Q# (Marks)
Performance Characteristics of Pelton Turbine: FP_1
Q1
What is the degree of reaction in Pelton wheel turbine:
(a) 0
(b) 0.5 (c) 1
(d) 1.5
Ans: (a) 0 Casing
Q2
What is the ratio of rotor to jet velocity in pelton wheel turbine for maximum efficiency?
Ans: 1/2
mical pumps in factories. Q3
Pelton wheel is used for:
(a) High head and high speed
(b) High head and low speed
(c) Low head and high speed
(d) Low head and low speed
Ans: (b) High head and low speed
Q4
What is the shape of Pelton wheel bucket?
Ans: Semi-ellipsoidal
Q5
A Pelton wheel turbine develops 10 MW power under a head of 500 m. Rotor of the wheel rotates at 450 rpm and rotor to jet velocity ratio is 0.5. For 85% hydraulic efficiency and 170 degree jet deflection angle, estimate:
(a) The diameter of the jet. (b) Specific speed of the turbine. (c) The diameter of rotor wheel. Ans: Given data : Power output = Po = 10 MW = 10x106 W, Input head = h = 500 m, Rotor speed = N = 450 rpm, Rotor to jet velocity = U/V1 = 0.5, Blade deflection angle = θ = 1700, so β = 1800-θ = 100, Efficiency = ղ = 0.85
Power input = power output / Efficiency => P = 10 x 106/0.85 = 11.764 x 106 W
Specific speed (Ns) = N P /H5/4 = 450
11.764 106 / 5005/4 = 652.81 m/sec (Ans b)
Efficiency = ղ = Po/Pi = Po / (ρQgh) 0.85 = (10x106) / (1000 x Q x 9.81 x 500) Q = 2.39 m3/sec Assuming no frictional losses in the nozzle Power input = Pi = Power at the turbine inlet = Power delivered by the nozzle to the wheel buckets => ρQgh = ρQ (V12/2) => V1 =
2 gh = 99.045 m/sec
Since, U/V1 = 0.5 => U = 49.52 m/sec Also, U = π D N/ 60
(where, D = wheel diameter)
=> D = 60xU/(π N) = 2.101 m (Ans c) Since, flow rate(Q) = (πd2/4) x V1 => d =
4Q = 0.17 m (Ans a) V
(where, d = jet diameter)
Q# (Marks) Q1
Performance Characteristics of Pelton Turbine: FP_2
What are the essential components of a Pelton Turbine? a. Buckets fitted on the periphery of disc b. Nozzle c. Spear rod d. Rotor shaft e. Breaking jet
Q2
List when Pelton Turbines are preferred for deployment? a.
High Head / Medium Head / Low Head(encircle your choice)
b. List the typical head __250 to 1800_ m Chemical pumps in factories. Q3
What is the formula for specific speed? Specific speed (Ns) = N P /H5/4 Jet pumps have no movable parts, need no mechanical drive and are designed in a simple structure.
Q4
What is the function of the splitter in the bucket of the Pelton Turbine?
Ans:
The splitter divide the jet in two equal streams, so that the side thrusts produced by the fluid in each half balance each other.It helps split the water flow equally on either side to avoid unbalanced axial thrust on the shaft and bearings.
Q5
Find the power produced by the Pelton Turbine if water flow is 0.1 m3/s and head at the nozzle inlet is 50 m.
Assume turbine efficiency to be 90%. What will be the
optimum speed of the buckets on the rotor? List any assumptions made. Ans: Efficiency = ղ = 0.9, Input head = h = 50 m, Flow rate = Q = 0.1 m3/sec
Power produced by the turbine = Po = Efficiency x Power available at the inlet => Po = ղ x ρQgh = 0.9x1000x0.1x9.81x50 = 44145 W
Assumption (i) Neglecting losses of head through nozzle Power available at nozzle inlet = power delivered by the nozzle => ρQgh = ρQ (V2/2) => V =
2 gh = 981 m/sec
Since, for maximum efficiency, Optimum speed of buckets (U) = V/2 => U = 981/2 = 490.5 m/sec
Q# (Marks) Q1
Performance Characteristics of Pelton Turbine: FP_3
To what pressure the water flowing through buckets of a Pelton wheel turbine is subjected?
Ans: Atmospheric pressure
Q2
Which of the following device is not used in a Pelton wheel turbine? (a) Venturimeter
(b) Draft tube
(c) Breaking jet
(d) None of the above
Ans: Draft tube
Q3 Which component is used to control the flow rate in a Pelton wheel turbine? Ans: Spear rod or, Spear valve
Q4
Pelton turbine is _________
(a) Tangential flow impulse turbine (b) Radial flow impulse turbine (b) Tangential flow reaction turbine (d) Mixed flow impulse turbine Ans: (a)
Q5 Which set of independent parameters are varied to obtain the characteristic curve of a hydraulic turbine? (a) Speed and Discharge
(b) Speed and Power output
(b) Head and Hydraulic efficiency
(c) Speed, Head and Discharge
Ans: (a) Speed and Discharge
Q6
A Pelton turbine with a wheel of 1.47 m diameter operates under an effective head of 675 m at nozzle inlet and uses 4 m3/s of water flow rate. Ratio of jet diameter to wheel diameter is 0.15. Assuming that the turbine operates at a blade speed to jet speed ratio of 0.47, determine [Please note: Blade deflection angle is missing from the question, Lets take it as 1650]
(a) Wheel rotational speed; (b) Power output (c) Hydraulic efficiency Ans: Given data: D = Wheel diameter = 1.47 m, Input head = h = 675 m, flow rate = Q = 4 m3/sec Jet diameter to wheel diameter ratio = d/D = 0.15 Blade speed to jet speed ratio = U/V = 0.47
d = 0.15 x D = 0.15 x 1.47 = 0.221 m Since, Q = (πd2/4) x V
(Where, V = jet speed)
=> V = 4Q/(πd2) = 104.75 m/sec Since, U/V = 0.47 => U = 0.47 x V = 49.23 m/sec Also, U = (πDN)/60 => N = (60 x U) /(πD) = 639.64 rpm (Ans a) Taking blade deflection angle, θ = 1650 => β = 1800 - 1650 = 150 Efficiency (ղ) = [2U (1+cosβ) (V-U)] / V2 => ղ = 0.97 (Ans c) Since, Power output = Po = ղ x Power input => Po = ղ x ρQgh = 0.97 x 1000 x 4 x 9,81 x 675 = 25.941 MW (Ans b)
Q# (Marks) Q1
Performance Characteristics of Pelton Turbine: FP_4
Name the instruments used to take the readings along with the parameters they measure.
Ans: (i) Orifice meter for volume flow rate, (ii) Tachometer for rotational speed, (iii) Spring mass balance for load applied, (iv) pressure gauge for input head
Q2
The hydraulic efficiency of Pelton turbine will be maximum when blade velocity is equal to -------------? (a) V/2
(b) V/4
(c) V/√2
(d) V×(√3/2)
Ans: (a)
Q3
What is the function of the spliter?
Ans: The splitter divide the jet in two equal streams, so that the side thrusts produced by the fluid in each half balance each other.I
Q4
Pelton turbine is _________ (a) Tangential flow
(b) Radial flow (c) Axial flow
(d) Mixed flow
Ans: (a)
Q5
Define Specific speed of turbine and write the formula.
Ans: Specific speed: It is the speed of a geometrically similar turbine which would produce one unit of power when working under a unit head. Mathematically, it is expressed as: Specific speed (Ns) = N P /H5/4
Q6
Calculate the force of water jet acting on semispherical blade of Pelton turbine. The turbine is rotating, rotational speed is 320rpm. Diameter of blade wheel D1 = 1.1 m. Discharge Q = 0.027 m3/s outflows from the nozzle with diameter 0.030 m.
Ans: Given data: N = wheel rotational speed = 320 rpm, D1 = wheel diameter = 1.1 m, Q = Discharge = 0.027 m3/sec, d = jet diameter = 0.030 m Since, Tangential velocity of bucket (U) = πD1N/60 => U = 18.43 m/sec Also, Q = (πd2/4) V1 => V1 = 4Q/(πd2) = 38.197 m/sec Since, Buckets are of semi-spherical shape. So, blade deflection angle (θ) = 1800 So, Angle between relative velocity at bucket exit and bucket tangential velocity (β) = 1800 - θ = 00 Force of fluid on the bucket, F = ρQ (Vw1 + Vw2) => F = ρQ (V1 + Vr2 cosβ - U) => F = ρQ (V1 + Vr2 - U) => F = ρQ x 2(V1 - U)
(Since, Vr = V1 - U)
=> F = 1000 x 0.027 x 2 x (38.197 - 18.43) => F = 1067.418 N