1. Calculate the energy transfer rate across 6 in. wall of firebrick with a temperature difference across the wall of 50 ˚C. The thermal conductivity of the firebrick is 0.65 BTU/hrft-˚F at the temperature interest. a. 285 W/m2

c. 112 W/m2

b. 369 W/m2

d. 429 W/m2

Solution Q A

kAΔT x

kΔT x

Where: ΔT

50

x

6in.

k

0.65

90˚F 0.5 ft. BTU hr ft ˚F

Then: Q A

0.65 90 BTU 0.50 hr ft ˚F

Q A

W 3.153 BTU m 117 hr ft 1 BTU ft hr

Thus: b

Q A

368.90

W m

2. At an average temperature of 100˚C, hot air flows through a 2.5 m long tube with an inside diameter of 50 mm. The temperature of the tube is ˚C along its entire length. Convective film coefficient is 20.1 W/m2-K. Determine the connective heat transfer from air to the tube.

a. 900 W

c. 624 W

b. 909 W

d. 632 W

Solution Q

h AΔT

Where: A

π d L

A

π 0.050 2.5

A

0.3927m

Q

20.1 0.3927 100

Thus;

d Q

20

631.46 W

3. Steam, initially saturated at 2.05 MPa, es through a 10.10 cm standard steel pipe for a total distance of 152 m. The steam line is insulated with 5.08 cm thickness of 85% magnesia. For an ambient temperature of 22˚C, what is the quality of the steam which arises at its destination if the mass flow rate is 0.125 kg steam per second? Properties of Steam: Pressure=2.05 MPa,

Temperature=213.67 ˚C

Enthalpy: hf=914.52 kJ/kg

hfg=1885.5 kJ/kg

hg=2800.00 kJ/kg

Note: k for 85% magnesia is 0.069 W/m-K and ho for still air is 9.36 W/m2-K a.93 %

c. 84 %

b.98 %

d. 76 %

Solution

From figure, r1=5.05 cm r2=10.13 cm Q

t r ln r 2πkL

t 1 A h

Where: A

2πr L

A

2π 10.13 152

A

96.746 m

Q

213.67 ln 0.1013/0.0505 2π 0.069 152

Q

16,427.4 W

Q

m h

22 1 96.746 9.36

16.43 kW

From:

16.43

h

0.125 2800

h

2668.6kJ/kg

h

h

h

Where:

2668 x

x 914.52

x 1885.5

0.92998 or 93 %

Thus; a 93%

4. The sun generates 1 kW/m2 when used as a source for solar collectors, a collector with an area of 1 m2 heat water. The flow rate is 3.0 liters per minute. What is the temperature rise in the water? The specific heat of water is 4,200 J/kg ˚C. a. 4.8 ˚C

c. 0.50 ˚C

b. 0.48 ˚C

d. 0.84 ˚C

Solution Q

mC ∆T Where: Q

1

kW 1m m Li min

m

3

C

4200

1kW

1

kg Li

J kg˚C

60

4.2

min s

0.05

kg s

kJ kg˚C

Thus; 1

0.05 4.2 ∆T

a ∆T

4.76 ˚C

5. The hot combustion gases of a furnace are separated from the ambient air and its surrounding which are a 25˚C, by a brick wall 0.15 m thick. The brick has a thermal conductivity of 1.2 W/m-K and a surface emissivity of 0.80. Under steady state conditions and outer surface temperature of 100 ˚C is measured. Free convection heat surface temperature of 100 ˚C is measured. Free convection heat transfer to the air ading this surface is characterized by a convection coefficient of 20 W/m2-K. What is the inner temperature in ˚C? a. 623.7

c. 461.4

b. 352

d. 256.3

Solution

Let, Q

Q

heat transmitted by convection Q

h t

Q

20 100

Q

1500

t 25

W m

heat transmitted by radiation Q

20,408.4

10 Fe T

Q

20,408.4

10

Q

1,872,793

Q

520

0.80

T 100

J hr 273

m 25

273

J hr

m

W m

Then; Q

Q

Q

Q

2020

W m

1500

520

Thus; Q

kA t

2020 b t

t x

1.2 t 100 0.15 352.5 ˚C

6. A 6 in. x 20 ft. uninsulated B.I. pipe conveys steam at 385 ˚F wit han average ambient temperature of 85˚F. If the cost of the fuel is P 250.00 per 106 BTU with the net energy conversion efficiency of 75%, what is the annual cost of the heat lost? a. P 60, 482.00

c. P 70, 482.00

b. P 65, 482.00

d. P 75, 482.00

Solution For 6 in. pipe schedule 80 D

6.625 in.

D

5.761 in.

For iron; k

52

k

30

W m

K

BTU hr ft ˚F

For the surface coefficients; h

1000

BTU ˚F hr ft

h

2

BTU ft ˚F

hr

Solving for Q: t D ln D 2πkL t

Q 1 Ah

1 A h

Where: A

πD L

A

πD L

30.16ft 34.69ft 385

Q 1 30.16 1000

85 6.625 ln 5.761 2π 30 20 Q

Then the annual cost of heat lost:

20,713

1 34.69 2 BTU hr

C

20,713 8760 250 0.75 10

Thus; a C

P 60,482.00

7. What is the external heating area in square feet of a tube with the following dimensions: tube inside diameter = 5 in. wall thickness = ½ in. length = 18 ft. a. 26.5

c. 19.25

b. 24.25

d. 28.26

Solution A

πD L

Where: D

5

6 12

18

2

1 2

6 in.

Thus; A

π

d A

28.27ft

8. Determine the vacuum efficiency of a surface condenser which operates at a vacuum of 635 mm Hg and exhaust steam enters the condenser at 45.81 ˚C, the barometric pressure is 760 mm Hg and the saturation pressure at 45.81 ˚C is 0.010 MPa.

Solution

a. 80.4%

c. 92.7%

b. 85.2%

d. 98.3%

P P

Vacuum Efficiency

P P

Where: P

101.325 kPa

P

0.010 MPa

P

760

P

125 mmHg

P

16.665kPa

10kPa

635 101.325kPa 760mmHg

Then; Vacuum Efficiency

101.325 16.665 101.325 10

Vacuum Efficiency

0.9270

Vacuum Efficiency

92.70%

Thus;

9. A heat exchanger was installed purposely to cool 0.50 kg of gas per second. Molecular weight is 28 and k=1.32. The gas is cooled from 150 ˚C to 80 ˚C. Water is available at the rate of 0.30 kg/s and at a temperature of 12 ˚C. Calculate the exit temperature of the water. a. 48

c. 46

b. 42

d. 44

Solution Q

Q

m C

∆T

m C

t

m C ∆T 12

m C

150

80

Where: C

1.2247

C

4.187

kJ kg K

kJ kg K

Then; 0.30 4.187 t

12

0.50 1.2247 150

80

Thus; c t

46.125 ˚C

10. An uninsulated steam pipe es through a room in which the air and walls are at 25 ˚C. The outside diameter of the pipe is 70 mm, and its surface temperature and emissivity are 200 ˚C and 0.80 respectively. If the coefficient associated with free convection heat transfer from the surface to the air is 15 W/m2-K, what is the rate of heat loss from the surface per unit length of pipe? a. 997.84 W/m

c. 797.84 W/m

b. 897.84 W/m

d. 697.84 W/m

Solution Q

Q

Q Where: Q

heat transmitted by convection

Q

h A t

Q

15 π 0.07 L 200

Q

577.27

Q

heat transmitted by radiation

t 25

W m

Q

20,408.4

10 A Fe T

Q

20,408.4

10

Q

1,514,032.40 L

Q

42,057 L

T

π 0.07 L 0.8

J hr 473

298

J hr

m

W m

Thus; Q

577.27L

Q

997.84 L

a

Q L

420.57L

997.84

W m

11. A heat exchanger is to be designed for the following specifications: H2O gas temperature = 1145 ˚C CO2 gas temperature = 45 ˚C Unit surface conductance on the hot side = 230 W/m2-K Unit surface conductance on the cold side = 290 W/m2-K Thermal Conductivity of the metal wall = 115 W/m-K Find the maximum thickness of the metal wall between the hot gas and cold gas so that the maximum temperature of the wall does not exceed 545 ˚C. a. 10.115 mm

c. 17.115 mm

b. 13.115 mm

d. 20.115 mm

Solution Q A

t 1 h

t x k

1 h

Solving for Q/A: Q A

h t

Q A

230 1145

Q A

138,000

t

545

W m

Then; 138,000

1.45 45 x 1 1 230 115 290

Thus; d x

20.115 mm

12. Calculate the heat transfer per hour through a solid brick wall 6m long, 2.9 m high, and 225 mm thick, when the outer surface is at 5 ˚C and the inner surface 17 ˚C, the coefficient of thermal conductivity of the brick being 0.5 W/m-K. a. 2,004.48 kJ

c. 2,400.48 kJ

b. 3,004.48 kJ

d. 3,400.48 kJ

Solution Q

kA∆T x

Q

0.60 6 2.9 17 0.225

Q

556.8 W

Q

556.8

Q

2,004.48

J 360J0s hr s kJ hr

5

1

kJ 1000J

Thus; a The heat transfer per hour is 2,004.48 kJ

13. A vertical furnace wall is made up of an inner wall of firebrick 20 cm thick followed by insulating brick 15 cm thick and an outer wall of steel 1 cm thick. The surface temperature of the wall adjacent to the combustion chamber is 1200 ˚C while that of the outer surface of steel is 50 ˚C. The thermal conductivities of the wall material in W/m-K are: firebrick, 10; insulating brick 0.26; and steel, 45. Neglecting the film resistances and resistance of ts, determine the heat loss per sq. m. of wall area. a. 1.93 W/m2

c. 1.55 W/m2

b. 2.93 W/m2

d. 2.55 W/m2

Solution Q A Q A

x k

t

x k

t

x k

1200 50 0.20 0.15 0.01 10 0.26 45

1.93

Thus; a

Q A

1.93

W m

14. A composite wall is made up of an external thickness of brickwork 110 mm thick inside which is a layer of fiberglass 75 mm thick. The fiber glass is faced internally by an insulating board 25 mm thick. The coefficients of thermal conductivity for the three are as follows: Brickwork

1.5 W/m-K

Fiberglass

0.04 W/m-K

Insulating board

0.06 W/m-K

The surface transfer coefficients of the inside wall is 3.1 W/m2-K while that of the outside wall is 2.5 W/m2-K. Take the internal ambient temperature as 10 ˚C and the external temperature is 27 ˚C. Determine the heat loss through such wall 6 m high and 10 m long. a. 330.10 W

c. 430.10 W

b. 230.10 W

d. 530.10 W

Solution Q

A∆T R

Where; R

1 h

x k

x k

R

1 3.1

0.110 1.5

R

3.09

m ˚C W

x k 0.075 0.04

1 h 0.025 0.06

1 2.5

Then; Q

6 10 27 3.09

10

Thus; a Q

330.10 W

15. One insulated wall of a cold-storage compartment is 8m long by 2.5 m high and consists of an outer steel plate 18 mm thick. An inner wood wall 22.5 mm thick, the steel and wood are 90 mm apart to form a cavity which is filled with cork. If the temperature drop across the extreme faces of the composite wall is 15 ˚C, calculate the heat transfer per hour through the wall and the temperature drop across the thickness of the cork. Take the coefficients of thermal conductivity for steel, cork and wood as 0.45, 0.045, and 0.18 W/m-K respectively. a. 408.24 kJ, 12.12 ˚C

c. 608.24 kJ, 13.12 ˚C

b. 708.24 kJ, 11.12 ˚C

d. 508.24 kJ, 14.12 ˚C

Solution Q

A∆T R

R

x k

R

0.018 45

R

2.125

x k

x x 0.09 0.045

0.09 0.045

0.0225 0.18

Then: Q

8 2.5 15 2.125

Q

141.176 W or

Q

508.24

J s

kJ hr

Thus, the heat transfer per hour is 508.24 kJ Solving for the temperature drop across the cork: Q

A∆T x k

141.176

∆T

20 ∆T 0.09 0.045

14.12 ˚C

Thus; d 508.24 kJ, 14.12˚C

16. A cubic tank of 2 m sides is constructed of metal plate 12 mm and contains water at 75˚C. The surrounding air temperature is 16 ˚C. Calculate the overall heat transfer coefficient

from water to air. Take the coefficient of thermal conductivity of the metal as 48 W/m-K, the coefficient of heat transfer of water is 2.5 kW/m2-K and the coefficient of heat transfer of the air is 16 W/m2-K. a. 15.84 W/m2-K

c. 16.84 W/m2-K

b. 14.84 W/m2-K

d. 13.84 W/m2-K

Solution Let U

overall heat transfer coefficient U

1 R

Where: 1

R

x x

h 1

R

2.5

1 h 0.012 48

10

1 16

0.063m ˚C /W

R Then: U

1 W 0.063 m ˚C

15.84

W m ˚C

Thus; a U

15.84

W m ˚C

17. Calculate the quantity of heat conducted per minute through a duralumin circular disc 127 mm diameter and 19 mm thick when the temperature drop across the thickness of the plate is 5 ˚C.Take the coefficient of thermal conductivity of duralumin as 150 W/m-K. a. 30 kJ

c. 35 kJ

b. 40 kJ

d. 45 kJ

Solution Q

Q

kA∆T x 150

π 0.127 4 0.019

Q

500.04 W

Q

30 kJ/min

5

Thus; a the quantity of heat conducted per minute is 30 kJ

18. A cold storage compartment is 4.5 m long by 4 m wide by 2.5 m high. The four walls, ceiling and floor are covered to a thickness of 150 mm with insulating material which has a coefficient of thermal conductivity of 5.8 x 10 -2 W/ m-K. Calculate the quantity of heat leaking through the insulation per hour when the outside and inside face temperatures of the material is 15 ˚C and -5 ˚C respectively. a. 2185.44 kJ

c. 3185.44 kJ

b. 1185.44 kJ

d. 4185.44 kJ

Solution Q

kA∆T x

Where: A

2 4.5 2.5

A

78.50 m

4 2.5

4.5 4

Then: Q

5.8

10

78.5 15 0.15

5

Q

607.07 Wor

Q

2185.44

J s

kJ hr

Thus: a the quantity of heat leaking through the insulation per hour is 2185.44

kJ . hr

19. Supplementary Problem A blower operating at 15000 rpm compresses air from 20°C and 1 atm to 1.6 atm. The design flow is 38 m3/min and at this point the power input is 60 kW. Determine the blower efficiency at the design flow. a. 65%

c. 59.81%

b. 64.91%

d. 60.01%

Solution: Blower Efficiency =

Isentropic Power Power Input

Solving for Isentropic Power: Pisen =

k-1 k -1

kPV (rp )

k-1

1.4-1

=

1.4(101.325)(38/60) (1.68/1) 1.4 -1 k-1

= 35.89 kW

then; eblower =

35.89 = 0.5981 60

thus; (a) eblower = 59.81% 20. Supplementary Problem A small blower handles 43.33 m3 of air per minute whose density is 1.169 kg/m3. The static and velocity heads are 16.38 and 1.22 cm WG (at 15.6°C) respectively. Local gravity acceleration is 9.741 m/s2. Find the power input to the air from the blower. a. 1.64 kW

c. 1.76 kW

b. 1.91 kW

d. 1.24 kW

Solution: P = Qh Where: h = 16.38 + 1.22 = 17.6 cm = 0.176 m Q = 43.33 m3/min = 0.72 m3/s then;

P = 9.741(0.72)(0.176) thus; (d) P = 1.24 kW 21. Supplementary Problem A fan can developed a static pressure head or 350 mm water gage through standard air condition. What is the new static pressure head if fan can operate at 95°C and 729 mm of Hg? a. 265 mm WG

c. 274 mm WG

b. 270 mm WG

d. 263 mm WG

Solution: ρ h2 = 2 h1 ρ1 Solving for ρ2 : ρ2 =

720(101.325/760) 0.287(95 + 273)

then; h2 0.91 = 350 1.2 thus; (a) h2 = 265.09 mm WG 22. Supplementary Problem A two-stage radial-type airplane supercharger is designed to deliver 4535 kg of air per hour at a pressure of 800 mmHg abs when operating at an altitude of 4570 m where the temperature is -15°Cand the pressure is 429 mm Hg abs. it rotates at 18,000 rpm and is to have an adiabatic over-all efficiency of 72 percent. It is to be tested at sea level (762 mm Hg abs and 26.67°C at a speed of 14,000 rpm. Considering that the efficiency at the design point does not change, determine for the design point under test conditions the volume of air taken m3/s. a. 2.27 m3/s b. 2.12 m3/s Solution: Q2 N2 = Q1 N1

c. 1.27 m3/s d. 1.12 m3/s

Solving for Q1: Q1 =

mRT1 P1

(4535/3600)(8.314/29)(-15 + 273) 429(101.325/760) = 1.63 m3/s

Then; 14,000 Q2 = 1.63 18,000 Thus; (a) Q2 = 1.27 m3/s 23. Supplementary Problem The fan has a total head of 190 m and a static pressure of 20 cm WG. If the air density is 1.2 kg/m3, what is the velocity of air flowing? a. 16.21 m/s

c. 16.67 m/s

b. 17.21 m/s

d. 17.766m/s

Solution: hv =

v2 2g

Solving for hv: h = hs + hv 190 = 0.20(1000/1.15) + hv hv = 16.09 m then; 16.09 =

V2 2(9.81)

V = 17.766 m/s thus; (d) 17.766 m/s 24. Supplementary Problem A sewerage aeration blower rotating at 3500 rpm is designed to deliver 567 m3/min of air from 20°C and 1 atm to a discharge of 158 kpa (abs) with an adiabatic efficiency of 65 %. During a summer the atmospheric temperature rises to 43°C but the barometric pressure does not change. It is desired to vary the

blower speed to maintain the same discharge pressure. Determine the discharge volume of standard air with the new speed. a. 8.20 m3/s

c. 8.64 m3/s

b. 9.10 m3/s

d. 9.74 m3/s

Solution: 2

Q1 h1 = h2 Q2 h1 : h2

Solving for

T2 h1 = h2 T1 =

43 + 273 20 + 273

= 1.08 Then; 567 Q2

1.08 =

2

Q2 = 545.98 m3/min Thus; (b) Q2 = 9.10 m3/s 25. Supplementary Problem A 40 in. diameter fan rated at 160,000 cfm standard air at 16 in. starting pressure is operating at 1200 rpm. Solve for the specific speed. a. 386,845.18 rpm

c. 384,845.18 rpm

b. 380,125.20 rpm

d. 392,865.28 rpm

Solution: Ns = specific speed Ns = =

N√Q 3

h

4

12000 160,000 3

( 4⁄3 )

4

Thus; (a) Ns = 386,845.18 rpm

26. Supplementary Problem A boiler requires 75,000 m3/hr of standard air. What is the motor power if it can deliver a total pressure of 145 mm or water gage. The mechanical efficiency of fan is 64 %. a. 40.30 kW

c. 42.45 kW

c. 46.30 kW

d. 43.69 kW

Solution: Pmotor =

Pair Pfan

Solving for Pair: Pair =γQH Where: h = 0.145

1000 1.2

= 120.83 m Then; Pair = [(1.2)(0.00981)]

75000 (120.83) 3600

= 29.63 kW Thus; Pmotor =

29.63 0.64

(a) pmotor = 46.30 kW 27. Supplementary Problem Calculate the required motor capacity needed to drive a forced-draft fan serving a stoker fired boiler using coal as fuel. Combustion data includes the following: Atmospheric Air

101.3 kPa ; 20 °C

Weight of fuel burned per hour

10 tons

Ultimate Analysis of fuel : C = 78 %

S=1%

H=3%

A=8%

O=3%

M=7%

ExcessAir

30 %

Fuel bed and air heater resistance

18 cm WG

Fan Efficiency

60 %

a. 87.84 kW

c. 84.87 %

b. 82.87 kW

d. 88.72 %

Solution: Pmotor =

Pair Pfan

Solving for Pair: Theoretical Air required for the combustion of coal. Wt =11.5C + 34.5

H -

O + 4.3S 8

= 11.5(0.78) + 34.5

0.03 -

0.03 +4.4(0.01) 8

= 9.92 kg air / kg fuel Actual weight of air supplied into the boiler: Wa = (1 + e) Wt = (1 + 0.30)(9.92 kg air/kg fuel)[(10(1000)kg fuel/hr] = 128,942 kg air/hr Volume of air demanded by the boiler from the forced draft fan: Q=

128,942 kg/hr =107,451.77 m3 /hr 1.2 kg/m3

= 29.85 m3/s Then; Pair = 1.2 0.00981

29.85 (0.18)

1000 1.2

= 52.71 kW Thus; Pmotor =

52.71 0.60

(a) Pmotor = 87.84 kW 28. Supplementary Problem The motor power needed to drive the fan is 75 kW and the volume flow of air delivered by fan is 23 m3/s and 20 cm water gage. The density of air is 1.2kg/m3. What is the fan efficiency? a. 60 %

c. 64 %

b. 62 %

d. 65 %

Solution: efan =

Pair Pmotor

Solving for Pair: Pair =[1.2 0.00981 23 ] 0.20

1000 1.2

= 45.126 kW Then; efan =

45.126 75

= 0.60168 Thus; (a) efan = 60.168 % 29. Supplementary Problem In a certain installation, a fan when driven by a 7.5 Hp motor at a speed of 600 rpm delivers 510 m3 of air per minute at a total pressure of 5 cm WC. If in the same installation, 6.5 cm WC pressure is required. What power and motor speed willthe fan be driven? Solution: Fan Drive Speed: N2 N1

2

=

h2 h1

N2 = N1 N2 = 600

h2 h1 6.5 5.0

N2 = 684.11 rpm Motor Power required:

P2 N2 = P1 N1

P2 = 7.5

3

684.11 600

3

P2 = 11.12 Hp Thus; (c) N2 = 684.11 rpm, P2 = 11.12 Hp

30. Supplementary Problem A fan is supplying forced draft into a boiler has the following specifications on its name plate: Capacity

280 m3/min

Air temp.

25°C

Total pressure

4 cm WC

Motor Rating

5 Hp; 1200 rpm

A tabular air heater is installed in line with the boiler, and the fan is now required to supply heated air for combustion at 90°C. What drive power is required and the new total pressure that this fan will operate if it is going to deliver the same volume of heated air at 1200rpm? a. 3.28 cm WC, 4.10 Hp

c. 3.28 cm WC, 5.50 Hp

b. 3.95 cm WC, 5.5 Hp

d. 3.95 cm WC, 4.10 Hp

Solution: Solving for the new head; ρ h1 = 2 h2 ρ1 Solving for

ρ2 : ρ1

ρ2 T1 = ρ1 T2 = Then;

25+273 90+273

h2 = 0.82 4 h2 = 3.28 cm WC

thus; (a) h2 = 3.28 cm WC & P2 = 4.10 Hp Note: when air is heated, its density decreases and the pressure needed to move the air to the combustion chamber will be lesser resulting to the decrease in the fan power requirement. 31. Supplementary Problem

A fan has a suction pressure of 5 cm water vacuum with air velocity of 5m/s. the discharge has 20 cm WG and discharge velocity of 10 m/s. Find the total head of the fan. Solution: h = hs + hv where: hs =

(hdw - hsw )ρw ρa

=

[(0.20 - (-0.05)]1000 1.2

= 208.33 m hv =

Vd 2 - Vs 2 2g

(10) - (5) 2(9.81)

2

2

= 3.28 m Thus; h = 208.11 + 3.28 (a) h = 212.15 m 32. Supplementary Problem An Air Handling Unit (AHU) for an air conditioning system has a centrifugal fan with backward curved blades mounted on a scroll housing driven by a motor at 750 rpm. The fan delivers 2000 cfm of air against 3 in. WC static pressure (including resistance of ducts, elbows, cooling coils, and outlet grills) and 0.80 in. WC velocity pressure. Calculate the tip speed of the wheel. a. 3100 fpm

c. 3586 fpm

b. 3000 fpm

d. 3500 fpm

Solution: The speed of the wheel: V=

2ghv

Solving for hv: ρa hva = hvw ρw (0.075)hva = (0.8/12)(62.4) hva = 55.46 ft of air

Note: ρair =0.075 lb/ft ρwater = 62.4 lb/ft

3

3

Then; V=

2(32.2)(55.46)

= 59.77 ft/s Thus; (c) V = 3586 ft/min. 33. Supplementary Problem A steam generator supplies 180,000 kg of steam per hour at 5.5 Mpa abs and 540°C with feedwater at 176°C. At this output, the thermal efficiency is 85% when burning 42,456 kJ/kg fuel oil at 15% excess air. The products of combustion with an average molecular weight of 30 are removed from the unit by a pair of duplicate induced-draft fans operating in parallel and the flue gas temperature for each fan suction is 150°C. Estimate the capacity of each fan using the following rule: “7.5 kg air required for perfect combustion for each 23,200 kJ per kg heat value of oil. The fan differential pressure is 190 mm WC. a. 69.57 kW

c. 59.75 kW

b. 89.75 kW

d. 76.57 kW

Solution: ρair = γQh = [1.2(0.00981)] Q h Where; h = 0.190 (1000/1.2) = 158.33 m Solving for Q: Mass of fuel burned per hour: mf =

ms hs - hv eboiler Qh

=

180,000(3520-746) 0.85(42,456)

= 13 836.33 kg/hr

Total mass of air used per kg of fuel burned: ma = 7.5

42,456 23,200

1.15 =17.58 kg air/kg fuel

Mass flow rate of air for combustion: ma = (15.78)(13,836.33) = 218,839.17 kg/hr Mass flow rate of gases leaving the boiler: mg = ma + mf = 218,839.17 + 13,836.33 = 232,225.5 kg/hr or mg = 116,112.75 kg/hr (each fan) Mass flow gas handled by each fan: Q=

(116,112.75)(8.314/30)(150 +273) 101.3

= 134,369.16 m3/hr = 37.32 m3/s Thus; (a) Pair = [ 1.2(0.00981) ] (37.32) (158.33) = 69.57 kW 34. ME Board Problem A fan listed as having the following performance with standard air: Volume discharge = 120 m3/s Speed = 7 rps Static pressure = 310 mm water gage Brake power = 620 kW The system duct will remain the same and the fan will discharge the same volume of 120 m3/s of air at 93°C and a barometric pressure of 735 mm Hg when its speed is 7 rps. Find the brake power input and the static pressure required. a. 482 kW, 241 mm WG

c. 482 kW, 256 mm WG

b. 492 kW, 241 mm WG

d. 492 kW, 256 mm WG

Solution: Brake power input=620 Solving for ρ2 : ρ2 =

735(101.325/760) 0.278(93+273)

= 0.9329 kg/m3

ρ2 ρ2 =620 ρ1 1.2

Solving for the static pressure, h2: 0.9392 h2 = 310 1.2 h2 = 241 mm of water gage then; Brakepower input= 620

0.9329 =482 kW 1.2

thus; (a) h2 = 241 mm & Pbrake = 482 kW 35. ME Board Problem Local coal with higher heating value of 5,500 kCal/kg is burned in a pulverized coal fired boiler with 25% excess air at the rate of 25.9 M.T. per hour when the steam generated is 220 M.T. per hour. This boiler is served by 2 forced-draft fans of equal capacity delivering the air at 305 mm of water to the furnace. Calculate the capacity of each fan in m3/hr if fan capacity is to be 110 percent of the maximum requirement. Ambient air is 100 kPa and 30°C. a. 31.75 m3/s

c. 28.87 m3/s

b. 63.50 m3/s

d. 57.73 m3/s

Solution: The theoretical weight of air to burn the fuel is given in an appropriate formula when the heat value ot the fuel is given: Wta = =

A HHV, kCal/kg = F t 745 5,500 745

= 7.382 kg air / kg fuel Mass flow rate of air required for combustion ma = (7.382)(1.25)[25.9(1000)] = 238,992.25 kg/hr Volume of Air needed at 100 kPa and 303 K ma Ra Ta P 238,992.25(0.287)(303) = 100 Q =

= 207 830.05 m3/hr Capacity of each fan at 110% of the maximum requirement: Q1 = Q2 =

207,830.05(1.10) 2

= 114,306.53 m3/hr = 31.75 m3/s Thus; (a) Q = 31.75 m3/s 36. Supplementary Problem Find the motor size needed provide the forced-draft service to a boiler that burns coal at the rate of 10 to per hour. The air requirements are 59,000 cfm, air is being provided under 6 in. water gage (WG) by the fan which has mechanical efficiency of 60 percent. Assume fan to deliver the total pressure of 6 in WG. a. 90 Hp

c. 97 Hp

b. 93 Hp

d. 99 Hp

Solution: Pmotor =

Pair 0.60

Solving for Pair: Pair = γQh Where: Q = 59,000 ft3/min. = 27.84 m3/s h = 6(1000/1.2) in. = 127 m then; Pair = [ 1.2(0.00981)) ](27.84)(127) Pair = 41.62 kW Thus; Pmotor =

41.62 =69.37 kW 0.60

(a) Pmotor = 92.99 Hp

37. Supplementary Problem A turbo-generator, 16 cylinder, Vee type diesel engine has an air consumption of 3000 kg/hr per cylinder at rated load and speed. This air is drawn in thru a filter by a centrifugal compressor direct connected to the exhaust gas turbine. The temperature of the air from the compressor is

145C and a counterflow air cooler



reduces the air temperature to 45 C before it goes to the engine suction header. Cooling water enter air cooler at mean temperature difference.

30C and leaves at 38C. calculate the arithmetic





a. 41 C b.

c. 61 C

51C

d,

71C

Solution:

t max  145  38  107C t min  45  30  15C t max  t min AMTD  2 107  15 AMTD  2

Thus; (c) AMTD  61  C

38. Supplementary Problem A pond is covered by a sheet of ice 2 cm thick (thermal conductivity

1.68W / mC ). The temperature of the lower surface of the ice is 0C and that of the upper surface is of the ice?

 10C. At what rate is heat conducted through each square meter

a. 840 W

c . 940 W

b. 740 W

d. 640 W

Solution:

kAt x (1.68)(1)(0  10) Q (0.02) Q

Thus;

t max

t min

(a) Q  840W

39. Supplementary Problem How much heat is

conducted

through

a

sheet

of

plates

glass,



k  0.0024Cal / s  m  C which is 2m by 3m and 5mm thick, when the   temperatures of the surfaces are 20 C and  10 C. 318,400 Cal/min b. 418,400 Cal/min

a.

c. 940 Cal/min d. 618,400 Cal/min

Solution:

Q

kAt x

Where:

k  0.0024 Cal/s- cm C A  2(3)  6 m 2

 60,000 cm 2

t  20  10  30  C L  5 mm  0.50 cm Then;

Q

0.002460,00030 0.50

= 8640 Cal/s Thus; (c)

Q  518,400 Cal/min

40. Supplementary problem A cooper rod whose diameter is 2 cm and length 50 cm has one end in boiling water, the other end in a jacket cooled by flowing water which enters at

10 C . The thermal conductivity of copper is 0.102 kCal / m  s   C. If 0.20 kg of water flows through the jacket in 6 min, by how much does the temperature of the water increase?

10.38  C  b. 9.38 C

a.

Solution:

Q Where:

kAt x

11.38  C  d. 12.38 C c.

x 0.022 4  3.14 10 4 m 2

A Then;

Q

0.1023.14  10 4 100  10

0.50 = 0.005765 kCal / s = 0.346 kCal / min 6 min  = 2.705 kCal

Heat required to raise the water temperature:

Q  m C t 2,075 Cal  200g (1 Cal / g C )(t ) Thus; 

(a) t  10.38 C

41. Supplementary Problem The thermal insulation of wooden glove may be regarded as being essentially

kCal / m  s   C. How 2 much heat does a person lose per minute from his hand of area 200 cm and skin   temperature 35 C on a winter day at  5 C. a layer of quiescent air 3 cm thick, of conductivity 5.7  10

6.12  C  b. 7.12 C

9.12  C  d. 8.12 C

a.

c.

Solution:

Q

kAt x

Where: k= 5.7  10

6

kCal / m  s   C

cm 2 2 = 0.02 m

A= 200

x=3 cm = 0.03 m

t  35  5  40  C Q

5.7 10 .0240 6

0.03

 1.52 10 4 kCal / s Thus;

6

(c) Q  9.12 Cal / min 42. Supplementary Problem The temperature directly beneath a 3 in. concentrate road is

5  F and the air



temperature is 20 F . Calculate the steady flow per square foot through the 

concrete. The thermal conductivity of the concrete is 0.50 Btu / ft  hr  F . a. 30 Btu / hr  ft

2

c. 50 Btu / hr  ft

2

b. 40 Btu / hr  ft

2

d. 60 Btu / hr  ft

2

Solution:

Q kt  x A 0.520  5  3 / 12 Thus; (a)

Q Btu  30 A hr  ft 2

43. Supplementary Problem At what rate does to sun lose energy by radiation? The temperature of the sun is about 6000 K and its radius is 6.95  10 km. . 5

3.48 10 26 W 26 b. 4.48  10 W

5.48  10 26 W 26 d. 6.48  10 W

a.

c.

Solution:

P  A T 4 Where:

  5.7 10 12 W / cm 2  K 4 A  4R 2  4 6.95  10 5   6.07  1012 km 2  6.07  10 22 m 2 Then;







P  5.7  10 12 6.07  10 22 6000 

Thus; (b)

P  4.48  10 26 W

4

44. Supplementary Problem How many watts will be radiated from a spherical block body 15 cm in 

diameter at a temperature of 800 C . a. 5.34 kW b. 4.34 kW

c. 6.34 kW d. 3.34 kW

Solution:

P  A T4 Where:

  5.7  10 12 W / cm 2  K 4 A  4 (7.5) 2  706.86 cm 2 T  800  273  1073K Then;





P  5.7  10 12 706.861073

4

Thus; (a) P  5,340 W  5.34 kW 45. Supplementary Problem Calculate the radiation in watts per square centimeter from a block of copper 



at 200 C and at 1000 C . The oxidized copper surface radiates at 0.60 the rate of a black body. a. 0.17 b. 0.27

c. 0.37 d. 0.07

Solution:

Q e AT4 Q e T4 A  0.60(5.7  10 12 )(300  273) 4  0.17 W / cm 2 Thus; (a)

Q  0.17 W / cm 2 A

46. Supplementary Problem A surface condenser serving a 50,000 kW steam turbo-generator unit receives exhaust steam at the rate of 196,000 kg/hr. Vacuum in condenser is 702 mm. Hg. Sea 



water for cooling enters at 29 and leaves at 37.5 C . For steam turbine condenser, manufacturers consider 950 Btu/lb of steam turbine condensed as heat given up to cooling water. Calculate the logarithmic mean temperature difference.





a. 4.57 C

c. 6.57 C

5.57C

b.

d.

7.57  C

Solution: Let: LMTD  Log mean temperature difference

LMTD 

t max  t min t max ln t min

Where:

  101.325kPa   P( condenser )  101.325  702 psi  760 psi    7.733 kPa(t sat  40.86  C )

(t ) max  40.86  29.5  11.36C (t ) min  40.86  37.5  3.36C Then;

LMTD 

11.36  3.36 11.36 ln 3.36

Thus; (c)

LMTD  6.57C

47. Supplementary Problem The stack gas from a chemical operation contains noxious vapors that must

315C to 35C . The gas flow rate is 0.70 m 3 / s . Water is available at 10  C at 1.26 kg / s . A two shell and 4 tube ,

be condensed by lowering its temperature from

conterflow heat exchanger will be used with a water flowing through the tubes. The gas has a specific heat of 1.10

kJ / kg  K and a gas constant of 0.26 kJ / kg  K .

Calculate the logarithmic mean temperature difference.

102.8  C  b. 110.9 C

120.8  C  d. 118.9 C

a.

c.

Solution: Average gas temperature

Density of gas

( ) :



315  35 2  175 C



P RT



101.325 (0.26)(175  273)

  0.867kg / m 3 Mass flow rate of gas:

mg  (0.7 m 3 / s )(0.867 kg / m 3 )  0.607 kg / s Heat gained by cooling water = Heat Lost by the gasses

mw C pw t w  m g C pg t g (1.26)(4.187)(t  10)  0.60(1.10)(315  35)

t  45.5  C Solving for

(t ) max and (t ) min :

(t ) max  315  45.5  269.5  C (t ) min  35  10  25  C Then;

LMTD 

(t ) max  (t ) min 269.5  25  (t ) max 269.5 ln ln 25 (T ) min

Thus; (a)

LMTD  102.8  C

48. Supplementary Problem Exhaust steam at 7 kPa at the rate of 75 kg/s enters a single condenser 2

containing 5,780 pcs copper tubes with a total surface area of 2950m . The steam has a moisture content of 10 % and the condense leaves saturated liquid at steam 

temperature. The cooling water flow rate is 4,413 liters per second entering at 20 C . Size of tubes, 25 mm O.D. by 3 min thick wall. Find the overall heat transfer coefficient. a. 5275 W / m b. 2275 W

2 K

c. 4274 W / m

/ m 2 K

d. 3225 W

Solution: Ethalpy of steam entering the condenser:

h1  (h f 1  xh fg ) 7 kPa  163.4  0.90(2572.5)  2478.65 kJ / kg Ethalpy and temperature of condensate:

2 K

/ m 2 K

h2  h f @ 7 kPa  163.4 kJ / kg , Tsat @ 7 kPa  29  C Ethalpy and temperature of the condensate:

Qwater  Qsteam m w C pw t w  m s (h1  h2 )

4,413L / s(1kg / L)(4.187)(t  20)  75(2478.65  163.4) t  29.40  C

(t ) max  39  20  19  C (t ) min  39  29.4  9.6  C LMTD 

19  9.6  13.77  C 19 ln 9 .6

From:

Q  A U (LMTD ) Where:

Q  m s (h1  h2 )

 75(2478.65  163.4)  173,643.75 kW Then:

173,643.75  2950 U 13.77 U  4.275

kW m2  K

Thus; (a) U  4275

W m K 2

49. Supplementary Problem What is the heat flow per hour through a brick and mortar wall 9 in. thick if the 

coefficient of thermal conductivity has been determined as 0.40 Btu / ft  hr  F and the wall is 10 ft high by 6 ft wide, the temperature on one side of the surface 



being 330 F and on the other 130 F . a. 6400 Btu / hr

c. 5400 Btu / hr

b. 7400 Btu / hr

d. 8400 Btu / hr

Soution:

kAt x (0.4)(10)(6)(330  130) 6 12

Q

then;

(a) Q  6,400 Btu / hr

50. Supplementary problem Water is flowing in a pipe with radius of 25.4 cm at a velocity of 5 m/s at the temperature in the pipe. The density and viscosity of the water are as follows:

density  997.9 kg / m 3 and vis cos ity  1.131Pa  s . What is the Reynolds number for this situation. a. 2241 b. 96.2

c. 3100 d.1140

Solution:

N RE 

VDp



Where:

V  5 m/s D  2(25.4)  50.8 cm  0.508 m   997.9kglm 3   1.131 Then;

N RE 

5(0.508)(9977.9) 1.131

Thus; (a) N RE  2241.08

51. Supplementary Problem A heat exchanger has an over-all

coefficient

of heat



transfer of

900W / m  K . The mean temperature difference is 20 C and the heat loss is 15,000 W. Calculate the heat transfer area. 2

a. 0.833 m

2

c. 0.933 m

2

b. 0.733 m

2

d. 0.633 m

2

Solution:

Q  AU  15,000  A(900)(20) A  0.833m 2 Thus; (a) A  0.833m

2

52. Supplementary Problem A complete furnace wall is made up of a 12 in. lining of magnesite refactory brick, a 5 in. thickness of 85% magnesia, and a steel casing 0.10 in. thick. Flue gas 



temperature is 2200 F and the boiler room is at 80 F .Gas side film coefficient is 15 

Btu / hr  ft 2   F air side is 4. Determine the thermal current Q / A . a. 187.41 Btu / hr  ft

2

c. 200.62 Btu / hr  ft

2

b. 197.41 Btu / hr  ft

2

d. 250.46 Btu / hr  ft

2

Solution:

Q t  A RT Where:

t  2200  80  2120  F 

hi  15 Btu / hr  ft 2 F 

ho  4 Btu / hr  ft 2 F 

k12  20.5 Btu / hr  ft 2 F (magnesite) 

k 23  0.04 Btu / hr  ft 2 F (magnesia) 

k 34  25 Btu / hr  ft 2 F ( steel ) RT 

1 x12 x 23 x34 1     hi k12 k 23 k 34 h0

RT  11.312 Thus;

2120 Q  A 11.312 (a)

Q  187.41 Btu / hr  ft 2 A

53. Supplementary Problem

 0.72 Btu / hr  ft   F ) 5 in. of insulating brick ( k  0.08) and 75 in. of red brick ( k  0.5) . The inner and outer A wall of a furnace is made up of 9 in. firebrick ( k

surface temperatures t1 and t 4 of the wall are

1500  F and 150  F respectively.

Neglecting the resistance of the mortar ts, compute the rate of heat flow through

1 ft 2 of the wall.

a. 80 Btu/hr b. 180 Btu/hr

c. 100 Btu/hr d. 200 Btu/hr

Solution:

Q

At RT

Where:

RT 

x12 x 23 x34   k12 k 23 k 34

RT 

9 / 12 5 / 12 2.5 / 12   0.72 0.08 0 .5

Then;

Q

1(1500  150) 7.5

Thus; (d) Q = 180 Btu/hr 54. Supplementary Problem What is the heat transfer in the glass surface area of 0.70 

m 2 having an inside



temperature of of 25 C and 13 C outside temperature. The thickness of the glass surface is 0.007 m. The thermal conductivity is 1.8 W/m-K. a. 4.16 kW b. 3.16 kW Solution:

kAt (1.8)(0.7)(25  13)  0.007 x  2160 W

Q Thus; (c)

Q  2160 W

c. 2.16 kW d. 1.16 kW

55. Supplementary Problem The interior of an oven is maintained at a temperature of 1500°F by means of a suitable control apparatus. The walls of the oven are 9 in. thick and constructed from a material having thermal conductivity of 0.18 Btu/hr-ft-°F. Calculate the heat loss for each square foot of wall surface per hour. Assume that the inside and outside wall temperatures are 1500°F and 400°F respectively. a. 264 Btu/hr-ft2

c. 164 Btu/hr-ft2

b. 364 Btu/hr-ft2

d. 64 Btu/hr-ft2

Solution:

0.18 1 1500 9 12

400

264

Thus; (a)

264

56. Past ME Board Problem Compute the amount condensate formed during 10 minutes warm-up of 150 m pipe conveys saturated steam with enthalpy of vaporization hfg = 1,947.8 kJ/kg. The minimum external temperature of pipe is 2°C and the final temperature is 195°C. The specific heat of pipe material is 0.6 kJ/kg-K and the specific weight is 28 kg/m. a. 249.69 kg

c. 124.85 kg

b. 499.38 kg

d. 62.42 kg

Solution:

ms (1947.8) = 4200(0.60)(195-2) Thus; (a) ms = 249.69 kg 57. Supplementary Problem A high pressure steam generator is to be fitted with convection type superheater having 72 elements in parallel. Steam at the rate of 70,000 kg/hr from the boiler drum enters the superheater inlet header at 8.3 Mpa and 485oC. Combustion products at 980oC enters the superheater proper at the rate of 160,000kg/hr. Superheater elements are made of 60 mm O.D. by 8 mm thick tubing of 30 m length. Assume that the flue gas has the same thermal properties of air. Calculate the heat transferred to the superheater tubes.

a. 12,152 kW

c. 10,152 kW

b. 11,512 kW

d. 13,152 kW

Solution: From mollier Chart: h1 = 2,715 kJ/kg @ 8.3 Mpa & 98% quality h2 = 3,340 kJ/kg @ 8.0 Mpa & 485oC Heat lost by flue gas = Heat gained by steam mgg tg = ms (h1 – h2) (160,000)(1.0)(980 – to) = 70,000 (3340 – 2715) to = 705oC The rate of heat transferred to the superheated tubes: Q = ms (h2 – h1) = 70,000 (3340 – 2715) = 43,750,000 kJ/hr = 12,152,000 kJ/S Thus: (a) Q = 12,152 kJ/s or kW 58. Supplementary Problem In a hot water heating system, water heated to 95oC and then is pumped at the rate of 4 L/min through a radiator where it is cooled to 35oC. If the water arrives at the radiator at at temperature of 85oC, how much heat does the radiator release each hour? a. 50,244 kJ/hr

c. 55,344 kJ/hr

b. 45,422 kJ/hr

d. 65,244 kJ/hr

Solution: Q = m t = 4(4.187)(85-35) = 837.4 kJ/min = 50,244 kJ/hr Thus: (a) Q = 50,244 kJ/hr 59. Supplementary Problem

Brine enters a cooler at the rate of 50 m3/hr at 15oC and leaves at 1oC. Specific heat and specific gravity of brine are 1.07 kJ/kg-K and 1.1 respectively. Calculate the heat transferred in kW. a. 158.21 kW

c. 258.21 kW

b. 228.86 kW

d. 128.86 kW

Solution: Q = m t Where: m = pV = 1.1(1000 kg/m3)(50m3/hr)(1hr/3600s) = 15.28 kg/s Thus: Q = (15.28)(1.07)(15-1) (b) Q = 228.86 kJ/s or kW 60. Supplementary Problem A metal rod is 10 cm long and has a diameter of 2 cm one end is in with steam at 100oC while the other end s a block of ice at 0oC. the cylindrical surface of the rod is carefully insulated so heat flows only from end to end. In a time of 20 minutes, 320 grams of ice melts. What is the thermal conductivity of metal? a. 0.28266 kJ/kgoC

c. 0.38266 kJ/kgoC

b. 0.18266 kJ/kgoC

d. 0.48266 kJ/kgoC

Solution: Qrod = Qice

0.01 100 0.10

0

20 60

0.320 3.3 10

K = 282.66 J/kgoC K = 0.28266 kJ/kgoC Thus: (a) k = 282.66 J/kgoC 61. Supplementary Problem A wall with an area of 10 m2 is made of 2 cm thickness of white pine (k = 0.113 W/moC) followed by 10 cm of brick (k = 0.649 W/moC). The pine is on the inside where the

temperature is 30oC while the outside temperature is 10oC. Assuming equilibrium conditions exist, what is the temperature at the interface between the two metals? a. 15.65oC

c. 18.21oC

b. 17.64oC

d. 19.31oC

Solution:

∆

Where: 0.02 . 113

0.10 0.649

RT = 0.331 m2oC/W Then;

10 30 10 0.331

Q = 604.23 W Solving for the temperature at the interface between the two materials: Q = Q12 604.23

604.23

.

.

Thus; (a) t2 = 19.31oC 62. Supplementary Problem In a hot air heating system, the furnace heats air from 60oF to 160oF. If the air is then circulated at the rate of 330 ft3/min by the blower, how much thermal energy is transferred per hour. Note: the specific heat of air at constant pressure is 0.250 Btu/lboF, the density of air is 0.0806 lb/ft3 at atmospheric pressure. a. 39,000 Btu/hr

c. 36,805 Btu/hr

b. 46,800 Btu/hr

d. 39,900 Btu/hr

Solution: Q = m∆t Where: m = pV = (0.0806 lb/ft3)(330 ft3/min) = 26.60 lb/min

Then; Q = (26.60 lb/min)(0.250 Btu/lboF)(160-60)oF Q = 665 Btu/min Thus; (d) Q = 39,900 Btu/hr 63. Supplementary Problem If 1000 liters of air at 27oC and pressure of 1 atm has a mass of 1.115kg and a specific heat at constant pressure of 1 x 103 J/kg – K, how much heat is required to raise the temperature of this gas from 27oC to 177oC at constant pressure? a. 176.25 kJ

c. 167.25 kJ

b. 157 kJ

d. 175 kJ

Solution: Q = m∆t = (1.115)(1)(177-27) Thus; (b) Q = 167.25 kJ 64. Supplementary Problem Calculate the amount of energy required to heat the air in a house 30 by 50 by 40 ft from 10 to 70oF. The density of air is about 0.08 lb/ft3, and its specific heat at constant pressure 0.24 Btu/lboF. a. 49,120 Btu

c. 69,120 Btu

b. 59,120 Btu

d. 79,120 Btu

Solution: Q = m∆t Where: m = pV = (0.08 lb/ft3)[(30)(40)(50)]ft3 = 4800 lb Then; Q = 4800(0.24)(70-10) Thus; (c) Q = 69,120 Btu

65. Supplementary Problem Water enters the condenser at 20oC and leaves at 35oC. What is the log mean temperature difference if the condenser temperature is 40oC. a. 16.37oC

c. 15.37oC

b. 13.37oC

d. 17.37oC

Solution:

∆

∆ ∆ ∆

Where: (∆t)max = 45 – 20 = 25oC (∆t)min = 45 – 35 = 10oC

25

10 25 10

Thus; (a) LMTD = 16.37oC 66. Supplementary Problem When 200 grams of aluminum at 100oC is dropped into an aluminum calorimeter (k = 0.909 x 103 J/kg-K) of mass 120 grams and containing 150 grams of kerosene at 15oC the mixture reaches a temperature of 50oC, what is the specific heat of kerosene? a. 1004 K/kg-k

c. 1110 K/kg-k

b. 1050 K/kg-k

d. 1080 K/kg-k

Solution: Heat loss by aluminum = heat gained by the kerosene and calorimeter QA = Qk + Qc maCa∆ta = mkCk∆tk + mcCc∆tc (0.20)(0.909x103)(100-50) = (0.15)(Ck)(50-15) + (0.12)(0.909x103)(50-15) Thus; (a) Ck = 1004.23 J/kg-K 67. Supplementary Problem A calorimeter contains 66 kg of turpentine at 10.6oC. When 0.147 kg of alcohol at 75oC is added, the temperature rises to 25.2oC. the specific heat of turpentine is 1.95x103 J/kg-oC

and the calorimeter is thermally equivalent to 30 grams of water. Find the specific heat of alcohol. a. 2.81745 kJ/kgoC

c. 0.81745 kJ/kgoC

b. 3.81745 kJ/kgoC

d. 1.81745 kJ/kgoC

Solution: Heat loss by alcohol = heat gained by the turpentine and calorimeter maCa∆ta = mtCt∆tt + mcCc∆tc (0.147)(Ca)(75-25.2) = (0.66)(1.95x103)(25.2-10.6) + (0.03)(4.187)(25.2-10.6) Ca = 2817.45 J/kgoC Thus; (a) Ca = 2.81745 J/KgoC 68. Supplementary Problem The temperature of a sample of molten lead near its temperature of solidification is falling at the rate of 6 K/min. If the lead continues to lose heat at this same rate and takes 35 min. to solidify completely, what is the heat of fusion of the lead? The specific heat of molten lead is 0.126 kJ/kg-K. a. 16.46 kJ/kg-K

c. 36.46 kJ/kg

b. 26.46 kJ/kg-K

d. 46.46 kJ/kg

Solution: Q = mC∆t = mLf Lf = C∆t = (0.126 kJ/kg-K)(6K/min)(35min) Thus; (b) Lf = 26.46 kJ/kg 69. Supplementary Problem A counterflow heat exchanger is designed to heat fuel oil from 45oC to 100oC while the heating fluid enters at 150oC and leaves at 115oC. Calculate the arithmetic mean temperature difference. a. 40oC

c. 60oC

b. 50oC

d. 70oC

Solution: Where:

∆

∆ 2

(∆t)max = 115 - 45 = 70oC (∆t)min = 150 - 100 = 50oC

70

50 2

Thus; (a) AMTD = 60oC 70. Supplementary Problem A fuel oil of 20oAPI is to be heated in a heater which makes two es thru heater tubes and the heating fluid makes one es but the flow is cross flow through the heater due to baffles inside the shell. Quantity of oil to be heated

3000L/hr

Temperature of oil entering heater tubes

21oC

Temperature of oil leaving heater tubes

95oC

Heating fluid, steam enter at 05oC and leaves as condesate at 105oC. Assume specific heat of oil to be 2.093 kJ/kg-K. Find the heating surface area if the over-all coefficient of heat transfer is taken as 140 W/m2oC. a. 24.76 m2

c. 23.75 m2

b. 30.75 m2

d. 32.54 m2

Solution: Q = UAs(LMTD) = UAs(θ) Where:

∆

∆ ∆ ∆

84

10 84 10

Θ = 43.77oC Q = mo∆to Solving for mo: . .

S.G. = 0.934 mo = (3000L/hr)(1kg/L)(0.934)

141.5 131.5

141.5 20 131.5

= 2802 kg/hr = 0.778 kg/s From: Q = (0.778)(2.093)(95-21) Q = 120.55 kW Q = 120,550 W Then: 120,550 – (140)(As)(34.77) Thus; (a) 24.76 m2 71. Supplementary Problem A 30 cm thick wall has an inside and outside surface temperatures of 300oC and 50oC respectively. If the thermal conductivity of the wall is 8 W/m-K. Calculate the heat transferred in kW/m2. a. 6.67

c. 7.67

b. 5.67

d. 8.67

Solution:

∆

8 300 50 0.30 = 6,666.67 W/m2 Thus; (a) Q/A = 6.67 kW/m2 72. Supplementary Problem A 4- low-pressure surface type feedwater heater is designed to heat 92,730 kg/hr of feedwater from 40oC initial to 80oC final temperature using steam bleed at 70 kPa abs. containing 2,645 kJ/kg enthalpy. Assume no subcooling of condensate, determine the effective length of 19 mm O.D. x 2 mm thick muntz metal tubes to be installed, if the water velocity inside the tubes is 1.22 m/s and U = 3000 W/m2K based on the external surface of the tubes. a. 2m

c. 3m

b. 4m

d. 5m

Solution:

A=

. .

A=

0.019

.

Solving for LMTD or θ:

∆

∆ ∆ ∆

50

10 50 10

= 24.85oC Solving for A: By: energy balance: ms(h1 – h2) = mww(t2 – t1) ms(2645 – 376.7) = 92,730(4.187)(80 – 40) ms = 6846.72 kg/hr Q = ms(h1 – h2) = 1.90(2645 – 376.7) Q = 4,314.01 kW Q = 4,314,010 W Then; 4,314,010 3000 24.85 = 57.87 m2 Solving for n: . 4 92,730 1000 3600 n = 477.91 pcs say 480 pcs Thus; A = 0.019 57.87 = 0.019

480

(a) L = 2.02 m tube length

0.015 4

1.22

4

73. Supplementary Problem Find the thermal conductivity of the 500 cm thick material with an area of 50,000 cm2 and a temperature difference of 10 K if the heat transmitted during 2 hours test is 2000 KJ. a. 0.014 W/m-K

c. 0.126 W/m-K

b. 0.025 W/m-K

d. 0.214 W/m-K

Solution: Q =

kA∆t x

10,000 k 10 2000 100 2 2 3600 0.500 Thus; (a) k = 0.014 W/m-K

74. Supplementary Problem A pipe with an outside diameter of 2.5 in. is insulated with a 2 in. layer of asbestos (ka = 0.396 Btu-in./hr-ft2-oF), followed by a layer of cork 1.5 in. thick (kc = 0.30 Btuin./hr-ft2-oF). If the the temperature of the outer surface of the cork is 90oF, calculate the heat lost per 100 ft of insulated pipe. a. 847.64 Btu/hr

c. 2847.42 Btu/hr

b. 3847.51 Btu/hr

d. 1847.14 Btu/hr

Solution: Q

Q

Thus;

t 1 ‐t 3 r r2 ln 3 ln r1 r2 2πk a L 2πk c L 290‐90 4.75 3.25 ln ln 3.25 1.25 0.396 0.30 100 2π 100 2π 12 12

c Q 2847.42 Btu/hr

75. Supplementary Problem At $ 0.25 per kW-hr, how much will it cost to maintain a temperature of 95oF for 24 hours in a box 2ft square on each side if the outside temperature is 72oF and the over-all heat transfer coefficient for the box is 0.10 Btu/hr-ft2-oF ? a. P 0.10

c. P 0.15

b. P 0.20

d. P 0.25

Solution: Area of One side: A 2 2 4 ft 2 Heat transferred through one side: Q AU t 1 ‐t 2 Q 4 0.1 96‐72 Q 9.6

Btu hr

Total Heat Transferred through 6 sides in 24 hours: Q 9.6 6 24 Q 1382.4 Btu The Cost to maintain: C

1382.4 $ 0.25 3412.75

Thus; a C $ 0.10


A steam pipe having a surface temperature of 300oC es through a room where the temperature is 25oC. The outside diameter of pipe is 100 mm and emissivity factor is 0.60. Determine the radiated heat loss for a 5 m pipe length. a. 5.34 kW

c. 3.34 kW

b. 4.34 kW

d. 6.34 kW

Solution: Q 20,408.4 x 10‐8 AFe T1 4 ‐T2 4 where: A πDL

T1 300 273 573 K

A π 0.10 5

T2 25 273 298 K

A 1.57 m2 Then; Q 20,408.4 x 10‐8 1.57 0.60 Q 19,208,138

573 4 ‐ 298

4

1hr J hr 3600s

J Q 5,335.59 or W s Thus; a Q 5.336 kW

77. Supplementary Problem An air-cooled condenser has an expected U value of 30 W/m2-K based on the air side area. The condenser is to transfer 60 kW with an airflow rate of 15 kg/s entering at 35oC. If the condensing temperature is to be 48oC, what is the required air-side area? a. 184 m2

c. 174 m2

b. 194 m2

d. 164 m2

Solution:

Q A U θ Solving for θ : θ

max ‐

∆t

∆t ln ∆t

max

∆t

from:

min

min

Q m ∆t 60 15 1 ∆t ∆t 4 K t 2 ‐ t 1 4 K t 2 39 K ∆t

max

48 ‐ 35 13°C

∆t

min

48 ‐ 39 9°C

Then; θ

13 ‐ 9 10.88°C 13 ln 9

60,000 A 30 10.88 Thus; a 183.82 m2

78. Supplementary Problem An air-cooled condenser is to reject 40 kW of heat from a condensing refrigerant to air. The condenser has an air-side area of 210 m2 and a U value based on this area of 0.037 kW/m2-K; it is supplied with 6.6 m3/s of air which has a density of 1.15 kg/m3. If the condensing temperature is to be limited to 55oC, what is the maximum allowable temperature of the inlet air? a. 30.7 oC

c. 50.7 oC

b. 40.7 oC

d. 20.7 oC

Solution: max ‐

∆t

∆t ln ∆t

max

∆t

θ

min

min

Solving for θ : Q A U θ 70 210 0.037 θ θ 9.018 K from: Q m ∆t 70 6.6 1.15

1.02 t 2 ‐ t 1

9.04

Then; 9.01

9.01

9.04 55‐ t 1 ln 55‐ t 2

9.04 55‐ t 1 ln 55‐ 9.04‐ t 1

Thus; b t 1 40.7 °C

79. Supplementary Problem Two walls are composed of 150 mm thick insulating material at the outer layer (k = 0.139 W/m-K) and 300 mm thick material at the inner layer (k = 1.111 W/m-K). Calculate the heat transmitted per m2 if the surface temperature of the cold side and hot side are 25oC and 300oC respectively. a. 203.79 W/m2

c. 254.65 W/m2

b. 303.79 W/m2

d. 354.65 W/m2

Solution: ∆t Q RT A where: RT

0.30 0.15 1.11 0.139

R T 1.35 Q 300‐25 1.35 A Thus; Q a 203.79 W/m2 A

80. Supplementary Problem The temperature of the flame in a furnace is 1277oC and the temperature of its surrounds is 277oC. Calculate the maximum theoretical quantity of heat energy radiated per minute per square meter to the surrounding surface area. a. 19,321.65 kJ

c. 17,321.65 kJ

b. 18,321.65 kJ

d. 16,321.65 kJ

Solution: Q 20408.4 x 10‐8 Fe T1 4 ‐ T2 4 where: Fe = 1 (if not given) T1 = 1277 + 273 = 1550 K T2 = 277 +273 = 550 K Q 20408.4 x 10‐8 1

1550 4 ‐ 550

Q 1.1593 x 109 J/m2-K thus;

4

a Q 19,321.65 kJ/m2-min

81. Supplementary Problem A small sphere has a radius of 3.50 cm and is maintained at a temperature of 360oC. Assuming it to be a black body surrounded by empty space, how much energy does it radiate each second? a. 30 J

c. 40 J

b. 35 J

d. 45 J

Solution: Q 20408.4 x 10‐8 Fe T1 4 ‐ T2 4 where: T1 = 360 + 273 = 633 K A = π( 0.035)2 A = 0.003848 m2 Q 20408.4 x 10‐8 1 0.003848 633

4

Q 126,083.68 J/hr Q 35.02 J/s Thus; b Q 35 J

82. Supplementary Problem The inner wall of a thermos bottle is at 0oC while the outer at 37oC. The space between the walls is evacuated and the walls are silvered so the emissivity is reduced to 0.10. It each wall has an area of 700 cm2, how much energy is transformed by radiation between the walls each second? a. 1.46 J

c. 16.5 J

b. 1.04 J

d. 17.03 J

Solution: Q 20408.4 x 10‐8 Fe T1 4 ‐ T2 4 where: Fe = 0.10 A = 700 cm2 = 0.07 m2 T1 = 37 + 273 = 310 K T2 = 0 + 273 = 273 K Substituting Values: Q = 5,258.116 J/hr Q = 1.46 J/s Thus; (a) Thus; the energy transferred by the radiation between the walls each second is 1.46 J

83. Supplementary Problem The hot gas temperature in a heat exchanger is 350oC (ho = 220 W/m2-K). What is the surface temperature on the wall if the heat transferred is 1500 W/m2? a. 3500C

c. 3430C

b. 3380C

d. 3580C

Solution: Q ho ∆t A 1500 = 220 (350 – t1) Thus; c t 1 343.18o C

84 ME Board Problem An oil heater heats 100 kg per minute of oil from 35oC to 100oC in a counterflow heat exchanger. The average specific heat of the oil is 2.5 kJ/kgoC. Exhaust gases used for heating enter the heater with an average specific heat of 1 kJ/kgoC, a mass flow rate of 250 kg/min and an initial temperature of 200oC. The over-all heat transfer coefficient is 75 W/m2oC. Determine the heating surface in m2. a. 36.110C

c. 32.720C

b. 41.720C

d. 25.340C

Solution: Q A U θ Solving for total heat transferred Q and θ : Qoil = Qgas moo ∆to = mgg ∆tg 100(2.5)(100 – 35) = 250(1)(200 – t) t = 1350C (∆t)max = 135 - 35 = 1000C (∆t)min = 200 – 100 = 1000C If (∆t)max = (∆t)min , then θ = average value = 1000C Q = m∆t Q

100 2.5 100‐35 60

Q 270.83 kW Thus; 270.83 = A (0.075) (100) a A 36.11 m2


If total resistance to heat flow of a composite wall is 3.0875 m2-K/W. What is the over-all transfer coefficient of the wall? a. 0.324 W/m2-K

c. 0.243 W/m2-K

b. 0.423 W/m2-K

d. 0.243 W/m2-K

Solution: U

1 1 3.0875 RT

Thus; (a) U = 0.324 W/m2-K

86. Supplementary Problem In a composite vertical furnace wall, the resistance due to insulating brick is 0.5769 oC/W.

What is the total resistance to heat flow if the percent of the total resistance

due to insulating brick is 96.80%. a. 0.597 0C/W

c. 0.975 0C/W

b. 0.795 0C/W

d. 0.957 0C/W

Solution: % brick

0.9680

R brick RT

0.5769 RT

Thus; (a) RT = 0.597 0C/W

87. Supplementary Problem A counterflow bank of boiler tubes has total area of 900 ft2 and its over-all efficiency of heat transfer is 13 Btu/hr-ft2-oF. Calculate the heat transferred if the log mean temperature difference is 1380oF. a. 16,146,000 Btu/hr

c. 18,148,000 Btu/hr

b. 17,147,000 Btu/hr

d. 15,145,000 Btu/hr

Solution: Q A U θ Q 900 13 1380 Thus; (a) Q = 16,146,000 Btu/hr

88. Supplementary Problem Calculate the quantity to be transferred to 3.25 kg of brass to raise its temperature from 30oC to 250oC taking the specific heat of the brass as 0.394 kJ/kg-K. a. 182 kJ

c. 151 kJ

b. 282 kJ

d. 251 kJ

Solution: Q m ∆t Q 3.25 0.394 250‐30 Thus; (b) Q = 281.712 kJ

89. Supplementary Problem The mass of the copper calorimeter is 0.28 kg and it contains 0.4 kg of water at 15oC. Taking the specific heat of copper as 0.39 kJ/kg-K, calculate the heat required to raise the temperature to 20oC? a. 6.92 kJ

c. 8.92 kJ

b. 7.92 kJ

d. 9.92 kJ

Solution: Water equivalent of calorimeter:

m 0.28

0.39 4.187

m 0.026 kg Heat received by the water and calorimeter Q m mw ∆t Q 0.026 0.40 4.187 20‐15 Thus; (c) Q = 8.92 kJ (d) 90. Supplementary Problem In an experiment to find the specific heat of lead, a 0.50 kg of lead shot at a temperature of 51oC is poured into an insulated calorimeter containing 0.25 kg of water at 13.5oC and the resultant temperature of the mixture is 15.5oC. If the water equivalent of the calorimeter is 0.020 kg, find the specific heat of the lead. a. 0.1278 kJ/kg-K

c. 0.01389 kJ/kg-K

b. 0.0278 kJ/kg-K

d. 0.0389 kJ/kg-K

Solution: Heat lost by the lead = Heat gained by the water and calorimeter mL CL ∆t L mw Cw ∆t w mC CC ∆t C 0.5 CL 51 ‐15.5 0.25 0.02 4.187 15.5‐13.5 thus; a CL 0.1278 kJ/kg-K

∆t

max

55 ‐ t 1

∆t

min

55 ‐ t 2

∆t

max ‐

∆t

min

t 2 ‐ t 1 9.04

91. Supplementary Problem The load on a water-cooled condenser is 90,000 Btu/hr. If the quantity of water circulated through the condenser is 15 gpm, determine the temperature rise of the water in the condenser. a) 12°F

c) 16°F

b) 14°F

d) 18°F

Solution: Q = mΔt 90,000

Btu hr

=m

Btu lb ° F

Δt

Where: m = ρV = 8.33

lb gal

= 7497

15

gal

60 min

min

hr

lb hr

Thus; 90,000 = 7497 1 Δt (a) Δt = 12°F

92. Supplementary Problem Thirty-six gallons of water per minute are circulated through a water-cooled condenser. If the temperature rise of the water in the condenser is 2°F, compute the load on the condenser in Btu/hr. a) 216,000 Btu/hr

c) 217,000 Btu/hr

b) 215,000 Btu/hr

d) 218,000 Btu/hr

Solution: Q = mΔt Where: m = 36

gal min

8.33

= 17992.8 lb/hr Thus:

60 min hr

Q =17992.8(1) (12) (a) Q = 215,913.6 Btu/hr

93. Supplementary Problem The load on an air-cooled condenser is 121,500 Btu/hr. If the desired temperature of the air in the condenser is 25°F, determine the air quantity in cfm that must be circulated over the condenser. a) 4500 cfm

c) 5500 cfm

b) 3500 cfm

d) 3000 cfm

Solution: Q = mΔt Where: Q = 121,500 Btu/hr = 0.24 Btu/°F Δt = 25°F Then: 121,500 = m(0.24)(25) m = 20,250 lb/hr the volume flow rate is: V=

m ρ

=

20,250 lb/hr 0.075 lb/ft³

V = 270,000 ft3/hr = 4500 ft3/min Thus; (a) V = 45000 cfm

94. Supplementary Problem Three thousand cubic feet per minute of air are circulated over an air-cooled condenser. If the load on the condenser iss 64,800 Btu/hr , compute the temperature rise of the air ing over the condenser. a) 16°F

c) 20°F

b) 18°F

d) 22°F

Solution: Q = mΔt Where: Q = 64, 800 Btu/hr m = ρV m = 0.075 lb/ft3 (3000 ft3/min) (60min/hr) m = 13,500 lb/hr = 0.24 Btu/lb °F Thus; 6400 = 13,500 (0.24) Δt (c) Δt = 20°F

95. Supplementary Problem The weight of ammonia circulated in a machine is found to be 21.8 lb/hr. If the vapor enters the compressor with a specific volume of 9.6 ft3/lb , calculate the piston displacement, assuming 80% percent volume efficiency. a) 261.6 ft3/hr

c) 281.8 ft3/hr

b) 271.6 ft3/hr

d) 291.6 ft3/hr

Solution: Actual volumetric efficiency =

Volume flow rate at suction Piston displacement

eva = V1/VD 0.80 =

21.8(9.6) VD

VD = 261.6 ft3/hr Thus; (a) VD = 261.6 ft3/hr

96. Supplementary Problem A single-stage ammonia compressor is producing 10 tons of refrigeration and the power consumed is 15 Hp. Suction pressure is 25 psi, condensing pressure is 180 psi. Brine temperature is 20°F off brine cooler. Determine the actual coefficient of performance.

a) 10.14

c) 12.14

b) 11.14

d) 13.14

Solution: COP = =

Refrigeration Capacity compressor power 10 3.516 15 0.746

= 13.14 Thus; (d) COP = 13.14

97. Supplementary Problem In an ammonia condensing machine (compressor plus condenser) the water used for condensing is 55°F and the evaporator is at 15°F. Calculate the ideal COP. a) 11.875

c) 10.875

b) 12.875

d) 13.875

Solution: COP =

T1 T2 ‐ T1

Where: T1 = 15 + 460 = 475 °R T1 = 55 + 460 = 515 °R Then; COP =

475 515 ‐ 475

Thus; (a) COP = 11.875

1. Supplementary Problem Calculate the specific volume of an air-vapor mixture in cubic meters pre kilogram of dry air when the following conditions prevail : t = 30°C, w = 0.015 kg/kg, and Pt = 90 kPa. 0.99 m3/kg

c) 0.79 m3/kg

b) 0.89 m3/kg

d) 0.69 m3/kg

a)

Solution: v=

RaT Pt ‐ Pv

Solving for Pv: w = 0.622

Pv Pt ‐ Pv

0.015 = 0.622

Pv 90 ‐ Pv

Pv = 2.12 kPa Thus; v=

(0.287) (30+273) 90 - 2.12

(a) v = 0.99 m3/kg

2. Supplementary Problem Compute the Humidity Ratio of air at 62 % relative humidity and 34°C when the barometric pressure is 101.325 kPa. a) 0.021 kgvapour/kgdry air

c) 0.041 kgvapour/kgdry air

b) 0.031 kgvapour/kgdry air

d) 0.051 kgvapour/kgdry air

Solution: w = 0.622

Pv Pt ‐ Pv

Solving for Pv: Psat @ 34°C = 5.32 kPa Pv = (RH) Psat

= (0.62) (5.32) = 3.30 kPa Then: w = 0.622

3.30 101.325 ‐ 3.30

Thus; (a) w = 0.021 kgvapour/kgdry air

3. Supplementary Problem A sample of air has dry-bulb temperature of 30°C and a wet-bulb temperature of 25°C. The barometric pressure is 101.325 kPa. Calculate the enthalpy of the air if it is adiabatically saturated. a) 75.94 kJ/kg

c) 79.54 kJ/kg

b) 70.94 kJ/kg

d) 74.09 kJ/kg

Solution: Enthalpy of air if adiabatically saturated: h = t + whg Solving for humidity ratio, w: w = 0.622

Pv Pt ‐ Pv

From steam table, at 25°C : Pv = 3.17 kPa, hg = 2547.2 kJ/kg w2 = 0.622

3.17 101.325‐ 3.17

= 0.02 kgvapour/kgdry air Thus; h2 = 1.0 kJ/kg-°C (25°C) + (0.02) (2547.2 kJ/kg) (a) h2 = 75.94 kJ/kg


An air-vapor mixture has dry bulb temperature of 30°C and a humidity ratio of 0.015. Calculate the enthalpy at 85 kPa barometric pressure. a) 68.34 kJ/kg

c) 72.45 kJ/kg

b) 54.35 kJ/kg

d) 67.45 kJ/kg

Solution: h = t + whg From steam table, at 30°C : h = 2556.3 kJ/kg then, h = (1 kJ/kg-°C) (30°C) + (0.015) (2556.3 kJ/kg) Thus; (a) h = 68.34 kJ/kg

5. Supplementary Problem In an air conditioning unit, 3.5 m3/s of air 27°C dry-bulb temperature 50 % relative humidity and standard atmospheric pressure enters the unit. The leaving condition of the air is 13°C dry-bulb temperature and 90% relative humidity. Using the properties from the psychrometric chart, calculate the refrigerating capacity in kW. a) 87.57 kW

c) 57.87 kW

b) 77.57 kW

d) 58.77 kW

Solution: Refrigerating Capacity QA: QA = m (h2 - h1) Solving for m: m= =

Volume Flow Rate Ave. Specific Volume 3.5 1/2(0.866 ‐ 0.822)

= 4.15 kg/s Thus; QA = 4.15 (55.3 - 34.2) (a) QA = 87.57 kJ or kW

6. Supplementary Problem A stream of outdoor air is mixed with a stream of return air in an air conditioning system that operates at 101 kPa pressure. The flow rate of outdoor system air is 2 kg/s and its condition is 35°C dry-bulb temperature and 25°C wet-bulb temperature. The flow rate of return air is 3 kg/s and its condition is 24°C and 50% relative humidity. Determine the enthalpy of the mixture. a) 91.56 kJ/kg

c) 91.56 kJ/kg

b) 91.56 kJ/kg

d) 91.56 kJ/kg

Solution: By Energy Balance: m1h1 + m2h2= m3h3 m1h1 + m2h2 = ( m1+ m2 ) h3 h3 =

2 (75.9) 3 (48) 2+3

Thus; (d) h3 = 59.16 kJ/kg


What is the specific volume of an air-vapor mixture at 30°C and a relative humidity of 45°C at 101.325 kPa. a) 0.578 m3/kg

c) 0.875 m3/kg

b) 0.785 m3/kg

d) 0.758 m3/kg

Solution: v=

RaT Pt ‐ Pv

Solving for Pv : Psat @ 30°C = 4.24 kPa Pv = (RH) Psat @ 30°C = 0.45 (4.24) = 1.91 kPa Thus; v=

0.287(30+273) 101.325 - 1.91

(c) v = 0.875 m3/kg

8. Supplementary Problem A mixture of dry-air and water vapor is at temperature of 21°C under a pressure of 101 kPa. The dew point temperature is 15°C. Calculate the relative humidity. a) 68.56 %

c) 56.68 %

b) 65.68 %

d) 58.66 %

Solution: RH =

Pv Psat @ 21°C

Where: PV = Psat @ 15°C = 1.7044 kPa Psat @ 21°C = 2.4861 kPa Thus;

RH =

1.7044 2.4861

= 0.6851

(b) RH = 68.56% 9. Supplementary Problem The density of air at 35°C and 101 kPa is 1.05 kg/m3. The humidity ratio is: a) 0.036 kgvapour/kgdry air

c) 0.36 kgvapour/kgdry air

b) 0.063 kgvapour/kgdry air

d) 0.63 kgvapour/kgdry air

Solution: PV = mRT P = mRT/V Pair = 1.08 ( 0.287 )( 35 + 273 ) Pair = 95.48 kPa Pt = Pair + Pvapor 101 = 95.48 + Pv Pv = 5.53 kPa Then; w = 0.622

= 0.622

Pv Pt ‐ Pv 5.53 101 ‐ 5.53

Thus; (a) w = 0.036 kgvapour/kgdry air


If the sensible heat ratio is 0.80 and the cooling load is 100 kW, what is the amount of sensible heating? a)

80 kW

c) 125 kW

b) 60 kW

d) 100 kW

Solution: SHR = sensible heat ratio =

Qs Qs QR

0.80 = Qs/100 Thus; (a) Qs = 80 kW

11. Supplementary Problem A 4m x 4m x 4m room has a relative humidity ratio of 80 %. The pressure in the room is 120 kPa and temperature is 35 °C (Psat = 5.628). What is the mass of vapor in the room. Use Rvapor = 0.4615 kN-m/kg-K. a) 3.03 kg

c) 4.03 kg

b) 2.03 kg

d) 5.03 kg

Solution: PvV = mvRvT Solving for Pv : RH = Pv / Psat 0.08 = Pv / 5.628 Pv = 4.5024 kPa Thus; 4.5024 [(4)(4)(4)] = mv (0.4615) (35 + 273) (a) mv = 2.027 kg

45. The bore and stroke of an air compressor are 276 mm and 164 mm respectively. If the piston displacement is 0.039 m /s, what is the operating speed of the compressor? a. 238.49 rpm

c. 338.49 rpm

b. 261.54 rpm

d. 361.54 rpm

Solution: Vd =

2

πd L N 4

0.039 =

2

π(0.276) 0.164 N 4

N = 3.975 rev/s = 238.49 Thus; (a) N = 238.49 rpm 46. A turbo-compressor is a gas turbine plant is used to compress 10 kg/s air from an initial pressure of 102 kpa to a discharge pressure of 622 kpa, with inlet and discharge temperature measured at 297 K and 527 K respectively. The compressor inlet pipe is 50 cm ID and the discharge pipe is 20 cm ID. Find the inlet discharge velocities of air. a. 42.56 m/s, 77.40 m/s

c. 45.45 m/s, 57.45 m/s

b. 34.72 m/s, 76.56 m/s

d. 54.45 m/s, 56.7 m/s

Solution: Velocity of air at suction: V

Q s= s As

Velocity of air in at discharge: V

d=

Qd Ad

solving for Q: Qs= V

mRT1 P1

s =

10 (.287)(297) 102 =8.357 m3 /s Qd= V

d=

=

mRT2 P2

10 (.287)(527) 622

2.43 m3 /s then; V

s=

8.357 m3 /s =42.56 m/s π 2 (.5) 4

V

2.43 m3 /s d= π =77.40 m/s 2 (.20) 4

Thus; (a) Vs=42.56m/s ,

Vd=77.40 m/s

47. The initial condition of air in an air compressor is 100 kpa and 25

and

discharges air at 450 kpa. The bore and stroke are 276 mm and 186 mm respectively with 8 % clearance running at 6 rev per second. Find the volume of air at suction. a. 203.39 m /hr

c. 261.25 m /hr

b. 303.39 m /hr

d. 361.25 m /hr

Solution: Solving for e : ev =1+c-c (

P1 1/n ) P2

=1+0.08-0.08 (

450 1/1.4 100

)

=0.84575 Vd= πD2 LN= π(0.276)2 (0.186)(6) 4

4

=0.0668 m3 /s Then; V1=0.84575(0.0668 ) =203.39 m3 /hr

48. If the power to drive shaft is 7 hp and the mechanical efficiency is 75 %, what is the actual compressor power? a. 5 hp

c. 2 hp

b. 3 hp

d. 4 hp

Solution: e

e

Compressor power Power to drive the shaft

0.74 =

Compressor power 7

Thus; (a) Compressor Power = 5.18 hp

49.A two stage compressor receives 0.35 kg/s of air at 100 kpa and 629 K and delivers it at 1000 kpa. Find the heat transferred in the intercooler? a. 70.49 kw

c. 90.49 kw

b. 80.49 kw

d. 100.49 kw

Solution: Q

m (Tx - T1

Solving for Tx : Px =

Tx T1

=(

Tx 269

100 5000 = 707.11 kpa

Px P1

=(

)

k-1 k

707.11 100

)

1.4-1 1.4

Tx =470.40 K

Thus; Q=0.35(1)(470.40 K- 269 K) Q=70. 49 kw 50.An air compressor which operates at 900 rpm has a piston displacement of 4500cm . Determine the mass flow rate of air standard density considering that the volume efficiency is 77 %. a. 224. 53 kg/hr

c. 314. 57 kg/hr

b. 324.35 kg/hr

d. 137.54 kg/hr

Solution: ev =

V1 ' VD

0.77=

V1 ' 4500

V'1 =3465 cm3 Then; V'1 =3465 cm3 900 =3,118,500 cm3 /min At standard temperature and pressure: γair =1.2 kg/m3 m=1.2

kg cm3 1m3 60 min 3118500 3 min (100)3 cm3 hr m

Thus; (a) m=224.53 kg/hr

1.A fuel is delivering 10 gallons per minute of oil with a specific gravity of 0.83. The total head is 9.14m; find how much energy the pump consumes in KJ per hour. a. 169

c. 189

b. 199

d. 179

Solution; P= γQH Where: γ=9.81 .083 =8.14 KN/m3 Q=10 gal/ min =2.27 m3 /hr H=9.14 m Then; P=8.14 2.27 9.14 Thus; (a) P=168.89

kj hr

2. A pump receives 8 kgs of water at 220 kpa and 110

and discharges it at 1100 kpa.

Compute the power required in kw. a. 8.126

c. 7.014

b. 5.082

d. 6.104

Solution: Let : P=power in kw P=Q(Pd - Ps ) Where: Q= 8

kg s

0.001

m3 = 0.008 m3 /s kg

Pd =1,100 kpa

P=(0.008)(1100-220) Thus; (b) P=7.04 kw

3.A pump lifts water at a rate of 283 liters per second from alake and force it into a tank 8 m above the level of the water at a pressure of 137 kpa. What is the power required in kw? a. 71

c. 61

b. 41

d. 51

Solution: P= γQH Where: γ=9.81 kN/m3 Q=

283L =0.283 m3 /s s

H=8+

137 9.81

H=21.97 Thus; P=9.81 0.283 21.97 (c) P=60.99

4.A pump discharges 150 liters per second of water to a height of 75 m. if the efficiency is 75 % and the speed of the pump is 1800 rpm, what is the torque in the N-m to which the drive shaft is subjected? a. 771

c. 791

b. 781

d. 681

Solution: Let: ep =pump efficiency P=

γQH 2πTN or P= ep 60

Where; N=1800 rpm P=

γQH 9.81(0.150)(75) = =147.15 kw ep 0.75

Then; 147.15=

2πT(1800) 60

T=0.781 kN-m Thus; (b) T=781 N-m

5.A centrifugal pump delivers 80 liters per second of water on test suction gauge reads 10 mm hg vacuum and 1.2 m below pump center line. Power input is 70 kw. Find the total dynamic head in meters. a. 66

c. 62

b. 60

d. 64

Solution:

γ=9.81 kN/m3 Q=0.80 m3 /s P=0.74 70 =51.80 Note: 74% is the usual pump efficiency used if not given. Thus; 51.80=9.81 0.80 H (a) H=66 m

6.A pump with a 400 mm diameter suction pipe and a 350 mm diameter discharge pipe is to deliver 20,000 liters per minute of 15.6

water. Calculate the pump head in meters

if suction gage is 7.5 cm below the pump centerline and reads 127 mm Hg vacuum and discharges gage is 45 cm above the pump centerline and reads 75 kpa. a. 15 m

c. 20 m

b. 5 m

d. 10 m

Solution: H=total dynamic head H=

Ps -Ps (Vd 2 -Vs 2 ) + +Zd -Zs 2g γ

Q=20,000

Vs =

Q 2

li =0.33 m/s min =

0.33 2

πd /4 π(0.4) /4

=2.63 m/s

Vd =

0.33 2

π(0.35) /4

=3.43 m/s

Thus:

H=

75-(-16.93) (3.432 -2.632 ) + +0.45+0.075 9.81 2(9.81) (d) H=10.14 m

7.A centrifugal pump delivers 300,000 liters per hour of water to a pressurized tank whose pressure is 284 kpa. The source of water is 5 meters below the pump. The diameter of the suction pipe is 300 mm and the discharge pipe is 250 mm. calculate the kw rating of the driving motor assuming the pump efficiency to be 72%. a. 41.75 kw

c. 43.28 kw

b. 35.23 kw

d. 38.16 kw

Solution: Let: Pbrake =brake input power Pbrake =

γQH ep

Where: Q=300,000

li =0.0833m3 /s hr

Solving for H: From: Bernoulli’s Equation:

H=


where: Vs =

V

0.0833 2

π(0.3) /4

=1.18 m/s

0.0833 π 0.25 /4

1.7 m/s

Then; 2

2

280-0 (1.70) -(1.18) + +5-0 H= 2(9.81) 9.81 H=33.62 m Thus; Pbrake =

9.81(0.0833)(33.62) 0.72

(d) Pbrake = 38.16 kw 8.A pump delivers 500 gpm of water against a total head of 200 ft and operating at 1770 rpm. Changes have increased the total head to 375 ft. at what rpm should the pump be operated to achieved the new head at the same efficiency. a. 2800 rpm

c. 3434 rpm

b. 3600 rpm

d. 2424 rpm

Solution: H1 N1 2 =( ) H2 N2 200 375

1770 N

Thus; (e) N2 =2423.67 rpm 9.The rate of flow of water in a pump installation is 60.6 kg/s. the intake static gage is 1.22 m below the pump centerline and reads 68.95 kpa gage ; the discharge static gage is 0.61 m below the pump centerline and reads 344.75 kpa gage. The gages are located lose to the pump as much as possible. The areas of the intake and discharge pipes are 0.093 m and 0.069 m respectively. The pump efficiency is 74 %. Take the density of water equals 1000 kg/m . What is the hydraulic power in kw? a. 17.0

c. 31.9

b. 24.5

d. 15.2

Solution: Pwater =Phydraulic =γQH Where: Q=

60.6 kg/s 1000 kg/m3

Q=0.0606 m3 /s Vs =

0.0606

Vd =

0.0606

0.093

0.069

=0.65 m/s =0.88 m/s

H=


H=

344.75-68.95 (0.88) -(0.65) ) + +(-0.61+1.22) 9.81 2(9.81)

2

H=28.74 m Thus the hydraulic power is: Phydraulic = 9.81 (0.0606) (28.74) Phydraulic =17.09 kW

2

10.It is desired to deliver 5 gpm at a head of 640 ft in a single stage pump having a specific speed not to exceed 40. If the speed is not exceeding 1352 rpm how many stages are required? a. 3

c. 5

b. 4

d. 2

Solution: Let; n=no. of stages h=head per stage then, h=

640 n

from; Ns =

40=

N√Q h3/4 1352√5 640 3/4 ( n )

Thus, n=2 stages 11. The power output is 30 Hp to a centrifugal pump that is discharging 900 gpm and which operates at 1800 rpm against a head H = 120 ft, 220 V, 3 phase, 60 Hertz. If this pump is modified to operate 1200 rpm, assuming its efficiency remains constant, determine its discharge in gpm, the theoretical head it imparts to the liquid and the power input to the pump. a. 600 gpm, 53.33 ft, 8.89 Hp

c. 500 gpm, 50.33 ft,7.89 Hp

b. 700 gpm, 63.33 ft, 9.89 Hp

d. 650 gpm, 53.33 ft, 8.95 Hp

Solution: Solution for the discharge, Q :

Q1 N1 = Q2 N2 900 1800 = Q2 1200 Q2 =600 gpm Solution for the theoretical head, H : H1 N1 2 =( ) H2 N2 120 1800 2 =( ) 1200 H2 H2 =53.33 ft Solving for the Power input, P : P1 N1 3 =( ) P2 N2 3

30 1800 =( ) P2 1200

P2 =8.89 Hp Thus; (a) Q2 =600gpm , H2 =53.33 ft , P2 =8.89 Hp

12. A pump operating at 1750 rpm delivering 500 gal/min against a total head of 150 ft. Changes in the piping system have increased the total head of 360 ft. At what rpm should the pump be operated to achieve this new head at the same efficiency? a. 2730 rpm

c. 2711 rpm

b. 2740 rpm

d. 2600 rpm

Solution:

H1 N1 2 =( ) H2 N2 2

150 1750 =( ) 360 N2 Thus;

N2 =2711.09 rpm

13. Water in the rural areas is often extracted from underground water source whose free surface is 60 m below ground level. The water is to be raised 5 m above the ground by a pump. The diameter of the pipe is 10 cm at the inlet and 15 cm at the exit. Neglecting any heat interaction with the surroundings and frictional heating effects. What is the necessary power input to the pump in kW for a steady flow of water at the rate of 15 li/s? Assume pump efficiency of 74 %. a. 9.54

c. 7.82

b. 5.54

d. 12.90

Solution: Input Power=

Input Power=

where:

Water Power Pump Efficiency

γQH ep

Q = 15 li/s Q = 0.05 m3 Vs =

0.015 2

π(0.10) /4

Vs =1.91 m/s Vd =

0.015 2

π(0.015) /4

Vd =0.85 m/s H=

Vd 2 -Vs 2 +Zd -Zs 2g

H=

(0.85) -(1.91) + 5-(-60) 2(9.81)

2

2

H = 64.85 Thus; Input power =

9.81(0.015)(64.85) 0.74

(d) Input Power = 12.896 kW

14. Past Board Problem Find the hydraulic horsepower and the mechanical efficiency of a rotary pump direct connected to a 5 Hp electrical motor operating at full load under the following conditions: Fluid handled

oil

Temperature

21°C

Specific gravity

0.85

Volume flow rate

20 li./sec

Total Head

175 kPa

a. 4.69 Hp , 94%

c. 5.69 , 74%

b. 3.69 Hp , 84%

d. 6.69 , 78%

Solution: Hydraulic Power of the Pump : Phydraulic = γ Q H where: γ = 9.81(0.85) = 8.34 kN/m3 Q = 20 li/s = 0.020 m3/s H = =

.

/

H = 20.98 m then ; Phydraulic = (8.34) (0.020) (20.98) = 3.50 kW Phydraulic = 4.69 Hp

ep =

.

=

ep = 0.94 = 94% thus; (a) Phydraulic = 4.69 Hp , ep = 94% 15. Past Board Problem Water from an open reservoir A at 8 m elevation is drawn by a motor driven pump to an open reservoir B at 70 m elevation. The inside diameter of the suction pipe is 200 mm and 150 mm for the discharge pipe. The suction line has a loss of head three times that of the velocity head in the 200 mm pipe. The discharge line has a loss of head twenty times that of the velocity head in the discharge pipeline. The pump centerline is at 4 m. Overall efficiency of the system is 78 %. For the discharge rate of 10 li/s, find the power input to the motor and the pressure gage readings installed just at the outlet and the inlet of the pump in kPag. a. 7.82 kW, 39 kPa, 650 kPa

c. 6.82 kW, 35 kPa, 550 kPa

b. 8.82 kW, 40 kPa, 680 kPa

d. 5.82 kW, 30 kPa, 500 kPa

Solution: Power input of the motor: Pinput =

.

=

.

where: Q = 0.010 m3/s Vd =

. .

/

Vd = 0.565884242 m/s .

Vs =

.

/

Vs = 0.318309886 m/s H=

+

+Z

Z

.

H=0+

.

+ 20

.

.

.

]+3

.

.

] + 64 – 4

H = 62.3530768 m thus; .

Pinput =

.

.

.

Pinput = 7.842098505 kW Pressure Gages Readings: Ps = γ Hs = 9.81(3.99) Ps = 39.14 kPa Pd = γ Hd = 9.81(66.34) Pd = 650.80 kPa

16. Supplementary Problem A pump adds 167.6 m of pressure head to 45.43 kg/s of water. What is the hydraulic power in kW. a. 64.69 kW

c. 66.54 kW

b. 74.69 kW

d. 76.54 kW

Solution: P=γQH where: Q=

.

/ /

= 0.04543 m3/s

P = 9.81 (0.04543) (167.6) thus; (b) P = 74.69 kW

17. Supplementary Problem A pump driven by an electrical motor moves 25 gal/min of water from a reservoir A to B, lifting the water to a total head of 245 ft. The efficiency of the pump is 64%. Neglecting velocity head, friction, and minor losses. What size motor is required? a. 2.64 Hp

c. 1.55 Hp

b. 2.55 Hp

d. 1.64 Hp

Solution: P=γQh = (8.33 lb/gal) (25 gal/min.) (245 ft) = 51,021.25

,

/

thus; (a) P = 1.55 Hp 18. Supplementary Problem A centrifugal pump is powered by a direct drive induction motor is needed to discharge 150 gal/min against a total head of 180 ft when turning at fully loaded speed of 3500 rpm. What type of pump should be selected? a. Radial

c. Mixed Flow

b. Francis

d. Propeller

Solution: Ns = =

/

√ /

= 872.286 rpm thus; (a) Radial turbine

19. Past ME Board Problem A boiler feed pump receives 40 L/s at 4 MPa and 180°C. It operates against a total head of 900 m with an efficiency of 60%. Determine the power output of the driving motor. a. 450.21 kW

c. 500.21 kW

b. 459.64 kW

d. 523.26 kW

Solution: Pmotor = Solving for Pwater: From Steam table: At 4 Mpa and 180°C h1 = 764.76 kJ/kg v1 = 0.00112484 m3/kg ρ=

= 889.015 kg/m3

.

= 8721.24 N/m3 = 8.721 kN/m3 Pwater = γ Q h = (8.721 kN/m3) (0.040 m3/s) (910 m) = 313 956 kW thus; Pmotor =

. .

(a) Pmotor = 523.26 kW

20. Past ME Board Problem A submersible, multi-stage, centrifugal deep well pump 260 gpm capacity is installed in a well 27 ft below the static water level. Drawdown when pumping at rated capacity is 10 ft. the pump deliverd that water into a 25,000 gallons capacity overhead storage tank. Totaldischarge head developed by pump including friction in piping is 243 ft. Calculate the brakepower required to drive the pump if pump efficiency is 70%. a. 15.86 Hp

c. 30.16 Hp

b. 21.22 Hp

d. 10.52 Hp

Solution: Pbrake = Solving for Pwater: Q = 260

.

= 34.76 ft3/min. H = 243 – (27 – 10) = 226 ft Pwater = (62.4

) (34.76

= 490,199.42 Pwater =

.

) (226 ft)

,

/

.

thus; Pwater = 21.22 Hp 21. Past ME Board Problem A boiler feed pump receives 45 li./s of water at 190°C and enthalpy of 839.33 kJ/kg. It operates against a head of 952 m with efficiency of 70%. Estimate the water leaving temperature assuming that the temperature rise is due to the inefficiency of the input energy.

a. 190.96 °C

c. 199.71 °C

b. 194.66 °C

d. 200 °C

Solution: m C

∆t

=

∆

- m ∆h

4.187 t – 190) =

.

(h – 839.33)

Solving for h : m (h - h ) = 0.00981 (952) h = 848.67 kJ/kg then; .

4.187 (t – 190) =

.

.

– (848.67 – 839.33)

(a) t = 190.96 °C 22. Past ME Board Problem A pump is driven by an electric motor moves 25 gal/min of water from reservoir A to reservoir B, lifting the water to a total of 245 ft. The efficiency of the pump and motor are 64% and 84% respectively. What size of motor in Hp is required? a. 3 Hp

c. 2 Hp

b. 5 Hp

d. 7 Hp

Solution: Pmotor = Solving for Pwater: Q = 25 gal/min = 0.0557 ft /s Pwater = (62.4 = 851.54 ft-lb/s

) (0.0557

/

) (245 ft)

Pwater = 1.55 Hp Pbrake =

=

. .

= 2.42 Hp thus; (a) Pmotor = 2.88 Hp or 3 Hp 23. Past ME Board Problem A vacuum pump is used to drain a flooded mine shaft at 20°C water. The pump pressure of water at this temperature is 2.34 kPa. The pump is incapable of lifting the water higher than 10.16 m. What is the atmospheric pressure? a. 90.21 kPa

c. 102.01 kPa

b. 96.02 kPa

d. 108.01 kPa

Solution: From Bernoulli’s Theorem:

= .

+

+Z =

=

+ . .

+

+Z

+ Z

Z

+ 0 + 10.16

(c) thus; P = 102.01 kPa 24. Past ME Board Problem Water reservoir is pumped over a hill through a pipe 45 mm in diameter and a pressure of 1 kg/cm (98.08 kPa) is maintained at the summit. Water discharge is 30 m above the reservoir. The quantity pumped is 0.50 m /s. Frictional losses in the discharge and suction pipe of the pump is equivalent to 1.5 m head loss. The speed of the pump is 800 rpm, what amount of energy must be furnished by the pump in kW? a. 206 kW

c. 250 kW

b. 210 kW

d. 245 kW

P=γQH Solving for H: H=

.

=

.

+

+

. / .

+Z .

Z + h

h

+ 30 +1.5

h = 42 m then; P = 9.81 (0.50) (42) thus; (a) P = 206 kW 25. Past ME Board Problem A pump is to deliver 80 gpm of water at 140°F with a discharge pressure of 150 psig. Suction pressure indicates 2 in. of mercury vacuum. The diameter of suction and discharge pipes are 5 in. and 4 in. respectively. The pump has efficiency of 70%, while the motor efficiency is 80%. Determine the power input to the drive motor. a. 12.59 Hp

c. 15.590 Hp

b. 10.59 Hp

d. 20.59 Hp

Solution: Pmotor = Pbrake = Solving for Pwater: Q = 80 gal/min = 0.1782 ft /s Vs = Vd =

. / / . / /

= 1.307 ft/s = 2.043 ft/s

From steam table: At 150 psig (164.7 psi) and 140°F: γ = 61.424 lb/ft h=

+

.

=

.

.

+

.

. .

= 354 ft then; Pwater = 61.424 (0.1782) (354) = 7.05 Hp

= 3874.80 =

P

.

= 10.07 Hp

.

Pmotor =

. .

thus; (a) Pmotor = 12.59 Hp 26. Past ME Board Problem Determine the water horsepower of a centrifugal water pump which has an input of 3.5 Hp if the pump has an 8 in. nominal size suction and 6 in. nominal size discharge and handles 150 gpm of water of 150°F. The suction line gage shows 4 in. Hg vacuum and the discharge gage shows 26 psi. The discharge gage is located 2 ft above the center of the discharge pipe line and the pump inlet and discharge lines are at the same elevation. a. 2.52 Hp

c. 2.78 Hp

b. 3.52 Hp

d. 3.78 Hp

Solution: P=γQH Solving for h:

Q = 150 gal/min = 0.334 ft /s

Vs =

= 0.957 ft/s

Vd =

= 1.701 ft/s

From steam table: At 150°F:

γ= H=

= 61.2 lb/ft

.

+

.

=

.

.

+

.

. .

= 67.83 ft thus; P = (61.2) (0.334) (67.83) = 1386.50 ft-lb/s (a) P = 2.52 Hp 27. Past ME Board Problem Water from an open reservoir A at 8 m elevation is drawn by a motor-driven pump to an open reservoir B at 70 m elevation. The inside diameter of the suction pipe is 200 mm and 150 mm for the discharge pipe. The suction line has a loss of head three times that of the velocity head in the 200 mm pipe. The discharge line has a loss of head 20 times that of the velocity head of the discharge pipeline. The pump centerline is at 4 m. Overall efficiency of the system is 78%. For a discharge rate of 10 li./s, find the power input to the motor. a. 10.06 kW

c. 6.12 kW

b. 4.80 kW

d. 7.85 kW

Solution: Pinput = :

Solving for P

Vs =

=

. .

= 0.318 m/s Vd =

=

. .

= 0.566 m/s .

=3

3

h

.

= 0.01546 m 20

h

= 20

.

+ Z

.

= 0.32642 m

H= =

+

.

. .

= 62.35 m

Z

+ h

h

+ (66 – 4) + (0.01546 + 0.32642)

then; = 9.81 (0.010) (62.35)

P

= 6.12 kW thus; Pinput =

. .

(d) Pinput = 7.85 kW 28. Past ME Board Problem A double suction, single stage, cenyrifugal pump delivers 900 m3/hr of seawater (S.G. = 1.03) from a source where the water level varies 2 m from high tide to low tide level. The pump centerline is located 2.6 m above the surface of the water at high tide level. The pump discharges into a surface condenser, 3 m above pump centerline. Loss of head due to friction in suction pipe is 0.80 m and the discharge side is 3 m. Pump is directly coupled to a 1750 rpm, 460 V, 3 phase, 60 Hz motor. Calculate the specific speed of pump in rpm. a. 5,149.20 rpm

c. 5,500 rpm

b. 6,149.20 rpm

d. 6,500 rpm

Solution: Ns =

/

Solving for H: hs = 2 + 2.6 + 0.8 = 5.4 m = 17.72 ft hd = 3 + 3 = 6 m = 19.69 ft H = 17.72 + 19.69 = 37.41 ft Q1 = Q2 = 900/2 = 450 m3/hr = 1981 gal/min then; Ns =

√ .

/

thus; (a) Ns = 5,149.20 rpm 29. Past ME Board Problem A DC driven pump running at 100 rpm delivers 30 liters per second of water at 40°C against a total pumping head of 27 m with a pump efficiency of 60%. Barometer pressure is 758 mm Hg abs. What pump speed and capacity would result if the pump rpm were increased to produce a pumping head of 36 m assuming no change in efficiency. a. 115.47 rpm , 34.64 L/s

c. 110.51 rpm , 34.46 L/s

b. 115.47 rpm , 38.68 L/s

d. 110.51 rpm , 38.68 L/s

Solution: New Speed required:

N N

H H

N 100

36 27

N = 115.47 rpm New Capacity Required: Q Q

N N

Q 30

115.47 100

Q = 34.64 L/s thus; (a) N = 115.47 rpm , Q = 34.64 L/s 30. Supplementary Problem A centrifugal pump discharged 20 L/s against a head of 17 m when the speed is 1500 rpm. The diameter of the impeller was 30 cm and the brake horsepower was 6.0. A geometrically similar pump 40 cm in diameter is to run at 1750 rpm. Assuming equal efficiencies, what brake horsepower is required?

a. 51.55 HP

c. 40.15 HP

b. 50.15 HP

d. 45.15 HP

Solution: New brake horsepower required: P

D N

P D N

6 0.30 1500

P 0.40 1750

thus; (c) P = 40.14 Hp 31. Supplementary Problem A two-stage centrifugal pump delivers 15,000 kg/hr of 110°C water against 76 m head at 3500 rpm. What is the specific speed of the pump? Solution: Ns =

.

/

Solving for Q: ,

Q= =

,

@

°

.

Q = 0.004383 m3/s H = 76/2 = 38 m then; Ns =

√ . .

/

= 780.39 rpm

thus; (a) Ns = 780.39 rpm

32. Supplementary Problem Calculate the impeller diameter of the centrifugal pump that requires 15m head to deliver if pump speed is 1500 rpm.

a. 218.43 mm

c. 345.75 mm

b. 300.75 mm

d. 276.45 mm

Solution: V= Solving for V: V= 2 = 2 9.81 15 = 17.15 m/s then; 17.5 = D = 0.21843 m thus; (a) D = 0.21843 m

33. Supplementary Problem A large centrifugal pump has a 254 mm diameter inlet and a 127 mm diameter outlet pipe. The measured flow rate is 51.6 L/s of cold water. The measured inlet pressure is 127 mmHg above atmospheric and discharge pressure measured at a point 1.22 m above the pump outlet is 212 kPa abs. The pump input is 10 Hp. Find the pump efficiency. a. 78.51 %

c. 74.54 %

b. 70.62 %

d. 76.77 %

Solution: epump = Pwater / Pinput Solving for Pwater : Vs = Q / As = 0.0516 / /4 0.254 Vs = 1.018 m/s Vd = Q / Ad =

. .

= 4.073 m/s H=

=

212‐101.325 ‐127

101.325 760

9.81

4.073 2 ‐ 1.018 2 2 9.81

1.22

= 11.57 m Pwater =9.81(0.516)(11.57) =5.86 Kw=7.85Hp thus; epump = 7.85/10 (a) epump = 0.785 = 78.51% 34. Supplementary Problem A closed tank contains liquefied butane gas whose specific gravity is 0.60. The tank pressure us 1.7 Mpa gage which is also the equilibrium vapor pressure of butane at the pumping temperature. Suction line losses is 1.5 m of gas and the static elevation gain is 4m. What is the Net Positive Suction Head available (NPSH)? a. 2.5 m

c. 3.5 m

b. 1.5 m

d. 1.7 m

Solution:

Use (+) for static elevation gain. NPSH =

.

= 0 + 4 – 1.5 Thus; (a)

NPSH= 2.5 m available

35. Supplementary Problem A condensate pump at sea level take water from a surface condenser where the vacuum is 15 in. of mercury. The friction and turbulence in the piping in the condenser hot well and the pump suction flange is assumed to be 6.5 ft. If the condensate pump to be installed has a required head of 9 ft, what would be the centerline of the pump to avoid cavitation? a. 2.5 ft

c. 18 ft

b. 15.5 ft

d. 5.5 ft

Solution: The Net Positive Suction Head required by the pump is 9. then; NPSH =

.

9 = 0 + S – 6.5 Thus; (b) S = 15.5 ft 36. Supplementary Problem A boiler feed pump receives 130 cfm of water with specific volume of 0.0025 m3 at a head of 800 m. If the pump efficiency is 64%, what is the output of the driving motor? a. 299 kW

c. 250.34 kW

b. 350.16 kW

d. 299.64 kW

Solution:

epump = Pwater / Pinput Solving for Pwater; Q = 130 ft3 = 0.061 m3/s =

1 0.0025

0.00981

3.9245 kN/m3

Pwater = 3.924(0.061)(800) = 191.49 Kw Then; Poutout = 191.49/0.64 Thus; a

Poutout = 299.21 kW

37. Supplementary Problem A dearator heater supplies 150 L/min of daerated feedwater into a booster pump at 115

pumping temperature, The heater pressure is maintained at 100 kPag by

bleed steam. Pump centerline is located 1m above the floor level. Suction line losses is 0.60 m. Determine the minimum height of water level in the heater that must be maintained above the centerline of the pump to avoid cavitation, if the pump to be installed has a reuired suction head 5.8 m. a. 9.05 m

c. 3.25 m

b. 2.25 m

d. 5.09 m

Solution: NPSH =

.

where; P = 100 kPag Pa = 101.325 kPa PV = Psat at 115

Pv = 172 kPa Vf at 15.6

= 0.001 m3 / kg

Vf at 11.5

= 0.001055 m3/ kg

S.G. =

0.001 0.001055

= .948

then; 5.8 =

100 101.325‐172 9.81 .948

+ S – 0.60

thus; (c) S = 3.25 38. Supplementary Problem A pump running at 1000 rpm delivers water against a head of 300 m. If the pump speed will increased to 1500 rpm, what is the change in head. a. 575 m

c. 675 m

b. 375 m

d. 475 m

Solution: H2 H1

N

=( ) N

H2 = 300(1500/1000)2 H2 = 675 m thus; (b) ∆H = 675 – 300 = 375 M

39. Supplementary Problem Oil is being pumped from a truck to a tank 10 ft higher than the truck through a 2 in. galvanized pipeline 100 ft long. If the pressure of the discharge side of the pump is 15 psi, at what rate in gpm is oil flowing through the pipe? The oil has a specific gravity of 0.92 at the temperature in the pipe.

a. 542.22 gpm

c. 642.44 gpm

b. 442.44 gpm

d. 742.44 gpm

Solution: Q = AV H=P/

15(144)

+ hs = 62.4(.92)

10

= 47.66 ft 2gh =

V=

2 32.2 47.66

= 55.40 fps A=

22 /

= 0.0218 ft2 then; Q = 0.0218 (55.40) = 1.209 ft3/s (7.489 gal/1ft3)(60 s/min) thus; (a) Q = 542.44 gal / min or gpm

40. Supplementary Problem A pump is operated by motor developing 30 Kw and is delivering water at a pressure of 200 psig. If the pump is drawing water from a lake 20 ft below the centerline and the mechanical and hydraulic efficiency may be assumed to be 97% and 65% respectively; estimate the amount of water discharge in gallons per minute. a. 358.26 gpm

c. 208.26 gpm

b. 258.26 gpm

d. 308.26 gpm

Solution: Pwater = QH

Solving for Pwater; Pwater = Pbrake (epump) = (30)[(.97)(.65)] = 18.915 Kw Solving for H; H=

200 144 62.4

20

= 481.54 ft = 146.766 m Then; 18.915 = 9.81 Q (146.766) Q = 0.01314 thus; (c) Q = 208.26 gpm 41. Supplementary Problem A 30 Hp centrifugal pump is used to deliver 70 cfm water. Calculate the number of stages needed if each impeller develops a 38 ft head. a. 4

c. 8

b. 6

d. 10

Solution: No. of stages = Total head / head per stage Solving for the total head, H; Q = 70 ft3/min Q = 0.033 m3/s P=

QH

30 / 0.746 = 9.81 (0.033) H

H = 124.22 n = 124.22 / 38 Thus; (a) n = 3.26 say 4 stages 42. Supplementary Problem In a test of a centrifugal pump driven by an electric motor, the suction pipe is 10 in. in diameter and its gage indicates a partial vacuum of 2.5 ft of water. The discharge pipe is 5 in. in diameter, is 2 ft higher than the suction gage and shows a pressure of 50 ft of water. if the pump efficiency is indicated assuming motor efficiency of 85%? a. 60%

c. 75%

b. 70%

d. 65%

Solution: Electrical Power Input = Water power / (epump)(emotor) Solving for water power: Vs = Q/AS = 1.6 / /4 10/12 2 = 2.935 fps Vd = Q/Ad = 1.6 / /4 5/12 2 = 11.74 fps H = (50 + 2.5) + 2 + [(11.74)2 – (2.935)2]/ 2(32.2) H = 56.51 ft Pwater = (62.4)(1.6)(56.51) = 5,641.96 ft-lb /s = 10.26 Hp then; 12/ .746 = 10.26 / (epump)(0.85) thus; (a) epumo = 0.7504 = 75.04%

43 Supplementary Problem A centrifugal pump is designed for 2000 rpm and head of 70 m. What is the speed if the head id increased to 100 m. a. 2100 rpm

c. 3000 rpm

b. 2390 rpm

d. 3010 rpm

Solution: )2 100/70 =(N2/2000)2 Thus; (b) N2 = 2,390.46 rpm 44. Supplementary Problem A water pump develops a total head of 200 ft. The pump efficiency id 80% and the motor efficiency is 87.5%. If the power rate is 1.5 cents per Kw-hr, hat is the power cost for pumping 100 gal? a. 34 cents per hour

c. 2.34 cents per hour

b. 1,34 cents per hour

d. 3,34 cents per hour

Solution: P=

QH

= (62.4 lb/ft3)[(1000 gal/hr)(1 ft3/ 7.48 gal)(1 hr / 16000 s)](200 ft) = 436.46 Hp P motor = 0.84 / 0.80(.875) Power Cost = 0.895 Kw ($ 0.015 / Kw- hr) Thus; (b) Power Cost = $ 0.0134 per hr or 1.34 cents per hr


A test on a centrifugal pump operating at 1150 rpm showed a total of 37.6ft at a capacity of 800 GPM. Estimate the total head and capacity if the pump were operated at 1750 rpm. Assume normal operation at point of maximum efficiency in each case. a. 87.07 ft., 1217.4 gpm

c. 97.07 ft., 1217.4 gpm

b. 87.07 ft., 1517.4 gpm

d. 97.07 ft., 1517.4 gpm

Solution: New Head Required:

New Capacity Required:

.

H2 = 87.07 ft

Q2 = 1217.4 gpm

Thus; (a) H2 = 87.07 ft; Q2 = 1217.4 gpm

46. Supplementary Problem A centrifugal pump operating at 1800 rpm develops total head of 200 ft at a capacity of 2500 gpm. What is the specific speed? a. 1590 rpm

c. 1650 rpm

b. 1690 rpm

d. 1550 rpm

Solution: Ns =

Thus:

/

√ /

(a) Ns = 1690 rpm 47. Supplementary Problem A centrifugal pump operating at 1150 rpm showed a total head of 40 ft at a capacity of 600 gpm. The impeller diameter is 10.5 in. Estimate the total head and capacity of a geometrically similar pump at 1150 rpm with an impeller diameter of 10 inches. a. 30. 25 ft.,513.8 gpm

c. 36.28 ft., 618.3 gpm

b. 36.28 ft., 518.3 gpm

d. 30.25 ft., 618.3 gpm

Solution: New Head Required:


.

H2 = 36.28ft

.

Q2 = 518.3gpm

Thus; (b) H2 = 36.28 ft; Q2 = 518.3 gpm

50. Supplementary Problem A pump delivers 20 cfm of water having a density of 62 lb/ft3. The suction and discharge gage reads 5 in. Hg vacuum and 30 psi respectively. The discharge gage is 5 ft above the suction gage. If pump efficiency is 70%, what is the motor power? Solution Pwater 0.70

Pmotor

30

₂

₁

₁

₂

Z₂ – Z₁

2

5 14.7/29.920 144 62

0

5

80.38

Pwater

62

ft3

20

3 min

80.38

99674.87

Pwater

1 33, 000

/

3.02 Hp

Thus; Pwater

3.02 0.70

4.31

49. Supplementary Problem A centrifugal pump-motor unit draws 100 li/min of water from a supply which has a level at the centreline of the pump. The discharge pressure is 28 kg/cm2 and the over-all unit efficiency is 67 %. What could be the required input to the electric motor in kW if the head is 280 m. a. 6.83 kW

c. 7.83 kW

b. 5.83 kW

d. 8.83 kW

Solution Pinput

9.81

.

Pwater 0.67 280

0.67

Pinput

6.83 kW

48. Supplementary Problem A double-suction centrifugal pump delivers 3 m3/s of water at a head of 15 m and running at 1200 rpm. Calculate the specific speed of the pump. Ns a. 9958.46 rpm

c. 16, 000 rpm

b. 12, 110.64 rpm

d. 17, 000 rpm

Solution Ns Where:

√ /

3 47, 556.14 2

15

Ns

23, 778.07

49.215

1200√23, 778.07 49.215 /

9958.56

Llamera, Kristine Joyce D.

10 – 19388

ME – 5206 Problems in Industrial Plant Engineering Part 4: Fans & Blowers 1. What horsepower is supplied to air moving at 20 fpm through a 2 x 3 ft duct under a pressure of 3 in. water gage? a. 0.786 Hp

c. 0.642 Hp

b. 0741 Hp

d. 0.0566 Hp

Solution: 

Solving for :

2 3 2



Solving for :

⁄

.

.

then, 2 

.

31.2

⁄

Thus, (d)

0.0567

2. A fan whose static efficiency is 40% has a capacity of 60,000

per hour at 60

and barometer of 30 in. Hg and gives a static pressure of 2 in. of water column on full delivery. What size electric motor shall be used to drive this fan? a. ½ Hp

c. 2 Hp

b. 1 Hp

d. 1 ½ Hp

1

Solution:

Brake (input) Power =



Solving for

:

where; ,

16.67 ⁄

.

.

then; .

16.67 173.37 0.315 

thus; .

0.788

.

(b) Use 1 Hp 3. Air is flowing in a duct with a velocity of 7.62 m/s and a static pressure of 2.16 cm water gage. The duct diameter is 1.22 m, the barometric pressure 99.4 kPa and the gage fluid temperature and air temperature are 30 . What is the total pressure of air against which the fan will operate in cm of water? a. 3.25

c. 3.75

b. 2.50

d. 1.25

Solution: 

Solving for velocity head,

:

. .

2.959 

Solving for velocity in of cm of water: 1000

2.959

where: . .

1.143

⁄ 2

then;

1000

2.959 1.143

0.0034 m of water 0.34 cm of water 

thus; h

2.16

(b) h

0.34

2.50 cm of water of air per second through a duct 91 cm in diameter

4. A ventilating fan discharge 4.4

against a static pressure of 22 mm water gauge. Barometric pressure is 730 mm Hg, the temperature of air is 29.44

and the gauge fluid density is 994 kg/

. If the power

input is 2.65 kW, determine the static efficiency. a. 35.62 %

c. 45.62 %

b. 25.62 %

d. 55 %

Solution: .

=

.

.

.

.

.

= 0.3562 

thus; (a) e = 35.62 %

5. A ventilation system includes a fan of 570

per minute. A capacity discharging thru

a 122 cm x 91 cm air duct against 30 mm static pressure. Air temperature is at 21 and barometer pressure is 730 mm Hg. (97 kPa). What input power will be required to give the fan an efficiency of 44.3%? a. 5 Hp

c. 3 Hp

b. 10 Hp

d. 7 Hp

Solution:

.

 Solving for

: @

.

= 995.85 kg/ 3

.

= 1.150 kg/  Static pressure in cm WG converted into m of air: 0.30 1.150

995.85 0.30 25.98 m of air

 Velocity of air at the fan outlet: / .

8.56 /

.

 Velocity pressure: . .

= 3.73 m of air  Total Pressure created by Fan: + = 5.98 + 3.73 = 29.71 m of air then; = [1.150(0.00981)(570/60)(29.71)] = 3.18 kW = 4.27 Hp  thus;

. .

9.64

10

6. The mechanical efficiency and static pressure of a fan are 40% and 20 m of air respectively. What is the static efficiency if the total pressure created by fan is 25 m of air? a. 30.44%

c. 35.44%

b. 33.44%

d. 37.44%

Solution:

= (0.443)(20/25) 

thus;

4

(a)

0.3544 or 35.44%

7. Air enters a fan through a duct at a velocity of 6.3 m/s and an inlet static pressure of 2.5 cm of water less than atmospheric pressure. The air leaves the fan through a duct at a velocity of 11.25 m/s and a discharge static pressure of 7.62 cm of water above the atmospheric pressure. If the specific weight of the air is 1.20 kg/ delivers 9.45

and the fan

/s, what is the fan efficiency when the power input to the fan is 13.75

kW at the coupling? a. 71.81%

c. 52.34%

b. 61.81%

d. 72.34%

Solution: Fan efficiency =



Solving for

: +

=

=

.

.

1000

.

. .

= 88.761 m of air then; = [1.2(0.00981)](9.45)(88.761) = 9.874 kW 

thus; . .

(a)

= 71.81%

8. The forced draft fan is in parallel with a capacity if 73.2

per second each supplying

combustion air to a steam generator. Air inlet is at 43.33 , a static pressure of 254 mm water gage is developed and the fan speed is 1200 rpm. The fan input is 257 kW each. Calculate the capacity of the fan for a speed increase of 20 percent. a. 77.86

/s

c. 89.46

/s

b. 87.84

/s

d. 59.49

/s

5

Solution: 1.21 = 1.21 (73.2) 

thus; (b)

87.84

/s

9. Calculate the air power of a fan that delivers 1200

/min of air through a 1 m by 1.5

m outlet. Static pressure is 120 mm WG and density of air is 1.18. a. 20.45 kW

c. 30.45 kW

b. 25.64 kW

d. 35.64 kW

Solution: 

Solving for h:

0.120

.

101.695

from: Q=AV 1 1.5 V = 13.33 m/s . .

= 9.06 m of air + = 101.695 + 9.06 = 110.756 m of air 

thus; 1.18 0.00981 (b)

10. A fan delivers 4.7

110.756

25.64 kW

/s at a static pressure of 5.08 cm of water when operating at a

speed of 400 rpm. The power input required is 2.963 kW. If 7.05

/s are desired in the

same fan and installation, finds the pressure in cm of water. a. 7.62

b. 17.14

6

c. 11.43

d. 5.08

Solution:

.



Solving for

:

600 rpm then; .



thus; (c)

11.43 cm of water

11. A fan described in a manufacturers table is rated to deliver 500

/min at a static

pressure (gage) of 254 cm of water when running at 250 rpm and requiring 3.6 kW. If the fan speed is changed to 305 rpm and the air handled were at 65

instead of

standard 21 , find the power in kW. a. 3.82

c. 4.66

b. 5.08

d. 5.68

Solution: 

Power Required at 305 rpm and 65 :



Solving for the Power required at 305 rpm and 21 :

.

6.5 kW then; .



thus; (d)

5.68 kW 7

12. What is the set efficiency of a fan if the fan efficiency is 45% and motor efficiency is 90%? a. 40.50%

c. 30.41%

b. 35.65%

d. 40.94%

Solution: Set of Efficiency = Fan Efficiency x Motor Efficiency = (0.45)(0.90) 

thus; 0.405 or 40.50%

(d)

13. A fan draws 1.42

/s of air at a static pressure of 2.54 cm of water through a duct 300

mm diameter and discharges it through duct of 275 mm diameter. Determine the static fan efficiency if total fan mechanical efficiency is 70% and air is measured at 25

and 760 mm Hg. a. 60%

c. 30%

b. 50%

d. 40%

Solution:



Solving for

:

.

=

=

.

= 1.18 kg/ 

Solving for

. .

= 21.52 cm of air

:

. .

. .

20.09 23.9 .

. .

8.54 m of air 

Solving for h: + = 21.52 + 8.54 8

= 30.06 m of air 

thus; 0.70 21.52/30.06 (b)

0.5011 or 50.11%

14. A fan manufacturer rates his fans at 152 mm water gage static pressure for 10

of

air per second at 21 , 1200 rpm and static efficiency of 69%. At what speed would these fans operate to develop 130 mm water gage when the temperature is 316 ? a. 1570 rpm

c. 1770 rpm

b. 1670 rpm

d. 1470 rpm

Solution: 

At 21 ,

1.52 

A change in temperature will effect a change in density:



With the same pressure;

:

1.52 2 2

13.09 

At 21

/

∶

.



thus; (a)

1570.8 rpm

15. A 12 Hp motor is used to drive a fan that has a total head of 20 m. If the fan efficiency is 70%, what is the maximum capacity in

/s?

a. 26.63

/s

c. 24.43

/s

b. 25.53

/s

d. 27.73

/s 9

Solution: 0.70 12

8.4

= 6.27 kW 6.27 = [(1.2)(0.00981)]Q(20) 

thus; /s

(a) Q =26.63

16. Find the air horsepower of an industrial fan that delivers 26 by 1.2 m; pressure is 127 mm of water; air temperature is 21

/s of air through a 1 m and barometric pressure

is 760 mm of mercury. a. 53.35 Hp

c. 46.45 Hp

b. 43.33 Hp

d. 56.45 Hp

Solution: 

Solving for h: 21.667 /

.

. .

= 1.2 kg/ @ 101.325

998.6

/

2 1.20 21.62 998.6 2 9.81

0.029 127

0.127

h = 0.127 + 0.029 = 0.156 

then; P = (9.81)(26)(0.156) = 39.79 kW



thus; (a) P = 53.34 Hp

10

17. A large forced – draft fan is handling air at 1 atm, 43.3

under a total head of 26.6

cm WG (at 43.3 ). The power input to the fan is 224 kW and the fan is 75 percent efficient. Compute the volume of air handled per minute. Local gravity of acceleration is g = 9.81m/ . a. 3,862.87

/

c. 2,862.87

/

b. 4,862.87

/

d. 4,567.97

/

Solution: Power Input to fan = Shaft Power

P= 224 =

.

.

.

Q = 64.38 = 3,862.87 

/s /

thus; (a) 3,862.87

/

18. A fan develops a brake power of 150 kW at 1.2 kg/ brake power of the fan if it operates at 100 kPa and 30

air density. What is the new at the same speed?

a. 163.75 kW

c. 133.86 kW

b. 143.75 kW

d. 173.86 kW

Solution:



Solving for

: .

= 1.15 

/

then; . .



thus; (b)

143.75 kW

11

18. Supplementary Problem A cold storage compartment is 4.5 m long by 4 m wide by 2.5 m high. The four walls, ceiling and floor are covered to a thickness of 150 mm with insulating material which has a coefficient of thermal conductivity of 5.8 x 10-2 W/m-K. Calculate the quantity of heat leaking through the insulation per hour when the outside and inside face temperatures of the material is 15°C and -5°C respectively.

a.

2185.44 kJ

c.

3185.44 J

b.

1185.44 kJ

d.

4185.44 kJ

Solution

Q

=

kAΔt x

where: A

=

2 [ (4.5)(2.5) + (4)(2.5) + (4.5(4) ]

=

78.5 m2

then: (5.8x10-2)(78.50)(15+5)

Q

=

Q

=

607.07 W or J/s

Q

=

2185.44 kJ/hr

0.15

thus; (a) the quantity of heat through the insulation per hour is 2185.44 kJ

19 Supplementary Problem A thin square steel plate, to 10 cm on a side, is heated in a blacksmiths forge to a temperature of 800°C. If the emissivity is 0.60, what is the total rate of radiation of energy?

a.

900 Watts

c.

300 Watts

b.

400 Watts

d.

700 Watts

Solution Q

=

A e σ T4

Q

=

(0.020m 2 ) (0.60) (5.67 x 10 -8

thus; (a) Q = 900 Watts

W m2K4

) (1073)4 K4

20. Supplementary Problem A furnace wall consist of 35 cm firebrick (k= 1.557 W/m-K), 12 cm insulating refractory (k=0.346) and 20cm common brick (k=1.692) covered with 7 cm steel plate (k=45). The temperature at the inner surface of the firebrick is 1230 degree C and at the outer face of the steel plate is 60 degree C. Atmosphere 27 degree C. What is the value of combined coefficient for convection and radiation from the outside wall? a. 31.13 W/m2-K

c. 41.3 W/m2-K

b. 30.13 W/m-K

d. 40.13 W/m2-K

Solution Q A

=

Δt RT

where: RT

=

=

=

k12

k23

+

X12

x23

0.35

x34

0.12

+

1.557

k34

+

0.346

+

+

k45 x45 0.2 0.692

m2K

0.862

W

then; Q A Q A

(1230-60)K

=

0.862

W

Q5-0

=

=

m2K

=

A

t5 - to 1 h0

1357.15

=

60-70 1 h0

h0

=

41.13

W m2K

1357.15 W/m2

+

0.07 45

21. Supplementary Problem A dry ice storage chest is a wooden box lined with glass fiber insulation 5cm thick. The wooden box (k= 0.069) is 2 cm thick and cubical 60 cm on an edge. The inside surface temperature is -76 degree C and the outside surface temperature is 18 degree C. Use k= 0.035 for fiber glass insulation. Determine the heat gain per day. a. 10211 kJ

c. 12211kJ

b. 11195 kJ

d. 9185 kJ

Solution Q

=

A∆t RT where: A

=

6[(0.60)(0.60)]

= RT

= =

2.16m2 k12 x12

+

k23 x23

0.05

+

0.035

=

0.02 0.069

0.1718

then; Q

= =

=

2.16(18+76) 1.718 (118.18)

10211.092

=

118.18

J S

(3600s)

(24hrs)

(1kJ)

(hr)

(Day)

(1000J)

kJ day

thus; (a) the heat gain per day is 10211.092kJ

22. Supplementary Problem One side of the refrigerated cold chamber is 6 m long by 3.17 m high and consists of 168mm thickness of cork between outer and inner walls of wood. The outer wood wall is 30 thick and its outside face temperature is 20 degree C, the inner wood wall is 35 mm thick and its inside face temperature is -3 degree C. Taking the coefficient of thermal conductivity of cork and wood as 0.42 and 0.20 W/m-K respectively, calculate the heat transfer per second per sq. m of surface area. a. 5.138 J

c. 6.318 J

b. 4.138 J

d. 3.318 J

Solution Q A

=

∆t RT

where: RT

=

0.03 0.2

+

0.168 0.042

+

0.035 0.2

=

4.325

then; Q A

= =

20+3

=

4.325 5.318

5.918W

J s

thus; (a) the heat transfer per second per sq.m of the surface is 5.318 J

23. Supplementary Problem Hot gases at 280 degree C flow on one side of a metal plate of 10 mm thickness and air at 35 degree C flows on the other side. The heat transfer coefficient of the gases is 31.5 W/m-K and that of the air 32 W/m-K. Calculate the overall transfer coefficient. a.

15.82 W/m2-K

c.

14.82 W/m2-K

b.

15.82 W/m2-K

d.

17.82 W/m2-K

Solution U

=

1 RT

where: RT

= =

F h1

+

k12 x12

+

1 x12

=

0.0632

thus; U

=

1 0.06032

(a) U=15.82

=

15.82 W

m2-K

1 31.5

+

0.01 50

+

1 32

24. Supplementary Problem The surface temperature of the hot side of the furnace wall is 1200 degree C. It is desired to maintain the outside of the wall at 38 degree C. A 152 mm of refractory silica is used adjacent to the combustion and 10 mm of steel covers the outside. What thickness of insulating bricks is necessary between refractory and steel, if the heat loss should be keep at 788 W/m2? Use k= 12.84 W/m-K for refractory silica; 0.15 for insulating brick, and 45 for steel. a. 220 mm

c. 260 mm

b. 240 mm

d. 280 mm

Solution RT

=

RT

=

x12 k12

+

0.152 31.84

x23 k23

+

x34 k34 x23

+

+

0.15

0.01 45

Solving for RT 788

=

788

=

RT

=

∆t RT (1200-38) RT

1.475

then; 1.475

=

0.152 13.48

thus; (a)

x23

=

0.22mm

x23

=

220mm

+

x23 0.15

+

0.01 45

25. Supplementary Problem How much heat will flow in 24 hours through a plaster wall that is 0.50 in thick and 8 ft x 14 ft in area if the temperature is 80 degree F on one side and 40 degree F on the other? Use = 3.25 Btu-in/hr-ft2-degree F a. 5.99 x 105 Btu

c. 7.99 x 105 Btu

b. 6.99 x 105 Btu

d. 4.99 x 105 Btu

Solution Q

=

=

kA∆T x 3.25[(8)(14)](80-40)(24)

thus; (b) Q = 6.99 x105 Btu

0.5

26. Supplementary Problem A hollow sphere has an outside radius of 1 m and is made of polystyrene foam with a thickness of 1 cm. A heat source inside keeps the inner surface 5.20 degree C hotter that the outside surface. How much power is produced by the heat source? Thermal conductivity of polystyrene foam is 0.033 W/M degree C. a. 200 W

c. 300 W

b. 216 W

d. 316 W

Solution Q

=

kA∆T x

where: A

= =

Q

=

4∏(1)2 12.56 m2 (0.033)(12.56)(5.20)

thus; (b) Q= 216 W

0.01

=

216 W

27. Supplementary Problem A glass window has an area of 1.60 m2 and a thickness of 4mm. If one side is at a temperature of 6.80 degree C and other is at -5 degree C, how much thermal energy flows through the window in a time of 24 hours? The thermal conductivity of glass is 1.89 x 10-4 Kcal/ m-s-degree C a. 26200 kCal

c. 40700 kCal

b. 58000 kCal

d. 77100 kCal

Solution Q

= = =

kA∆T x (1.89 x 10-4)(1.60)(6.80+5)(3600)(24) 0.004 77100 kCal

thus; (d) Q = 77100 kCal

28. Supplementary Problem The wall of a cold room consist of a layer of cork sandwiched between outer and inner walls of wood, the wood walls being each 30 mm thick. The inside atmosphere of the room is maintained at -20 degree C when the external atmospheric temperature is 25 degree C, and the heat loss through the wall is 42 W/m2. Taking the thermal conductivity of wood and cork as 0.20 W/m-K and 0.05 W/m-k respectively, and the rate of heat transfer between each exposed wood surface and their respective atmospheres as 15 W/m2 –K, calculate the thickness of the cork. a. 31.90 mm

c. 41.90 mm

b. 21.90 mm

d. 51.90 mm

Solution Q

k∆T

=

A

x

where: Q A

=

42

W m2 W

k

=

0.05

x

=

30mm

m-K =

0.30m

from the heat through each wood wall: 42

=

∆t

=

0.20∆t 0.03 6.30◦C

Interface temperature of outer wood and cork: =

22.2 - 6.3

=

15.6◦C

Interface temperature of inner wood and cork: =

-17.2

+

6

=

10.9◦C

Temperature difference acros cork: =

15.9-(-10.9)

=

26.8◦C

From heat flow through cork: 42

=

x

=

0.05(26.8) x 0.0319m

thus; (a) x = 31.90 mm

29. Supplementary Problem A slab of material has an area of 2m2 and is 1mm thick. One side is maintained at a temperature of 0 degree C while the other is at 12 degree C. It is determined the 6820 J of heat flows through the material in a time of 10 minutes. What is the thermal conductivity of the material? a. 4.74 x 10-4 W/m◦C

c. 2.66 x 10-4 W/m◦C

b. 5.74 x 10-4 W/m◦C

d. 9.79 x 10-4 W/m◦C

Solution Q

=

kA∆T x

6820 J

=

10(60) s k

=

k(2)(12-0) 0.001 4.74 x 10- 4

thus; (a) k = 4.74 x 10-4

W m◦C

J s-m-◦C

30. Supplementary Problem An insulated steam pipe located where the ambient temperature is 32°C, has an inside diameter of 50 mm with 10 mm thick wall. The outside diameter of the corrugated asbestos insulation is 125 mm and the surface coefficient of still air, h0 = 12 W/m2-K. Inside the pipe is steam having a temperature of 150°C with film coefficient hi = 6000 W/m2-K. Thermal conductivity of pipe and asbestos insulation are 45 and 0.12 W/m2-K respectively. Determine the heat loss per unit length of pipe. a. 110 W

c. 130 W

b. 120 W

d. 140 W

Solution: Q=

∆

Where: RT =

.

RT =

.

.

RT = 0.98345 / L Then; Q=

.

/

Thus; (b) Q/L = 120 W per meter length

.

31. Supplementary Problem A pipe 200 mm outside diameter and 20 m length is covered with a layer, 70 mm thick of insulation having a thermal conductivity of 0.05 W/m2-K and a thermal conductance of 10 W/m2-K at the outer surface. If the temperature of the pipe is 350°C and the ambient temperature is 15°C, calculate the external surface temperature of the lagging. a. 32.6°C

c. 42.6°C

b. 22.6°C

d. 53.6°C

Solution: Q(for lagging) = Q(for surface film) .

0.340 10

T2 = 32.6°C

15

32. Supplementary Problem Dry and saturated steam at 6 Mpa abs. enters a 40 m length of 11.5 cm O.D. steel pipe at a flow rate of 0.12 kg/s. The pipe is covered with 5 cm thick asbestos insulation (k= 0.022 W/m-k). The pipe is located in a tunnel with stagnant air temperature of 27 degree C. The unit outside convective coefficient is 10 W/m2-K. Neglecting steam film and pipe wall resistances, determine the mass of steam. a. 4.86 kg/hr

c. 5.86 kg/hr

b. 3.86 kg/hr

d. 6.86 kg/hr

Solution:

The temperature of the outer surface of the pipe is equal to that of the steam since the resistance of metal pipe and vapor are negligible. Vapour temperature t1 =t2 =tsat @ 6Mpa = 275 degree C. RT

= = =

Q

=

R2

+

R0

ln(10.75/5.75)

+

2∏(40)(0.022) 0.11687 ∆t RT

=

=

2112 W

=

2.122kW

1 ∏(0.215)(40)(10)

◦C 275-27 0.11687

The heat necessary to condense steam at 6Mpa is hfg @ 6Mpa is equal to 1571 kJ/kg. Since there are 2.122 kW of heat lost from the steam, then

Steam Condensed

= = =

2.122kJ/s 1577kJ/s 0.00135 kg/s 4.86 kg/hr

33. Supplementary Problem Calculate the heat loss per linear ft from 2 in. nominal pipe (2.375 in. outside diameter ) covered with 1 in. of an insulating material having an average thermal conductivity of 0.0375 Btu/ hr- ft- degree F. Assume that the inner and outer surface temperatures of the insulation are 380 degree F and 80 degree F respectively. a. 110 Btu/ hr-ft

c. 120 Btu/ hr-ft

b. 116 Btu/ hr-ft

d. 126 Btu/ hr-ft

Solution Q

=

∆t RT

=

∆t ln( d2/d1) 2∏kL

=

380 ln(4.375/2.375) 2∏(0.0375)

Q

=

116

Btu hr-ft

34. Supplementary Problem Calculate the heat loss per linear foot from a 10 in. normal pipe ( outside diameter = 10.75 in. ) covered with a composite pipe insulation consisting of 1 ½ in of insulation I placed next to the pipe and 2 in. of insulation II placed upon insulation I. assume that the inner and outer surface temperatures of the composite insulation are 700o F and 100oF respectively, and that the thermal conductivity of material I is 0.05 Btu/hr-ft-oF and for material II is 0.039 Btu/hr-ft-oF. a. 423.13 Btu/hr-ft

c. 120 Btu/hr-ft

b. 123.13 Btu/hr-ft

d. 126 Btu/hr-ft

Solution Q = Where: .

RT =

/ .

= 1.826 Then: Q =

.

Thus; (d) Q = 323.13

.

+

.

/ .

.

35. Supplementary Problem A steam pipe carrying a steam at 380 kPa pressure for a distance of 120 m in a chemical plant is not insulated. Estimate the saving in steam cost that would be made per year if this 8 cm steam line were covered with 85% magnesia pipe covering 5 cm thick. Take room temperature to be 25˚C, the cost of steam is 65 cents per 1000 kg. thermal conductivity of magnesia k = 0.0745 W/m-K, unit convective coefficient of room air, ho = 12 W/m2-K. a. $ 305

c. $ 505

b. $ 405

d. $ 605

Solution Steam temperature = 142o C Latent heat hfg = 2139.4 kJ/kg Heat Loss from the bare pipe: Q1 =

Where: Ro = = =

.

0.00276 oC/W

Then; Q1 =

.

= 42,343.64 W =

42,344 kW

Total Resistance from the insulated pipe: R1 = R2 + Ro =

=

/

+

/ .

+

.

R1 = 0.01113 ˚C/W Heat loss from the insulated pipe:

Q2 =

=

.

= 10,512.13 W Heat Saved = Q1- Q2 = 42,34364 – 10.51213 = 31,831 kW Amount of steam saved due to condensation (m2):

m =

.

/ . /

= 0.014878 kg/yr

36. Supplementary Problem A liquid to liquid counter flow heat exchanger is sued to heat a cold fluid from 120 310 . Assuming that the hot fluid enters at 500F and leaves at 400F, calculate the log mean temperature difference for the heat exchanger a. 132F

c. 332F

b. 232F

d. 432F

Solution

LMTD = log mean temperature difference LMTD =

∆

∆ ∆ ∆

Where:

Thus, LMTD =

= 232

∆

400

120

280

∆

500

310

190

37. Supplementary Problem A blower with the inlet open to the atmosphere delivers 300 cfm of air at a pressure of 2in. WG trough a duct 11 in. diameter, the manometer being attached to the discharge duct at the blower. Air temperature is 70ºF, and the barometer pressure is 30.22 in Hg. Calculate the horsepower. a. 1.54 Hp

c. 3.54 Hp

b. 2.54 Hp

d. 0.75 Hp

Solution: Pair = ɣgh where: Q = 3000 cfm Solving for h: Air density at the following condition by correcting the standard density. ρair = 0.075 (

30.2

29.92

hs =

hv =

2

(

62.4

12 0.076

V 2g

=[

) = 0.076 lb/ft3

) = 136.84 ft of air

3000/60 11 2 ) 12

π/4(

2(32.2)

] = 89.13 ft

then; P = 0.075 (3000/60(136.84 + 89.13) P = 847.39

ft - lb s

thus; (a) P = 1.54 Hp

(

1 Hp 550 ft-lb/s

)

38. Supplementary Problem A certain fan delivers 12,000 cfm at a static pressure of 1in. WG when operating at a speed of 400 rpm and requires an input of 4 Hp. If the same installation 15,000 cfm are desired, what will be the new speed, and the new power needs? a. 450 rpm, 6.81 Hp

c. 500 rpm, 6.81 Hp

b. 500rpm, 7.81 Hp

d. 450 rpm, 7.81 Hp

Solution: New Speed Required: N2 N1

=

N2 400

The New Power Required:

Q2

P2

Q1

P1

=

15000

P2

12000

4

N2 = 500 rpm

= ( = (

N2 N1 500 400

P2 = 7.81 hp

thus; (b) 500 rpm, 7.81 Hp

39. Supplementary Problem A certain fan delivers 12,000 cfm at 70ºF and normal barometric pressure at a static pressure of 1 in. WG when operating at 400 rpm and requires 4 Hp. If the air temperature is increased to 200ºF (density 0.06018 lb/ft3) and the speed of the fan remains the same, what will be the new static pressure and power? a. 0.81 in. WG, 3.21 Hp

c. 0.71 in. Wg, 3.24 Hp

b. 0.81 in. WG, 2.24 Hp

d. 0.71 in. WG, 2.24 Hp

Solution: New Static Pressure Required: h2 h1

=

New Power Required:

ρ2

P2

ρ1

P1

=

ρ2 ρ1

h2 1

=

0.06018

P2

0.075

4

h2 = 0.80 in. WG

=

0.06018 0.075

P2 = 3.21 Hp

thus; (a) 0.81 in. WG, 3.21 Hp

40. Supplementary Problem If the speed of the fan in problem 39 is increased so as to produce a static pressure of 1 in. WG at 200ºF. What will be the new speed and new capacity? a. 446.54 rpm, 13,396.33 cfm

c. 457.45 rpm, 12,457.45 cfm

b. 454.34 rpm, 15,345.17 cfm

d. 745.54 rpm, 11,345.34 cfm

Solution: New Speed Required: N2 400

=


0.075

Q2

0.06018

12000

N2 = 446.54 rpm

=

0.075 0.06018

Q2 = 13,396.33 cfm

thus; (a) 446.54 rpm, 13,396.33 cfm

41. Supplementary Problem If the speed of the fan of the previous examples (Problem 39-40) is increased so as to deliver the same weight of air at 200ºF as at 70ºF. What will the new speed, new capacity, new static pressure and new power? a. 498.50 rpm, 14,955.14 cfm, 1.25 in. WG, 6.21 Hp b. 646.54 rpm, 15,396.33 cfm, 2.25 in. WG, 7.21 Hp c. 464.54 rpm, 15,396.33 cfm, 3.25 in. WG, 5.21 Hp d. 546.54 rpm, 12,396.33 cfm, 4.25 in. WG, 4.21 Hp

Solution: New Speed Required: N2 400

=

New Static Pressure Required:

0.075

h2

0.06018

1

N2 = 498.50 rpm

New Capacity Required: Q2 12000

=

=

0.075 0.06018

h2 = 1.25 in. WG

New Power Required:

0.075

P2

0.06018

4

Q2 = 14,955.135 cfm

=(

0.075

0.06018

P2 = 6.21 Hp

thus; a. 498.50 rpm, 14,955.14 cfm, 1.25 in. WG, 6.21 Hp

42. Supplementary Problem A fan discharges 10,000 cfm of air through a duct 2 ft by 2 ft against a static pressure of 0.90 in. of water. The gage fluid density is 62lb/ft3, air temperature is 85ºF and the barometric pressure is 28.7 in. Hg. If the power input to the fan is measured as 3.6 Hp, what is the over-all mechanical efficiency of the fan? a. 50 %

c. 60 %

b. 56 %

d. 65 %

Solution: em =

Pair 3.6

Solving for Pair : V=

Q A

h = hs + hv h = 19.05 + 8.22 = 27.27 m Q = 10000 ft3/min

V = 2500 ft/min = 41.67 ft/s

Q = 4.72 m3/s

V=

then;

2gh

41.67 =

2(32.2) hv

hv = 29.96 ft = 8.22 m hs = hw (

ρw

)

1000 1.2

Pair = [1.2(0.00981)](4.72)(27.27) Pair = 1.515 kW

ρair

hs = 0.9 (

Pair = ɣgh

) = 750 in.

hs = 19.05 m

Pair = 2.03 Hp thus; (a) em =

2.03 3.6

= 0.564 = 56.4 %

43. Supplementary Problem A ventilation system includes a fan with a mechanical efficiency of 45% against a static pressure of 30 cm WG. If the total pressure created by fan is 300m of air, what is the static efficiency? a. 37.5 %

c. 40.5 %

b. 35 %

d. 45 %

Solution: es = em

thus; hs h

)

(a) es = 0.375 or 37.5 %

Solving for hs: hs = 0.30

1000 1.2

)

hs = 250 m then; es = 0.45

250 300

)

44. Supplementary Problem It requires 55 Hp to compress 1000 cfm of air at 60ºF and 14.7 psi to a pressure of 10 psig. The temperature of the air leaving the blower is 184ºF. What is the flow in cfm from the blower discharge?

a. 852.64 cfm

c. 737.06 cfm

b. 801.62 cfm

d. 700.91 cfm

Solution: P1 V1 T1 P1 Q

1

T1

P2 V

= =

2

T2 P2 Q

2

T2

(14.7)(1000) 60 + 460

=

(10 + 47)Q2 184 + 460

thus; (a) Q2 = 737.06 cfm

45. Supplementary Problem A blower draws 3000 cfm of air through a duct of 12 in. in diameter with a suction of 3 in. of water. The air is discharged through a duct 10 in. in diameter against a pressure of 2in. of water. The air is measured at 70ºF and 30.2 in. Hg. Calculate the air horsepower. Use specific weight of 62.34lb/ft3. a. 2.82 Hp

c. 3.87 Hp

b. 2.87 Hp

d. 1.75 Hp

Solution: P = ɣgh Solving for h: h = Zd – Zs +

Pd - Ps ɣ

+

Vd2 – Vs2 2g

where: 3000

Vs = π6010

3

ft /s 2

2

( ) ft

= 91.67 ft/s

4 12

Vd =

3000 3 ft /s 60 π 2 2 (1) ft 4

= 63.56 ft/s

Ps = ɣh = (62.34)(2/12) = 0.072 psi Pd = (62.34)(-3/12) = -0.108 psi Zs = Zd The density of Air @ 30.2 in. Hg and 70ºF

ɣ=

P RT

14.7

=

30.2(29.92)(144) 53.34(70 + 460)

ɣ = 0.0756 lb/ft3 then; h=

(0.072 + 0.108)(144)lb/ft3 0.0756 lb/ft3

+

2

2

(91.67) -(63.66) 2(32.2)

h = 410. 42 ft of air thus; Pair = ɣgh = 0.0756(3000/60)(410.42) ft-lb/s Pair = 1551.39 ft-lb/s (

1 Hp

)

550 ft-lb/s

(a) Pair = 2.82 Hp

46. Supplementary Problem A blower operating at 15,000 rpm, compresses air from 68F and 14.7 psia to 10 psig. The design flow is 1350 cfm and at this point the BHp is 80. Determine the efficiency of the blower at the design point. a. 48.41 Hp

c. 55.62 Hp

b. 40.54 Hp

d. 57.65 Hp

Solution: Pair = ɣgh Solving for the density of air at 68ºF and 14.7 psia ρ=

P RT

=

(14.7)(144) 53.34 (68 + 460)

ρ = 0.07516 lb/ft3 Solving for h: h=

h=

kRT1 k - 1

P

k-1

[( 2 ) k -1 P1

1.4(53.34)(68 + 460) 1.4 - 1

24.7 1.4-1

[(

14.7

) 1.4 -1

h = 15,743.7 ft

then; Pair = 0.07516(

1350 60

(15,743.7)

Pair = 26,624.17 ft-lb/s thus; (a) Pair = 48.41 Hp

47. Supplementary Problem A fan running at 2000 rpm delivers 16,000 cfm against 3 in. static pressure, thereby consuming 15BHp. If the rpm is increased to 2200 rpm so that the rpm ratio is 1:1:1. What is the new cfm? a. 17,500

c. 17,600

b. 16,600

d. 16,500

Solution: N2 N1

=

Q2 Q1

1.1 =

Q2 16000

thus; (a) Q2 = 17,600 cfm

48. Supplementary Problem A 0.70 m vane axial fan is running at 2000 rpm delivers 7.5 m3/s against 0.08 m static pressure thereby consuming 12 BkW. If the fan wheel diameter is increased from 0.70 m to 0.76 m, so that the diameter ratio 1.10:1, what is the new static pressure? a. 0.10 m

c. 0.13 m

b. 0.09 m

d. 0.15 m

Solution: h2 h1 h2 0.8

=

D2 D1

= (1.10

thus; (a) h2 = 0.097 m

49. Supplementary Problem At standard air density a fan delivers 8m3/s against 0.05 m static pressure consuming 12 BkW. If it will operate in Baguio City where due to high altitude the air density is only 82.4% of the standard air density, what is the new BkW? a. 7.86 BkW

c. 9.89 BkW

b. 15 BkW

d. 13 BkW

Solution: P2 P1 P2 12

=

ρ2 ρ1

= (0.824)

thus; (a) P2 = 9.89 BkW

50. Supplementary Problem What horsepower is supplied to air moving at 20 fpm through a 2x3 ft duct under a pressure of 3in. WC? a. 0.057 Hp

c. 0.123 Hp

b. 0.043 Hp

d. 0.241 Hp

Solution: Pair = ɣgh where: Q = AV Q = (20 ft/min)[ (2)(3) ft2 ] Q = 120 ft3/min = 2 ft3/s h = 3 in. (

62.4

0.075

h = 2496 in. h = 208 ft of air then; Pair = 0.075(2)(208) = 31.2 ft-lb/s thus; (a) Pair = 0.057 Hp

51. Supplementary Problem A fan whose static efficiency is 40% has a capacity of 60,000 ft3/hr at 60ºF and barometer of 3 in. Hg and give a static pressure of 2 in. WC on full delivery. What size of electric motor be used to drive this fan? a. ½ Hp

c. 2 Hp

b. 1 Hp

d. 1.5 Hp

Solution: Pair

efan =

Pmotor

Solving for Pair: Q = 60,000 ft3/hr Q = 16.67 ft3/s h = 2(

62.4

0.075

)

h = 1664 in. of air h = 138.67 ft of air then; Pair = ɣgh Pair = 0.075(16.67)(138.67) Pair = 173.37 ft-lb/s Pair = 0.315 Hp thus; 0.40 =

0.315 Hp Pmotor

(b) Pmotor = 0.788 Hp say 1 Hp

52. Supplementary Problem A fan delivers 4.7 m3/s at a static pressure of 5.08 cm of water when operating at speed of 400 rpm. If the power input required is 2.963 kW and if on the same installation 7.05 m3/s is desired, find the static pressure in cm of water. a. 12.43 cm of H2o

c. 10.43 cm of H2o

b. 9.43 cm of H2o

d. 11.43 cm of H2o

Solution: h2 h1

Solving for N2: N2 2

N2

N1

N1

=(

h2 5.08

)

=

N2 2

N2

400

400

=(

)

Q2 Q1

=

7.05 4.7

N2 = 600 rpm thus; h2 5.08

600 2

=(

400

)

(d) h2 = 11.43 cm of H2o

53. Supplementary Problem What horsepower is supplied to air moving at 7m/min through a 70 cm x 90 cm duct under pressure of 8 cm of H20? a. 0.049 Hp

c. 0.077 Hp

b. 0.058 Hp

d. 0.066 Hp

Solution: Pair = ɣgh where: Q = AV Q = [ (0.70)(0.90) ](7/60) m3/s Q = 0.0735 m3/s h = 0.08(1000/1.2) h = 66.67 m of air then; Pair = [ 1.2(0.00981) ] (0.0735) (66.67) Pair = 0.058 kW

thus; (c) Pair = 0.077 Hp

54. Supplementary Problem What horsepower is supplied of air moving at 25 fpm through an air duct 2 ft x 3 ft? Fan pressure is 4 in. of water column. a. 0.0945 Hp

c. 0.0495 Hp

b. 0.495 Hp

d. 0.849 Hp

Solution: Pair = ɣgh where: Q = [ (2)(3) ](25/60) ft3/s = 2.5 ft3/s h = 4(62.4/0.075) = 3328 in. of air = 277.33 ft of air then; Pair = (0.075)(2.5)(277.33) Pair = 52 ft-b/s thus; (a) Pair = 0.0945 Hp

55. Supplementary Problem A fan delivers 1.42 m3/s air at static pressure head of 2.54 cm of water through a duct 300 mm in diameter and discharge it through a duct 275 mm in diameter. Determine the static fan efficiency if the total fan mechanical efficiency is 70% and air is measured at 25ºC and 760 mm of Hg. a. 60 %

c. 30 %

b. 40 %

d. 50 %

Solution: estatic = efan (

hs h

)

Solving for h: V1 = Q/A1 = 1.42/( /4)(0.3)2 V1 = 20.089 m/s V2 = Q/A2 = 1.42/( /4)(0.275)2 V2 = 23.907 m/s

hv =

V22 – V12 2g

=

(23.907)2 – (20.089 2(9.81)

hv = 8.562 m of air hs = 0.0254 (1000/1.2) hs = 21.17m of air h = hs + hv h = 21.17 + 8.562 h = 29.73 m of air thus; estatic = 0.70 (21.17/29.73) estatic = 0.4984 (d) estatic = 49.84 %

2

19. Supplementary Problem A 95 tons refrigeration system has a compressor power of 90 Hp. Find the coefficient of performance, COP. a. 3.85

c. 4.77

b. 2.77

d. 1.99

Solution:

where: RE = 95 tons = 334.02 kW = 90 hp = 120.64 kW thus; 334.02 120.64 (b) COP = 2.77 20. Past ME Board Problem A refrigeration system operates on the reversed Carnot cycle. The minimum and maximum temperatures are -25

and 72 , respectively. If the heat rejected at the

condenser is 6000 kJ/min, find the power input required. a. 1 663.38 kJ/min

c. 1 686.83 kJ/min

b. 1 888.38 kJ/min

d. 1 886.83 kJ/min

Solution:

∆

where:

T 72 25

273

345

273

345 K 3

2

248

solving for ∆ : ∆

4

2

∆ = 17.39 kJ/min-K thus, 345 (c)

248 17.39

1 686.83 /

248 K

1

S

21. Supplementary Problem In a

0 refrigerating plant, the specific enthalpy of the refrigerant as it leaves the

condenser is 135 kJ/kg and as it leaves the evaporator it is 320 kJ/kg. If the mass flow the refrigerant is 5 kg/min, calculate the refrigerating effect per hour. a. 55 500 kJ/hr

c. 65 500 kJ/hr

b. 35 500 kJ/hr

d. 45 500 kJ/hr

Solution: where: m = 5 kg/min = 300 kg/hr 320 / 135 / thus; RE = 300 ( 320 – 135 ) (a) RE = 55 500 kJ/hr 22. Supplementary Problem A refrigerator is 2 m high, 1.2m wide and 1m deep. The over-all heat transfer coefficient is 0.532 W/

. How many kilograms of 0

temperature is maintained at 10

ice will melt per hour if the inside

while the surrounding air temperature is at 35 ?

a. 1.60 kg

c. 2.60 kg

b. 1.80 kg

d. 2.80 kg

Solution: Q=m Solving for Q: Q=UA∆ where: A = 2 2 1.2

1.2 1

2 1

A = 11.2 COP = 0.532

11.2

Q = 148.96 W = 0.14896 kW

)(35 – 10)

then; 0.14896 = m ( 335 ) m = 4.4466 x 10

kg/s

thus; ice melted per hour is: (a) m = 1.60 kg/hr 23. Supplementary Problem The power requirement of a Carnot refrigerator in maintaining a low temperature region at 300 K is 1.5 kW per ton. Find the heat rejected. a. 4.02 kW

c. 5.02 kW

b. 7.02 kW

d. 6.02 kW

Solution: ∆ where: =

+1

Solving for COP: .

COP =

.

COP = 2.34 then; =

.

+ 300

427.99

∆ = ∆

=

. .

0.012

thus; 427.99 0.012 (c)

5.02

24. Supplementary Problem A simple saturated refrigeration cycle for R-12 system operates at an evaporating temperature of -5

and a condensing temperature of 40 . Determine the volume

flow rate for a refrigerant capacity of 1 kW. Properties of R-12: At -5 ,

349.3

0.06496

At 40 ,

238.5

a. 0.0005866

/

c. 0.05865

/

b. 0.005866

/

d. 0.00005866

/

/

Solution: Solving for m: 1 / 349.3 238.5 /

m = 0.00903 kg/s thus; = (0.00903)(0.06496) (a)

0.0005866

/

25. Past ME Board Exam A vapor compression refrigeration system is designed to have a capacity of 150 tons of refrigeration. It produces chilled water from 22

to 2 . Its actual coefficient of

performance is 5.86 and 35% of the power supplied to the compressor is lost in the form of friction and cylinder cooling losses. Determine the condenser cooling water required for a temperature rise of 10 . a. 14.75 kg/s

c. 18.65 kg/s

b. 15.65 kg/s

d. 13.76 kg/s

Solution: By Energy Balance: ∆

where: RE = 150 ( 3.516 ) RE = 527.4 kW from: COP = .

5.86 =

= 90 kW then: 527.4

90

617.4 =

617.4

∆

617.4 =

417.4 10

thus; (a)

14.75

/

26. Past ME Board Exam An air conditioning system of a high rise building has a capacity of 350 kW of refrigeration, uses R-12. The evaporating and condensing temperatures are 0 35

respectively. Determine the mass of refrigeration 12 circulated per second.

Properties of R-12 At 0

At 35 0.05339

/

200 /

233.5 / 351.48 /

a. 2.97 kg/s

c. 4.57 kg/s

b. 3.57 kg/s

d. 1.97 kg/s

Solution:

m= where:

35

350 0

351.48 /

0

and

35

233.5 /

thus; m= (a)

.

.

m = 2.97 kg/s

27. Supplementary Problem Cold salt brine at an initial temperature of 0 from 40

to 5

is used in a packing plant to chill beef

in 18 hours. Determine the volume of brine in liters per minute required

to cool 1000beeves of 250 kg each, if the final temperature of brine is 3 . Specific heat of brine is 3.76 kJ/kg

and S.G. = 1.05. Specific heat of beef is 3.15 kJ/kg .

a. 37.59 kg/s

c. 38.79 kg/s

b. 39.67 kg/s

d. 35.67 kg/s

Solution Volume of brine circulated: = where: = 1

1.05

/

:

Solving for the

(3.14

= 424 then; ∆

1.05

424

3.76 3

0

37.59

/

thus; (a) 37.59 kg/s

424

)(40 – 5)

28. Past ME Board Exam A simple vapor compression cycle develops 13 tons of refrigeration. Using ammonia as refrigerant and operating at a condensing temperature of 24

and evaporating

temperature of -18 ; and assuming that the compressions are isentropic and that the gas leaving the condenser is saturated. Find the power requirement. a. 13 kW

c. 12 kW

b. 8.79 kW

d. 9.79 kW

Properties of R-12: At 24 :

312.87 /

h @ 974 kPA ( P at 24 and

974 kPA

1657 /

At -18 :

-18

1439.94 / 0.5729

/

Solution where: @ 947

&

= 1657 kJ/kg @

18

= 1439.94 kJ/kg from: 13tons ( 3.516 kW/ton) = m ( 1439.94 – 312.87 ) kJ/kg m = 0.0415 kg/s thus; 0.0405 1567 (b)

24

= 8.79 kW

1439.94

29. Past ME Board Exam A belt driven compressor is used in refrigeration system that will cool 10 liter per second of water from 13

to 1 . The belt efficiency is 98%, motor efficiency is 85%

and the input of the compressor is 0.7 kW per ton of refrigeration. Find the actual coefficient of performance of over-all efficiency is 65%. a. 4.44

c. 6.44

b. 5.44

d. 3.44

Solution COP = where: ∆ = [ 10(1)] ( 4.187 ) ( 13 – 1 ) = 502.44 kW = 142.90 TOR Solving for compressor work, .

= 0.65 =

.

.

. .

:

. .

78.055 thus; COP =

. .

(c) COP = 6.44 30. Supplementary Problem Four thousand liters per hour of distillates are to be cooled from 21

to 12

of wax by weight is separated out at 15 . The specific heat of oil is 2 kJ/kg

and 12% and S.G. is

0.87. The specific heat of the wax is 2.5 and the latent heat of fusion is 290 kJ/kg. The specific heat of the wax is 2.5 and the latent heat is 290 kJ/kg. Allow 10% for the losses, find the capacity of the refrigerating machine. a. 20 TOR

c. 40 TOR

b. 51.08 TOR

d. 31.08 TOR

Solution

Distillate

Oil

21

15

Oil -12

Wax

15

Wax -12

where: m = mass of distillate chilled per hour ( 0.87 ) 1

m = 400

m = 3,480 kg/hr = 0.97 kg/s and; = 0.97

2

(21 – 15)

= 11.64 kW

= (0.12)(0.97)

(290)

= 33.76 kW

= (0.12)(0.97)

(2.5)

(15+12)

= (0.88)(0.97)

(2)

(15+27)

33.76

7.86

0.10 11.64

= 7.86 kW = 46.09 kW

46.09

9.935

thus; 109.285 (d)

31.08

31. Supplementary Problem A 50 ton vapor compression system using Ammonia as refrigerant operates between 20

condenser and -16

evaporator temperature. If simple saturation cycle with

isentropic compression is assumed, determine the piston displacement of the reciprocating compressor to be used in the system operating at 600 rpm.

Properties of Ammonia: At 20

At -16 274.9 /

1424.4 /

@ 857.12 &

@20

0.5296 /

1640 /

a. 5000

c. 7000

b. 6000

d. 8000

Solution Piston displacement of the compressor: = Where:

m=

974 kPA

20

m=

-16

.

m=

.

.

m = 0.513 kg/s then; =

.

.

0.0081

8000

thus; (a)

8000

32. Past ME Board Problem A simple vapor compression cycle develops 15 tons of refrigeration using Ammonia as refrigerant and operating at condensing temperature of 24 temperature of -18

and assuming compression are isentropic and that the gas

leaving the condenser is saturated, find the power per ton. Properties of Ammonia At 24

At -16

312.87 /

1439.94 /

1665 /

and evaporating

a. 0.702 kW/ton

c. 0.602 kW/ton

b. 0.802 kW/ton

d. 0.502 kW/ton

Solution

= =

where: Solving for m: 15(3.516) =

1439.94

312.87

M = 0.04688 kg/s then; 0.04688 1665

1439.94

10.531 thus;

=

(a)

.

0.702

/

33. Past ME Board Problem In an Ammonia refrigerator the pressure in the evaporator is 267,58 kPa and the ammonia at entry is 0.12 dry while at exit is 0.91 dry. During compression the work done per kg of ammonia is 17,033 kg-m. Calculate the coefficient of performance. If the rate of ammonia circulation is 5.64 kg, calculate the volume of vapor entering the compressor per minute. The compressor is single acting, its volumetric efficiency is 80% and it runs at 120 rpm. The ratio of stroke to bore is 1. Calculate the bore and stroke. It is given that the latent enthalpy and specific volume of ammonia a 267.58 kPa are 320 kCal/kg and 0.436

/kg respectively.

a. 2.235

/

c. 1.457

/

b. 4.357

/

d. 3.567

/

Solution Properties of Ammonia: 267.58 @ 267.58

320

/

0.436

/

0.91 320 291.2

/

0.12 320 38.4

/

Solving for COP:

COP =

=

where: RE = 201.2 -

=

38.4

= 252.8 kCal/kg = 1058.47 kJ/kg = 17 033

/

( 0.00981 kN/

then; COP =

. .

= 6.335

Solving for the volume of vapor entering the compressor per minute:

where: m = 5.64 kg/min

= 0.001527 + 0.91(0.436) = 0.3983

/

then; = 5.61

0.3983

= 2.235

/

Solving for the bore and stroke: from: =

LN

= . .

LN

=

(D)(120)

Note:

L=D

thus; (a) L = D = 0.31 m = 31 cm 34. Supplementary Problem Freon-12 leaves the condenser of a refrigerating plan as a saturated liquid at 5.673 bar. The evaporator pressure is 1.509 bar and the refrigerant leaves the evaporator at this pressure and at a temperature of -5 . Calculate the refrigerating effect per kg. a. 132.88 kJ/kg

c. 160.91 kJ/kg

b. 123.77 kJ/kg

d. 123.86 kJ/kg

Solution Properties of Freon-12 At 5.673 bar

At 1.509 bar ( t = - 20

54.87 /

)

17.82 / 178.73 / 160.91 /

Since, the saturation temperature at 1.509 is -20

and the refrigerant at this pressure

leaves the evaporator at -5 , it is superheated by -5 , it is superheated by 15 @ 1.509 bar superheated 15 then; RE = RE = 187.75 – 54.87 thus; (a) RE = 132.88 kJ/kg

= 187.75 kJ/kg

35. Past ME Board Exam An ammonia compressor operates at an evaporator pressure of 316 kPa and a condenser pressure of 1514.2 kPa. The refrigerant is subcooled 5

and is superheated

8 . A twin cylinder compressor with bore to stroke ratio of 0.85 is to be used at 1200 rpm. The mechanical efficiency is 76%. For a load of 87.5 kW, determine the size of the driving motor. a. 24.26 kW

c. 34.26 kW

b. 25.26 kW

d. 35.26 kW

Solution Properties of Ammonia @ 316

& 0

1472 /

@ 316

& 0

0.41

@ 34

361.2 /

@ 1514.2

= = =

1715 /

P

=

/

5

34

3

1514 kPa

316 kPa

8

Solving for m: Q=

87.5 = m (1472 – 361.2) m = 0.079 kg/s =

.

.

thus; (b)

= 25.26 kW

0

36. Supplementary Problem A refrigeration system having a 30 kW capacity requires 10 Hp compressors. Find the COP of the system. a. 2.78

c. 4.02

b. 3.78

d. 5.02

Solution COP = =

.

thus; (c) COP = 4.02

37. Supplementary Problem A refrigerating machine uses ammonia as the working fluid. It leaves the compressor as dry saturated vapor at 8.57 bar es the condenser at this pressure and leaves as saturated liquid. The pressure in the evaporator is 1.902 bar and the ammonia leaves the evaporator 0.96 dry. If the rate of flow of the refrigerant through the circuit is 2 kg/min, calculate the volume taken into the compressor in m³/min, and the refrigerating effect in kJ/min. a. 1.198 m³/min, 2,183 kJ/min   b. 2.198 m³/min, 3,183.38 kJ/min

c.   1.198 m³/min, 3,183 kJ/min d.   2.198 m³/min, 2,183 kJ/min

Solution:

Properties of Ammonia:

    At 8.57 bar





Solving for the volume taken into compressor per minute:

Solving for the Refrigerating effect in kJ/min:

RE= m(h1 – h2)

                          89.8 0.96 1330.2                                                                                                                      1336.79 /                               275.1 /

Then;

    At 1.902 bar

=  275.1 kJ/kg



=  1462.6 kJ/kg



=  1420 kJ/kg

.

= 0.6237 m³/kg



= 1330.2 kJ/kg



Where:

2

/ min 1366.79

2,183.38

275.1 /



Thus; (a)



= 1.198 m³/min



.

2

=  89.8 kJ/kg

.

,

,

.

/



38. Supplementary Problem: The water enters the condenser at 30°C and leaves at 50°C. If the heat rejected in the condenser is 500 kW, determine the volume of water needed to cool the refrigerant. a. 5.69 kg/s b. 4.69 kg/s

4.187 50

30

c.    6.69 kg/s d.    7.69 kg/s

Solution:

∆

500

5.97

/

Then; the volume of water is: .

/

/

Thus; .



/

39. Supplementary Problem: A 500 kW refrigeration system is used to produce cooled water from 24°C to 3°C. Calculate the mass flow rate of water in kg/s. a. 5.69 kg/s b. 4.69 kg/s

c.    6.69 kg/s d.    7.69 kg/s

Solution:

∆

500

4.187 24

(a)

.

3

Thus;

/

40. Supplementary Problem: A vapor compression refrigeration system is designed to have a capacity of 100 TOR. It produces chilled water from 22°C to 2°C. Its actual coefficient of performance is 5.86 and 35% of the power supplied to the compressor is lost in the form of friction and cylinder cooling looses.

Determine the size of the electric motor required to drive the compressor in kW and the volume flow rate of chilled water is L/s.   a. 92.31 kW, 4.199 L/s   b. 90.71 kW, 5.277 L/s

c.  93.75 kW, 5.724 L/s d.   91.75 kW, 7.575 L/s

Solution:

.

5.86



6



.35

0.65



60

.

.

Thus;

92.31

Solving for the volume flow of chilled water:

∆

100 3.516

4.187 22

4.199

2

/

Thus; .

/ /

4.199 /

Thus; (a)

.

,

.

/

41. Supplementary Problem: A four cylinder, single‐acting, V‐type compressor with 8 cm and 10 cm stroke operates at 600 rpm. It used in a Freon‐12 vapor compression system with condenser and evaporator pressure

of 725.5 kPa and 189.5 kPa respectively. If the compression is dry and isentropic, the clearance is 2 percent and the there is no subcooling or superheating (before compression) of the refrigerant, determine the refrigerating capacity of the compressor in tons. a. 7.31 TOR b. 8.54 TOR

c.  7.54 TOR d.   8.31 TOR

Solution:

Properties of Freon‐12

@

14°

345.365 /



@

14°

0.0878951

/

@ 29° @ 725.5

227.557 /

@29°

368 /

And



345.365







∶

Solving for the refrigerating Capacity



227.557

Solving for m :



0.0878951

.

0.10 4

     Where;

1

1.02

0.9544

.

0.02

.

.



Then;

0.0878951

0.2183

0.9544 /

.

0.10 4



Thus;   0.2183 345.365

(a)

25.72

227.557

.

42. Supplementary Problem:

The dryness fractions of the entering and leaving the evaporator of a refrigerating plant are of at the 0.28 and 0.92 respectively. If the specific enthalpy of the evaporation ( evaporator pressure is 290.7 kJ/kg, Calculate the mass of ice at ‐5°C that would theoretically be through the machine is 0.5 kg/s.                                  made per day from water at 14°C when the mass flow of   Note:    Specific heat of water = 4.2 kJ/kg‐K                 Specific heat of fusion of ice = 2.04 kJ/kg‐K         Enthalpy of fusion = 335 kJ/kg‐K                   a. 17.89 tons/day b. 18.89 tons/day

c.  19.89 tons/day   d.   20.89 tons/day

Solution: through evaporator:

Specific enthalpy gain of



0.92

0.28



0.64 290.7

186.05 /

0.92

0.28







Heat to be extracted from water to make 1 kg of ice:

                  1 4.2 14





404 /



0.50

0

335



Then;



1 2.04 0

5

/







404

0.5 186.05

0.23

/

Thus; the mass of ice in tons per day:

0.23



.

(b)

/





43. Past ME Board Problem A vapor compression refrigeration system has a 30 kW motor driving the compressor. The compressor inlet pressure and temperature are 64.17 kPa and ‐20°C respectively and discharge pressure of 960 kPa. Standard liquid enters the expansion valve. Using Freon‐12 as refrigerant, determine the capacity of the unit in tons of the refrigeration. a.   17.145 TOR b. 18.145 TOR

c.   19.145 TOR d.   20.145 TOR

Solution:


345 /



398 /



238.5 /



Solving for the RefrigeratingCapacity:





345

          238.5

Solving for m:



   30

398



0.566



0.566 345



60.28

          345

Thus; 238.5           17.145



(a) 17.145 TOR 44. Past ME Board Problem A refrigerating system operates on the reversed Carnot Cycle. The higher temperature of the refrigerant in the system is 120°F and the lower is 10°F. The capacity is 20 tons. Neglect losses. Determine the network in Btu/min. a. 935.21 Btu/min b. 457.57 Btu/min

c.   745.71 Btu/min d.   765.81 Btu/min

Solution:

∆



Where;



120



10

460

580°

460

470°

Solving for ∆ :

  ∆

∆

∆



/

8.511



/

°

Thus;



580

470 8.511 .

(a)

/

45. Supplementary Problem: What is the coefficient of performance of a vapor compression refrigeration system with the following properties: Enthalpy at suction is 190 kJ/kg; enthalpy after compression is 210 kJ/kg. The enthalpy after condensation is 60 kJ/kg. a. 4.5 b. 5.5

Solution:

c.   6.5 d.   3.5



                Thus;

.

(c)

46.  Supplementary Problem:

A refrigerating machine is driven by a motor of output power 2.25 kW and 2.5 tons of ice at ‐7 °C made per day from water at 18 °C. Calculate the coefficient of performance of the machine and express its capacity in  of tons of ice per 24 hours from and 0 °C, taking the following values:

Specific heat of water = 4.2 kJ/kg‐K                 Specific heat of fusion of ice = 2.04 kJ/kg‐K         Enthalpy of fusion = 335 kJ/kg‐K                   a. 5.476, 3.17 tons/day b. 4.476, 3.17 tons/day

c.   5.476, 4.17 tons/day   d.   4.476, 4.17 tons/day

Solution;

Where;

2.25

Solving for

;

.

4.2 18

0.029 424.88

0

12.32

Then;

. .

5.476

Solving for m in tons per day;





2.5

424.88

335

2.04 0

7

2.5

335

424.88

Then;   3.17

/

Thus; .

(a)

&

.

/

47. Supplementary Problem: Determine the heat extracted from 2000 kg of water from 25°C to ice at ‐10°C. a. 621, 150 kJ b. 721, 150 kJ

c.   821, 150 kJ d.   921, 150 kJ

Solution:

Where;

0

209,350    2000 335

2000 4.187 25

Thus;

670, 000

2000 2.09 0

10

41,800 209, 350

(d)

,

670, 000

41, 800

48. Supplementary Problem: A single acting, twin cylinder, ammonia compressor with bore equal to stroke is driven by an engine at 250 rpm. The machine is installed in a chilling plant to produce 700 kW of refrigeration at ‐18°C evaporating temperature. At this temperature the cooling effect per kg mass is 1160 kJ.

The specific volume of the vapor compressor is 0.592 m³ per kilogram. Assume 85% volumetric efficiency, determine the bore in mm. a. 400 mm b. 300 mm



c.   450 mm d.   500 mm

Solution:

2

6.545

Solving for the piston displacement,



. .

.

:

From:

700

          1160

0.603

/

0.42 ³/

Then; Thus;

0.42

6.545 ³

(a) D = 0.40 m = 400 mm 49.  Past ME Board Problem Saturated vapor Freon‐12 refrigerant at 219.12 kPa leaves the evaporator and enters the compressor at ‐5°C. The refrigerant leaves the condenser as saturated liquid at 25°C and enters the expansion valve at 22°C. Heat rejected from the condenser amount to 74 kW. The work to the compressor is 55.5kJ/kg while the heat lost from the compressor is 4.2kJ/kg. If 1.15 kJ/kg 0f heat are lost in the piping between the compressor and condenser, calculate the refrigeration capacity in tons. a. 15.06 TOR b. 17.76 TOR Solution:

c.   14.57 TOR d.   12.75 TOR


@219.12

@ 25°

5°

223.65 /

@ 22°

4.2

401.30 /



347.13 /

By energy balance in the compressor:





220.75 /

@ 219.12

350 /

55.5

By energy balance in the piping from the compressor to condenser;

1.15

400.15 /

By energy balance in the condenser;

223.65

400.15

176.5 /

Solving for the mass flow rate: 0.419

/

.

/

/

Thus; the refrigerating capacity:

                      (a)



0.419 347.13 .

50.    Past ME Board Problem

220.75        .

The mass flow of water entering the condenser is 20 kg/s. If the temperature difference between the entrance and exit temperature is 20°C, determine the rejected heat in the condenser. a. 1, 674.80 kW b. 1, 774.80 kW Solution:

c.   1, 574.80 kW d.   1, 884.80 kW

∆

20 4.187 20

Thus; (a)

,

.

51. Supplementary Problem: The cooling load of a small walk‐in freezer has been calculated to be 1.10 tons of refrigeration at ‐30°C. A compressor and motor must selected to handle the load. The following conditions are given: Refrigerant   F‐12 Compressor, rpm 600 Motor, rpm 1800 Compressor discharge pressure 800kPa   Liquid receiver temperature 20°C Assume dry and isentropic compression, compressor volumetric efficiency of 80%, mechanical efficiency of 85%, and power transmission efficiency of 90%. Calculate the displacement of the compressor in cm³. a. 641.40 b. 651.40

c.   661.40 d.   671.40

Solution:

Properties of Freon‐12                                                                                                                                                     @ 800 33

338.143 /

375 /





218.321 / 0.159375 ³/



Solving for

:

Where: .

. 322

Then:

.

.

.

.



.

0.006414

.

/



.

0.0006414

Thus;   (a)

.

³/

52. Supplementary Problem: The refrigerant leaves the compressor and enters the condenser of a Freon‐12 refrigerating plant at 5.673 bar and 50 and leaves the condenser as saturated liquid at the same pressure. At compressor suction the pressure is 1.826 bar and temperature 0 . Calculate the coefficient of performance. a. 3.09 b. 4.09

c.   5.09   d.   6.09

Solution:


At 5.673 bar, Sat. temp = 20 Thus, At 50 refrigerant is superheated by 30 , and the compressor   216.75 / discharge, At 1.826 bar, Sat. temp. = ‐15    thus, At 0 refrigerant is superheated by 15 190.15 / 54.87 /    54.87 /

Then;

.

.

.

.

Thus;

.

(c)

53. Supplementary Problem: An industrial plant requires 10 kg/s to cool water from 30 to 1 . Find the tons of refrigeration required.   a. 345.34 b. 245.34

c.   145.34 d.   445.34

Solution:

∆

10 4.187 30

1, 214.23

345.34

1

Thus;   .

(a)

54. Supplementary Problem: Air flowing at a rate of 2.5 kg/s is heated in a heat exchanger from ‐10 to 30 . What is the rate of heat transfer?   a. 100 kW b. 150 kW

c.   200 kW d.   50 kW

Solution:

∆

2.5

100

1.0

30

10



Thus;

(a)

73. Supplementary Problem To cool farm products, 300 kg of ice at -4.4 0C are placed in bunker. Twenty four hours later, the ice have melted into water at 7.2 0C. What is the average rate of cooling provided by the ice in kJ/hr? a. 2679.28 kJ/hr

c. 3679.8 kJ/hr

b. 5679.8 kJ/hr

d. 4679.28 kJ/hr

Solution

2.09 0

4.4

335

4.187 7.2

0

thus; (d)

4679.28 /

74. Supplementary Problem The combined loads of an ice and cold storage are 25 tons of ice per day and 137,000 kJ/hr, respectively. Refrigeration required per ton of ice is 1.925. Ammonia compressor carrying these combined loads operates between -14 0C and 42 0C liquefaction. Determine the number of units of 7 multiple tube condenser each unit made up of 200 mm pipe shell where there are 7 pieces extra strong 50 mm tubes inside each pipe shell. Length is 6 m condensing water enters at 29 0C and leaves at 38 0C. U = 539 W/m2 – K, LMTD = 39.22 0C and cross flow factor = 0.75. a. 3 units

c. 5 units

b. 4 units

d. 6 units

Solution Properties of Ammonia: h1 = 1427.7 kJ/kg h2 = 1714.0 kJ/kg

h3 = h4 = 383.5 kJ/kg

Let: n = no. of units Q = heat gained by cooling water Q1 = heat transferred per unit then; Solving for Q: kW refrigeration for cold storage = 137,000 kJ/hr (1hr/3600s) = 38.06 kW kW refrigeration for ice plant

= 25(3.516)(1.925) = 169.21 kW

QT = Total kW = 38.06 + 169.21 QT = 207.27 From: m= m=

. .

.

m = 0.1985 kg/s then; Q = m (h2 – h1) = 0.1985 (1714 – 383.5) = 264.12 kW

Solving for Q1: Area of each unit, A: A = π (O.D.) L N = π (0.06) (6) (7) = 7.92 m2 then; Q1 = A U F (LMTD) = (7.92)(539)(0.75)(39.22) = 125,569.11 W = 125.57 kW/units thus; n=

=

. .

/

n = 2.10 units ( or n must be 3 units ) (a) n = 3

75. Past ME Board Problem Calculate the power required by a system of one compressor serving two evaporators. One evaporator carries a load of 35 kW at 10 0C and the other a load of 70 kW at -5 0C. a back pressure valve reduces the pressure in the 10 0C evaporator to that of the -5 0C evaporator. The condensing temperature is 37 0C. if the refrigerant is ammonia , then what is the COP. a. 4.33

c. 6.33

b. 5.33

d. 3.33

Properties of Ammonia: h3 = h4 = h7 @ 37 0C = 375.9 kJ/kg h5 = h6 = hg @ 10 0C = 1471.6 kJ/kg h8 = hg @ -5 0C = 1456.2 kJ/kg h2 = h @ 1432 kPa (Psat @ 37 0C) and S2 = S1 = 1665 kJ/kg then;

COP =

.

COP = Solving for WC: m4 =

=

.

.

= 0.0319 kg/s m7 =

=

.

= 0.0648 kg/s m1 = m6 + m8 = m4 + m7

.

.

= 0.0967 kg/s By Energy Balance: m1h1 = m6h6 + m8h8 0.0967h1 = 0.0319(1471.6) + 0.0648(1456.2) h1 = 1461.3 kJ/kg then; WC = m (h2 – h1) = 0.0967(1665 – 1461.3) = 19.7 kW thus; COP =

.

(a) COP = 5.33

76. Past ME Board Problem Twenty pounds of water at an initial temperature of 80 0F are heated until the temperature is increased to 190 0F. Compute the quality of heat energy supplied. a. 2200 BTU

c. 2400 BTU

b. 2300 BTU

d. 2500 BTU

Solution Q = m∆t = (20lb)(1 thus; (a) Q = 2200 Btu

)(190-80)0F

77. Supplementary Problem Suppose that 30 gpm of water are removed from 60 0F to 40 0F. Calculate the heat energy removed in Btu per hour. a. -299,880 Btu/hr

c. -199,880 Btu/hr

b. -399,880 Btu/hr

d. -499,880 Btu/hr

Solution Q = m∆t = m(1)(40-60) Solving for m: m = ρV = (8.33

)(30

)(60

= 14,994

thus; Q = (14,994)(1)(-20) (a) Q = -299,880

78. Supplementary Problem If the latent heat of water is 144 Btu/lb, determine the quantity of latent heat given up by 10 lb of water at 32 0F when it freezesinto ice at 32 0F. a. 1550 Btu

c. 2880 Btu

b. 1440 Btu

d. 3100 Btu

Solution Q = mLf = 10 lb (144 thus;

)

(b) Q = 1440 Btu

79. Supplementary Problem Compute the cooling rate (energy flow rate in Btu/hr) produces by ice melting at the rate of 150 lb/hr. a. 30000 Btu/hr

c. 21,600 Btu/hr

b. 10,530 Btu/hr

d. 15,000 Btu/hr

Solution Q = mLf = (150 )(144

)

thus; (c) Q = 21,600 Btu/hr

80. Supplementary Problem Twenty kilograms of water at initial temperature of 25 0C are heated until the temperature is increased to 80 0C. Compute the quantity of heat energy supplied. a. 4,605.7 kJ

c. 2,000.1 kJ

b. 4,000.2 kJ

d. 2,302.85 kJ

Solution Q = m∆t = (20 kg)(4.187 thus; (a) Q = 4,605 kJ

)(80-25) K

81. Supplementary Problem One-tenth m3 of water is cooled from 39 0C to 2 0C. Determine the quantity of heat energy rejected by the water. a. 15,491.90 kJ

c. 17,321.90 kJ

b. 14,591.90 kJ

d. 18,231.80 kJ

Solution Q = m∆t =

1000

4.187 35

10

thus; (a) Q = 15,491.90 kJ

82. Supplementary Problem Suppose that 30 kg/s of water are cooled from 35 0C to 10 0C. Compute the required energy flow rate in kJ/s. a. 3140.25 kW

c. 3457.75 kW

b. 3240.25 kW

d. 3567.25 kW

Solution Q = m∆t = 30

4.187

35

10

thus; (a) Q = 3140.25 kJ/s or 3140.25 kW

83. Supplementary Problem Compute the cooling rate produced by ice melting at the rate of 150 kg/hr. a. 737.5 kW

c. 937.5 kW

b. 837.5 kW

d. 637.5 kW

Solution Q = mLf = m 335 where: m = 1.50 kg/hr = 2.50 kg/s Q = 2.5

335

thus; (b) Q = 837.5 kJ/s or kW

84. Supplementary Problem Seventy-five hundred pounds of fresh beef enter a chilling cooler at 102 0F and are chilled to 45 0F each day. Compute the product load in Btu per 24 hours. The specific heat of beef above freezing is 0.75 Btu/lb-0F. a. 320,600 Btu

c. 220,600 Btu

b. 420,600 Btu

d. 520,600 Btu

Solution Production Load = (7500)(0.75)(102-45) = 320,600 Btu per 24 hrs (a) Product Load = 320,600 Btu per 24 hrs

85. Supplementary Problem Calculate the piston displacement of a two cylinder compressor rotating at 1450 rpm if the diameter of the cylinder is 2.5 in. and the length of stroke. a. 16.48 ft3/min

c. 14.48 ft3/min

b. 15.48 ft3/min

d. 17.48 ft3/min

Solution VD =

.

=

2 1450 2

= 28,470.68 in.3/min thus; (a) VD = 16.48 ft3/min

86. Supplementary Problem Calculate the compression ratio of an R-12 compressor when the suction temperature is 20°F and the condensing temperature is 100°F. a. 4.68

c. 2.68

b. 1.68

d. 3.68

Solution: Compression =

R= Where: At 20°F, Ps = 35.75 psi At 100°F, Pd = 131.6 psi

Thus; R=

. .

(d) R = 3.68

87. Supplementary Problem Determine the shaft power required by the compressor if the theoretical power is 2.66 Hp and the overall efficiency of the compressor is 80%. a. 2.22 Hp

c. 4.44 Hp

b. 3.33 Hp

d. 1.11 Hp

Solution: Let eo = overall efficiency Wc = compressor theoretical Ws = shaft power

Then: eo = 0.80 =

.

Thus; (b) Ws = 3.33

88. Supplementary Problem A refrigeration compressor having a 10 in. flywheel is driven by a four-pole, alternating current motor. If the diameter of the motor pulley is 4 in., determine the speed of the compressor. a. 700 rpm

c. 600 rpm

b. 500 rpm

d. 800 rpm

Solution: N1D1 = N2D2 Where: N1 = speed of the compressor D1 = diameter of the compressor flywheel N2 = speed of the compressor driver D2 = diameter of the driver pulley Note: If the compressor driver is a four-pole, alternating current motor operating on 60 cycle power, the approximate driver speed is 1750 rpm. For a two-pole, alternating motor the approximate speed is 3500 rpm. N(10) = (1750)(4) Thus; (a) 700 rpm

89. Supplementary Problem Determine the estimated condenser load for an open-type compressor having a cooling capacity of 16,500 BTU/hr and a heat rejection factor of 1.32. a. 22,280 BTU/hr

c. 21,780 BTU/hr

b. 20,780 BTU/hr

d. 19,780 BTU/hr

Solution: Condenser Load = Compressor Capacity x Heat rejection factor = (16,500)(1.32) = 21,780 BTU/hr Thus;

(c) Condenser Load = 21,780 BTU/hr

90. Supplementary Problem If the load on the water-cooled condenser in 150,000 BTU/hr and the temperature rise of the water in the condenser is 10°F, what is the quantity of water circulated in gpm. a. 30

c. 20

b. 40

d. 50

Solution: Q = mΔt 150,000 = m(1)(10) m = 15,000 lb/hr Solving for V in gpm: V=

=

, .

/ /

V = 1800.72 gal/hr = 30.01 gal/min Thus; (a) V = 30 gpm

CHAPTER VI

12. Past ME Board Problem A coil has an inlet temperature of 60 and outlet of 90 . If the mean temperature of the coil is 110 , find the by factor of the coil. a. 0.20 b. 0.30

c. 0.40 d. 0.50

Solution:

By factor =

BF = Thus; (a) BF = 0.40

13. Past ME Board Problem If the latent and sensible heat loads are 20 kW and 80 kW respectively, what is the sensible heat ratio? a. 0.80 b. 0.60

c. 0.70 d. 0.90

Solution: Let: SHR = sensible heat ratio SHR =

Thus; (a) SHR = 0.80

=

14. A room being air conditioned is being held at 25 dry bulb and 50% relative humidity. A flow rate of 5 ⁄ of supply air at 15 dry bulb and 80% RH is being delivered to the room to maintain that steady condition at 100 kPa. What is the sensible heat absorbed from the room air in kW? a. 50.8 b. 60.8

c. 40.5 d. 70.9

Solution: =m

(

)

Solving for m: PV = mRT 100(5) = m (0.287)(15 + 273) m = 6.049 kg/s thus; = (6.049)(1.003)(25 – 25) 60.80 kW

(a)

15. Supplementary Problem Compute the pressure drop of 30 air flowing with a mean velocity of 8 m/s in a circular sheet-metal duct 300 mm in diameter and 15 m long. Use a friction factor, f = 0.02 and 1.1644 / . a. 37.26 Pa b. 25.27 Pa

c. 29.34 Pa d. 30.52 Pa

Solution: ∆ = ∆ = Thus;

.

. .

(a) ∆

37.26 Pa

16. Supplementary Problem A pressure difference of 350 Pa is available to force 20oC air through a circular sheetmetal duct 450 mm in diameter and 25 m long. At 20oC, 1.204 / and take friction factor, f = 0.016. Determine the velocity. a. 25.57 m/s b. 27.55 m/s

c. 28.54 m/s d. 24.85 m/s

Solution: ∆ = 350 =

.

. .

Thus; (a) V = 25.57 m/s

17. Supplementary Problem A duct 0.40 m high and 0.80 m wide suspended from the ceiling in a corridor, makes a right angle turn in the horizontal plane. The inner radius is 0.2 m and the outer radius is 1.0 m measured from the same center. The velocity of air in the duct is 10 m/s. Compute the pressure drop in this elbow. Assuming; f = 0.3, 1.204 / and L = 10 m. a. 341 Pa b. 441 Pa Solution: ∆ = Where: =

(for rectangular duct)

c. 143 Pa d. 144 Pa

.

=

.

. .

= 0.53 m Thus; ∆ = (a) ∆

.

. .

341 Pa

18. Supplementary Problem A rectangular duct has a dimension of 0.25 m by 1 m. Determine the equivalent diameter of the duct. a. 0.40 m b. 0.80 m

c. 0.70 m d. 0.30 m

Solution: = =

. .

Thus; 0.40 m

(a)

19. Supplementary Problem A 0.30 x 0.40 m branch duct leaves a 0.30 x 0.60 main duct at an angle of 60°. The air temperature is 20 . The dimensions of the main duct remain constant following branch. ⁄ . What is the pressure downstream in the main duct. The flow rate upstream is 2.7 Note: at 20 , 1.2041 / . a. 346 Pa b. 436 Pa Solution: Pressure Loss in the main duct:

c. 634 Pa d. 643 Pa

0.4 1

=

and

Solving for

:

=

=

1.3

2.7 = ⁄

= 1.4

=

.

=

.

.

= 7.78 m/s

=

=

. .

.

= 15 m/s

=

=

. .

.

= 10.43 m/s Then; =

.

.

.

1

.

= 3.38 From: Bernoulli Equation:

2 9.81 1.2041 Thus;

250 3.38 9.81 1.02041

15

7.78 2 9.81

345. 64

(a)

20. Supplementary Problem A sudden enlargement in a circular duct measures 0.20 m diameter upstream and 0.40 m downstream. The upstream pressure is 150 Pa, downstream pressure is 200 Pa. What is the flow rate of 20˚C air through the fitting? Use p = 1.02041 kg/m3. a. 0.49 m3/s

c. 0.38 m3/s

b. 0.83 m3/s

d. 0.94 m3/s

Solution: Q = AuVu Solving for Vu: 1

Ploss = where: =(

.

2

.

= 0.25

then: (200-150) =

.

1

0.25

= 12.15 m/s Thus; Q=

Л

.

(12.15)

(a)Q = 0.38 m3/s

21. Past ME Board Exam Water at 55oC is cooled in a cooling tower which has an efficiency of 65%. The temperature of the surrounding air is 32oC dry bulb and 70% relative humidity. The heat

dissipated from the condenser is 2,300,000 kJ/hr. Find the capacity in liters per second of the pump used in the cooling tower. a. 8.50 L/s

c. 7.60 L/s

b. 6.80 L/s

d. 6.70 L/s

Solution: Pump Capacity = m ѵf@t4: Solving for m: e= from psychometric chart: At 32oC and 70% RH: = 27.40oC 0.65 =

.

= 37.06oC Using energy balance is the condenser: m(t3 – t4) =

,

m(4.187)(55 – 37.06) =

,

,

,

m = 8.51 kg/s From steam table at t4 = 37.06oC: Ѵf = 1.0068 L/kg Thus; Pump Capacity = (8.51 kg/s) (1.0068 L/kg) (a) Pump Capacity = 8.57 L/s

22. Past ME Board Exam An atmospheric cooling tower is to provide cooling for the jacket water of a four stroke, 800 kW Diesel generator. The cooling tower efficiency is 60% at a temperature approach of 10oC. If the ambient air has a relative humidity of 70% and dry bulb temperature of 32oC, determine the cooling tower supplied to the diesel engine in liters per hour. Generator efficiency is 97% useful work = 30% and cooling loss = 25%. a. 39,800 L/hr

c. 45,700L/hr

b. 35,700 L/hr

d. 49,800 L/hr

Solution: Volume of water = m Ѵf at t4 : Solving for m and Ѵf : At tdb1 = 32oC and RH = 70% twb = 27.45oC tapproach = t4 – 27.45 10 = t4 – 27.45 t4 = 37.45oC

Brake power of engine = Power input to generator =

.

Heat supplied to Engine, QA:

= 824.74 kW

QA =

. .

= 2749.14 kW

Heat absorbed by Cooling water, Qw: Qw = 0.25 (2749.14) = 687.285 mww(t3 – t4) = 687.285 mw(4.187)(52.4 – 37.45) = 687.285 mw = 10.98 kg/s = 39,527.14 kg/hr Specific Volume of water at 37.45oC, Ѵf = 1.007 L/kg Thus; Vol. of cooling water, Vw: Vw = 39,527.14(1.007) L/hr (a) Vw = 39,803.83 L/hr

23. Past ME Board Exam Fifty gallons per minute of water enters a cooling tower at 46oC. Atmospheric air at 16oC db and 55% RH enters the tower at 2.85 m3/s and leaves at 32oC saturated. Determine the volume of water that leaves the tower. a. 4.10 L/s

c. 2.10 L/s

b. 3.10 L/s

d. 5.10 L/s

Solution: Volume water leaving the tower, V4: V4 = m4(Ѵf at t4) Solving for m4 and Ѵf: At tdb1 = 16oC and 55%RH Ѵ1 = 0.828 m3/kg

At 32oC and 100% RH ὼ2 = 0.0308 kg/kg

ὼ1 = 0.0056 kg/kg

h2 = 110.9 kJ/kg

h1 = 32 kJ/kg

h3 = hf at 46oC

ma =

=

ѵ

.

= 192.62 kJ/kg

.

Ѵ3 = 0.0010103 m3/kg

= 3.44 kg/s m3 =

. .

= 3.12 kg/s

By mass balance: m3 – m4 = ma(W2 – W1) 3.12 – m4 = 3.44 (0.0308 – 0.0056) (a) m4 = 3.09 kg/s By energy balance: m3h3 – m4h4 = ma(h2 – h1) 3.12(192.62) – 3.09h4 = 3.44(110.9 – 32) h4 = 106.65 kJ/kg From Steam table 1: t4 = 25.42oC Ѵ4 = 1.0031 L/kg Thus; volume of water leaving, Vw: Vw = 3.09(1.0031) (a) Vw = 3.10 L/s

24. Past ME Board Exam A 250,000 kg/hr of water 35oC enters a cooling tower where it is to be cooled to 17.5oC. The energy is to be exchanged with atmospheric air entering the units at 15oC and leaving the unit at 30oC. The air enters at 30% RH and leaves at 85% RH. If all process

are assumed to occur at atmospheric pressure, determine the percentage of total water flow that is make up water. a. 2.22%

c. 4.44%

b. 3.335

d. 1.11%

Solution: Percentage make-up water =

,

Solving for mass of make-up water: At 15oC and 30% RH: h1 = 23.02 kJ/kg w1 = 0.0033 kg/kg At 30oC and 85% RH: h2 = 89.01 kJ/kg w1 = 0.0233 kg/kg Heat lost by water = Heat gained by air mww∆ = ma(h2 – h1) 250,000(4.187)(35 – 17.5) = ma(89.01 – 23.02) ma = 277,589.41 kg/hr then; the mass of make-up water, ms: m5 = ma(w2 – w1) m5 = 277,589.41(0.0233 – 0.0033) = 5,551.79 kg/hr Thus; the percentage make-up water, %make-up =

,

. ,

(a) %make –up = 0.0222 or 2.22%

25. Past ME Board Exam How much refrigeration capacity is required to cool 56.67 m3 of air per minute from 29oC to 21oC. Assume that the cooled air is saturated. a. 4.76 TOR

c. 3.76 TOR

b. 1.76 TOR

d. 2.76 TOR

Solution: Refrigeration Capacity, QA: QA = ma(h1 – h2) Solving for ma: From psychometric chart: At 21oC db and 100% RH h1 = 70 kJ/kg ѵ1 = 0.875 m3/kg At constant SH intersecting 29oC db: h2 = 70 kJ/kg

ma = =

. .

/

/

ma = 64.766 kg/min = 1.079 kg/s Then; QA = 1.079(70-61) = 9.715 kW Thus; (a) 2.76 tons of refrigeration

26. Supplementary Problem Find the refrigeration capacity required to cool 29 cubic meter per minute from 29oC to 18oC if air from the outside has an RH of 90%. a. 2.9 TOR

c. 4.9 TOR

b. 3.9 TOR

d. 5.9 TOR

Solution: Qn = ma(h1 – h2) Solving for ma: From psychometric chart: At 29oC db and 90% RH h1 = 88.45 kJ/kg At 218oC db and 100% RH h2 = 50.45 kJ/kg

ѵ1 = 0.886 m3/kg ma =

.

= 32.73 kg/min

ma = 0.546 kg/s then; QA = 0.546(88.45 – 50.45) = 20.75 kW Thus; (a) QA = 5.9 tons of refrigeration 27. Past ME Board Exam The temperature of the air in a dryer is maintained constant by the use of steam coils within the dryer. The product enters the dryer at the rate of one metric ton per hour. The

initial moisture content is 3 kg moisture per kg of dry solid and will be dried to moisture content of 0.10 kg moisture per kg dry solid. Air enters the dryer with a humidity ratio of 0.016 kg moisture per kg of dry air and leaves with a relative humidity of 100% while the temperature remains constant at 60 . If the total pressure of the air is 101.3 kPa, ⁄ determine the capacity of the forced draft far to handle this air in . a. 85.75

⁄

c. 55.87

⁄

b. 87.55

⁄

d. 58.75

⁄

Capacity of fan = :

Solving for m and At point 1: .

=

.

0.16 =

.

2.54 1 101.3

2.54 1 = 0.287 (60 + 273) 1

⁄

0.968

At point 2: = 19.94 kPa

@

= RH (

@

= (1)(19.94) = 19.94 kPa = =

.

.

.

. .

= 0.1524 kg/kg At point 3:

)

= 250 kg At point 4: .

Moisture content =

.

= 0.0909 or 9.09% = 0.0909 = 0.0909 = 275

250 /

By mass balance in the dryer:

=

= =

.

.

Thus; Capacity of Fan = 5315.25 (0.968) ⁄

= 5142.16

(a) Capacity of the Fan = 85.75

⁄

Alternate Solution: Fan Capacity = From psychrometric chart: At 60

and 0.016 humidity ratio: ⁄

= 0.968

From steam table at 60 , =

.

.

.

0.1524

.

/

Moisture removed = m (

19.94

3 (350) – (0.10)(250) = m (0.154- 0.1016) M = 5315.25 kg/hr Thus; Fan Capacity = 5315.16 (0.968) ⁄

= 5145.16

⁄

(a) Fan Capacity = 82.75

28. Past ME Board Exam Wet material containing 215% moisture (dry basis) is to be dried at the rate of 1.5 kg/s in a continuous dryer to give a product containing 5% moisture (wet basis). The drying medium consist of air heated to 373 K and containing water vapor equivalent to a partial pressure of 1.40 kPa. The air leaves the dryer at 310 K and 70% saturated. Calculate how much air will be required to remove the moisture. a. 50 kg/s

c. 60 kg/s

b. 55 kg/s

d. 65 kg/s

Solution: Amount of moisture removed = amount of moisture absorbed by air. Let m = rate of flow of dried product .

(1.5) = 0.95 m m = 0.501 kg/s

Amount of moisture removed = 1.5 – 0.501 = 0.999 kg/s Solving for W1: W1 =

.

=

.

. .

.

= 0.00871 kg/kg From psychrometric chart, W2 = 0.0289 kg/kg

Then, ma(0.0289 – 0.00871) = amount of moisture removed ma(0.0289 – 0.00871) = 0.999 thus; (a) ma = 49.48 kg/s

29. Past ME Board Exam One hundred fifty cubic meters of air per minute at 35oC dry bulb and 25oC wet bulb temperature are to be cooled to 21oC. Determine the refrigeration capacity. a. 10 TOR

c. 12 TOR

b. 11 TOR

d. 13 TOR

Solution: Refrigeration Capacity, QA: QA = ma(t1 – t2) Solving for ma: From psychrometric chart: 1

= 0.892 m3/kg,

ma =

.

then; QA = 2.8 (1) (35 - 21) = 39.24 kW Thus; (a) QA = 11.16 TOR

2

= 0.855 m3/kg

= 168.16 kg/min

30. Past ME Board Exam In an auditorium maintained at a temperature not to exceed 24oC and relative humidity not to exceed 60%, a sensible heat load of 132 kW and 78 kg of moisture per hour to be remove. Air is supplied to the auditorium at 18oC. How many kilograms of air must be supplied per hour? a. 79,200 kg/hr

c. 72,900 kg/hr

b. 97,200 kg/hr

d. 92,700 kg/hr

Solution: Qs = ma(t2 – t1) 132 = ma(1)(24 - 18) ma = 22 kg/s thus; (a) ma = 79,200 kg/hr

31. Supplementary Problem Eleven thousand three hundred kilograms per hour of water enters a cooling tower at 45oC. Atmospheric air at 16oC and 55 percent relative humidity enters the tower at the rate of 10,200 m3/hr and leaves at 32oC and saturated. Determine the mass of water evaporated per hour during the cooling process? a. 2,912.53 kg/hr

c. 1,292.53 kg/hr

b. 2,219.53 kg/hr

d. 1,912.53 kg/hr

Solution: Mass of water evaporated, mW: mW = ma (SH2 – SH1) From Psychrometric chart: Entering air at 16oC and 55% RH: V1 = 0.83 m3/kg; SH1 = 0.07 kg/kg Leaving air at 32oC and 100% RH (Saturated) SH2 = 0.307 kg/kg Mass of air entering the tower: 3

ma =

10,200 m /hr 0.83 m3 /kg

= 12,289.16 kg/hr then; mW = 12,289.16 (0.307 – 0.07) thus; (a) mW = 2,192.53 kg/hr 32. Supplementary Problem Water at 55 ̊C is cooled in a cooling tower which has an efficiency of 65%. The temperature of the surrounding air is 32 ̊C dry bulb and relative humidity of 70%. The heat dissipated from the condenser is 2,300,000 kJ/hr . Find the capacity in liters per second of the pump used in the tower. a. 8.66 L/s

c. 4.76 L/s

b. 8.76 L/s

d. 7.26 L/s

Solution: From Psychrometric Chart: At 32oC db and 70% RH

twb1 = 27.5oC The temperature of water leaving the tower can be determined by tower efficiency equation: Tower eff. = = 0.65 =

Actual cooling range Theoretical Cooling range ta -tb ta -twb1 55-tb 55-27.5

tb = 37.125 oC By Energy Balance in the condenser: QR = mw w ( ta – tb ) 2,300,000 = mw (4.187) (55-37.125) mw = 30,731.15 kg/hr Density of water at 55oC: ρw =

1 vf@55o C

=

1 0.0010146

= 985.6 kg/m3 Then; the capacity of the pump to be used in the cooling tower: Pump capacity =

kg hr kg 935.6 hr

30,731

(1000 (3600

L ) m3 s ) hr

thus; (a) Pump Capacity = 8.66 L/s 33. Supplementary Problem A dryer is to deliver 1000kg/hour of palay with a final moisture content of 10%. The initial moisture content in the feed is 15% at atmospheric condition with 32 ̊C dry bulb and 21 ̊C wet bulb. The dryer is maintained at 45 ̊ C while the relative humidity of the hot humid air from the dryer is 80%. If the steam pressure supplied to the heater is 2 Mpa, determine the air supplied to the dryer in mᵌ/hr. a. 1332.25 mᵌ/hr

c. 1223.25 mᵌ/hr

b. 1233.25 mᵌ/hr

d. 1523.13 mᵌ/hr

Solution: Amount of moisture removed = Amount of moisture absorbed; Let; m = amount of palay in wet feed

Solid in wet feed = solid in dried product 0.85 = 0.90 (1000) m = 1,058.83 kg/hr Amount of moisture removed: m = 1,058.83 -1000 = 58.28 From psychrometric chart: W1 = W2 = 0.0111 kg/kg; v2 = 0.915 m3/kg W3 = 0.0515 kg/kg then; the amount of moisture absorbed; = ma (W3 – W2) 58.823 = ma (0.0515 – 0.0111) ma = 1456.015 kg/hr Va = 1456.015 kg/hr (0.915 m3/kg) thus; (a) Va = 1332.25 m3/hr 34. Supplementary Problem Copra enters a dryer containing 60% water and 40% of solids and leaves with 5% water and 95% solids. Find the weight of water removed based on each pound of original product. a. 0.58 lb

c. 0.47 lb

b. 0.40 lb

d. 0.67 lb

Solution: Let; m = weight of original product per lb of wet feed Solid in wet feed = solid in dried product 0.95 = 0.40 (1) m = 0.42 lb thus; Weight of water removed = 1 – 0.42 (a) Weight of water removed = 0.58 35. Supplementary Problem In an air conditioning system, If the re-circulated air is three times the outside the mass of supply air is 20 kg/s, what is the mass of the outside air? a. 3 kg/s

c. 5 kg/s

b. 4 kg/s

d. 6 kg/s

Solution: mo + mt = ms mo + 3mo = 20 4mo = 20 thus; (a) mo = 5 kg/s 36. Supplementary Problem An auditorium is to be maintained at a temperature of 26 °C dry bulb and 50% RH. Air is to be supplied at a temperature not lower than 15 °C dry bulb. The sensible heat gain is 110 kW and the latent gain is 37.5 kW. Take ventilating air as 25% by weight of the air from the room, and is at 35 °C dry bulb and 60% RH. Determine refrigerating capacity in tons. a. 43.45

c. 63.28

b. 54.23

d. 76.34

Solution: Refrigeration Capacity = ms (h4 – h1) From psychrometric chart: h3 = 90.49 KJ/kg h2 = 53 KJ/kg Solving for ms: Qs = ms (t2 – t1) 110 = ms (1.0) (26 – 15) ms = 10 kg/s Solving for h3: QT = ms (h2 – h1)

110 + 37.5 = 10 (53 – h1) h1 = 38.25 KJ/kg Solving for h4: by mass balance: mo + mr = ms 0.25mr + mr = 10 mr = 8 kg/s by Heat Balance moh3 + mrh2 = msh4 [0.25(8)] (90.49) + 8 (53) = 10 h4 h4 = 60.50 KJ/kg thus; Refrigerating Capacity = 10 (60.50 – 38.25) = 222.48 kW Refrigerating Capacity = 63.28 Tons of Refrigeration 37. Supplementary Problem An assembly hall was to have an air conditioning unit installed which would be maintained at 26 °C dry bulb and at 50% RH. The unit delivers air at 15 °C dry bulb temperature and the calculated sensible heat load is 150 kW and the latent heat is 51.3 kW. Twenty percent by weight of extracted air is made up of outside air at 34 °C dry bulb and 60% RH, while 80% is extracted by the air conditioner from the assembly hall. Determine the air conditioners refrigeration capacity in tons of refrigeration and its ventilation load in kW. a. 83.22 TOR, 37.47 TOR

c. 89.56 TOR, 45.77 TOR

b. 76.43 TOR, 57.34 TOR

d. 56.78 TOR, 47.68 TOR

Solution: Refrigeration Capacity, QA: QA = ms (h4 – h1) Ventilation load, QV: QV = mo (h3 – h1) Solving for ms:

Qs = ms (t2 – t1) 150 = ms (1.0) (26 – 15) ms = 13.64 kg/s From psychrometric chart: h3 = 86.5 KJ/kg h2 = 53 KJ/kg Solving for h1: QT = ms (h2 – h1) 150 + 51.3 = 13.64 (53 – h1) h1 = 38.24 KJ/kg Solving for mr and mo: mr = 0.80 (13.64) mr = 10.91 kg/s mo = 0.20 (13.64) = 2.73 kg/s By heat balance: moh3 + msh4 2.736(86.5) + 10.91(53) = 13.64h4 h4 = 59.69 kJ/kg thus; QA = 13.64 (59.69 – 38.24) = 292.61 kW = 83.22 tons of refrigeration Qv = 2.73(86.5 – 38.24) = 131.75 kW = 37.47 tons of refrigeration (a) QA = 83.22 TOR, QV = 37.47 TOR 38. Supplementary Problem An air conditioned theater is to be maintained at 80 °F dry bulb temperature and 50% RH. The calculated total sensible heat load in the theater is 620,000 BTU/hr, and the latent heat load is 210,000 Btu/hr. The air mixture at 84 °F and 59 °F wet bulb temperature by chilled water cooling coils and delivered as supply air to the theater. Calculate the tons refrigeration required. a. 100.65 TOR

c. 142.67 TOR

b. 124.67 TOR

d. 112.60 TOR

Solution: Conditioner Capacity, QA: QA = ms (h4 – h1) Solving for ms: QT = ms (h2 – h1) From Psychrometric Chart: h4 = 35.82 Btu/lb h1 = 25.78 Btu/lb h2 = 31.35 Btu/lb then; 620,000 + 210,000 = ms (31.35 – 25.78) ms = 149,012.57 lb/hr thus; QA =

149,012.57 35.82 - 25.78 Btu/hr 12,000 Btu/hr TOR

(a) QA = 124.67 Tons of Refrigeration 39. Supplementary Problem Determine the quantity of heat required to raise 20 m ᵌ /min of air 20 °C and 80 percent relative humidity to 35 °C. a. 5 kW

c. 7 kW

b. 6 kW

d. 8 kW

Solution: From Psychrometric Chart: At tdb1 = 20oC and 80% RH h1 = 50kJ/kg v1 = 0.85 m3/kg h2 = 65.5 kJ/kg Q = m (h2 – h1) Solving for m: m=

V v

=

20 0.847

= 23.61 kg/min = 0.394 kg/s thus; Q = 0.394 (65.5 – 50) (a) Q = 6.10 kJ/s or kW 40. Supplementary Problem Determine the partial pressure of water vapor if the barometric pressure is 101.325 kPa and the humidity ratio is 0.05. a. 7.54 kPa

c. 5.74 kPa

b. 4.75 kPa

d. 5.47 kPa

Solution: W = 0.622

Pv Pt -Pv

0.05 = 0.622

Pv 101.325-Pv

thus; (a) 7.54 kPa 41. Supplementary Problem The evaporative condenser of an ammonia refrigeration plant has a water flow rate 226 kg/s and enters a natural draft cooling tower at 40 °C. The water is cooled to 29 °C by air entering at 38 °C db and 24 °C wb. The air leaves the tower as saturated at 40 °C db. Calculate the make-up water required in kg/hr. Water properties: At 49oC; hf = 167.48 kJ/kg

Air Properties: At 38oC db and 42oC wb

At 29oC; hg = 121.43 kJ/kg

h = 72.5 kJ/kg w = 0.013 kg/kg

At 40oC db saturated; h = 166 kJ/kg, w = 0.0488 kg/kg a. 8977

c. 8055

b. 8055

d. 8388

Solution: m = ma (W2 – W1) Solving for mass of air, ma: Heat absorbed by air = heat rejected by water ma (h2 – h1) = mw Cw ∆t ma (166 – 72.5) = 126 (4.187) (40 – 29) ma = 62.07 kg/s then; m = 62.07 (0.0488 – 0.013) = 2.22 kg/s thus; (a) m = 7999.08 kg/hr 42. Supplementary Problem Determine the absolute humidity (vapor density) of an air sample that has a dew point temperature of 45 °F if the value of the gas constant R for low pressure water vapor is 85.66 ft-lbm °R. The vapor pressure corresponding to a saturation temperature of 45 °F is 0.1475 psia. a. 0.000491 lb/ftᵌ

c. 0.000149 lb/ftᵌ

b. 0.000941 lb/ftᵌ

d. 0.000194 lb/ftᵌ

Solution: PV = mRT m V =

=ρ=

P RT

0.1475 (144) 85.66 (45+460)

thus; ρ = 0.000491 lb/ft3

43. Supplementary Problem A certain sample of the air has a temperature of 70 °F (partial pressure of 0.36 psia) and a dew point temperature of 50°F. The partial pressure of the water is

vapor corresponding to a 50 °F dew point temperature is 0.178 psia. Determine the relative humidity RH. a. 49.44%

c. 39.44%

b. 59.44%

d. 69.44%

Solution RH = =

Actual partial pressure partial pressure at saturation

0.178 0.36

x 100

x 100

thus; (a) RH = 49.44 % 44. Supplementary Problem Air at normal atmospheric pressure has a temperature of 70 °F and a dew point temperature of 50 ° F. Determine the saturation ratio of the air. The humidity ratios corresponding to dew point temperatures of 50 °F and 70 °F respectively are 0.00763 lb/lb and 0.01576 lb/lb respectively a. 58.51%

c. 38.31%

b. 28.21%

d. 48.41%

Solution: Saturation ratio = =

Wactual Wsaturation 0.00763 0.01576

x 100

x 100

thus; (c) Saturation ratio = 48.41 45. Supplementary Problem Determine the sensible heat of 5 lb of air having a dry bulb temperature of 70 °F and a humidity ratio of 0.0092 lb/lb. The latter corresponding to a dew point temperature of approximately 55°F. a. 94 BTU

c. 84 BTU

b. 48 BTU

d. 49 BTU

Solution: Qs = m (0.24 DB) = 5 (0.24)(70) thus; Qs = 84 BTU 46. Supplementary Problem If the total heat removed per pound of dry air is 0.10 Btu and the sensible heat removed per pound of dry air is 6 BTU. Compute the sensible heat factor (SHR). a. 0.50

c. 0.70

b. 0.60

d. 0.40

Solution: SHR =

QR QS

=

6 10

thus; (a) SHR = 0.60 47. Supplementary Problem Determine the approximate load on a cooling tower if the entering and leaving temperatures are 96 °F and 88 °F, respectively and the flow rate of the water over the tower is 30 gpm. a. 2500 Btu/min

c. 3000 Btu/min

b. 2000 Btu/min

d. 3500 Btu/min

Solution: Tower load = 8.33 V ∆t Btu/min = 8.33 (30) (96 – 88) thus; (b) Tower load = 2000 Btu/min

48. Supplementary Problem Determine the equipment standard air volume for 150 mᵌ/s of air having a dry bulb temperature of 15 °C.

a. 150 mᵌ/s

c. 170 mᵌ/s

b. 160 mᵌ/s

d. 180 mᵌ/s

Solution: Vs =

150 (21+273) (15+273)

thus; (a) Vs = 153 m3/s 49. Supplementary Problem Determine the quantity of water lost by bleed off if the water flow rate over the tower is 30 gpm and the range is 10 °F. Percent bleed-off required is 33%. a. 0.077 gpm

c. 0.099 gpm

b. 0.088 gpm

d. 0.066 gpm

Solution: The quantity of water lost by bleed-off, Vw = 0.33 (30) thus; (a) Vw = 0.099 gpm

50. Supplementary Problem The mass of an outside air is 50°C in an air conditioning unit is 60 kg. find the temperature after mixing if the outside air mixed with 40 kg with recirculated air at 35 °C. a. 44 °C

c. 52 °C

b. 39 °C

d. 47 °C

Solution: moto + mrtr = msts 60(50) + 40(35) = (60 + 40)ts thus; (a) ts = 44oC

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