Question 1 of 8: Hydro Turbines

Question 1: Hydro Turbines (10 marks)

Part I — Hydro turbine plant (Kaplan, 4 MW)

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. From FIG. 602 the water surfaces are: reservoir normal high WL 45.54 m, normal low WL 42.97 m; tailrace full-load WL 27.30 m. Turbine-generator max power 4 MW, η = 100%, frictionless. The 18.52 m vertical dimension on the drawing sets the drawing scale, from which the steel penstock bore scales to ≈ 2.8 m.

Find. Flow rate and penstock velocity at the design (low-water) head, and the velocity at the higher (normal-high-water) head.

Approach. With 100% efficiency and no friction the gross head between reservoir and tailrace surfaces is the head across the machine; get Q from P = ρgQH, then V = Q/A with the scaled penstock area.

1. Operating head at low water. Gross head between reservoir normal-low and tailrace full-load surfaces: $$H_a = 42.97-27.30 = \boxed{15.67\ \text{m}}$$
2. Flow rate at maximum power. From $P=\rho g Q H$ with $\eta=1$: $$Q_a=\frac{P}{\rho g H_a}=\frac{4\times10^{6}}{1000\cdot 9.81\cdot 15.67}=\boxed{26.0\ \text{m}^3/\text{s}}$$
3. Penstock diameter from the drawing. Scaling the penstock bore against the 18.52 m reference gives $D\approx 2.8\ \text{m}$, so $A=\tfrac{\pi}{4}D^{2}=\tfrac{\pi}{4}(2.8)^2=6.16\ \text{m}^2$.
4. Penstock velocity (b). $$V_b=\frac{Q_a}{A}=\frac{26.0}{6.16}=\boxed{4.2\ \text{m/s}}$$
5. Effect of higher reservoir level (c). At normal high WL the head rises to $H_c=45.54-27.30=18.24\ \text{m}$. Because the power is fixed at 4 MW, a larger head requires a smaller flow ($Q=P/\rho g H$), hence a lower penstock velocity.
6. Velocity at high water (d). $Q_c=\dfrac{4\times10^{6}}{1000\cdot 9.81\cdot 18.24}=22.4\ \text{m}^3/\text{s}$, so $$V_d=\frac{22.4}{6.16}=\boxed{3.6\ \text{m/s}}$$

Part II — Kaplan turbine efficiency (Mactaquac)

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.

Find. Hydraulic (water) power delivered across the machine, the electrical output, and the overall turbine-generator efficiency.

Approach. Apply the steady-flow energy equation between the inlet and outlet gauges to get the total head H across the machine (pressure + velocity + elevation heads); water power is ρgQH; efficiency is electrical output over water power.

1. Throat velocities. $V=\dfrac{Q}{\frac{\pi}{4}D^2}$: $$V_{in}=\frac{354}{\frac{\pi}{4}(6.4)^2}=11.0\ \text{m/s},\qquad V_{out}=\frac{354}{\frac{\pi}{4}(7.0)^2}=9.2\ \text{m/s}$$
2. Head across the machine. With $p_{in}/\rho g = 226000/(1000\cdot 9.81)=23.04\ \text{m}$, $p_{out}=-4.5\ \text{m}$ (given as head), and the inlet 5.0 m above the outlet: $$H=\Big(\tfrac{p_{in}}{\rho g}+\tfrac{V_{in}^2}{2g}+z_{in}\Big)-\Big(\tfrac{p_{out}}{\rho g}+\tfrac{V_{out}^2}{2g}+z_{out}\Big)$$$$H=23.04+6.17+5.0-(-4.5)-4.31=\boxed{34.4\ \text{m}}$$
3. (a) Hydraulic power. $$P_{hyd}=\rho g Q H=1000\cdot 9.81\cdot 354\cdot 34.4=\boxed{119.4\ \text{MW}}$$
4. (b) Electrical output. Given directly: $P_{elec}=\boxed{110\ \text{MW}}$.
5. (c) Overall efficiency. $$\eta=\frac{P_{elec}}{P_{hyd}}=\frac{110}{119.4}=\boxed{92.1\%}$$