---
title: "Design pipes PQ, PR and RS, C = 110"
subject: "Water Supply Engineering"
chapter: 7
key: "semester_5/water_supply_engineering:pd-s71"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/pd-s71
site: BCT Engineering
---

# Design pipes PQ, PR and RS, C = 110

Water Supply Engineering · Chapter 7: Reservoirs and Distribution System

- Asked in: 2071 Shrawan, 10 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/water_supply_engineering/pd-s71
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-7

## Question

Design pipes PQ, PR and RS. Minimum pressures have to be maintained at 1 kg/cm$^2$ in all taps. Take Hazen William constant C = 110.

[Figure: Source at P, RL of source = 950 m. Pipe PR, length 150 m, carries 0.7 lps to R (RL 860 m, Tap 1); from R, pipe RS, length 160 m, carries 0.3 lps to S (RL 845 m, Tap 2); pipe PQ, length 180 m, carries 0.25 lps to Q (RL 930 m, Tap 3).]

## Short answer

Use Hazen-Williams equation with C=110. Pipe PQ (0.25 lps, 180 m) adopts 25 mm, pipe PR (0.7 lps, 150 m) adopts 20 mm, pipe RS (0.3 lps, 160 m) adopts 20 mm. Residual heads exceed 10 m at all taps.

## Answer

Residual pressure = 1 kg/cm² = 10 m at all taps; C = 110. HGL at source P = 950 m. The flows given for each pipe are taken as the discharge carried by that pipe (PR = 0.7 lps, RS = 0.3 lps, PQ = 0.25 lps).

#### Method
Use the Hazen-Williams equation in SI units:
$$
h_f = \frac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}
$$
with $Q$ in m³/s, $L$ and $D$ in m. For each pipe the available head is (HGL at upstream end − ground RL of the tap − required residual head). The gradient $S = H/L$ gives the calculated diameter, the next larger commercial size is adopted, and the real head loss and residual head are then checked. Commercial sizes (nominal bore): 15, 20, 25, 32, 40, 50, 65, 80, 100, 125, 150, 200 mm and so on. Constant C = 110.

#### Pipe PR
- Discharge $Q = 0.700$ lps, length $L = 150$ m
- HGL at P = 950.00 m; ground RL at R = 860 m; residual head required = 10 m
- Available head $H = 950.00 - 860 - 10 = 80.00$ m
- Gradient $S = H/L = 80.00/150 = 0.53333$
- From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 19.5$ mm, adopt commercial size **20 mm**
- Actual $h_f = 71.565$ m, HGL at R = 878.43 m, residual head = 18.43 m (≥ 10 m, OK)
- Velocity $v = Q/A = 2.23$ m/s

#### Pipe RS
- Discharge $Q = 0.300$ lps, length $L = 160$ m
- HGL at R = 878.43 m; ground RL at S = 845 m; residual head required = 10 m
- Available head $H = 878.43 - 845 - 10 = 23.43$ m
- Gradient $S = H/L = 23.43/160 = 0.14647$
- From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 18.5$ mm, adopt commercial size **20 mm**
- Actual $h_f = 15.894$ m, HGL at S = 862.54 m, residual head = 17.54 m (≥ 10 m, OK)
- Velocity $v = Q/A = 0.95$ m/s

#### Pipe PQ
- Discharge $Q = 0.250$ lps, length $L = 180$ m
- HGL at P = 950.00 m; ground RL at Q = 930 m; residual head required = 10 m
- Available head $H = 950.00 - 930 - 10 = 10.00$ m
- Gradient $S = H/L = 10.00/180 = 0.05556$
- From $h_f = \dfrac{10.67\,L\,Q^{1.852}}{C^{1.852}D^{4.87}}$: $D_{calc} = 21.0$ mm, adopt commercial size **25 mm**
- Actual $h_f = 4.303$ m, HGL at Q = 945.70 m, residual head = 15.70 m (≥ 10 m, OK)
- Velocity $v = Q/A = 0.51$ m/s

#### Summary
| Pipe | Q (lps) | L (m) | Calc. D (mm) | Adopted D (mm) | h_f (m) | Residual (m) | v (m/s) |
|---|---|---|---|---|---|---|---|
| PR | 0.7 | 150 | 19.5 | 20 | 71.57 | 18.43 | 2.23 |
| RS | 0.3 | 160 | 18.5 | 20 | 15.89 | 17.54 | 0.95 |
| PQ | 0.25 | 180 | 21.0 | 25 | 4.30 | 15.70 | 0.51 |

Static heads: Q = 950 − 930 = 20 m, R = 90 m, S = 105 m. Pipes PR and RS must therefore be of a class for at least 105 m; if the high pressure is a problem, a break pressure tank can be provided near R.

**Answer: PQ = 25 mm, PR = 20 mm, RS = 20 mm.** (Velocity in PR is 2.23 m/s, still within the 3 m/s limit.)

## Check yourself

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

1. Head available for PR with HGL 950, R at 860 and residual 10 m?
   A. 80 m
   B. 90 m
   C. 70 m
   D. 100 m
   Answer: A. 950 - 860 - 10 = 80 m. (Key point: Available head PR)

2. RS begins from which HGL?
   A. 950 m
   B. 878.43 m, the HGL at R after loss in PR
   C. 860 m
   D. 862.54 m
   Answer: B. Head left after PR's loss (71.57 m) gives HGL at R = 878.43 m. (Key point: Next pipe start)

3. PR has v = 2.23 m/s. Is it acceptable?
   A. No, must be below 1 m/s
   B. Yes, within the 0.6-3 m/s limit
   C. No, must exceed 3 m/s
   D. Velocity is not checked
   Answer: B. 2.23 m/s lies between 0.6 and 3 m/s. (Key point: Velocity check)

## Related in Water Supply Engineering

- [Design pipes AB and BC, source RL 1050 m](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apd-c76) — Water Supply Engineering, chapter 7
- [Design pipes AB and BC, source RL 180 m](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apd-as67) — Water Supply Engineering, chapter 7
- [Design pipes AB, BC and AC, C = 110](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apd-j66) — Water Supply Engineering, chapter 7
- [Design pipes AB, BC, AD (min pressure 1.1 kg/cm2)](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apipe-design-ab-bc-ad-110) — Water Supply Engineering, chapter 7, asked 2×
- [Design pipes AB and BC, A RL 210 m, 44 and 18 lps](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apd-c73) — Water Supply Engineering, chapter 7
- [Design main transmission pipes PQ and QR](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Apd-c70o) — Water Supply Engineering, chapter 7

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