Question 3 of 8: Hydro Turbine Design

Question 3: Hydro Turbine Design (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. A Francis (radial inward-flow) turbine at its design point:

Find. The power output, the volume flow, the runner diameter, the inlet velocity triangle and its radial component, and the runner height (the gate width) at inlet.

Approach. The dimensionless specific speed on page 15 contains the power as its only unknown, so it yields the output directly; the hydraulic efficiency then converts that power into a flow rate. The tip-speed ratio fixes the peripheral speed and hence the diameter, Euler's equation supplies the inlet whirl, and continuity through the cylindrical inlet surface gives the gate height.

1. Part (a) — angular velocity. Converting the required speed, $$\omega=\frac{2\pi N}{60}=\frac{2\pi\times 750}{60}=78.54\ \text{rad/s}$$
2. Power from the specific speed. The page-15 turbine specific speed is $N_s=\omega P^{1/2}/\left[\rho^{1/2}(gH)^{5/4}\right]$, which rearranges to $$P=\left[\frac{N_s\,\rho^{1/2}\,(gH)^{5/4}}{\omega}\right]^{2} =\left[\frac{0.9\times 31.62\times (9.81\times 160)^{1.25}}{78.54}\right]^{2}$$ With $(gH)^{1.25}=(1569.6)^{1.25}=9.888\times 10^{3}$, $$\boxed{P=12.8\ \text{MW}}$$ The attachment chart on page 10 shows a Francis runner peaking near $N_s\approx 0.9$ at about 91 % overall efficiency, so this machine sits exactly at the top of the Francis band — a reassuring consistency check on the choice of machine type.
3. Part (b) — volume flow rate. The hydraulic power delivered to the runner is $P=\eta_h\,\rho g Q H$, so $$Q=\frac{P}{\eta_h\,\rho g H}=\frac{12.82\times 10^{6}}{0.94\times 1000\times 9.81\times 160}$$ $$\boxed{Q=8.69\ \text{m}^3/\text{s}}$$
4. Part (c) — peripheral speed and runner diameter. The free-jet velocity corresponding to the net head is $V_{jet}=\sqrt{2gH}=\sqrt{2\times 9.81\times 160}=56.03$ m/s, so the runner tip speed is $$U_1=0.7\,V_{jet}=0.7\times 56.03=39.22\ \text{m/s}$$ and since $U_1=\omega D/2$, $$D=\frac{2U_1}{\omega}=\frac{2\times 39.22}{78.54}$$ $$\boxed{D=1.00\ \text{m}}$$ The same answer follows from $D=60U_1/(\pi N)$, and landing on a round metre is a good sign that the paper was designed backwards from this value.
5. Part (d) — inlet whirl from Euler's equation. A well-designed reaction runner discharges with no swirl, so the whole hydraulic head is absorbed at inlet: $$\eta_h\,gH=U_1V_{w1}\qquad\Longrightarrow\qquad V_{w1}=\frac{0.94\times 9.81\times 160}{39.22}=37.62\ \text{m/s}$$
6. Radial flow velocity. The absolute velocity makes 18° with the tangent, so its radial component is the whirl component times the tangent of that angle: $$V_{f1}=V_{w1}\tan\alpha_1=37.62\times\tan 18^\circ$$ $$\boxed{V_{f1}=12.2\ \text{m/s}}$$ The absolute velocity itself is $V_1=\sqrt{V_{w1}^2+V_{f1}^2}=39.6$ m/s, and the relative velocity entering the blade makes $\beta_1=\tan^{-1}\!\left[V_{f1}/(V_{w1}-U_1)\right]=97.5^\circ$ with the tangent — the blade leading edge leans very slightly backwards, which is typical of a Francis runner whose whirl is a little below the runner speed. Figure 3.1 is the required sketch.
7. Part (e) — runner height at inlet. All of the flow crosses the cylindrical surface of diameter D and height b at the radial velocity, so $$Q=\pi D\,b\,V_{f1}\qquad\Longrightarrow\qquad b=\frac{Q}{\pi D V_{f1}}=\frac{8.687}{\pi\times 0.9987\times 12.22}$$ $$\boxed{b=0.227\ \text{m}\ (227\ \text{mm})}$$ The height-to-diameter ratio is $b/D=0.23$, which is the narrow, tall-headed proportion expected of a Francis runner in the middle of its specific-speed range; a low-head machine would be much wider.
8. Closure check. Feeding the answers back through continuity, $\pi\times 0.9987\times 0.2265\times 12.22=8.69$ m³/s, reproduces the part (b) flow, and $\tan^{-1}(12.22/37.62)=18.0^\circ$ reproduces the given guide-vane angle. The design is self-consistent.