---
title: "Manometer reading in pipe with minor losses"
subject: "Hydraulics"
chapter: 2
key: "semester_4/hydraulics:manometer-reading-pipe-minor-losses"
source: https://www.bctengineer.com/q/semester_4/hydraulics/manometer-reading-pipe-minor-losses
site: BCT Engineering
---

# Manometer reading in pipe with minor losses

Hydraulics · Chapter 2: Simple pipe flow problems and solution

- Asked in: 2070 Magh, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/hydraulics/manometer-reading-pipe-minor-losses
- Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-2

## Question

Calculate the magnitude and direction of the manometer reading when water is flowing with velocity of 4.5 m/s for figure below. Consider minor losses also. [Figure: flow enters at A; a 75 mm diameter pipe (f = 0.02) 3 m long runs from A to B; a mercury (Hg) manometer connects the pipe to a point C; vertical dimensions of 2.4 m (from the tank water surface) and 0.9 m are marked between B and C]

## Short answer

Energy equation yields p_A/γ=3.226 m, p_C/γ=3.3 m. The difference 0.074 m? No, friction head 0.8257 m water creates pressure difference. Manometer reading h = 0.8257 / 12.6 ≈ 0.0655 m (66 mm Hg), mercury higher on C side.

## Answer

**Reading of the figure.** Water flows along the horizontal 75 mm pipe from A to B (3 m, $f = 0.02$) and discharges at B into a large tank whose water surface is 2.4 m above the pipe axis. The mercury manometer joins the pipe at A to the tank wall at C, which is 0.9 m below B. Velocity in the pipe $V = 4.5$ m/s. Minor loss: exit (sudden enlargement into the tank), $K = 1.0$.

#### Pressure at A
$$
\frac{V^2}{2g} = \frac{4.5^2}{2(9.81)} = 1.0321\ \text{m},\qquad h_f = f\frac{L}{D}\frac{V^2}{2g} = 0.02\left(\frac{3}{0.075}\right)(1.0321) = 0.8257\ \text{m},\qquad h_{exit} = \frac{V^2}{2g} = 1.0321\ \text{m}
$$
Energy equation from A (pipe axis, datum) to the tank water surface (2.4 m above the axis, velocity zero, atmospheric pressure):
$$
\frac{p_A}{\gamma} + \frac{V^2}{2g} = 2.4 + h_f + h_{exit}
\;\Rightarrow\; \frac{p_A}{\gamma} = 2.4 + 0.8257 + 1.0321 - 1.0321 = 3.2257\ \text{m}
$$
The exit loss cancels the velocity head, so only the friction loss changes the pressure.

#### Pressure at C
C is in the quiet water of the tank, so the pressure is hydrostatic: $\dfrac{p_C}{\gamma} = 2.4 + 0.9 = 3.3$ m.

#### Manometer reading
Piezometric heads (datum at the pipe axis):
$$
\left(\frac{p_A}{\gamma} + z_A\right) - \left(\frac{p_C}{\gamma} + z_C\right) = 3.2257 - (3.3 - 0.9) = 0.8257\ \text{m of water}
$$
For a mercury-water differential manometer: $\left(S_{Hg} - 1\right)h = 12.6\,h = 0.8257$
$$
h = \frac{0.8257}{12.6} = 0.0655\ \text{m} = 65.5\ \text{mm of mercury}
$$

**Direction.** The piezometric pressure at A is greater than at C, so mercury is pushed down in the limb connected to A and rises in the limb connected to C. The mercury level in the C limb is higher than in the A limb by about 65.5 mm (the reading equals the friction head $h_f$, converted to mercury).

**Answer:** Manometer reading $\approx 66$ mm of Hg, with mercury higher on the C side (the A side is at the higher pressure).

## Check yourself

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

1. Why does only the friction loss change the pressure at A?
   A. The pipe is horizontal and smooth
   B. The exit loss (1.0 V^2/2g) cancels the velocity head, leaving p_A/gamma = 2.4 + h_f
   C. Velocity head is zero in the pipe
   D. Minor losses are ignored
   Answer: B. The V^2/2g on the left and the exit loss term on the right are equal, so p_A/gamma = 2.4 + 0.8257. (Key point: Pressure at A)

2. How is the pressure at C found?
   A. Energy equation along the pipe
   B. Hydrostatic: depth 2.4 + 0.9 = 3.3 m in the quiet tank
   C. From velocity 4.5 m/s
   D. Equal to atmospheric
   Answer: B. C is in still water of the tank, so pressure is hydrostatic. (Key point: Pressure at C)

3. How is the piezometric head difference of 0.8257 m of water converted to mercury reading h?
   A. h = 0.8257/13.6
   B. h = 0.8257/(13.6 - 1) = 65.5 mm
   C. h = 0.8257 x 13.6
   D. h = 0.8257/12.6 x 13.6
   Answer: B. A differential water-mercury manometer gives (S_Hg - 1) h = piezometric difference. (Key point: Converting to manometer reading)

4. Which side shows higher mercury?
   A. A side, since A is higher in pressure
   B. C side, because A has the higher piezometric pressure and pushes mercury down
   C. Both equal
   D. Depends on flow rate only
   Answer: B. Higher piezometric pressure at A pushes mercury down in its limb, raising it in the C limb. (Key point: Direction)

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