22-Mec-B3 Energy Conversion and Power Generation · May 2018
Question 5 of 8: Hydro Power Plant
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 16-Mec-B3
Energy Conversion and Power Generation, National Examinations, May 2018.
Three hours, closed book. Two sections: Section A is calculative
(Questions 1–5) and Section B is descriptive (Questions 6–8).
Candidates answer four questions from Section A and two from Section B; six
questions of 10 marks each constitute a complete paper (60 marks). Reference
data for individual questions are bound in as attachments on pages 9–15,
reference formulae and constants on pages 16–19, and Granet &
Bluestein steam tables are supplied. All eight questions are solved
below, because the set is a study resource rather than a timed attempt.
Reference texts.
Granet, I. and Bluestein, M., Thermodynamics and Heat Power, 6th ed.
— the steam tables issued with this paper, and the vapour-cycle and
gas-cycle chapters.
Çengel, Y. A. and Boles, M. A., Thermodynamics: An Engineering
Approach, 9th ed. — Brayton and Rankine cycles, combined cycles,
isentropic component efficiencies.
El-Wakil, M. M., Powerplant Technology — heat recovery steam
generators and pinch behaviour, condensers and their off-design response,
boiler heat-absorption surfaces, load scheduling.
Lamarsh, J. R. and Baratta, A. J., Introduction to Nuclear
Engineering, 4th ed. — the pressurised water reactor, primary loop
heat removal, once-through steam generators.
Dixon, S. L. and Hall, C. A., Fluid Mechanics and Thermodynamics of
Turbomachinery, 7th ed. — hydraulic turbines, net head and the energy
accounting of a waterway.
Wherever the paper's own attachments carry a value that duplicates a computed
result — the Koeberg terminal temperature difference and back pressure on
page 10, the gas-turbine output quoted in the preamble to Question 2,
the published rating of the Oconee unit — that printed value is used as an
independent check and the agreement is quoted in the answer.
Given. Water passes from a reservoir surface at 1170.5 m through a 7.0 m penstock to the turbine inlet at 1091.5 m, leaves through a 5.0 m draft tube at 1086.5 m and discharges into a tailrace whose surface stands at 1094.7 m, at a volumetric flow of 200 m3/s.
$\rho = 1000\ \text{kg/m}^3,\ g = 9.81\ \text{m/s}^2$
Find. The velocities at points 2 and 3, the head lost between 1 and 2 and between 3 and 4, the potential power of the site and the hydraulic power actually delivered to the runner.
Question 5 — the Vanderkloof waterway from reservoir to tailrace, with the four numbered stations of the question. The gross head is the 75.8 m between the reservoir surface and the minimum tailwater level; the penstock and the draft tube each consume part of it.
Approach. Use continuity for the two velocities, then write the total head $z + p/\rho g + V^2/2g$ at each numbered station. The head loss in a passage is the drop in total head across it, and each power is $\rho g \dot{V}$ times the appropriate head difference.
Part (a)(i) — velocities from continuity. The flow areas are $$A_2 = \frac{\pi}{4}(7.0)^2 = 38.48\ \text{m}^2,\qquad A_3 = \frac{\pi}{4}(5.0)^2 = 19.63\ \text{m}^2,$$ so with $V = \dot{V}/A$, $$V_2 = \frac{200}{38.48} = \boxed{5.197\ \text{m/s}},\qquad V_3 = \frac{200}{19.63} = \boxed{10.19\ \text{m/s}} .$$ The draft tube is deliberately the smaller passage at the runner exit and then flares, so that the kinetic energy leaving the runner is recovered as suction rather than thrown away.
Part (a)(ii) — total head at each station. Working in metres of water, the pressure heads are $p_2/\rho g = 700\,000/(1000\times9.81) = 71.36\ \text{m}$ and $p_3/\rho g = 65\,000/(1000\times9.81) = 6.63\ \text{m}$, and the velocity heads are $V_2^2/2g = 5.197^2/19.62 = 1.377\ \text{m}$ and $V_3^2/2g = 10.19^2/19.62 = 5.288\ \text{m}$. At the two free surfaces both the gauge pressure and the velocity are negligible, so $$H_1 = 1170.5\ \text{m},\qquad H_4 = 1094.7\ \text{m},$$ $$H_2 = 1091.5 + 71.36 + 1.377 = 1164.24\ \text{m},\qquad H_3 = 1086.5 + 6.63 + 5.288 = 1098.42\ \text{m} .$$
Head losses in the two passages. Each loss is simply the fall in total head across the passage: $$h_{L,1-2} = H_1 - H_2 = 1170.5 - 1164.24 = \boxed{6.27\ \text{m}},$$ $$h_{L,3-4} = H_3 - H_4 = 1098.42 - 1094.7 = \boxed{3.71\ \text{m}} .$$ Together they consume 9.98 m of the 75.8 m available, about 13 % — a high but believable figure for a short, large-diameter waterway operating at minimum tailwater, and dominated by the exit loss from the draft tube, which alone throws away the 5.29 m of velocity head at point 3.
Part (b)(i) — potential power of the site. The gross head is the difference between the two free surfaces, $z_1 - z_4 = 1170.5 - 1094.7 = 75.8\ \text{m}$, so $$P_{\text{pot}} = \rho g \dot{V} (z_1 - z_4) = 1000 \times 9.81 \times 200 \times 75.8 = \boxed{148.7\ \text{MW}} .$$ This is what the site would deliver if the waterway were frictionless and the machine perfect; it is the yardstick against which the plant is judged.
Part (b)(ii) — hydraulic power developed in the turbine. The runner sees only the net head between its own inlet and outlet flanges, $$H_2 - H_3 = 1164.24 - 1098.42 = 65.82\ \text{m},$$ $$P_{\text{hyd}} = \rho g \dot{V} (H_2 - H_3) = 1000 \times 9.81 \times 200 \times 65.82 = \boxed{129.1\ \text{MW}} .$$ The two answers must differ by exactly the two head losses, and they do: $\rho g \dot{V}(6.27 + 3.71) = 19.6\ \text{MW}$, and $148.7 - 19.6 = 129.1\ \text{MW}$. That closure is the check to run before writing either figure down.
What the numbers mean for the plant. The waterway efficiency is $65.82/75.8 = 86.8$ %; multiplying by a runner efficiency of about 93 % and a generator efficiency of about 98 % would put the terminal output near 118 MW, which is the right order for a Vanderkloof machine. Note also that the tailwater level given is the minimum: at a higher tailwater the gross head, and therefore the output, would fall.
