Question 2 of 6: Hydro Power Plants — waterway losses, potential power and turbine efficiency

Question 2: Hydro Power Plants — waterway losses, potential power and turbine efficiency (15 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. Two independent hydro problems worked in head form, with ρ = 1000 kg/m³ and g = 9.81 m/s² from page 14 of the paper.

Find. Part I: the two pipe velocities, the head lost in the penstock and in the draft tube and tailrace, the gross potential power of the site and the hydraulic power actually delivered to the runner. Part II: the efficiency implied by the Mactaquac test measurements.

Approach. Both parts are the steady-flow energy equation written in head form. Continuity gives the velocities; the total head $H=z+p/\rho g+V^{2}/2g$ is then evaluated at each measuring point, differences in total head between points outside the machine are friction losses, and the difference across the machine multiplied by $\rho g Q$ is the power passing through it.

1. Part (a)(i) — velocities from continuity. With $A=\pi D^{2}/4$ and $V=Q/A$:$$A_2=\dfrac{\pi (7.0)^{2}}{4}=38.48\ \text{m}^{2}\qquad V_2=\dfrac{200}{38.48}=\boxed{5.20\ \text{m/s}}$$$$A_3=\dfrac{\pi (5.0)^{2}}{4}=19.64\ \text{m}^{2}\qquad V_3=\dfrac{200}{19.64}=\boxed{10.19\ \text{m/s}}$$The draft tube is the smaller pipe, so the water leaves the runner nearly twice as fast as it enters — which is precisely why a draft tube is fitted, to recover that kinetic energy before the flow reaches the tailrace.
2. Part (a)(ii), step 1 — total head at each measuring point. Taking gauge pressures throughout and the still reservoir and tailrace surfaces as having zero velocity head,$$H_1=z_1=1170.5\ \text{m}\qquad H_4=z_4=1094.7\ \text{m}$$$$H_2=z_2+\dfrac{p_2}{\rho g}+\dfrac{V_2^{2}}{2g}=1091.5+\dfrac{700\,000}{1000\times 9.81}+\dfrac{5.20^{2}}{2\times 9.81}=1091.5+71.36+1.38=1164.23\ \text{m}$$$$H_3=z_3+\dfrac{p_3}{\rho g}+\dfrac{V_3^{2}}{2g}=1086.5+6.63+5.29=1098.41\ \text{m}$$
3. Part (a)(ii), step 2 — the two friction losses. No work crosses the boundary between 1 and 2 or between 3 and 4, so the fall in total head over each of those reaches is entirely loss:$$h_{L,1\to 2}=H_1-H_2=1170.5-1164.23=\boxed{6.27\ \text{m}}$$$$h_{L,3\to 4}=H_3-H_4=1098.41-1094.7=\boxed{3.71\ \text{m}}$$The draft tube and tailrace loss is smaller than the penstock loss even though the velocity there is twice as high, because the reach is far shorter.
4. Part (b)(i) — potential power of the whole plant. The gross head is the difference between the two free water surfaces, and the potential power is what the site would deliver with no losses anywhere:$$P_{pot}=\rho g Q\,(z_1-z_4)=1000\times 9.81\times 200\times 75.8=\boxed{148.7\ \text{MW}}$$
5. Part (b)(ii) — hydraulic power developed in the turbine. Between points 2 and 3 the machine extracts the fall in total head, and the question asks for that quantity with no internal losses charged against it:$$H_{turb}=H_2-H_3=1164.23-1098.41=65.82\ \text{m}$$$$P_{hyd}=\rho g Q\,H_{turb}=1000\times 9.81\times 200\times 65.82=\boxed{129.1\ \text{MW}}$$The two answers close on each other exactly, which is the check worth writing down: $148.7-\rho g Q\,(6.27+3.71)/10^{6}=148.7-19.6=129.1$ MW. Just over 13 % of the site potential is lost in the waterways before the runner sees it.
6. Part II, step 1 — velocities at the two Mactaquac tappings. The same continuity calculation on the measured flow gives$$V_{in}=\dfrac{354}{\pi (6.4)^{2}/4}=\dfrac{354}{32.17}=11.00\ \text{m/s}\qquad V_{out}=\dfrac{354}{\pi (7.0)^{2}/4}=\dfrac{354}{38.48}=9.20\ \text{m/s}$$
7. Part II, step 2 — net head across the machine. Add the pressure, elevation and velocity contributions between the two tappings. The outlet pressure is already quoted as a head, and it is negative because the draft tube runs under vacuum:$$\dfrac{p_{in}-p_{out}}{\rho g}=\dfrac{226\,000}{9810}-(-4.5)=23.04+4.5=27.54\ \text{m}$$$$\Delta z=5.0\ \text{m}\qquad \dfrac{V_{in}^{2}-V_{out}^{2}}{2g}=\dfrac{121.09-84.61}{19.62}=1.86\ \text{m}$$$$H=27.54+5.0+1.86=\boxed{34.40\ \text{m}}$$
8. Part II, step 3 — water power and efficiency. The power arriving in the water is $\rho g Q H$, and the efficiency is the measured electrical output divided by it:$$P_{water}=1000\times 9.81\times 354\times 34.40=119.5\ \text{MW}$$$$\eta=\dfrac{P_e}{P_{water}}=\dfrac{110.0}{119.5}=\boxed{0.921\ (92.1\ \%)}$$This is the combined turbine-and-generator efficiency, because the numerator is measured at the generator terminals. If the generator is taken as 98.5 % efficient, the runner alone is at 93.5 % — both figures sit squarely in the range expected of a large Kaplan machine. As a cross-check on the machine type, the specific speed $N_s=N\sqrt{P}/H^{1.25}=112.5\sqrt{110\,000}/34.40^{1.25}=448$, which is in the axial-flow Kaplan band and consistent with the low head.