--- title: "Francis overall efficiency with draft tube" subject: "Hydropower Engineering" chapter: 7 key: "semester_7/hydropower_engineering:francis-draft-tube-efficiency-32mw" source: https://www.bctengineer.com/q/semester_7/hydropower_engineering/francis-draft-tube-efficiency-32mw site: BCT Engineering --- # Francis overall efficiency with draft tube Hydropower Engineering · Chapter 7: Hydro-electric Machines - Asked in: 2081 Bhadra, 5 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/hydropower_engineering/francis-draft-tube-efficiency-32mw - Whole chapter: https://www.bctengineer.com/md/semester_7/hydropower_engineering/chapter-7 ## Question Determine the overall efficiency of a Francis turbine developing 32 MW of power under a net head of 54 m. It is provided with a draft tube, which has an inlet diameter of 3 m and is set 2.2 m above the tailrace level. A vacuum gauge connected to the draft tube indicates a reading of 4.8 m of water. Assume the efficiency of the draft tube is 78%. ## Short answer From draft tube efficiency 78% and vacuum 4.8 m, V1²/(2g) = (4.8 − 2.2)/0.78 = 3.33 m gives V1 = 8.09 m/s, discharge 57.2 m³/s, overall efficiency 105.7%. This exceeds 100%, indicating inconsistent data; method uses ηo = P/(γQH). ## Answer #### 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 $\eta_o = 64.4$ %.) ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does Bernoulli between draft tube inlet and tailrace surface give? A. V1^2/2g = 4.8 m B. (V1^2 - V2^2)/2g - hf = 4.8 - 2.2 = 2.6 m C. (V1^2 - V2^2)/2g - hf = 4.8 + 2.2 m D. V1 = V2 Answer: B. With p1/gamma = -4.8 m and the inlet 2.2 m above tailwater, the regained head is 4.8 - 2.2 = 2.6 m. (Key point: Bernoulli at draft tube) 2. How is V1^2/2g found from the draft tube efficiency? A. Add 0.78 to 2.6 B. Take 2.6 m directly C. Multiply 2.6 by 0.78 D. Divide the regained head by eta_d: 2.6/0.78 = 3.333 m Answer: D. eta_d = (regained head)/(V1^2/2g), so V1^2/2g = 3.333 m and V1 = 8.087 m/s. (Key point: Draft tube efficiency) 3. How is the discharge obtained? A. Q = V1 x 3 B. Q = (pi/4) d1^2 V1 = 57.16 m3/s C. Q = P/(gamma H) D. Q = eta_d P Answer: B. Flow through the 3 m inlet at V1 = 8.087 m/s gives Q = 57.16 m3/s. (Key point: Discharge) 4. What does the computed overall efficiency of 105.7% indicate? A. Cavitation is absent B. The draft tube is inefficient C. The given data are inconsistent, since efficiency cannot exceed 100% D. A very efficient turbine Answer: C. Efficiency above 100% is impossible, so one of the printed values is inconsistent, though the method stands. 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