---
title: "Pipe outlet minimum water level, 50 lps"
subject: "Irrigation and Drainage Engineering"
chapter: 7
key: "semester_6/irrigation_and_drainage_engineering:pipe-outlet-min-water-level-50lps"
source: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/pipe-outlet-min-water-level-50lps
site: BCT Engineering
---

# Pipe outlet minimum water level, 50 lps

Irrigation and Drainage Engineering · Chapter 7: Regulating Structures

- Asked in: 2072 Magh, 1+4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/irrigation_and_drainage_engineering/pipe-outlet-min-water-level-50lps
- Whole chapter: https://www.bctengineer.com/md/semester_6/irrigation_and_drainage_engineering/chapter-7

## Question

Making a suitable sketch compute the minimum water level required in the distributary to convey a flow as 50 lps through a 10 m long, free-discharging pipe outlet (n = 0.016) of 20 cm dia to a water course with FSL at 100.0 m.

## Short answer

For Q=0.05 m³/s, d=0.20 m, L=10 m, n=0.016, compute velocity 1.59 m/s, friction loss 0.352 m, entry loss 0.065 m, velocity head 0.129 m. Total head required = 0.546 m, so minimum distributary water level = 100.55 m.

## Answer

A free-discharging pipe outlet needs a head equal to the sum of the entry loss, the friction loss and the velocity head at exit. The minimum water level in the distributary is the outlet level plus this head. The pipe delivers into the water course at its FSL 100.0 m; the pipe centre at the exit is taken at 100.0 m.

```text
 distributary ~~~~ H.W.L (min) ~~~~
                     H = ?
               ---- pipe, L = 10 m, d = 20 cm --> free fall
                                     water course FSL 100.0 m
```

#### Data
$Q = 50$ l/s $= 0.05$ m³/s, $L = 10$ m, $d = 0.20$ m, Manning's $n = 0.016$, entry loss coefficient 0.5 (assumed).

#### Calculation
$$
\begin{aligned} A &= \frac{\pi}{4}(0.2)^2 = 0.03142\ \text{m}^2, \qquad v = \frac{Q}{A} = 1.592\ \text{m/s} \\ \frac{v^2}{2g} &= 0.1291\ \text{m}, \qquad R = \frac{d}{4} = 0.050\ \text{m} \\ h_f &= \frac{n^2 v^2 L}{R^{4/3}} = \frac{0.016^2\times1.592^2\times10}{0.050^{4/3}} = 0.3520\ \text{m} \\ h_e &= 0.5\frac{v^2}{2g} = 0.0646\ \text{m} \end{aligned}
$$

Total head required (entry loss + friction + velocity head at free discharge):
$$
H = h_e + h_f + \frac{v^2}{2g} = 0.0646 + 0.3520 + 0.1291 = 0.546\ \text{m}
$$

Minimum water level in the distributary $= 100.0 + 0.546 = 100.546$ m.

**Answer:** minimum water level in the distributary $\approx 100.55$ m (head required 0.55 m).

## Check yourself

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

1. Which losses make up the head required for a free-discharging pipe outlet?
   A. Friction loss only
   B. Entry loss, friction loss and the velocity head at exit
   C. Bend loss and silt loss
   D. Entry loss and gate loss only
   Answer: B. For free discharge H = h_e + h_f + v²/2g, since the exit velocity head is lost as a free jet. (Key point: Head components)

2. What is the velocity in the 20 cm pipe at 50 l/s?
   A. 0.25 m/s
   B. 1.59 m/s
   C. 5.0 m/s
   D. 0.05 m/s
   Answer: B. A = π/4 x 0.2² = 0.03142 m²; v = 0.05/0.03142 = 1.59 m/s. (Key point: Velocity)

3. Which formula gives the friction loss with Manning's n?
   A. h_f = n v L
   B. h_f = v²/2g
   C. h_f = n² v² L / R^(4/3) with R = d/4
   D. h_f = L/d
   Answer: C. Using Manning, h_f = n²v²L/R^(4/3) with R = 0.05 m gives 0.352 m. (Key point: Friction loss)

4. What minimum distributary water level results?
   A. 100.00 m
   B. 99.45 m
   C. 101.50 m
   D. About 100.55 m, i.e. 100.0 + 0.546 m head
   Answer: D. H = 0.0646 + 0.352 + 0.1291 = 0.546 m, added to the exit level 100.0 m. (Key point: Minimum water level)

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