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Absolute and gauge pressure, mercury manometer

Short answer

Absolute pressure is relative to vacuum, gauge to atmosphere. Gauge pressure at G is 26.85 kPa, and the mercury manometer height h is 201 mm.

Define absolute and gauge pressure. Determine (i) the gauge pressure reading on the pressure gauge and (ii) the height h, of the mercury manometer. Take liquid density = 800 kg/m³, vapour pressure = 120 kPa (abs) and atmospheric pressure = 101 kPa (abs). [Figure: closed vessel with vapour over liquid; pressure gauge G at 1 m below the liquid surface level; the vessel bottom is connected to a mercury U-tube manometer open to atmosphere with reading h]

Definitions

  • Absolute pressure is measured from absolute zero (complete vacuum): pabs=pgauge+patmp_{abs}=p_{gauge}+p_{atm}.
  • Gauge pressure is measured relative to the local atmospheric pressure: pgauge=pabs−patmp_{gauge}=p_{abs}-p_{atm}. It is negative for vacuum.

Numerical

Reading of the figure: gauge GG and the liquid-mercury interface of the U-tube are at the same level, 1 m below the liquid surface; the mercury stands hh above that interface in the open limb. ρl=800 kg/m3\rho_l=800\ \text{kg/m}^3, pv=120p_v=120 kPa (abs), patm=101p_{atm}=101 kPa.

(i) Gauge reading. Absolute pressure at G:

pG=pv+ρlg(1)=120+800×9.81×11000=127.85 kPa absp_G=p_v+\rho_l g(1)=120+\frac{800\times9.81\times1}{1000}=127.85\ \text{kPa abs} pG,gauge=127.85−101=26.85 kPap_{G,gauge}=127.85-101=26.85\ \text{kPa}

(ii) Mercury height hh. The same absolute pressure is balanced by atmosphere plus the mercury column:

pG=patm+γHgh ⇒ h=26.8513.6×9.81=0.2012 mp_G=p_{atm}+\gamma_{Hg}h\ \Rightarrow\ h=\frac{26.85}{13.6\times9.81}=0.2012\ \text{m}

Answer: (i) gauge pressure =26.85=26.85 kPa; (ii) h≈201h\approx201 mm of mercury.

✓ Answer checked by Claude Opus 5.5 (high effort)

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