Skip to main content

Use of manometer and inverted manometer

Short answer

An inverted U-tube manometer measures small pressure differences using a light fluid at the top. Pressure difference is pA − pB = ρ1 g h1 − ρ2 g h2 − ρ_s g h. With air on top the light-liquid term is negligible, giving precise readings for low differentials.

What is the use of manometer? Show the working principle of inverted manometer with neat sketch.

Use of a manometer

A manometer measures pressure at a point (simple manometer) or the difference of pressure between two points (differential manometer) by balancing it against a column of liquid. Used for pipes, venturimeters, orifice meters and laboratory flow measurement; it is simple, cheap and accurate.

Inverted U-tube differential manometer

It is an inverted U-tube whose top part contains a light fluid (air or a light oil) lighter than the liquid in the pipes. It is used to measure small pressure differences, where a heavy liquid like mercury would give too small a reading.

        .---------------.
        | light liquid  |  (air or oil, rho_s)
  X ----+------.        |
        |      |  h     |
        |      '--------+---- right interface
   h_1  | liquid        |
        | rho_1         | h_2   liquid rho_2
        o A             |
                        o B

Let X–X be the left (higher) interface, A be h1h_1 below it, the right interface be hh lower than X–X, and B be h2h_2 below the right interface.

Working principle

Pressure at level X–X in the left limb:

pA−ρ1gh1p_A - \rho_1 g h_1

Pressure at level X–X in the right limb (go up h2h_2 of pipe liquid from B, then up hh of light liquid):

pB−ρ2gh2−ρsghp_B - \rho_2 g h_2 - \rho_s g h

Equating:

pA−pB=ρ1gh1−ρ2gh2−ρsghp_A - p_B = \rho_1 g h_1 - \rho_2 g h_2 - \rho_s g h

If air is used at the top, ρsgh≈0\rho_s g h \approx 0.

Example: water in both pipes, oil S=0.8S = 0.8 at top, h1=0.30h_1 = 0.30 m, h2=0.20h_2 = 0.20 m, h=0.12h = 0.12 m: pA−pB=9.81(0.30−0.20−0.8×0.12)=0.039p_A - p_B = 9.81(0.30 - 0.20 - 0.8\times0.12) = 0.039 kN/m², a very small difference that an inverted manometer can read well.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Hydro Power chapter 1 questions →