---
title: "Equivalent length of pipeline losses"
subject: "Hydraulics"
chapter: 2
key: "semester_4/hydraulics:equivalent-length-of-minor-losses"
source: https://www.bctengineer.com/q/semester_4/hydraulics/equivalent-length-of-minor-losses
site: BCT Engineering
---

# Equivalent length of pipeline losses

Hydraulics · Chapter 2: Simple pipe flow problems and solution

- Asked in: 2076 Bhadra, 4 marks
- Video: [Module 1: 20. Pressure Losses and Equivalent Pipe Length](https://www.youtube.com/watch?v=DThcSkL40Jk) — Educate-Learn
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/equivalent-length-of-minor-losses
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-2

## Question

A pipe line system consists of the following sources of head losses. (i) Entrance loss in 200 mm diameter (ii) Friction loss in 500 m of 200 mm diameter pipe with f = 0.02 (iii) Sudden expansion from 200 mm to 250 mm diameter (iv) Exit loss from 250 mm diameter pipe. Obtain the equivalent length of 200 mm diameter pipe with f = 0.02.

## Short answer

Sum of loss coefficients expressed in 200 mm pipe velocity heads is 51.04. Equivalent length of 200 mm pipe with f=0.02 is L_e = (51.04 × 0.2)/0.02 = 510.4 m.

## Answer

**Idea.** Express every loss in terms of the velocity head in the 200 mm pipe, $\dfrac{V_1^2}{2g}$, and find the length of 200 mm pipe ($f = 0.02$) that has the same total loss: $f\dfrac{L_e}{D_1} = \sum K$.

Areas: $\dfrac{A_1}{A_2} = \left(\dfrac{200}{250}\right)^2 = 0.64$, so $V_2 = 0.64V_1$.

| Loss | Expression | In terms of $V_1^2/2g$ |
|---|---|---|
| (i) Entrance | $0.5\dfrac{V_1^2}{2g}$ | 0.5 |
| (ii) Friction, 500 m of 200 mm | $f\dfrac{L}{D}\dfrac{V_1^2}{2g} = 0.02\left(\dfrac{500}{0.2}\right)\dfrac{V_1^2}{2g}$ | 50 |
| (iii) Sudden expansion | $\dfrac{(V_1 - V_2)^2}{2g} = \left(1 - \dfrac{A_1}{A_2}\right)^2\dfrac{V_1^2}{2g}$ | $(1 - 0.64)^2 = 0.1296$ |
| (iv) Exit from 250 mm pipe | $\dfrac{V_2^2}{2g} = (0.64)^2\dfrac{V_1^2}{2g}$ | 0.4096 |
| **Total** | | **51.0392** |

Equivalent length of 200 mm pipe:
$$
f\frac{L_e}{D_1} = 51.0392 \;\Rightarrow\; L_e = \frac{51.0392\times 0.2}{0.02} = 510.39\ \text{m}
$$

**Answer:** Equivalent length $\approx 510.4$ m of 200 mm pipe (with $f = 0.02$). The minor losses (entrance, expansion, exit) alone are equivalent to about 10.4 m of pipe.

## Check yourself

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

1. What does the equivalent length of 200 mm pipe mean here?
   A. Length whose friction loss equals the total loss: f Le/D1 = sum of K
   B. Length equal to 500 m
   C. Length with the same velocity
   D. Length of the 250 mm pipe
   Answer: A. All losses are expressed in V1^2/2g and equated to f Le/D1. (Key point: Equivalent length idea)

2. What is the sudden expansion loss in terms of V1^2/2g?
   A. (1 - A1/A2)^2 = 0.1296
   B. (1 - A1/A2) = 0.36
   C. 0.5
   D. (A1/A2)^2 = 0.41
   Answer: A. Loss = (V1 - V2)^2/2g with V2 = 0.64 V1, giving (0.36)^2. (Key point: Sudden expansion loss)

3. How is the exit loss expressed in terms of V1^2/2g?
   A. 1.0
   B. (0.64)^2 = 0.4096 since V2 = 0.64 V1
   C. 0.64
   D. 0.5
   Answer: B. Exit loss is V2^2/2g and V2 = 0.64 V1. (Key point: Exit loss from 250 mm pipe)

4. Total K = 51.04. What is Le?
   A. 51.04 m
   B. 500 m
   C. 10.4 m
   D. 510.4 m
   Answer: D. Le = 51.0392 x 0.2/0.02 = 510.4 m. (Key point: Total and result)

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