---
title: "Pelton wheel power with penstock losses"
subject: "Fluid Machines"
chapter: 3
key: "semester_5/fluid_machines:pelton-wheel-numerical-penstock"
source: https://www.bctengineer.com/q/semester_5/fluid_machines/pelton-wheel-numerical-penstock
site: BCT Engineering
---

# Pelton wheel power with penstock losses

Fluid Machines · Chapter 3: Water Turbine

- Asked in: year not given, 8 marks
- Video: [Pelton Wheel Problem Example 1](https://www.youtube.com/watch?v=j5t8H74Kb7Q) — TutorialsPoint
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/fluid_machines/pelton-wheel-numerical-penstock
- Whole chapter: https://www.bctengineer.com/md/semester_5/fluid_machines/chapter-3

## Question

A Pelton wheel is supplied with 0.5 m³/s of water through a penstock 2000 m long and 0.6 m in diameter from a reservoir whose surface is 450 m above the nozzle. Take the Darcy friction factor of the penstock as 0.016 and the minor losses as negligible. The nozzle velocity coefficient is 0.98, the speed ratio is 0.46, the wheel (pitch circle) diameter is 1.4 m, the buckets deflect the jet through 165 degrees, and the relative velocity is reduced by 5 % while passing over the bucket. Mechanical efficiency is 97 %. Calculate (a) the net head, (b) the jet diameter, (c) the speed of the wheel, (d) the power developed by the runner and its hydraulic efficiency, and (e) the shaft power.

## Short answer

Net head H = H_g − f L V_p²/(2g D) = 441.50 m. Jet velocity V₁ = C_v√(2gH) = 91.21 m/s, jet diameter d = 83.5 mm. Bucket speed u = 0.46√(2gH) = 42.81 m/s, N = 584 rpm. Euler power P = ρQ u (V_w₁ − V_w₂) = 1986.7 kW, η_h = 91.7%, shaft power = 1927.1 kW.

## Answer

**Given:** $H_g=450$ m, $L=2000$ m, $D=0.6$ m, $f=0.016$, $Q=0.5$ m³/s, $C_v=0.98$, $\phi=0.46$, $D_w=1.4$ m, deflection $165^\circ$, $k=0.95$ ($V_{r2}=kV_{r1}$), $\eta_m=0.97$.

#### (a) Net head

Velocity in penstock: $V_p=\dfrac{Q}{\frac{\pi}{4}D^2}=\dfrac{0.5}{0.2827}=1.768$ m/s.

$$
h_f=\frac{fLV_p^2}{2gD}=\frac{0.016\times2000\times1.768^2}{2\times9.81\times0.6}=8.50\ \text{m}
$$

$$
H = H_g - h_f = 450 - 8.50 = 441.50\ \text{m}
$$

#### (b) Jet diameter

$$
V_1 = C_v\sqrt{2gH} = 0.98\times\sqrt{2\times9.81\times441.50} = 0.98\times93.07 = 91.21\ \text{m/s}
$$

$$
a=\frac{Q}{V_1}=5.482\times10^{-3}\ \text{m}^2,\qquad d=\sqrt{\frac{4a}{\pi}}=83.5\ \text{mm}
$$

#### (c) Speed of wheel

$$
u=\phi\sqrt{2gH}=0.46\times93.07=42.81\ \text{m/s},\qquad N=\frac{60u}{\pi D_w}=\frac{60\times42.81}{\pi\times1.4}=584\ \text{rpm}
$$

#### (d) Power developed and hydraulic efficiency

Velocity triangles: $V_{w1}=V_1=91.21$ m/s; $V_{r1}=V_1-u=48.40$ m/s; $V_{r2}=0.95V_{r1}=45.98$ m/s.

The bucket turns the jet by $165^\circ$, so the relative exit velocity makes $180-165=15^\circ$ with the direction opposite to the bucket motion.

$$
V_{w2}=u-V_{r2}\cos15^\circ=42.81-44.41=-1.60\ \text{m/s (backwards)}
$$

$$
V_{w1}+|V_{w2}|=91.21+1.60=92.81\ \text{m/s}
$$

$$
\begin{aligned}
P&=\rho Q\,u\,(V_{w1}+|V_{w2}|)=1000\times0.5\times42.81\times92.81=1986.7\ \text{kW}\\
\text{Water power}&=\rho g Q H=2165.6\ \text{kW}\\
\eta_h&=\frac{1986.7}{2165.6}=91.7\%
\end{aligned}
$$

#### (e) Shaft power

$$
P_s=\eta_m P=0.97\times1986.7=1927.1\ \text{kW}
$$

**Answer:** (a) 441.50 m; (b) 83.5 mm; (c) 584 rpm; (d) 1986.7 kW, $\eta_h=91.7\%$; (e) 1927.1 kW.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Net head is obtained by:
   A. Gross head only
   B. Gross head divided by C_v
   C. Gross head plus friction loss
   D. Gross head minus penstock friction loss hf = fLV^2/2gD
   Answer: D. H = 450 - 8.50 = 441.50 m. (Key point: Net head)

2. The jet velocity is:
   A. sqrt(2gH)
   B. Q/A of penstock
   C. phi sqrt(2gH)
   D. Cv sqrt(2gH) = 91.21 m/s
   Answer: D. V1 = 0.98 x 93.07 = 91.21 m/s. (Key point: Jet velocity)

3. Bucket speed u is found from:
   A. u = phi sqrt(2gH) = 42.81 m/s
   B. u = V1/2 exactly
   C. u = V1
   D. u = Cv V1
   Answer: A. u = 0.46 x 93.07 = 42.81 m/s, so N = 60u/(pi D) = 584 rpm. (Key point: Bucket speed)

4. The relative velocity at exit is:
   A. Zero
   B. Same as inlet
   C. 0.95 times Vr1 = 45.98 m/s, and makes 15 degrees with the reverse direction
   D. Equal to u
   Answer: C. With 5% reduction and 165 degree deflection, Vw2 = u - Vr2 cos15 = -1.60 m/s. (Key point: Exit relative velocity)

## Related in Fluid Machines

- [Design of Pelton wheel for given power and head](https://www.bctengineer.com/question.md?key=semester_5%2Ffluid_machines%3Apelton-wheel-design-numerical) — Fluid Machines, chapter 3
- [Pelton wheel: components and working](https://www.bctengineer.com/question.md?key=semester_5%2Ffluid_machines%3Apelton-wheel-working-components) — Fluid Machines, chapter 3

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