Dja3032 - Chapter 2-1 6x2s13
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(DJA3032) INTERNAL COMBUSTION ENGINE
CHAPTER 2: PISTON ENGINE PROCESS ANALYSIS by MOHD SAHRIL BIN MOHD FOUZI, Grad. IEM (G 27763) DEPARTMENT OF MECHANICAL ENGINEERING
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
General objective: To understand the structure and various types of cycle engines. Specific objectives: At the end of this unit you should be able to: define the air standard cycle. define constant pressure () and constant volume (cv). draw p-v diagram of Otto cycle, Diesel cycle and Combined cycle. explain Otto cycle, Diesel cycle and Combine/dual cycle.
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Introduction : In this unit we are to discuss the meaning of standard cycle, heat supplied at constant volume and heat supplied at constant pressure. An internal combustion engine can be classified into three different cycles and they are Otto cycle, Diesel cycle and Dualcombustion cycle. Qin Pressure (P) P3
pv cons tan t
Qin P2
Pressure, P
3
2
2
P2 = P3
3
4
4
P4
P4
pv cons tan t
Qout
Qout P1
P1
1
1
V2
V1
Figure 2.1: P-V Diagram (Otto Cycle)
Volume ( v )
V2
V3
V1= V4
V Volume
Figure 2.2: P-V Diagram (Diesel Cycle) © MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
P
Qin 3
P3=p4 Qin
4 p.v conts
2 5
Qout
1 V2=V3
V1=V5
V
Figure 2.3: P-V Diagram (Dual/Combined Cycle)
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Air Standard Cycles The air standard cycle is a cycle followed by a heat engine which uses air as the working medium. Since the air standard analysis is the simplest and most idealistic, such cycles are also called ideal cycles and the engine running on such cycles are called ideal engines. In order that the analysis is made as simple as possible, certain assumptions have to be made. These assumptions result in an analysis that is far from correct for most actual combustion engine processes, but the analysis is of considerable value for indicating the upper limit of performance. The analysis is also a simple means for indicating the relative effects of principal variables of the cycle and the relative size of the apparatus.
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Assumptions: 1. The working medium is a perfect gas with constant specific heats and molecular weight corresponding to values at room temperature. 2. No chemical reactions occur during the cycle. The heat addition and heat rejection processes are merely heat transfer processes. 3. The processes are reversible. 4. Losses by heat transfer from the apparatus to the atmosphere are assumed to be zero in this analysis. 5. The working medium at the end of the process (cycle) is unchanged and is at the same condition as at the beginning of the process (cycle).
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
T-S Diagram
Figure 2.4: T-S Diagram for Otto Cycle
Figure 2.5: T-S Diagram for Diesel Cycle
Figure 2.6: T-S Diagram for Combined/Dual Cycle
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Compression ratio To give direct comparison with an actual engine the ratio of specific volume, v 1 / v2, is taken to be the same as the compression ratio of the actual engine, Compression ratio, rv = Pressure (P) P3
Clearance volume
P2
v1 v2
=
sweptvolum e clearencev olume clearencev olume
3 pv cons tan t
2
Clearance volume 4
TDC
P4 P1
V2 Minimum volume
BDC
1
Swept volume
V1
Figure 2.7: P-V Diagram for Otto Cycle
Swept volume
Volume ( v )
Maximum volume Figure 2.8: Animated 4 Stroke Engine © MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Example 2.1 An Otto cycle in a petrol engine with a cylinder bore of 55mm, a stroke of 80mm, and a clearance volume of 23.3 cm3 is given. Find the compression ratio of this engine. Solution: Cylinder bore, B = 55mm = 5.5 cm Stroke, S = 80mm = 8.0 cm Clearance Volume = 23.3 cm 3
swept volume = = =
Abstract the data from the question. Convert the data into centimetre unit.
Compression ratio, rv = v1 x B² x S 4 v2 x (5.5)² x 8 = sweptvolum e clearencev olume 4 clearencev olume 3 190.07 cm 3 = (190.07 + 23.3) cm 23.3 cm 3
= 9.16
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Self-Exercise Exercise 2.1 A petrol engine with a cylinder bore of 73 mm, a stroke of 95 mm, and a clearance volume of 26.3 cm 3 is given. Find the compression ratio of this engine. [Ans: 16.12] Exercise 2.2 An air engine is operated with cylinder bore 65 mm with the stroke of 73 mm. The clearance volume for this engine is 1/10 of swept volume. Calculate the compression ratio for this air engine. [Ans: 11.00] Exercise 2.3 Given compression ratio for a petrol engine is 10.5.The cylinder bore and stroke length for this engine are 69 mm and 83 mm. Calculate the clearance volume for this engine. [Ans: 32.66 cm3 ] Exercise 2.4 An otto cycle engine with compression ratio 9.5 operating with clearrance volume 23.45 cm 3. The stroke length for this engine is 83 mm, find the cylinder bore for this engine. [Ans: 5.53 cm]
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Otto Cycle Analysis Pressure (P) P3
Otto Cycle Process:
3
1 to 2 is isentropic compression with compression ratio v1 / v2, or rv .
pv cons tan t
Qin 2
P2
2 to 3 is reversible constant volume heating, the heat supplied Q1
4
P4
Qout P1
1 Volume ( v )
V1 V2 Figure 2.9: P-V Diagram for Otto Cycle
3 to 4 is isentropic expansion, v4 /v3 is similar to v1 /v2. 4 to 1 is reversible constant volume cooling, the heat rejected Q2.
Formula to find the thermal efficiency for Otto Cycle P1V1 T1V1
= P2V2
1
=
T2V2
Qin, Q1 = Cv(T3 ─ T2 )
1
= 1-
Qout, Q2 = Cv(T4 ─ T1 )
© MSF @ POLITEKNIK UNGKU OMAR
Qout, Q2 Qin, Q1
(DJA3032) INTERNAL COMBUSTION ENGINE Example 2.2 One petrol engine is working at a constant volume, the compression ratio is 8.5:1. Pressure and temperature at a beginning compression process is 101 kN/m 2 and 840 C. Temperature at the beginning of an expand process is14960 C. Calculate the temperature and pressure at the important points based on the Otto cycle. Solution: Pressure (P) P3
3
rv
pv cons tan t
Qin P2
V1 V4 = 8.5 V2 V3
*T1 = 84°C + 273K = 357 K
2
4
P1 = 101 kN/m²
P4
Qout P1
1
V2
V1
Volume ( v )
*T3 = 1496°C + 273K = 1769 K * Temperature must be convert into Kelvin
P-V Diagram for Otto Cycle
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Point 2 (1 to 2 is isentropic compression ) P1V1
= P2V2
T1V1
P2 = [ V1 ] x P1 V2 P2 = [8.5]
1.4
x 101 kN / m²
P2 = 2020.73 kN / m²
1
=
1
T2V2
1
V1 T2 = [ V2 ] T2 = [8.5]
x T1
0.4
x 357 K
T2 = 840.30 K
Point 3 (2 to 3 is reversible constant volume heating ) Constant volume
P2V2 = P3V3 T2 T3
P3 = T3 x P2 T2 P3 = 1769 x 2020.73 kN / m² 840.30 P3 = 4254.04 kN / m²
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Point 4 (3 to 4 is isentropic expansion ) P3V3
= P4V4
T3V3
P4 = [ V3 ] x P3 V4 P4 = [
1 8.5
1.4
]
x 4254.04 kN / m²
P4 = 212.63 kN / m²
1
=
T4V4
1
1
V3 T4 = [ V4 ]
x T3
0.4 T4 = [ 1 ] x 1769 K
8.5
T4 = 751.55 K
Qin, Q1 = Cv (T3 ─ T2 ) = 0.718 kJ/kg.K x ( 1769 K – 840.30 K) Qin, Q1 = 666.81 kJ/kg Qout, Q2 = Cv (T4 ─ T1 ) = 0.718 kJ/kg.K x ( 751.55 K – 357 K) Qout, Q2 = 283.29 kJ/kg
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Exercise 2.5 A four cylinder engine operates in Otto cycle, the volume is constant and the compression ratio is 9:1, beginning pressure is 105 KN/m2 , temperature is 83 ° C and final temperature is 1520 °C. Draw a p-v diagram and find the temperature and pressure for each point. Lastly calculate the efficiency of Otto cycle. Exercise 2.6 One petrol engine is working at a constant volume, the compression ratio is 8.5:1. Pressure and temperature at a beginning compression process is 101 kN/m 2 and 84 0C. Temperature at the beginning of an expand process is14960 C. Calculate the temperature and pressure at the important points based on the Otto cycle. Hence, calculate thermal efficiency for this engine.
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Solution for Exercise 2.5 V V4 rv 1 = 9 V2 V3
Pressure (P) P3
3 pv cons tan t
Qin
T1 = 83°C + 273K = 356 K
P2
2
4
P4
Qout
P1 = 105 kN/m²
P1
1
T3 = 1520°C + 273K = 1793 K
V1
V2
P-V Diagram for Otto Cycle
Point 2 (1 to 2 is isentropic compression ) P1V1
= P2V2
T1V1
1
=
T2V2
1
P2 = [ V1 ] x P1 V2
V1 T2 = [ V2 ]
P2 = [9]
T2 = [9]
1.4
x 105 kN / m²
P2 = 2275.77 kN / m²
1
0.4
x T1
x 356 K
T2 = 857.33 K © MSF @ POLITEKNIK UNGKU OMAR
Volume ( v )
(DJA3032) INTERNAL COMBUSTION ENGINE
Solution for Exercise 2.5 Point 3 (2 to 3 is reversible constant volume heating ) Constant volume
P2V2 = P3V3 T2 T3
P3 = T3 x P2 T2 P3 = 1793 x 2275.77 kN / m² 857.33 P3 = 4759.49 kN / m²
Point 4 (3 to 4 is isentropic expansion ) P3V3
= P4V4 P4 = [ V3 ] x P3 V4 P4 = [
1 9
1.4
]
x 4759.49 kN / m²
P4 = 219.59 kN / m²
T3V3
1
1
=
T4V4
T4 = [
V3 V4 ]
x T3
T4 = [
1 9
x 1793 K
1
0.4
]
T4 = 744.53 K © MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Solution for Exercise 2.5 Qin, Q1 = Cv (T3 ─ T2 ) = 0.718 kJ/kg.K x ( 1793 K – 857.33 K) Qin, Q1 = 671.81 kJ/kg Qout, Q2 = Cv (T4 ─ T1 ) = 0.718 kJ/kg.K x ( 744.53 K – 356 K) Qout, Q2 = 278.96 kJ/kg Thermal Efficiecy for Otto Cycle
th, Otto
= 1-
Qout, Q2 Qin, Q1
= 1 - 278.96 kJ.K/kg 671.81 kJ.K/kg
= 0.585 @ 58.5 %
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Diesel Cycle Analysis
Diesel Cycle Process: 1 to 2 is isentropic compression with compression ratio v1 / v2, or rv .
Qin Pressure,
P2 = P3
P 2
3
pv cons tan t
2 to 3 is reversible constant pressure heating, the heat supplied Q1, v3 / v2, cut off ratio,
4
P4
Qout P1
3 to 4 is isentropic expansion,
1
V2
V3
V1= V4
rc
V Volume
4 to 1 is reversible constant volume cooling, the heat rejected Q2.
Figure 2.10: P-V Diagram for Diesel Cycle
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Formula to find the thermal efficiency for Diesel Cycle Point 1 Get the data from the question. Point 2 (1 to 2 is isentropic compression )
= P2V2 1 1 = T2V2 T1V1 P1V1
To get pressure To get temperature (Kelvin)
Point 3 (2 to 3 is reversible constant pressure heating ) Constant pressure
P2V2 = P3V3 T2 T3
V3 = T 3 = V2 T2
rc
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Qin
Point 4 (3 to 4 is isentropic expansion ) P3V3
= P4V4
P2 = P3
P4 = [ V3 ] x P3 V4
[ VV34 ] [ VV34 ] Put
= =
Pressure,
2
3
pv cons tan t
1 4
P4
[ VV32 ] x [ VV21 ] [ rc ] x [ r1v ]
P
Qout P1
1
2 V2
V3
V1= V4
2 into 1
r c P4 = [ ] x rv 1
T3V3
= T4V4
T4 = [ V3 ] V4
P3
1
1
x T3
3 © MSF @ POLITEKNIK UNGKU OMAR
V Volume
(DJA3032) INTERNAL COMBUSTION ENGINE Qin
[ VV34 ] =
[ VV32 ] x
[ VV34 ]
[ rc ] x [ r1v ]
Put
=
Pressure,
[ VV21 ]
P2 = P3
2
3
pv cons tan t
4 4
P4
4 into 3
Qout P1
1
T4 = [ r c ] rv
P
1
x T3
Qin, Q1 = (T3 ─ T2 ) Qout, Q2 = Cv(T4 ─ T1 )
V2
th, Diesel =
1-
V3
Qout, Q2 Qin, Q1
V1= V4
Thermal Efficiency
© MSF @ POLITEKNIK UNGKU OMAR
V Volume
(DJA3032) INTERNAL COMBUSTION ENGINE Example 2.3 Diesel engine has an inlet temperature and a pressure at 15°C and 1 bar respectively. The compression ratio is 12/ 1 and the maximum cycle temperature is 1100°C. Calculate the air standard thermal efficiency based on the diesel cycle. Solution
rv =
12
T1 = 15°C + 273K = 288 K P1 = 1 bar T3 = 1100°C + 273K = 1373 K
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Point 2 (1 to 2 is isentropic compression ) P1V1
= P2V2
T1V1
1
=
T2V2
1
1
P2 = [ V1 ] x P1 V2
V1 T2 = [ V2 ]
P2 = [12] x 1 bar
T2 = [12] x 288 K
P2 = 32.42 bar
T2 = 778.15 K
1.4
x T1
0.4
Point 3 (2 to 3 is reversible constant pressure heating ) Constant pressure
P2V2 = P3V3 T2 T3
V3 = T3 = rc = 1373 K = 1.764 V2 T2 778.15 K
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Point 4 (3 to 4 is isentropic expansion ) P3V3
= P4V4
P4 = [ V3 ] x P3 V4
[ VV34 ]
=
[ VV32 ] x [ VV21 ]
[ VV34 ]
=
[ rc ] x [ r1v ]
1
T3V3
= T4V4
Put 1
=
[ rc ] x [ r1v ]
P3
P4 = [ 1.764 ] x 32.42 bar 12 2
P4 = 2.21 bar
Put
1
[ VV34 ]
r c P4 = [ ] x rv 1.4
1
T4 = [ V3 ] x T3 V4 [ VV34 ] = [ VV32 ] x [ VV21 ]
2 into 1
3
4
4 into 3
T4 = [ r c
1
rv ]
x T3 0.4
T4 = [ 1.764 ] x 1373 K 12 T4 = 637.67 K © MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Thermal Efficiecy for Diesel Cycle Qin, Q1 = (T3 ─ T2 ) = 1.005 kJ/kg.K x ( 1373 K – 778.15 K) Qin, Q1 = 597.82 kJ/kg Qout, Q2 = Cv (T4 ─ T1 ) = 0.718 kJ/kg.K x ( 637.67 K – 288 K) Qout, Q2 = 251.06 kJ/kg
th, Diesel
= 1-
Qout, Q2 Qin, Q1
= 1 - 251.06 kJ/kg 597.82 kJ/kg
= 0.580 @ 58 %
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Exercise 2.7 One engine operates in diesel cycle, the beginning temperature is 18 °C, and pressure is 1 bar. The compression ratio is 14:1 , maximun temperature is 1320°C. Given = 1.005 KJ/kg.K, R=287 KJ/kg.K. Calculate the temperature and r c and mechanical efficiency. Exercise 2.8 A diesel cycle’s engine running with the beginning of compression process temperature 17 °C and pressure 1.1 bar. The compression ratio for this engine is 13.5 and the temperature for beginning expansion process is 1290 °C. Calculate thermal efficiency for this diesel cycle engine. Exercise 2.9 An engine was operated with diesel cycle process with the beginning temperature and pressure at 16.8 °C and 1.05 bar. The compression ratio for this engine is 12.5 and cut of ratio for this engine is 1.75. Calculate thermal efficiency for this engine. © MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Solution for Exercise 2.7
rv =
14
T1 = 18°C + 273K = 291 K P1 = 1 bar T3 = 1320 °C + 273K = 1593 K Point 2 (1 to 2 is isentropic compression ) P1V1
= P2V2
T1V1
1
=
T2V2
1
1
P2 = [ V1 ] x P1 V2
V1 T2 = [ V2 ]
P2 = [14] x 1 bar
T2 = [14] x 291 K
P2 = 40.23 bar
T2 = 836.27 K
1.4
x T1
0.4
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Solution for Exercise 2.7 Point 3 (2 to 3 is reversible constant pressure heating ) Constant pressure
P2V2 = P3V3 T2 T3
V3 = T3 = rc = 1593 K = 1.905 V2 T2 836.27 K
Point 4 (3 to 4 is isentropic expansion ) P3V3
= P4V4
P4 = [ V3 ] x P3 V4
[ VV34 ]
=
[ VV32 ] x [ VV21 ]
[ VV34 ]
=
[ rc ] x [ r1v ]
Put 1
2 into 1
r c P4 = [ ] x rv 1.4
2
P3
P4 = [ 1.905 ] x 40.23 bar 14 P4 = 2.47 bar
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Solution for Exercise 2.7 1
T3V3
= T4V4
1
Put
1
T 4 = [ V3 ] x T3 V4 [ VV34 ] = [ VV32 ] x [ VV21 ]
[ VV34 ]
=
3
4 into 3
T 4 = [ rc
1
rv ]
x T3 0.4
T4 = [ 1.905 ] x 1593 K 14
[ rc ] x [ r1v ]
4
T4 = 717.34 K
Thermal Efficiecy for Diesel Cycle Qin, Q1 = (T3 ─ T2 ) = 1.005 kJ/kg.K x ( 1593 K – 836.27 K) Qin, Q1 = 760.51 kJ/kg Qout, Q2 = Cv (T4 ─ T1 ) = 0.718 kJ/kg.K x ( 717.34 K – 291 K) Qout, Q2 = 306.11 kJ/kg
th, Diesel
= 1-
Qout, Q2 Qin, Q1
= 1 - 306.11 kJ/kg 760.51 kJ/kg
= 0.597 @ 59.7% © MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Solution for Exercise 2.8
rv =
13.5
T1 = 17°C + 273K = 290 K P1 = 1.1 bar T3 = 1290 °C + 273K = 1563 K Point 2 (1 to 2 is isentropic compression ) P1V1
= P2V2
T1V1
1
=
T2V2
1
1
P2 = [ V1 ] x P1 V2
V1 T2 = [ V2 ]
x T1
P2 = [13.5] x 1.1 bar
T2 = [13.5]
x 291 K
P2 = 42.06 bar
T2 = 821.36 K
1.4
0.4
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Solution for Exercise 2.8 Point 3 (2 to 3 is reversible constant pressure heating ) Constant pressure
P2V2 = P3V3 T2 T3
V3 = T3 = rc = 1563 K = 1.903 V2 T2 821.36 K
Point 4 (3 to 4 is isentropic expansion ) P3V3
= P4V4
P4 = [ V3 ] x P3 V4
[ VV34 ]
=
[ VV32 ] x [ VV21 ]
[ VV34 ]
=
[ rc ] x [ r1v ]
Put 1
2 into 1
r c P4 = [ ] x rv 1.4
2
P3
P4 = [ 1.903 ] x 40.23 bar 13.5 P4 = 2.71 bar
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Solution for Exercise 2.8 1
T3V3
= T4V4
1
Put
1
T 4 = [ V3 ] x T3 V4 [ VV34 ] = [ VV32 ] x [ VV21 ]
[ VV34 ]
=
3
4 into 3
T 4 = [ rc
1
rv ]
x T3 0.4
T4 = [ 1.903 ] x 1563 K 13.5
[ rc ] x [ r1v ]
4
T4 = 713.84 K
Thermal Efficiecy for Diesel Cycle Qin, Q1 = (T3 ─ T2 ) = 1.005 kJ/kg.K x ( 1563 K – 821.36 K) Qin, Q1 = 745.35 kJ/kg Qout, Q2 = Cv (T4 ─ T1 ) = 0.718 kJ/kg.K x ( 713.84 K – 290 K) Qout, Q2 = 304.32 kJ/kg
th, Diesel
= 1-
Qout, Q2 Qin, Q1
= 1 - 304.32 kJ/kg 745.35 kJ/kg
= 0.592 @ 59.2% © MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Solution for Exercise 2.9
rv = rc =
12.5 1.75
T1 = 16.8°C + 273K = 289.8 K P1 = 1.05 bar
Point 2 (1 to 2 is isentropic compression ) P1V1
= P2V2
T1V1
1
=
T2V2
1
1
P2 = [ V1 ] x P1 V2
V1 T2 = [ V2 ]
x T1
P2 = [12.5] x 1.05 bar
T2 = [12.5]
x 289.8 K
P2 = 36.05 bar
T2 = 795.91 K
1.4
0.4
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Solution for Exercise 2.9 Point 3 (2 to 3 is reversible constant pressure heating ) Constant pressure
P2V2 = P3V3 T2 T3
V3 = T3 = rc = T3 = 1.75 V2 T2 795.91 K T3 = 1392.84 K
Point 4 (3 to 4 is isentropic expansion ) P3V3
= P4V4
P4 = [ V3 ] x P3 V4
[ VV34 ]
=
[ VV34 ]
=
Put 1
[ VV32 ] x [ VV21 ] [ rc ] x [ r1v ]
2
2 into 1
r c P4 = [ ] x rv P4 = [ 1.75 12.5
P3
1.4
]
P4 = 2.30 bar
© MSF @ POLITEKNIK UNGKU OMAR
x 36.05 bar
(DJA3032) INTERNAL COMBUSTION ENGINE
Solution for Exercise 2.9 1
T3V3
= T4V4
1
Put
1
T 4 = [ V3 ] x T3 V4 [ VV34 ] = [ VV32 ] x [ VV21 ]
[ VV34 ]
=
3
4 into 3
T 4 = [ rc
1
rv ]
x T3 0.4
T4 = [ 1.75 ] x 1392.84 K 12.5
[ rc ] x [ r1v ]
4
T4 = 634.38 K
Thermal Efficiecy for Diesel Cycle Qin, Q1 = (T3 ─ T2 ) = 1.005 kJ/kg.K x ( 1392.84 K – 795.91 K) Qin, Q1 = 599.91 kJ/kg Qout, Q2 = Cv (T4 ─ T1 ) = 0.718 kJ/kg.K x ( 634.38 K – 289.8 K) Qout, Q2 = 247.41 kJ/kg
th, Diesel
= 1-
Qout, Q2 Qin, Q1
= 1 - 247.41 kJ/kg 599.91 kJ/kg
= 0.588 @ 58.8 % © MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Dual/Combined Cycle Analysis P
P3=P4
Dual/Combined Cycle Process: 1 to 2 is isentropic compression with compression ratio v1 / v2, or rv ,
Qin 3
Qin
2 to 3 is reversible constant volume heating, the heat supplied Q1
4
3 to 4 is reversible constant pressure
p.v conts
heating, the heat supplied Q1, v4/ v3, cut off
2
ratio,
5
Qout
1 V2=V3
V1=V5
V
rc
4 to 5 is isentropic expansion, 5 to 1 is reversible constant volume cooling, the heat rejected Q2.
Figure 2.11: P-V Diagram for Dual/Combined Cycle
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Formula to find the thermal efficiency for Dual/Combined Cycle Point 1 Get the data from the question. Point 2 (1 to 2 is isentropic compression )
= P2V2 1 1 = T2V2 T1V1 P1V1
To get pressure To get temperature (Kelvin)
Point 3 (2 to 3 is reversible constant volume heating ) Constant volume
P2V2 = P3V3 T2 T3
P3 = T3 x P2 T2 T3 = P3 x T2 P2
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Assume the heat added at constant volume is equal to the heat added at constant pressure; cv(T3 – T2 ) = (T4 – T3 ) T4 = cv(T3 – T2 ) + . (T3) Point 4 (3 to 4 is reversible constant pressure heating ) Constant pressure
P3V3 = P4V4 T3 T4
V4 = T 4 = V3 T3
rc
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE P
Qin
Point 5 (4 to 5 is isentropic expansion ) P4V4
= P5V5
P3=P4
P5 = [ V4 ] x P4 V5 =
[ VV45 ] = Put
4
Qin
[ VV45 ]
3
1
p.v conts
2
[ VV43 ] x [ VV21 ] 1
[ rc ] x [ rv ]
2
Qout
1 V2=V3
2 into 1
r c P5 = [ ] x rv
5
P4
© MSF @ POLITEKNIK UNGKU OMAR
V1=V5
V
(DJA3032) INTERNAL COMBUSTION ENGINE
1
T4V4
= T5V5
P
1
Qin
1
T 5 = [ V4 ] x T4 V5 [ VV45 ] = [ VV43 ] x [ VV21 ]
[ VV45 ] = Put
1
[ rc ] [ rv ] x
3
3 P3=P4
4
Qin 4
p.v conts
2 5
4 into 3 T 5 = [ rc
1
1
rv ]
x T4
Qin, Q1 = Cv(T3 ─ T2 ) + (T4 ─ T3 ) Qout, Q2 = Cv(T5 ─ T1 )
Qout
V1=V5
V2=V3
th, Dual
= 1-
Qout, Q2 Qin, Q1
© MSF @ POLITEKNIK UNGKU OMAR
Thermal Efficiency
V
(DJA3032) INTERNAL COMBUSTION ENGINE
Example 2.4 An oil engine takes in air at 1.01 bar, 200C and the maximum cycle pressure is 69 bar. The compression ratio is 18/1. Draw the p-v diagram and calculate the air standard thermal efficiency based on the dual combustion cycle. Assume that the heat added at constant volume is equal to the heat added at constant pressure. Solution
rv =
18
T1 = 20°C + 273K = 293 K P1 = 1.01 bar P3 = 69 bar
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Point 2 (1 to 2 is isentropic compression ) P1V1
= P2V2
T1V1
1
=
T2V2
1
1
P2 = [ V1 ] x P1 V2
V1 T2 = [ V2 ]
P2 = [18] x 1.01 bar
T2 = [18] x 293 K
P2 = 57.77 bar
T2 = 931.06 K
1.4
x T1
0.4
Point 3 (2 to 3 is reversible constant volume heating ) Constant volume
P2V2 = P3V3 T2 T3
P3 = T3 x P2 T2 T3 = P3 x T2 P2 69 bar x 931.06 K 57.77 bar T3 = 1112.05 K T3 =
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Assume the heat added at constant volume is equal to the heat added at constant pressure; cv(T3 – T2 ) = (T4 – T3 ) T4 = cv(T3 – T2 ) + . (T3) T4 =
0.718 kJ/kg.K (1112.05 K – 931.06 K) + 1.005 kJ/kg.K(1112.05 K) 1.005 kJ/kg.K
T4 = 1241.30 K Point 4 (3 to 4 is reversible constant pressure heating ) Constant pressure
P3V3 = P4V4 T3 T4
V4 = T4 = V3 T3
rc 1241.30 K 1112.05 K
rc
=
rc
= 1.116
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Point 5 (4 to 5 is isentropic expansion )
P4V4
= P5V5
P5 = [ V4 ] x P4 V5
[ VV45 ]
=
[ VV43 ] x [ VV21 ]
[ VV45 ]
=
[ rc ] x [ r1v ]
1
T4V4
= T5V5
1
r c P5 = [ ] x rv
P4
1.4
P5 = [ 1.116 ] x 69 bar 18 2
P5 = 1.41 bar
Put
1
=
2 into 1
1
T 5 = [ V4 ] x T4 V5 [ VV45 ] = [ VV43 ] x [ VV21 ]
[ VV45 ]
Put
[ rc ] x [ r1v ]
3
4
4 into 3
T 5 = [ rc
1
rv ]
x T4 0.4
T5 = [ 1.116 ] x 1241.30 K 18 T5 = 408.16 K © MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Thermal Efficiecy for Dual/Combined Cycle Qin, Q1 = Cv(T3 ─ T2 ) + (T4 ─ T3 ) Qin, Q1 = 0.718 kJ/kg.K(1112.05 K ─ 931.06 K ) + 1.005 kJ/kg.K (1241.30 K ─1112.05 K) Qin, Q1 = 259.85 kJ/kg Qout, Q2 = Cv (T5 ─ T1 ) Qout, Q2 = 0.718 kJ/kg.K x (408.16 K– 293 K) Qout, Q2 = 82.68 kJ/kg
th, Dual
= 1-
Qout, Q2 Qin, Q1
= 1 - 82.68 kJ/kg 259.85 kJ/kg
= 0.682 @ 68.2 %
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Self –Assessment Exercise 2.10 The pressure and temperature of air standard dual combustion cycle are given below, i. ii. iii. iv. v. vi.
T1 = 290 K P1 = 1.01 bar T2 = 871.1K T3 = 1087.5 K T4 = 1236.3 K T5 = 429.3 K
and ratio of compression is 16:1. Calculate the thermal efficiency for this dual cycle engine. Exercise 2.11 In a dual combustion cycle, the maximum pressure is 64 bar. Calculate the thermal efficiency when the pressure and temperature at the start of the compression are 1.01 bar and 18 0 C respectively. The compression ratio is 17/1. © MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Exercise 2.12 In a dual combustion cycle, the maximum pressure is 62 bar. Calculate the thermal efficiency when the pressure and temperature at the start of the compression are 1.01 bar and 16.7 °C respectively. The compression ratio is 17.7. Exercise 2.13 In a combine cycle, the maximum pressure is 67 bar. Calculate the thermal efficiency when the pressure and temperature at the start of the compression are 1.03 bar and 17 °C respectively. The compression ratio is 16.9. Exercise 2.14 An engine is operating in dual combustion cycle, the maximum pressure is 66 bar. Calculate the thermal efficiency for this engine when the pressure and temperature at the beginning of the compression process are 1.03 bar and 20.3 °C respectively. The compression ratio for this engine is 17.9.
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(DJA3032) INTERNAL COMBUSTION ENGINE
Answer for Exercise 2.11 T1 = 291 K P1 = 1.01 bar
T2 = 903.80 K P2 = 53.33 bar
Qin = 259.75 kJ/kg Qout = 84.61 kJ/kg
T3 = 1084.68 K P3 = 64 bar
T4 = 1213.91 K P4 = 64 bar rc = 1.119
T5 = 408.85 K P5 = 1.42 bar
T4 = 1069.25 K P4 = 62 bar rc = 1.064
T5 = 347.30 K P5 = 1.21 bar
ɳth, dual = 0.674 @ 67.4 %
Answer for Exercise 2.12 T1 = 289.7 K P1 = 1.01 bar
T2 = 914.40 K P2 = 56.43 bar
Qin = 129.70 kJ/kg Qout = 41.35 kJ/kg
T3 = 1004.72 K P3 = 62 bar
ɳth, dual = 0.681 @ 68.1 %
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Answer for Exercise 2.13 T1 = 290 K P1 = 1.03 bar
T2 = 898.57 K P2 = 53.94 bar
Qin = 312.54 kJ/kg Qout = 102.25 kJ/kg
T3 = 1116.22 K P3 = 67 bar
T4 = 1271.71 K P4 = 67 bar rc = 1.139
T5 = 432.40 K P5 = 1.54 bar
T4 = 1135.68 K P4 = 66 bar rc = 1.082
T5 = 369.62 K P5 = 1.30 bar
ɳth, dual = 0.673 @ 67.3 %
Answer for Exercise 2.14 T1 = 293.3 K P1 = 1.03 bar
T2 = 929.94 K P2 = 58.46 bar
Qin = 172.33 kJ/kg Qout = 54.79 kJ/kg
T3 = 1049.94 K P3 = 66 bar
ɳth, dual = 0.682 @ 68.2 %
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
© MSF @ POLITEKNIK UNGKU OMAR
CHAPTER 2: PISTON ENGINE PROCESS ANALYSIS by MOHD SAHRIL BIN MOHD FOUZI, Grad. IEM (G 27763) DEPARTMENT OF MECHANICAL ENGINEERING
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
General objective: To understand the structure and various types of cycle engines. Specific objectives: At the end of this unit you should be able to: define the air standard cycle. define constant pressure () and constant volume (cv). draw p-v diagram of Otto cycle, Diesel cycle and Combined cycle. explain Otto cycle, Diesel cycle and Combine/dual cycle.
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(DJA3032) INTERNAL COMBUSTION ENGINE
Introduction : In this unit we are to discuss the meaning of standard cycle, heat supplied at constant volume and heat supplied at constant pressure. An internal combustion engine can be classified into three different cycles and they are Otto cycle, Diesel cycle and Dualcombustion cycle. Qin Pressure (P) P3
pv cons tan t
Qin P2
Pressure, P
3
2
2
P2 = P3
3
4
4
P4
P4
pv cons tan t
Qout
Qout P1
P1
1
1
V2
V1
Figure 2.1: P-V Diagram (Otto Cycle)
Volume ( v )
V2
V3
V1= V4
V Volume
Figure 2.2: P-V Diagram (Diesel Cycle) © MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
P
Qin 3
P3=p4 Qin
4 p.v conts
2 5
Qout
1 V2=V3
V1=V5
V
Figure 2.3: P-V Diagram (Dual/Combined Cycle)
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Air Standard Cycles The air standard cycle is a cycle followed by a heat engine which uses air as the working medium. Since the air standard analysis is the simplest and most idealistic, such cycles are also called ideal cycles and the engine running on such cycles are called ideal engines. In order that the analysis is made as simple as possible, certain assumptions have to be made. These assumptions result in an analysis that is far from correct for most actual combustion engine processes, but the analysis is of considerable value for indicating the upper limit of performance. The analysis is also a simple means for indicating the relative effects of principal variables of the cycle and the relative size of the apparatus.
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(DJA3032) INTERNAL COMBUSTION ENGINE
Assumptions: 1. The working medium is a perfect gas with constant specific heats and molecular weight corresponding to values at room temperature. 2. No chemical reactions occur during the cycle. The heat addition and heat rejection processes are merely heat transfer processes. 3. The processes are reversible. 4. Losses by heat transfer from the apparatus to the atmosphere are assumed to be zero in this analysis. 5. The working medium at the end of the process (cycle) is unchanged and is at the same condition as at the beginning of the process (cycle).
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
T-S Diagram
Figure 2.4: T-S Diagram for Otto Cycle
Figure 2.5: T-S Diagram for Diesel Cycle
Figure 2.6: T-S Diagram for Combined/Dual Cycle
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(DJA3032) INTERNAL COMBUSTION ENGINE
Compression ratio To give direct comparison with an actual engine the ratio of specific volume, v 1 / v2, is taken to be the same as the compression ratio of the actual engine, Compression ratio, rv = Pressure (P) P3
Clearance volume
P2
v1 v2
=
sweptvolum e clearencev olume clearencev olume
3 pv cons tan t
2
Clearance volume 4
TDC
P4 P1
V2 Minimum volume
BDC
1
Swept volume
V1
Figure 2.7: P-V Diagram for Otto Cycle
Swept volume
Volume ( v )
Maximum volume Figure 2.8: Animated 4 Stroke Engine © MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Example 2.1 An Otto cycle in a petrol engine with a cylinder bore of 55mm, a stroke of 80mm, and a clearance volume of 23.3 cm3 is given. Find the compression ratio of this engine. Solution: Cylinder bore, B = 55mm = 5.5 cm Stroke, S = 80mm = 8.0 cm Clearance Volume = 23.3 cm 3
swept volume = = =
Abstract the data from the question. Convert the data into centimetre unit.
Compression ratio, rv = v1 x B² x S 4 v2 x (5.5)² x 8 = sweptvolum e clearencev olume 4 clearencev olume 3 190.07 cm 3 = (190.07 + 23.3) cm 23.3 cm 3
= 9.16
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Self-Exercise Exercise 2.1 A petrol engine with a cylinder bore of 73 mm, a stroke of 95 mm, and a clearance volume of 26.3 cm 3 is given. Find the compression ratio of this engine. [Ans: 16.12] Exercise 2.2 An air engine is operated with cylinder bore 65 mm with the stroke of 73 mm. The clearance volume for this engine is 1/10 of swept volume. Calculate the compression ratio for this air engine. [Ans: 11.00] Exercise 2.3 Given compression ratio for a petrol engine is 10.5.The cylinder bore and stroke length for this engine are 69 mm and 83 mm. Calculate the clearance volume for this engine. [Ans: 32.66 cm3 ] Exercise 2.4 An otto cycle engine with compression ratio 9.5 operating with clearrance volume 23.45 cm 3. The stroke length for this engine is 83 mm, find the cylinder bore for this engine. [Ans: 5.53 cm]
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Otto Cycle Analysis Pressure (P) P3
Otto Cycle Process:
3
1 to 2 is isentropic compression with compression ratio v1 / v2, or rv .
pv cons tan t
Qin 2
P2
2 to 3 is reversible constant volume heating, the heat supplied Q1
4
P4
Qout P1
1 Volume ( v )
V1 V2 Figure 2.9: P-V Diagram for Otto Cycle
3 to 4 is isentropic expansion, v4 /v3 is similar to v1 /v2. 4 to 1 is reversible constant volume cooling, the heat rejected Q2.
Formula to find the thermal efficiency for Otto Cycle P1V1 T1V1
= P2V2
1
=
T2V2
Qin, Q1 = Cv(T3 ─ T2 )
1
= 1-
Qout, Q2 = Cv(T4 ─ T1 )
© MSF @ POLITEKNIK UNGKU OMAR
Qout, Q2 Qin, Q1
(DJA3032) INTERNAL COMBUSTION ENGINE Example 2.2 One petrol engine is working at a constant volume, the compression ratio is 8.5:1. Pressure and temperature at a beginning compression process is 101 kN/m 2 and 840 C. Temperature at the beginning of an expand process is14960 C. Calculate the temperature and pressure at the important points based on the Otto cycle. Solution: Pressure (P) P3
3
rv
pv cons tan t
Qin P2
V1 V4 = 8.5 V2 V3
*T1 = 84°C + 273K = 357 K
2
4
P1 = 101 kN/m²
P4
Qout P1
1
V2
V1
Volume ( v )
*T3 = 1496°C + 273K = 1769 K * Temperature must be convert into Kelvin
P-V Diagram for Otto Cycle
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Point 2 (1 to 2 is isentropic compression ) P1V1
= P2V2
T1V1
P2 = [ V1 ] x P1 V2 P2 = [8.5]
1.4
x 101 kN / m²
P2 = 2020.73 kN / m²
1
=
1
T2V2
1
V1 T2 = [ V2 ] T2 = [8.5]
x T1
0.4
x 357 K
T2 = 840.30 K
Point 3 (2 to 3 is reversible constant volume heating ) Constant volume
P2V2 = P3V3 T2 T3
P3 = T3 x P2 T2 P3 = 1769 x 2020.73 kN / m² 840.30 P3 = 4254.04 kN / m²
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Point 4 (3 to 4 is isentropic expansion ) P3V3
= P4V4
T3V3
P4 = [ V3 ] x P3 V4 P4 = [
1 8.5
1.4
]
x 4254.04 kN / m²
P4 = 212.63 kN / m²
1
=
T4V4
1
1
V3 T4 = [ V4 ]
x T3
0.4 T4 = [ 1 ] x 1769 K
8.5
T4 = 751.55 K
Qin, Q1 = Cv (T3 ─ T2 ) = 0.718 kJ/kg.K x ( 1769 K – 840.30 K) Qin, Q1 = 666.81 kJ/kg Qout, Q2 = Cv (T4 ─ T1 ) = 0.718 kJ/kg.K x ( 751.55 K – 357 K) Qout, Q2 = 283.29 kJ/kg
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Exercise 2.5 A four cylinder engine operates in Otto cycle, the volume is constant and the compression ratio is 9:1, beginning pressure is 105 KN/m2 , temperature is 83 ° C and final temperature is 1520 °C. Draw a p-v diagram and find the temperature and pressure for each point. Lastly calculate the efficiency of Otto cycle. Exercise 2.6 One petrol engine is working at a constant volume, the compression ratio is 8.5:1. Pressure and temperature at a beginning compression process is 101 kN/m 2 and 84 0C. Temperature at the beginning of an expand process is14960 C. Calculate the temperature and pressure at the important points based on the Otto cycle. Hence, calculate thermal efficiency for this engine.
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Solution for Exercise 2.5 V V4 rv 1 = 9 V2 V3
Pressure (P) P3
3 pv cons tan t
Qin
T1 = 83°C + 273K = 356 K
P2
2
4
P4
Qout
P1 = 105 kN/m²
P1
1
T3 = 1520°C + 273K = 1793 K
V1
V2
P-V Diagram for Otto Cycle
Point 2 (1 to 2 is isentropic compression ) P1V1
= P2V2
T1V1
1
=
T2V2
1
P2 = [ V1 ] x P1 V2
V1 T2 = [ V2 ]
P2 = [9]
T2 = [9]
1.4
x 105 kN / m²
P2 = 2275.77 kN / m²
1
0.4
x T1
x 356 K
T2 = 857.33 K © MSF @ POLITEKNIK UNGKU OMAR
Volume ( v )
(DJA3032) INTERNAL COMBUSTION ENGINE
Solution for Exercise 2.5 Point 3 (2 to 3 is reversible constant volume heating ) Constant volume
P2V2 = P3V3 T2 T3
P3 = T3 x P2 T2 P3 = 1793 x 2275.77 kN / m² 857.33 P3 = 4759.49 kN / m²
Point 4 (3 to 4 is isentropic expansion ) P3V3
= P4V4 P4 = [ V3 ] x P3 V4 P4 = [
1 9
1.4
]
x 4759.49 kN / m²
P4 = 219.59 kN / m²
T3V3
1
1
=
T4V4
T4 = [
V3 V4 ]
x T3
T4 = [
1 9
x 1793 K
1
0.4
]
T4 = 744.53 K © MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Solution for Exercise 2.5 Qin, Q1 = Cv (T3 ─ T2 ) = 0.718 kJ/kg.K x ( 1793 K – 857.33 K) Qin, Q1 = 671.81 kJ/kg Qout, Q2 = Cv (T4 ─ T1 ) = 0.718 kJ/kg.K x ( 744.53 K – 356 K) Qout, Q2 = 278.96 kJ/kg Thermal Efficiecy for Otto Cycle
th, Otto
= 1-
Qout, Q2 Qin, Q1
= 1 - 278.96 kJ.K/kg 671.81 kJ.K/kg
= 0.585 @ 58.5 %
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Diesel Cycle Analysis
Diesel Cycle Process: 1 to 2 is isentropic compression with compression ratio v1 / v2, or rv .
Qin Pressure,
P2 = P3
P 2
3
pv cons tan t
2 to 3 is reversible constant pressure heating, the heat supplied Q1, v3 / v2, cut off ratio,
4
P4
Qout P1
3 to 4 is isentropic expansion,
1
V2
V3
V1= V4
rc
V Volume
4 to 1 is reversible constant volume cooling, the heat rejected Q2.
Figure 2.10: P-V Diagram for Diesel Cycle
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Formula to find the thermal efficiency for Diesel Cycle Point 1 Get the data from the question. Point 2 (1 to 2 is isentropic compression )
= P2V2 1 1 = T2V2 T1V1 P1V1
To get pressure To get temperature (Kelvin)
Point 3 (2 to 3 is reversible constant pressure heating ) Constant pressure
P2V2 = P3V3 T2 T3
V3 = T 3 = V2 T2
rc
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Qin
Point 4 (3 to 4 is isentropic expansion ) P3V3
= P4V4
P2 = P3
P4 = [ V3 ] x P3 V4
[ VV34 ] [ VV34 ] Put
= =
Pressure,
2
3
pv cons tan t
1 4
P4
[ VV32 ] x [ VV21 ] [ rc ] x [ r1v ]
P
Qout P1
1
2 V2
V3
V1= V4
2 into 1
r c P4 = [ ] x rv 1
T3V3
= T4V4
T4 = [ V3 ] V4
P3
1
1
x T3
3 © MSF @ POLITEKNIK UNGKU OMAR
V Volume
(DJA3032) INTERNAL COMBUSTION ENGINE Qin
[ VV34 ] =
[ VV32 ] x
[ VV34 ]
[ rc ] x [ r1v ]
Put
=
Pressure,
[ VV21 ]
P2 = P3
2
3
pv cons tan t
4 4
P4
4 into 3
Qout P1
1
T4 = [ r c ] rv
P
1
x T3
Qin, Q1 = (T3 ─ T2 ) Qout, Q2 = Cv(T4 ─ T1 )
V2
th, Diesel =
1-
V3
Qout, Q2 Qin, Q1
V1= V4
Thermal Efficiency
© MSF @ POLITEKNIK UNGKU OMAR
V Volume
(DJA3032) INTERNAL COMBUSTION ENGINE Example 2.3 Diesel engine has an inlet temperature and a pressure at 15°C and 1 bar respectively. The compression ratio is 12/ 1 and the maximum cycle temperature is 1100°C. Calculate the air standard thermal efficiency based on the diesel cycle. Solution
rv =
12
T1 = 15°C + 273K = 288 K P1 = 1 bar T3 = 1100°C + 273K = 1373 K
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Point 2 (1 to 2 is isentropic compression ) P1V1
= P2V2
T1V1
1
=
T2V2
1
1
P2 = [ V1 ] x P1 V2
V1 T2 = [ V2 ]
P2 = [12] x 1 bar
T2 = [12] x 288 K
P2 = 32.42 bar
T2 = 778.15 K
1.4
x T1
0.4
Point 3 (2 to 3 is reversible constant pressure heating ) Constant pressure
P2V2 = P3V3 T2 T3
V3 = T3 = rc = 1373 K = 1.764 V2 T2 778.15 K
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Point 4 (3 to 4 is isentropic expansion ) P3V3
= P4V4
P4 = [ V3 ] x P3 V4
[ VV34 ]
=
[ VV32 ] x [ VV21 ]
[ VV34 ]
=
[ rc ] x [ r1v ]
1
T3V3
= T4V4
Put 1
=
[ rc ] x [ r1v ]
P3
P4 = [ 1.764 ] x 32.42 bar 12 2
P4 = 2.21 bar
Put
1
[ VV34 ]
r c P4 = [ ] x rv 1.4
1
T4 = [ V3 ] x T3 V4 [ VV34 ] = [ VV32 ] x [ VV21 ]
2 into 1
3
4
4 into 3
T4 = [ r c
1
rv ]
x T3 0.4
T4 = [ 1.764 ] x 1373 K 12 T4 = 637.67 K © MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Thermal Efficiecy for Diesel Cycle Qin, Q1 = (T3 ─ T2 ) = 1.005 kJ/kg.K x ( 1373 K – 778.15 K) Qin, Q1 = 597.82 kJ/kg Qout, Q2 = Cv (T4 ─ T1 ) = 0.718 kJ/kg.K x ( 637.67 K – 288 K) Qout, Q2 = 251.06 kJ/kg
th, Diesel
= 1-
Qout, Q2 Qin, Q1
= 1 - 251.06 kJ/kg 597.82 kJ/kg
= 0.580 @ 58 %
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Exercise 2.7 One engine operates in diesel cycle, the beginning temperature is 18 °C, and pressure is 1 bar. The compression ratio is 14:1 , maximun temperature is 1320°C. Given = 1.005 KJ/kg.K, R=287 KJ/kg.K. Calculate the temperature and r c and mechanical efficiency. Exercise 2.8 A diesel cycle’s engine running with the beginning of compression process temperature 17 °C and pressure 1.1 bar. The compression ratio for this engine is 13.5 and the temperature for beginning expansion process is 1290 °C. Calculate thermal efficiency for this diesel cycle engine. Exercise 2.9 An engine was operated with diesel cycle process with the beginning temperature and pressure at 16.8 °C and 1.05 bar. The compression ratio for this engine is 12.5 and cut of ratio for this engine is 1.75. Calculate thermal efficiency for this engine. © MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Solution for Exercise 2.7
rv =
14
T1 = 18°C + 273K = 291 K P1 = 1 bar T3 = 1320 °C + 273K = 1593 K Point 2 (1 to 2 is isentropic compression ) P1V1
= P2V2
T1V1
1
=
T2V2
1
1
P2 = [ V1 ] x P1 V2
V1 T2 = [ V2 ]
P2 = [14] x 1 bar
T2 = [14] x 291 K
P2 = 40.23 bar
T2 = 836.27 K
1.4
x T1
0.4
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Solution for Exercise 2.7 Point 3 (2 to 3 is reversible constant pressure heating ) Constant pressure
P2V2 = P3V3 T2 T3
V3 = T3 = rc = 1593 K = 1.905 V2 T2 836.27 K
Point 4 (3 to 4 is isentropic expansion ) P3V3
= P4V4
P4 = [ V3 ] x P3 V4
[ VV34 ]
=
[ VV32 ] x [ VV21 ]
[ VV34 ]
=
[ rc ] x [ r1v ]
Put 1
2 into 1
r c P4 = [ ] x rv 1.4
2
P3
P4 = [ 1.905 ] x 40.23 bar 14 P4 = 2.47 bar
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Solution for Exercise 2.7 1
T3V3
= T4V4
1
Put
1
T 4 = [ V3 ] x T3 V4 [ VV34 ] = [ VV32 ] x [ VV21 ]
[ VV34 ]
=
3
4 into 3
T 4 = [ rc
1
rv ]
x T3 0.4
T4 = [ 1.905 ] x 1593 K 14
[ rc ] x [ r1v ]
4
T4 = 717.34 K
Thermal Efficiecy for Diesel Cycle Qin, Q1 = (T3 ─ T2 ) = 1.005 kJ/kg.K x ( 1593 K – 836.27 K) Qin, Q1 = 760.51 kJ/kg Qout, Q2 = Cv (T4 ─ T1 ) = 0.718 kJ/kg.K x ( 717.34 K – 291 K) Qout, Q2 = 306.11 kJ/kg
th, Diesel
= 1-
Qout, Q2 Qin, Q1
= 1 - 306.11 kJ/kg 760.51 kJ/kg
= 0.597 @ 59.7% © MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Solution for Exercise 2.8
rv =
13.5
T1 = 17°C + 273K = 290 K P1 = 1.1 bar T3 = 1290 °C + 273K = 1563 K Point 2 (1 to 2 is isentropic compression ) P1V1
= P2V2
T1V1
1
=
T2V2
1
1
P2 = [ V1 ] x P1 V2
V1 T2 = [ V2 ]
x T1
P2 = [13.5] x 1.1 bar
T2 = [13.5]
x 291 K
P2 = 42.06 bar
T2 = 821.36 K
1.4
0.4
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Solution for Exercise 2.8 Point 3 (2 to 3 is reversible constant pressure heating ) Constant pressure
P2V2 = P3V3 T2 T3
V3 = T3 = rc = 1563 K = 1.903 V2 T2 821.36 K
Point 4 (3 to 4 is isentropic expansion ) P3V3
= P4V4
P4 = [ V3 ] x P3 V4
[ VV34 ]
=
[ VV32 ] x [ VV21 ]
[ VV34 ]
=
[ rc ] x [ r1v ]
Put 1
2 into 1
r c P4 = [ ] x rv 1.4
2
P3
P4 = [ 1.903 ] x 40.23 bar 13.5 P4 = 2.71 bar
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Solution for Exercise 2.8 1
T3V3
= T4V4
1
Put
1
T 4 = [ V3 ] x T3 V4 [ VV34 ] = [ VV32 ] x [ VV21 ]
[ VV34 ]
=
3
4 into 3
T 4 = [ rc
1
rv ]
x T3 0.4
T4 = [ 1.903 ] x 1563 K 13.5
[ rc ] x [ r1v ]
4
T4 = 713.84 K
Thermal Efficiecy for Diesel Cycle Qin, Q1 = (T3 ─ T2 ) = 1.005 kJ/kg.K x ( 1563 K – 821.36 K) Qin, Q1 = 745.35 kJ/kg Qout, Q2 = Cv (T4 ─ T1 ) = 0.718 kJ/kg.K x ( 713.84 K – 290 K) Qout, Q2 = 304.32 kJ/kg
th, Diesel
= 1-
Qout, Q2 Qin, Q1
= 1 - 304.32 kJ/kg 745.35 kJ/kg
= 0.592 @ 59.2% © MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Solution for Exercise 2.9
rv = rc =
12.5 1.75
T1 = 16.8°C + 273K = 289.8 K P1 = 1.05 bar
Point 2 (1 to 2 is isentropic compression ) P1V1
= P2V2
T1V1
1
=
T2V2
1
1
P2 = [ V1 ] x P1 V2
V1 T2 = [ V2 ]
x T1
P2 = [12.5] x 1.05 bar
T2 = [12.5]
x 289.8 K
P2 = 36.05 bar
T2 = 795.91 K
1.4
0.4
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Solution for Exercise 2.9 Point 3 (2 to 3 is reversible constant pressure heating ) Constant pressure
P2V2 = P3V3 T2 T3
V3 = T3 = rc = T3 = 1.75 V2 T2 795.91 K T3 = 1392.84 K
Point 4 (3 to 4 is isentropic expansion ) P3V3
= P4V4
P4 = [ V3 ] x P3 V4
[ VV34 ]
=
[ VV34 ]
=
Put 1
[ VV32 ] x [ VV21 ] [ rc ] x [ r1v ]
2
2 into 1
r c P4 = [ ] x rv P4 = [ 1.75 12.5
P3
1.4
]
P4 = 2.30 bar
© MSF @ POLITEKNIK UNGKU OMAR
x 36.05 bar
(DJA3032) INTERNAL COMBUSTION ENGINE
Solution for Exercise 2.9 1
T3V3
= T4V4
1
Put
1
T 4 = [ V3 ] x T3 V4 [ VV34 ] = [ VV32 ] x [ VV21 ]
[ VV34 ]
=
3
4 into 3
T 4 = [ rc
1
rv ]
x T3 0.4
T4 = [ 1.75 ] x 1392.84 K 12.5
[ rc ] x [ r1v ]
4
T4 = 634.38 K
Thermal Efficiecy for Diesel Cycle Qin, Q1 = (T3 ─ T2 ) = 1.005 kJ/kg.K x ( 1392.84 K – 795.91 K) Qin, Q1 = 599.91 kJ/kg Qout, Q2 = Cv (T4 ─ T1 ) = 0.718 kJ/kg.K x ( 634.38 K – 289.8 K) Qout, Q2 = 247.41 kJ/kg
th, Diesel
= 1-
Qout, Q2 Qin, Q1
= 1 - 247.41 kJ/kg 599.91 kJ/kg
= 0.588 @ 58.8 % © MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Dual/Combined Cycle Analysis P
P3=P4
Dual/Combined Cycle Process: 1 to 2 is isentropic compression with compression ratio v1 / v2, or rv ,
Qin 3
Qin
2 to 3 is reversible constant volume heating, the heat supplied Q1
4
3 to 4 is reversible constant pressure
p.v conts
heating, the heat supplied Q1, v4/ v3, cut off
2
ratio,
5
Qout
1 V2=V3
V1=V5
V
rc
4 to 5 is isentropic expansion, 5 to 1 is reversible constant volume cooling, the heat rejected Q2.
Figure 2.11: P-V Diagram for Dual/Combined Cycle
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Formula to find the thermal efficiency for Dual/Combined Cycle Point 1 Get the data from the question. Point 2 (1 to 2 is isentropic compression )
= P2V2 1 1 = T2V2 T1V1 P1V1
To get pressure To get temperature (Kelvin)
Point 3 (2 to 3 is reversible constant volume heating ) Constant volume
P2V2 = P3V3 T2 T3
P3 = T3 x P2 T2 T3 = P3 x T2 P2
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Assume the heat added at constant volume is equal to the heat added at constant pressure; cv(T3 – T2 ) = (T4 – T3 ) T4 = cv(T3 – T2 ) + . (T3) Point 4 (3 to 4 is reversible constant pressure heating ) Constant pressure
P3V3 = P4V4 T3 T4
V4 = T 4 = V3 T3
rc
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE P
Qin
Point 5 (4 to 5 is isentropic expansion ) P4V4
= P5V5
P3=P4
P5 = [ V4 ] x P4 V5 =
[ VV45 ] = Put
4
Qin
[ VV45 ]
3
1
p.v conts
2
[ VV43 ] x [ VV21 ] 1
[ rc ] x [ rv ]
2
Qout
1 V2=V3
2 into 1
r c P5 = [ ] x rv
5
P4
© MSF @ POLITEKNIK UNGKU OMAR
V1=V5
V
(DJA3032) INTERNAL COMBUSTION ENGINE
1
T4V4
= T5V5
P
1
Qin
1
T 5 = [ V4 ] x T4 V5 [ VV45 ] = [ VV43 ] x [ VV21 ]
[ VV45 ] = Put
1
[ rc ] [ rv ] x
3
3 P3=P4
4
Qin 4
p.v conts
2 5
4 into 3 T 5 = [ rc
1
1
rv ]
x T4
Qin, Q1 = Cv(T3 ─ T2 ) + (T4 ─ T3 ) Qout, Q2 = Cv(T5 ─ T1 )
Qout
V1=V5
V2=V3
th, Dual
= 1-
Qout, Q2 Qin, Q1
© MSF @ POLITEKNIK UNGKU OMAR
Thermal Efficiency
V
(DJA3032) INTERNAL COMBUSTION ENGINE
Example 2.4 An oil engine takes in air at 1.01 bar, 200C and the maximum cycle pressure is 69 bar. The compression ratio is 18/1. Draw the p-v diagram and calculate the air standard thermal efficiency based on the dual combustion cycle. Assume that the heat added at constant volume is equal to the heat added at constant pressure. Solution
rv =
18
T1 = 20°C + 273K = 293 K P1 = 1.01 bar P3 = 69 bar
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Point 2 (1 to 2 is isentropic compression ) P1V1
= P2V2
T1V1
1
=
T2V2
1
1
P2 = [ V1 ] x P1 V2
V1 T2 = [ V2 ]
P2 = [18] x 1.01 bar
T2 = [18] x 293 K
P2 = 57.77 bar
T2 = 931.06 K
1.4
x T1
0.4
Point 3 (2 to 3 is reversible constant volume heating ) Constant volume
P2V2 = P3V3 T2 T3
P3 = T3 x P2 T2 T3 = P3 x T2 P2 69 bar x 931.06 K 57.77 bar T3 = 1112.05 K T3 =
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Assume the heat added at constant volume is equal to the heat added at constant pressure; cv(T3 – T2 ) = (T4 – T3 ) T4 = cv(T3 – T2 ) + . (T3) T4 =
0.718 kJ/kg.K (1112.05 K – 931.06 K) + 1.005 kJ/kg.K(1112.05 K) 1.005 kJ/kg.K
T4 = 1241.30 K Point 4 (3 to 4 is reversible constant pressure heating ) Constant pressure
P3V3 = P4V4 T3 T4
V4 = T4 = V3 T3
rc 1241.30 K 1112.05 K
rc
=
rc
= 1.116
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE Point 5 (4 to 5 is isentropic expansion )
P4V4
= P5V5
P5 = [ V4 ] x P4 V5
[ VV45 ]
=
[ VV43 ] x [ VV21 ]
[ VV45 ]
=
[ rc ] x [ r1v ]
1
T4V4
= T5V5
1
r c P5 = [ ] x rv
P4
1.4
P5 = [ 1.116 ] x 69 bar 18 2
P5 = 1.41 bar
Put
1
=
2 into 1
1
T 5 = [ V4 ] x T4 V5 [ VV45 ] = [ VV43 ] x [ VV21 ]
[ VV45 ]
Put
[ rc ] x [ r1v ]
3
4
4 into 3
T 5 = [ rc
1
rv ]
x T4 0.4
T5 = [ 1.116 ] x 1241.30 K 18 T5 = 408.16 K © MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Thermal Efficiecy for Dual/Combined Cycle Qin, Q1 = Cv(T3 ─ T2 ) + (T4 ─ T3 ) Qin, Q1 = 0.718 kJ/kg.K(1112.05 K ─ 931.06 K ) + 1.005 kJ/kg.K (1241.30 K ─1112.05 K) Qin, Q1 = 259.85 kJ/kg Qout, Q2 = Cv (T5 ─ T1 ) Qout, Q2 = 0.718 kJ/kg.K x (408.16 K– 293 K) Qout, Q2 = 82.68 kJ/kg
th, Dual
= 1-
Qout, Q2 Qin, Q1
= 1 - 82.68 kJ/kg 259.85 kJ/kg
= 0.682 @ 68.2 %
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Self –Assessment Exercise 2.10 The pressure and temperature of air standard dual combustion cycle are given below, i. ii. iii. iv. v. vi.
T1 = 290 K P1 = 1.01 bar T2 = 871.1K T3 = 1087.5 K T4 = 1236.3 K T5 = 429.3 K
and ratio of compression is 16:1. Calculate the thermal efficiency for this dual cycle engine. Exercise 2.11 In a dual combustion cycle, the maximum pressure is 64 bar. Calculate the thermal efficiency when the pressure and temperature at the start of the compression are 1.01 bar and 18 0 C respectively. The compression ratio is 17/1. © MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Exercise 2.12 In a dual combustion cycle, the maximum pressure is 62 bar. Calculate the thermal efficiency when the pressure and temperature at the start of the compression are 1.01 bar and 16.7 °C respectively. The compression ratio is 17.7. Exercise 2.13 In a combine cycle, the maximum pressure is 67 bar. Calculate the thermal efficiency when the pressure and temperature at the start of the compression are 1.03 bar and 17 °C respectively. The compression ratio is 16.9. Exercise 2.14 An engine is operating in dual combustion cycle, the maximum pressure is 66 bar. Calculate the thermal efficiency for this engine when the pressure and temperature at the beginning of the compression process are 1.03 bar and 20.3 °C respectively. The compression ratio for this engine is 17.9.
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Answer for Exercise 2.11 T1 = 291 K P1 = 1.01 bar
T2 = 903.80 K P2 = 53.33 bar
Qin = 259.75 kJ/kg Qout = 84.61 kJ/kg
T3 = 1084.68 K P3 = 64 bar
T4 = 1213.91 K P4 = 64 bar rc = 1.119
T5 = 408.85 K P5 = 1.42 bar
T4 = 1069.25 K P4 = 62 bar rc = 1.064
T5 = 347.30 K P5 = 1.21 bar
ɳth, dual = 0.674 @ 67.4 %
Answer for Exercise 2.12 T1 = 289.7 K P1 = 1.01 bar
T2 = 914.40 K P2 = 56.43 bar
Qin = 129.70 kJ/kg Qout = 41.35 kJ/kg
T3 = 1004.72 K P3 = 62 bar
ɳth, dual = 0.681 @ 68.1 %
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
Answer for Exercise 2.13 T1 = 290 K P1 = 1.03 bar
T2 = 898.57 K P2 = 53.94 bar
Qin = 312.54 kJ/kg Qout = 102.25 kJ/kg
T3 = 1116.22 K P3 = 67 bar
T4 = 1271.71 K P4 = 67 bar rc = 1.139
T5 = 432.40 K P5 = 1.54 bar
T4 = 1135.68 K P4 = 66 bar rc = 1.082
T5 = 369.62 K P5 = 1.30 bar
ɳth, dual = 0.673 @ 67.3 %
Answer for Exercise 2.14 T1 = 293.3 K P1 = 1.03 bar
T2 = 929.94 K P2 = 58.46 bar
Qin = 172.33 kJ/kg Qout = 54.79 kJ/kg
T3 = 1049.94 K P3 = 66 bar
ɳth, dual = 0.682 @ 68.2 %
© MSF @ POLITEKNIK UNGKU OMAR
(DJA3032) INTERNAL COMBUSTION ENGINE
© MSF @ POLITEKNIK UNGKU OMAR
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