Francis overall efficiency with draft tube

Data

$P = 32$ MW, $H = 54$ m, draft tube inlet $d_1 = 3$ m, inlet at $H_s = 2.2$ m above tailrace level, vacuum at inlet $= 4.8$ m of water, $\eta_d = 0.78$.

Energy regained in the draft tube

Apply Bernoulli between the inlet (1) and the tailrace surface (2), using gauge pressures ($p_1/\gamma = -4.8$ m, tailrace surface 0):

$$-4.8 + \frac{V_1^2}{2g} + 2.2 = \frac{V_2^2}{2g} + h_f$$ $$\frac{V_1^2 - V_2^2}{2g} - h_f = 4.8 - 2.2 = 2.6\ \text{m}$$

By definition of draft tube efficiency,

$$\eta_d = \frac{\frac{V_1^2 - V_2^2}{2g} - h_f}{V_1^2/2g} \;\Rightarrow\; \frac{V_1^2}{2g} = \frac{2.6}{0.78} = 3.333\ \text{m}$$ $$V_1 = \sqrt{2 \times 9.81 \times 3.333} = 8.087\ \text{m/s}$$

Discharge and overall efficiency

$$Q = \frac{\pi}{4}(3)^2 \times 8.087 = 57.16\ \text{m}^3/\text{s}$$

Water power $= \gamma Q H = 9.81 \times 57.16 \times 54 = 30{,}282$ kW.

$$\eta_o = \frac{P}{\gamma Q H} = \frac{32{,}000}{30{,}282} = 1.057$$

Answer: the data as printed give $Q = 57.2\ \text{m}^3/\text{s}$ and $\eta_o = 1.057$ (105.7 %). This is above 100 %, which is physically impossible, so one of the given values (32 MW, 4.8 m vacuum or 2.2 m setting) is inconsistent. The method above is the required one: find $V_1$ from the draft tube efficiency and the vacuum reading, then $Q$, then $\eta_o = P/\gamma QH$. (For example, if the inlet were 2.2 m below tailwater instead, the regained head would be $4.8 + 2.2 = 7.0$ m, giving $Q = 93.8\ \text{m}^3/\text{s}$ and $\eta_o = 64.4$ %.)