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Piezometer levels in layered fluids

Short answer

Using hydrostatic balance from the 1.5 kPa sensor, gasoline (γ = 6671 N/m³) rises to Z_B ≈ 2.72 m and glycerin (γ = 12 361 N/m³) to Z_C ≈ 1.93 m.

In Fig. below, sensor A reads 1.5 kPa (gage). All fluids are at 20°C. Determine the elevations Z in meters of the liquid levels in the open piezometer tubes B and C. [Figure: closed tank with air (2 m deep) on top, gasoline (1.5 m) and glycerin (1 m) below; sensor A on the air; piezometer tube B connected at the gasoline layer and tube C at the glycerin layer; z = 0 at the tank bottom]

Fluid properties at 20 °C (White's table): gasoline ρ=680 kg/m3\rho=680\ \text{kg/m}^3, γg=680×9.81=6671 N/m3\gamma_g=680\times9.81=6671\ \text{N/m}^3; glycerin ρ=1260 kg/m3\rho=1260\ \text{kg/m}^3, γgl=12 361 N/m3\gamma_{gl}=12\,361\ \text{N/m}^3. Air weight is neglected.

Reading of the figure: z=0z=0 at the tank bottom; glycerin 1 m deep (z=0z=0 to 1), gasoline 1.5 m deep (z=1z=1 to 2.5), air above; pA=1.5p_A=1.5 kPa gauge acts on the gasoline surface at z=2.5z=2.5 m. Piezometer B joins the gasoline layer, C the glycerin layer.

   A (1.5 kPa)  air
  ------------------ z = 2.5 m
   gasoline  <- tube B
  ------------------ z = 1.0 m
   glycerin  <- tube C
  ------------------ z = 0

Piezometer B (gasoline)

The pressure at the top of the gasoline is 1500 Pa. The gasoline rises in tube B to a height where the pressure is zero:

ZB=2.5+15006671=2.5+0.225=2.725 mZ_B=2.5+\frac{1500}{6671}=2.5+0.225=2.725\ \text{m}

Piezometer C (glycerin)

Pressure at the gasoline-glycerin interface (z=1z=1 m):

p=1500+6671(1.5)=11 507 Pap=1500+6671(1.5)=11\,507\ \text{Pa}

The glycerin in tube C rises above z=1z=1 m until the pressure is zero:

ZC=1+11 50712 361=1+0.931=1.931 mZ_C=1+\frac{11\,507}{12\,361}=1+0.931=1.931\ \text{m}

Answer: ZB≈2.72Z_B\approx2.72 m and ZC≈1.93Z_C\approx1.93 m.

✓ Answer checked by Claude Opus 5.5 (high effort)

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