---
title: "Conical draft tube inlet pressure and efficiency"
subject: "Hydro Power"
chapter: 7
key: "semester_6/hydro_power:conical-draft-tube-numerical"
source: https://www.bctengineer.com/q/semester_6/hydro_power/conical-draft-tube-numerical
site: BCT Engineering
---

# Conical draft tube inlet pressure and efficiency

Hydro Power · Chapter 7: Hydro-Electric Machines

- Asked in: 2074 Bhadra, 6 marks
- Video: [Problem 2 on Draft tube/ Conical draft tube/   Fluid mechanics](https://www.youtube.com/watch?v=FvrEB1V3PGI) — Edutech Guru Engineering Learning-By Priyanka ma'am
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/conical-draft-tube-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-7

## Question

A conical draft tube having inlet and outlet diameters 1.5 m and 2.1 m discharges water at outlet with a velocity of 3 m/s. The total length of the draft tube is 7.5 m and 1.5 m of length of draft tube is immersed in water. If loss of head due to friction in draft tube is 0.2 × velocity head at the tube outlet, determine pressure head at inlet of the draft tube. Also find the efficiency of the draft tube.

## Short answer

Using continuity, inlet velocity V1=5.88 m/s. Bernoulli equation with friction loss hf=0.092 m and outlet pressure head 1.5 m yields inlet pressure head -7.21 m gauge (3.09 m absolute). Efficiency ηd = (V1^2−V2^2)/2g − hf divided by V1^2/2g = 68.8%.

## Answer

**Given:** inlet diameter $D_1 = 1.5$ m, outlet diameter $D_2 = 2.1$ m, outlet velocity $V_2 = 3$ m/s, total length 7.5 m, length immersed in tailwater $y = 1.5$ m, friction loss $h_f = 0.2\,V_2^2/2g$. Take atmospheric pressure head $= 10.3$ m of water.

```text
  (1) inlet  D1 = 1.5 m
          /  \          ^
         /    \         | 6.0 m above TWL
 ~~~~~~~/      \~~~~~~~~|~~~~~ TWL
       /        \       | 1.5 m below TWL
  (2) outlet D2 = 2.1 m, V2 = 3 m/s
```

Height of inlet above tailwater $H_s = 7.5 - 1.5 = 6.0$ m.

#### Velocity at inlet (continuity)
$$
\begin{aligned}
V_1 &= V_2\left(\frac{D_2}{D_1}\right)^2 = 3 \times \left(\frac{2.1}{1.5}\right)^2 = 5.88\ \text{m/s}
\end{aligned}
$$

#### Velocity heads and loss
$$
\begin{aligned}
\frac{V_1^2}{2g} &= \frac{5.88^2}{19.62} = 1.762\ \text{m} \\
\frac{V_2^2}{2g} &= \frac{3^2}{19.62} = 0.459\ \text{m} \\
h_f &= 0.2 \times 0.459 = 0.092\ \text{m}
\end{aligned}
$$

#### Pressure head at inlet
Bernoulli between 1 and 2 (datum at outlet, gauge pressures; $p_2/\rho g = 1.5$ m):

$$
\begin{aligned}
\frac{p_1}{\rho g} + \frac{V_1^2}{2g} + 7.5 &= \frac{p_2}{\rho g} + \frac{V_2^2}{2g} + h_f \\
\frac{p_1}{\rho g} + 1.762 + 7.5 &= 1.5 + 0.459 + 0.092 \\
\frac{p_1}{\rho g} &= -7.21\ \text{m (gauge)}
\end{aligned}
$$

Absolute pressure head $= 10.3 - 7.21 = 3.09$ m of water (abs).

#### Efficiency of the draft tube
$$
\begin{aligned}
\eta_d &= \frac{\dfrac{V_1^2 - V_2^2}{2g} - h_f}{\dfrac{V_1^2}{2g}} = \frac{1.762 - 0.459 - 0.092}{1.762} \\
&= \frac{1.211}{1.762} = 0.688
\end{aligned}
$$

**Answer:** pressure head at inlet $= -7.21$ m of water (vacuum), i.e. **3.09 m of water absolute**; efficiency of draft tube $= 68.8\%$.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What is the velocity at the draft tube inlet (D1 = 1.5 m, D2 = 2.1 m, V2 = 3 m/s)?
   A. 4.29 m/s
   B. 5.88 m/s
   C. 3.00 m/s
   D. 8.24 m/s
   Answer: B. V1 = V2 (D2/D1)^2 = 3 x 1.96 = 5.88 m/s. (Key point: Continuity for V1)

2. With total length 7.5 m and 1.5 m immersed, how high is the inlet above tailwater?
   A. 7.5 m
   B. 1.5 m
   C. 9.0 m
   D. 6.0 m
   Answer: D. Hs = 7.5 - 1.5 = 6.0 m. (Key point: Height of inlet above TWL)

3. What is the gauge pressure head at the inlet?
   A. +7.21 m
   B. -7.21 m
   C. -3.09 m
   D. -1.5 m
   Answer: B. p1/rho g = 1.5 + 0.459 + 0.092 - 1.762 - 7.5 = -7.21 m (vacuum). (Key point: Pressure head at inlet)

4. What is the efficiency of the draft tube?
   A. 68.8%
   B. 26.1%
   C. 100%
   D. 31.2%
   Answer: A. eta = [(V1^2 - V2^2)/2g - hf]/(V1^2/2g) = (1.762 - 0.459 - 0.092)/1.762 = 0.688. (Key point: Draft tube efficiency)

## Related in Hydro Power

- [Draft tube functions and inlet pressure](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Adraft-tube-pressure-inlet) — Hydro Power, chapter 7, asked 2×
- [Head loss with enlarged central pipe length](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Apipe-replaced-central-length) — Hydro Power, chapter 1
- [Discharge and head loss in horizontal pipe](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Apipe-discharge-head-loss) — Hydro Power, chapter 1
- [Head loss at sudden enlargement](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Asudden-enlargement-loss) — Hydro Power, chapter 1
- [Why draft tube is used in Francis turbine](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Adraft-tube-francis) — Hydro Power, chapter 7
- [Draft tube function and turbine selection criteria](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Adraft-tube-turbine-selection) — Hydro Power, chapter 7

## Related in other subjects

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