---
title: "Inverted manometer differential level"
subject: "Fluid Mechanics"
chapter: 2
key: "semester_3/fluid_mechanics:inverted-manometer-differential-level"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/inverted-manometer-differential-level
site: BCT Engineering
---

# Inverted manometer differential level

Fluid Mechanics · Chapter 2: Pressure and Head

- Asked in: 2078 Bhadra, 8 marks
- Video: [Problem on Inverted U-Tube Manometer - Pressure and Pressure Measurement - Fluid Mechanics 1](https://www.youtube.com/watch?v=CTlMHkDKnNM) — Ekeeda
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/inverted-manometer-differential-level
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-2

## Question

An inverted manometer is connected to the pipe and tank as shown in figure. What will be the differential level on U-tube connecting pipe? [Figure: water tank connected to a pipe through an inverted manometer with fluid S = 0.83, dimensions 0.5 m, 0.2 m, 0.15 m, 0.25 m, R = 0.15 m, with a mercury (S = 13.6) U-tube]

## Short answer

Balance pressures stepwise: add γh going down, subtract going up, then equate to γHg h. For the assumed dimensions, differential mercury level is about 42 mm.

## Answer

The figure dimensions are not fully legible, so this reading is used and stated. Specific weight of water $\gamma_w=9.81$ kN/m³, $g=9.81$.

**Assumed reading:** the pipe centre lies 0.5 m below the free surface in the tank, so the pipe is under a static head of 0.5 m of water. Left leg of the inverted U (water) rises 0.25 m from the pipe to the lighter fluid ($S=0.83$); the manometer reading $R=0.15$ m (left interface is higher than the right one); the right leg water goes down 0.20 m to a mercury U-tube whose other limb is open. We find the differential mercury level $h$.

```text
        ______ oil S=0.83 ______
       |                        |
 water |                  R     | water 0.20
 0.25  |                        |
 pipe  P                        Hg  |--h--| open
```

Pressure at pipe (gauge):

$$
p_P=\gamma_w(0.5)=4.905\ \text{kPa}
$$

At the left oil-water interface (rise of 0.25 m in water):

$$
p_{L}=4.905-9.81(0.25)=2.453\ \text{kPa}
$$

Across the lighter fluid to the right interface (it is $R=0.15$ m lower):

$$
p_R=2.453+0.83(9.81)(0.15)=3.674\ \text{kPa}
$$

Down 0.20 m of water to the mercury surface:

$$
p_{Hg}=3.674+9.81(0.20)=5.636\ \text{kPa}
$$

This is balanced by a mercury column $h$ in the open limb:

$$
h=\frac{p_{Hg}}{\gamma_{Hg}}=\frac{5.636}{13.6\times9.81}=0.0422\ \text{m}
$$

**Answer: differential mercury level $h\approx42$ mm** (for this reading). The method is the same for other dimensions: add $\gamma h$ going down, subtract going up, and equate to $\gamma_{Hg}h$.

## Check yourself

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

1. Going up 0.25 m in water from the pipe to the inverted U, what happens to pressure?
   A. Increases by 2.45 kPa
   B. Stays constant
   C. Decreases by 2.45 kPa
   D. Becomes zero
   Answer: C. Moving up subtracts gamma h: 4.905 - 2.453 = 2.453 kPa. (Key point: Climbing the water leg)

2. Going down R = 0.15 m through the lighter fluid (S = 0.83) to the right interface, how does pressure change?
   A. Increases by 0.83 x 9.81 x 0.15
   B. Decreases by 9.81 x 0.15
   C. Increases by 9.81 x 0.15
   D. No change
   Answer: A. Going down in the light fluid adds its gamma h = 1.22 kPa. (Key point: Crossing the lighter fluid)

3. How is the mercury level difference h found from the pressure p_Hg?
   A. h = p_Hg x gamma_Hg
   B. h = p_Hg / gamma_Hg
   C. h = p_Hg / gamma_w
   D. h = gamma_Hg / p_Hg
   Answer: B. The open mercury limb balances p_Hg, so h = 5.636/(13.6 x 9.81) = 0.042 m. (Key point: Mercury differential)

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