---
title: "Branch pipe diameters from a single feeder (f 0.025)"
subject: "Hydraulics"
chapter: 3
key: "semester_4/hydraulics:reservoir-feeds-two-reservoirs-branch-diameters"
source: https://www.bctengineer.com/q/semester_4/hydraulics/reservoir-feeds-two-reservoirs-branch-diameters
site: BCT Engineering
---

# Branch pipe diameters from a single feeder (f 0.025)

Hydraulics · Chapter 3: Three reservoirs problem and Pipe networks

- Asked in: 2073 Bhadra, 10 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/reservoir-feeds-two-reservoirs-branch-diameters
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-3

## Question

A reservoir A feeds two lower reservoirs B and C through a single pipe 10 km long, 750 mm diameter having a downward slope of $2.2\times10^{-3}$. This pipe then divides into two branch pipes, one 5.5 km long laid with a downward slope of $2.75\times10^{-3}$ (going to B), the other 3 km long having a downward slope of $3.2\times10^{-3}$ (going to C). Calculate the necessary diameters of the branch pipes so that the steady flow rate in each shall be 0.24 m³/s when the level in each reservoir is 3 m above the end of the corresponding pipe. Neglect all losses except pipe friction and take f = 0.025 throughout.

## Short answer

Main pipe friction loss h_f = 8fLQ²/(π²gD⁵) = 20.06 m, so H_J = 4.94 m. Available heads: to B 17.07 m, to C 11.54 m. Branch D = (8fLQ²/(π²g h_f))^(1/5) yields D_B ≈ 521 mm, D_C ≈ 499 mm.

## Answer

**Data.** Main pipe AJ: $L = 10$ km, $D = 0.75$ m, slope $2.2\times10^{-3}$. Branch JB: $L = 5.5$ km, slope $2.75\times10^{-3}$. Branch JC: $L = 3$ km, slope $3.2\times10^{-3}$. $Q_B = Q_C = 0.24$ m³/s, so the main pipe carries $Q = 0.48$ m³/s. $f = 0.025$. Only friction loss: $h_f = \dfrac{8fLQ^2}{\pi^2gD^5}$. Each reservoir level is 3 m above the end of its pipe (A: above the start of the main pipe; B and C: above the ends of the branches).

#### Elevations (datum: level of junction J)
- The main pipe falls $10\,000(2.2\times10^{-3}) = 22$ m from A to J, so the level of A is $22 + 3 = 25$ m above J.
- The branch to B falls $5500(2.75\times10^{-3}) = 15.125$ m, so the level of B is $-15.125 + 3 = -12.125$ m (12.125 m below J).
- The branch to C falls $3000(3.2\times10^{-3}) = 9.6$ m, so the level of C is $-9.6 + 3 = -6.6$ m.

#### Head loss in the main pipe and the head at J
$$
h_{f,AJ} = \frac{8(0.025)(10\,000)(0.48)^2}{\pi^2(9.81)(0.75)^5} = 20.06\ \text{m}
$$
Piezometric head at J (above J's level): $H_J = 25 - 20.06 = 4.94$ m.

#### Head available for each branch
- To B: $h_B = H_J - (-12.125) = 17.07$ m
- To C: $h_C = H_J - (-6.6) = 11.54$ m

#### Branch diameters
$$
D = \left(\frac{8fLQ^2}{\pi^2g\,h_f}\right)^{1/5}
$$
$$
D_B = \left(\frac{8(0.025)(5500)(0.24)^2}{\pi^2(9.81)(17.07)}\right)^{1/5} = 0.5209\ \text{m} \approx 521\ \text{mm}
$$
$$
D_C = \left(\frac{8(0.025)(3000)(0.24)^2}{\pi^2(9.81)(11.54)}\right)^{1/5} = 0.4989\ \text{m} \approx 499\ \text{mm}
$$

**Answer:** Diameter of the branch to B $\approx 521$ mm; diameter of the branch to C $\approx 499$ mm (head at J is 4.94 m above the junction level).

## Check yourself

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

1. Each branch carries 0.24 m^3/s. What does the main pipe carry?
   A. 0.12 m^3/s
   B. 0.72 m^3/s
   C. 0.48 m^3/s
   D. 0.24 m^3/s
   Answer: C. By continuity the main pipe carries the sum of the two branches. (Key point: Discharge in main pipe)

2. The main pipe falls 22 m and the reservoir is 3 m above its start. How high is A above J?
   A. 3 m
   B. 25 m
   C. 22 m
   D. 19 m
   Answer: B. A is 22 + 3 = 25 m above the junction level. (Key point: Level of A above J)

3. With main pipe friction loss 20.06 m, what is H_J above J's level?
   A. 4.94 m
   B. 45.06 m
   C. 20.06 m
   D. 25 m
   Answer: A. H_J = 25 - 20.06 = 4.94 m. (Key point: Head at J)

4. How is each branch diameter found?
   A. D = sqrt(4Q/(pi V)) with any V
   B. D from continuity only
   C. D = (h_f/(8fLQ^2))^5
   D. D = (8fLQ^2/(pi^2 g h_f))^(1/5), with h_f = H_J minus the reservoir level
   Answer: D. Head available to B is 17.07 m and to C is 11.54 m; solving gives 521 mm and 499 mm. (Key point: Branch diameter formula)

## Close variants set in other exams

- [Branch pipe diameters from a single feeder (f 0.024)](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Areservoir-feeds-two-reservoirs-branch-f024) — Hydraulics, chapter 3, asked 2×

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