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22-Mec-A6 Fluid Machinery · December 2016

Question 3 of 8: Hydro Turbine Design (Francis)

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format: National Examination 07-Mec-A6-1 Fluid Machinery, December 2016 — closed book, 3 hours. Section A (Calculative, Q1–Q5) and Section B (Descriptive, Q6–Q8); candidates do four of A and two of B for 60 marks. All eight questions are solved as a study resource. Values marked “from the figure” are read from the examination attachment drawings.

Reference texts: Dixon & Hall, Fluid Mechanics and Thermodynamics of Turbomachinery (7th ed.); Cohen, Rogers & Saravanamuttoo, Gas Turbine Theory (6th ed.); Turton, Principles of Turbomachinery; Çengel & Boles, Thermodynamics (9th ed.); Fox & McDonald, Introduction to Fluid Mechanics (9th ed.). Constants from the paper: $g=9.81\ \text{m/s}^2$, $c_p=1.005\ \text{kJ/kg\,K}$, $k=1.4$, $R=0.287\ \text{kJ/kg\,K}$, $\rho_{water}=1000\ \text{kg/m}^3$, $p_{atm}=100\ \text{kPa}$, $p_{vap}=2.34\ \text{kPa}$.

Question 3: Hydro Turbine Design (Francis) (10 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Francis turbine, electrical output $P_e=200\ \text{MW}$, net head $H=110\ \text{m}$, dimensionless power specific speed $\Omega_{sp}=0.9$, $\alpha_1=22^\circ$, $V_{abs}/V_{jet}=0.77$, $V_{blade}/V_{jet}=0.6583$, with $V_{jet}=\sqrt{2gH}$.

Find. Turbine efficiency and hydraulic input power, rotational speed and runner diameter, the runner-inlet velocity triangle and radial velocity, and the flow rate and runner-inlet height.

u1 = 30.6 m/s (blade)V1 = 35.8 m/s (alpha1=22deg)Vr1 = 13.4 m/s (radial)whirl Vw1 = 33.2 m/sVw1 (33.2) exceeds u1 (30.6), so the relative inlet velocity W1 (red) has forward whirl.
Fig. 3.1 — Runner-inlet velocity triangle: absolute velocity V1 at 22° to the tangent (blade speed u1), resolved into whirl Vw1 and radial Vr1.

Approach. Read $\eta$ from the Francis efficiency chart to get hydraulic power; invert the dimensionless specific speed for $\omega$; get the blade speed from the velocity ratio to size the runner; resolve the inlet triangle for the radial velocity; and close with continuity for the flow and inlet height.

  1. Efficiency and hydraulic power (a). From the Page-13 chart, the Francis line at $\Omega_{sp}=0.9$ reads $\eta\approx0.94$. Hence $$P_{hyd}=\frac{P_e}{\eta}=\frac{200}{0.94}=\boxed{212.8\ \text{MW}}$$
  2. Rotational speed (b). From $\Omega_{sp}=\dfrac{\omega\sqrt{P}}{\rho^{1/2}(gH)^{5/4}}$ with $P=P_e$: $$\omega=\frac{\Omega_{sp}\,\rho^{1/2}(gH)^{5/4}}{\sqrt{P}}=\frac{0.9\times31.62\times(1079.1)^{1.25}}{\sqrt{200\times10^6}}=12.45\ \text{rad/s}$$ $$N=\frac{60\omega}{2\pi}=\boxed{118.8\ \text{rev/min}}$$
  3. Runner diameter (b). Jet velocity $V_{jet}=\sqrt{2gH}=\sqrt{2(9.81)(110)}=46.46\ \text{m/s}$; blade speed $U=0.6583V_{jet}=30.58\ \text{m/s}$. From $U=\omega D/2$: $$D=\frac{2U}{\omega}=\frac{2(30.58)}{12.45}=\boxed{4.91\ \text{m}}$$
  4. Inlet velocity diagram & radial velocity (c). Absolute inlet velocity $V_1=0.77V_{jet}=35.77\ \text{m/s}$ at $\alpha_1=22^\circ$ to the tangent: $$V_{r1}=V_1\sin\alpha_1=35.77\sin22^\circ=\boxed{13.4\ \text{m/s}},\qquad V_{w1}=V_1\cos22^\circ=33.2\ \text{m/s}$$
  5. Flow rate and inlet height (d). From $P_{hyd}=\rho g Q H$: $$Q=\frac{P_{hyd}}{\rho g H}=\frac{212.8\times10^6}{1000(9.81)(110)}=\boxed{197\ \text{m}^3/\text{s}}$$ and with $Q=\pi D\,b\,V_{r1}$ over the runner periphery, $$b=\frac{Q}{\pi D V_{r1}}=\frac{197}{\pi(4.91)(13.4)}=\boxed{0.95\ \text{m}}$$
Question 3 results
QuantityValue
Efficiency (chart)0.94
Hydraulic power input212.8 MW
Rotational speed $N$118.8 rev/min
Runner diameter $D$4.91 m
Radial inlet velocity $V_{r1}$13.4 m/s
Flow rate $Q$197 m³/s
Runner-inlet height $b$0.95 m