---
title: "Parallel pipeline to increase discharge"
subject: "Hydro Power"
chapter: 1
key: "semester_6/hydro_power:parallel-pipeline-diameter"
source: https://www.bctengineer.com/q/semester_6/hydro_power/parallel-pipeline-diameter
site: BCT Engineering
---

# Parallel pipeline to increase discharge

Hydro Power · Chapter 1: Introduction to Basic Fluid Mechanics and Hydraulics

- Asked in: 2071 Bhadra, 8 marks
- Video: [Problem 3 on flow through parallel pipes/ fluid mechanics](https://www.youtube.com/watch?v=FmGhE2_dG_0) — 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/parallel-pipeline-diameter
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-1

## Question

A pipeline supplies 0.09 m³/s water from a reservoir to a building. The pipe is 2 km long and 0.15 m diameter with friction factor 0.025. It is desired to increase the discharge by 25% by installing another pipeline having same friction factor in parallel with this over half the length. Recommend a suitable diameter of the pipe to be installed.

## Short answer

Existing pipe needs 440.68 m head. For 0.1125 m³/s flow, old pipe carries 0.0595 m³/s in parallel half, new pipe must take 0.053 m³/s under same head loss, requiring diameter 0.143 m; recommend standard 150 mm.

## Answer

The head available (reservoir to building) stays the same. Adding a parallel pipe over the first half reduces resistance there, so more flow passes. Darcy–Weisbach with friction factor $f$ is used: $h_f = \dfrac{fLV^2}{2gD} = \dfrac{8fLQ^2}{\pi^2 g D^5}$.

#### Step 1: Available head (existing pipe)
$$
\begin{aligned}
h &= \frac{8 \times 0.025 \times 2000 \times 0.09^2}{\pi^2 \times 9.81 \times 0.15^5} = 440.68\ \text{m}
\end{aligned}
$$
(This large value is simply the head the original pipe needs; it stays fixed.)

#### Step 2: New layout
```text
           Pipe A (0.15 m, 1000 m), Q1
 Res ==+=================================+===== 0.15 m, 1000 m ====> Bldg
       +---------------------------------+        Q = 0.1125
           Pipe B (D2, 1000 m), Q2
```
New discharge $Q = 1.25 \times 0.09 = 0.1125$ m³/s.

Loss in the second (single) half:
$$
h_2 = \frac{8 \times 0.025 \times 1000 \times 0.1125^2}{\pi^2 \times 9.81 \times 0.15^5} = 344.28\ \text{m}
$$
Loss allowed in the parallel half:
$$
h_p = 440.68 - 344.28 = 96.40\ \text{m}
$$

#### Step 3: Flow in old pipe of the parallel portion
$$
\begin{aligned}
96.40 &= \frac{8 \times 0.025 \times 1000\,Q_1^2}{\pi^2 \times 9.81 \times 0.15^5} \\
Q_1 &= 0.0595\ \text{m}^3/\text{s}
\end{aligned}
$$
So the new pipe must carry $Q_2 = 0.1125 - 0.0595 = 0.0530$ m³/s.

#### Step 4: Diameter of new pipe
Same $f$, same length and same head loss $h_p$, so $Q \propto D^{2.5}$:
$$
\begin{aligned}
D_2 &= 0.15\left(\frac{Q_2}{Q_1}\right)^{0.4} = 0.15\left(\frac{0.0530}{0.0595}\right)^{0.4} = 0.143\ \text{m}
\end{aligned}
$$

**Answer:** A parallel pipe of about 0.143 m diameter is needed. Recommend the next standard size, **150 mm**, which gives slightly more than the required 25% increase.

## Check yourself

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

1. What stays constant when the discharge is increased by laying a parallel pipe?
   A. The friction factor of the total line only
   B. The head available between reservoir and building
   C. The velocity in the old pipe
   D. The discharge in each pipe
   Answer: B. The total head available is fixed; the parallel pipe reduces resistance so more flow passes for the same head. (Key point: Fixed head)

2. How is the allowed head loss in the parallel portion found?
   A. Total head minus the loss in the single 1000 m half at the new Q
   B. Half of the total head
   C. Total head divided by 1.25
   D. Equal to the loss in the single half at the old Q
   Answer: A. hp = 440.68 - 344.28 = 96.40 m, using the new Q = 0.1125 m^3/s in the single half. (Key point: Head split)

3. Once the old 0.15 m pipe carries Q1 = 0.0595 m^3/s under the parallel head loss, how is the new pipe flow found?
   A. Q2 = Q1
   B. Q2 = 0.1125 - Q1 = 0.0530 m^3/s
   C. Q2 = 0.09 - Q1
   D. Q2 = 1.25 x Q1
   Answer: B. Parallel flows add, so the new pipe carries the remainder of the total 0.1125 m^3/s. (Key point: Flow split)

4. For same f, length and head loss, how does discharge vary with diameter?
   A. Q is proportional to D
   B. Q is proportional to D^2
   C. Q is proportional to D^5
   D. Q is proportional to D^2.5
   Answer: D. From hf = 8fLQ^2/(pi^2 g D^5), Q is proportional to D^2.5, giving D2 = 0.15(Q2/Q1)^0.4 = 0.143 m. (Key point: Diameter relation)

## Related in Hydro Power

- [Flow between reservoirs (15 cm and 25 cm pipe)](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Areservoirs-pipeline-15-25cm) — 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
- [Maximising discharge with same lining](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Arectangular-channel-max-discharge) — Hydro Power, chapter 1
- [Laminar/turbulent flow and friction factor](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Alaminar-turbulent-friction-factor) — Hydro Power, chapter 1
- [Discharge over overflow spillway](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Aoverflow-spillway-discharge) — Hydro Power, chapter 6
- [Water hammer in 200 mm pipe](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Awater-hammer-200mm-pipe) — Hydro Power, chapter 1

## Related in other subjects

- [Percentage increase with parallel pipe at midpoint](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aparallel-pipe-added-midpoint-increase) — Hydraulics, chapter 2
- [Second parallel pipe to increase flow](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Atwo-reservoirs-second-pipe-increase-flow) — Hydraulics, chapter 2
- [Parallel pipe added in second half](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Aparallel-pipe-added-second-half) — Hydraulics, chapter 2
- [Branch pipe diameters from a single feeder (f 0.025)](https://www.bctengineer.com/question.md?key=semester_4%2Fhydraulics%3Areservoir-feeds-two-reservoirs-branch-diameters) — Hydraulics, chapter 3, asked 2×