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Gauge pressure in cleaning fluid tank

Short answer

pA = pB – γw×0.055 m = 12.86 kPa gauge. Then h = (pC – pA)/(0.8×γw) = 0.0713 m = 71.3 mm.

A cylindrical tank contains water at a height of 55 mm as shown in figure below. Inside a smaller open cylinder tank containing cleaning fluid (S = 0.8) at a height h. If PBP_B = 13.4 kPa and PCP_C = 13.42 kPa gauge, what are gauge pressure PAP_A and height h of cleaning fluid? Assume that the cleaning fluid kerosene is prevented from moving to the top of the tank. [Figure: closed tank with air at the top (gauge PAP_A), water 55 mm deep, a smaller open cylinder of kerosene (height h) inside; gauges PBP_B and PCP_C at the bottom]

Reading of the figure: gauge B is at the bottom of the main tank (under 55 mm of water). Gauge C is at the bottom of the small cylinder (under a depth hh of cleaning fluid, S=0.8S=0.8). The air above both liquids is at gauge pressure pAp_A.

Gauge pressure pAp_A. From B:

pB=pA+γw(0.055) ⇒ pA=13.4−9.81(0.055)=12.86 kPap_B=p_A+\gamma_w(0.055)\ \Rightarrow\ p_A=13.4-9.81(0.055)=12.86\ \text{kPa}

Height hh. From C:

pC=pA+γfh,γf=0.8×9.81=7.848 kN/m3p_C=p_A+\gamma_f h,\qquad \gamma_f=0.8\times9.81=7.848\ \text{kN/m}^3 h=pC−pAγf=13.42−12.8607.848=0.0713 mh=\frac{p_C-p_A}{\gamma_f}=\frac{13.42-12.860}{7.848}=0.0713\ \text{m}

Answer: pA≈12.86p_A\approx12.86 kPa (gauge) and h≈71.3h\approx71.3 mm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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