Question 1 of 8: Hydro Turbines

Question 1: Hydro Turbines (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.

Part I — Mactaquac efficiency

Given. Flow and gauge readings at the machine inlet and outlet of a Kaplan set.

Find. Water (hydraulic) power input, electrical output, and overall turbine-generator efficiency.

Approach. Apply the steady-flow energy equation between the inlet and outlet gauge sections to get the head extracted, multiply by $\rho g Q$ for water power, then form efficiency against the stated output.

1. Section velocities from continuity. $V=\dfrac{Q}{\tfrac{\pi}{4}D^{2}}$ gives $V_{in}=\dfrac{354}{\tfrac{\pi}{4}(6.4)^{2}}=11.00\ \text{m/s}$ and $V_{out}=\dfrac{354}{\tfrac{\pi}{4}(7.0)^{2}}=9.20\ \text{m/s}$.
2. Head across the machine. With the inlet gauge as head $p/\rho g=226000/(1000\cdot9.81)=23.04\ \text{m}$ and the outlet gauge already in metres of water ($-4.5$ m), the energy equation gives $H=\left(23.04+\dfrac{11.00^{2}}{2g}+5.0\right)-\left(-4.5+\dfrac{9.20^{2}}{2g}\right)$. Substituting the velocity heads (6.17 m and 4.31 m) yields $\boxed{H=34.40\ \text{m}}$.
3. Hydraulic (water) power. $P_{hyd}=\rho g Q H=1000\cdot9.81\cdot354\cdot34.40=119.4\ \text{MW}$.
4. Efficiency. The electrical output is the stated 110 MW, so $\eta=\dfrac{P_{elec}}{P_{hyd}}=\dfrac{110}{119.4}=0.921$.

The energy equation is essential here: the suction (−4.5 m) draft-tube reading is a vacuum that adds to the head recovered, and the velocity-head difference between the unequal inlet and outlet bores is a genuine (if small) contribution. A single overall efficiency of about 92 % is exactly what a well-set large Kaplan unit achieves at its design point.

Part II — plant flow and penstock velocity

Given. The plant cross-section (Attachment page 11) fixes the head-water and tail-water levels; the penstock bore is scaled from the 18.52 m dimension on the drawing.

Find. Penstock flow and velocity at normal-low head, and how the velocity changes when the reservoir rises.

Approach. With negligible system friction the net head is the level difference; the required flow follows from $P=\eta\rho g Q H$, and the penstock velocity from continuity.

1. Operating head (normal low reservoir). $H_a=42.97-27.30=15.67\ \text{m}$.
2. Flow at 4 MW. $Q_a=\dfrac{P}{\eta\rho g H_a}=\dfrac{4\times10^{6}}{0.90\cdot1000\cdot9.81\cdot15.67}=\boxed{28.9\ \text{m}^3/\text{s}}$.
3. Penstock velocity. With $A=\tfrac{\pi}{4}(2.8)^{2}=6.16\ \text{m}^2$, $V_b=\dfrac{Q_a}{A}=\dfrac{28.9}{6.16}=4.70\ \text{m/s}$.
4. Higher reservoir (part c). At normal high water $H_c=45.54-27.30=18.24\ \text{m}$. For the same 4 MW the flow falls, $Q_c=\dfrac{4\times10^{6}}{0.90\cdot1000\cdot9.81\cdot18.24}=24.84\ \text{m}^3/\text{s}$, so the velocity is less.
5. Velocity at high water (part d). $V_d=\dfrac{24.84}{6.16}=4.03\ \text{m/s}$, confirming the reduction.

The result in (c) is often counter-intuitive: a higher reservoir gives a greater head, but because the power is fixed at 4 MW the machine simply passes less water to deliver it, and the penstock velocity drops from 4.70 to 4.03 m/s. Selecting a realistic efficiency is legitimate exam practice under the "state your assumptions" rubric.