---
title: "Head loss ratio, parallel versus series"
subject: "Hydraulics"
chapter: 2
key: "semester_4/hydraulics:parallel-vs-series-head-loss-ratio"
source: https://www.bctengineer.com/q/semester_4/hydraulics/parallel-vs-series-head-loss-ratio
site: BCT Engineering
---

# Head loss ratio, parallel versus series

Hydraulics · Chapter 2: Simple pipe flow problems and solution

- Asked in: 2069 Bhadra, 8 marks
- Video: [Mastering Parallel Pipe Flow Systems | Fluid Mechanics Explained](https://www.youtube.com/watch?v=B4pysiBSBWM) — Engineering Key Tutorials
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/parallel-vs-series-head-loss-ratio
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-2

## Question

Two pipes have a length L each. One of them has diameter $D_1$ and the other has diameter $D_2$. If the pipes are arranged in parallel, the loss of head when a total quantity of water Q flows through them is $H_1$. If the pipes are arranged in series and the same quantity Q flows through them, the loss of head is $H_2$. If $D_2 = D_1/2$, find the ratio of $H_1$ to $H_2$, neglecting minor losses and assuming same f.

## Short answer

For D2 = D1/2, the head loss in parallel is only about 2.2% of that in series: ratio H1/H2 = 1/45.7 (≈0.0219).

## Answer

**Method.** Use $h_f = \dfrac{8fLQ^2}{\pi^2gD^5}$, i.e. $h_f = rQ^2$ with $r \propto D^{-5}$ for the same $L$ and $f$. Let $D_2 = D_1/2$.

#### Parallel arrangement (head loss $H_1$)
Both pipes have the same head loss $H_1$, so $Q_i = \sqrt{H_1/r_i}$, which gives $Q_i \propto D_i^{5/2}$:
$$
Q = Q_1' + Q_2' = \sqrt{\frac{\pi^2gH_1}{8fL}}\left(D_1^{5/2} + D_2^{5/2}\right)
\;\Rightarrow\; H_1 = \frac{8fLQ^2}{\pi^2g\left(D_1^{5/2} + D_2^{5/2}\right)^2}
$$
With $D_2 = D_1/2$: $D_2^{5/2} = (0.5)^{2.5}D_1^{5/2} = 0.17678\,D_1^{5/2}$, so $\left(D_1^{5/2} + D_2^{5/2}\right)^2 = (1 + 0.17678)^2D_1^5 = 1.3848D_1^5$:
$$
H_1 = \frac{8fLQ^2}{\pi^2g}\cdot\frac{1}{1.3848D_1^5}
$$

#### Series arrangement (head loss $H_2$)
The same $Q$ passes through both pipes and the losses add:
$$
H_2 = \frac{8fLQ^2}{\pi^2g}\left(\frac{1}{D_1^5} + \frac{1}{D_2^5}\right) = \frac{8fLQ^2}{\pi^2g}\cdot\frac{1 + 32}{D_1^5} = \frac{8fLQ^2}{\pi^2g}\cdot\frac{33}{D_1^5}
$$

#### Ratio
$$
\frac{H_1}{H_2} = \frac{1/1.3848}{33} = 0.02188 \approx \frac{1}{45.7}
$$

**Answer:** $H_1 : H_2 = 1 : 45.7$, i.e. $H_1/H_2 \approx 0.0219$ (the parallel arrangement loses only about 2% of the head lost in series).

## Check yourself

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

1. For pipes in parallel with equal L and f, how does the flow in each pipe vary with D?
   A. Q proportional to D^2
   B. Q proportional to D^(5/2)
   C. Q proportional to D^5
   D. Q proportional to D^(1/2)
   Answer: B. h_f = rQ^2 with r proportional to D^-5, so Q = sqrt(h_f/r) is proportional to D^(5/2). (Key point: Discharge-diameter relation in parallel)

2. Which expression gives H1 for parallel pipes?
   A. H1 proportional to Q^2/(D1^5 + D2^5)
   B. H1 proportional to Q^2/(D1^(5/2) + D2^(5/2))^2
   C. H1 proportional to Q^2 (1/D1^5 + 1/D2^5)
   D. H1 proportional to Q/(D1 + D2)
   Answer: B. Both pipes have the same loss and Q is the sum of the two branch flows, each proportional to D^(5/2). (Key point: Parallel head loss)

3. For D2 = D1/2 in series, what is H2 in units of 8fLQ^2/(pi^2 g D1^5)?
   A. 1
   B. 2
   C. 33
   D. 32
   Answer: C. Losses add: 1/D1^5 + 1/D2^5 = (1 + 32)/D1^5 = 33/D1^5. (Key point: Series head loss)

4. What is H1/H2?
   A. About 0.0219, i.e. 1:45.7
   B. About 0.5
   C. About 33
   D. About 0.30
   Answer: A. H1/H2 = (1/1.3848)/33 = 0.0219. (Key point: Ratio H1:H2)

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## Related in other subjects

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