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Pressure difference using compound manometer

Short answer

Go from A to B through each column: add ρgh going down, subtract going up. The pressure difference is pA − pB ≈ 41.4 kN/m², with A at higher pressure.

Calculate the pressure difference between the points A and B. Specific gravity of oil is 0.8. [Figure: pipes A and B both contain water and are connected by two mercury U-tubes joined at the top by an inverted U-tube containing oil; marked heights are 90 cm (centre of A above the bottom datum), 60 cm and 40 cm (mercury columns on the left), 50 cm (mercury column on the right) and 80 cm (centre of B above the right mercury level).]

Go from A to B through each column: add ρgh\rho g h going down, subtract going up.

Assumed reading of the figure (heights above the bottom datum): centre of A = 0.90 m. In the left mercury U-tube, the mercury surface is at 0.40 m in A's limb and 0.60 m in the other limb. Oil fills the inverted U-tube, down to the right U-tube's mercury surface, taken at 0.40 m. The right U-tube's mercury in B's limb is at 0.50 m. B is 0.80 m above that, i.e. at 1.30 m.

Given: Soil=0.8S_{oil} = 0.8, SHg=13.6S_{Hg} = 13.6, water S=1S = 1, ρwg=9.81\rho_w g = 9.81 kN/m³.

StepColumnHeight (m)Pressure change (kN/m²)
A down to Hg (left)water0.50+9.81(0.50)=+4.905+9.81(0.50) = +4.905
Up in Hg to 0.60mercury0.20−13.6(9.81)(0.20)=−26.683-13.6(9.81)(0.20) = -26.683
Down through oil to 0.40oil0.20+0.8(9.81)(0.20)=+1.570+0.8(9.81)(0.20) = +1.570
Up in Hg to 0.50mercury0.10−13.6(9.81)(0.10)=−13.342-13.6(9.81)(0.10) = -13.342
Up in water to Bwater0.80−9.81(0.80)=−7.848-9.81(0.80) = -7.848

Manometer equation

pA+4.905−26.683+1.570−13.342−7.848=pBp_A + 4.905 - 26.683 + 1.570 - 13.342 - 7.848 = p_B pA−pB=41.398 kN/m2p_A - p_B = 41.398\ \text{kN/m}^2

Answer: pA−pB≈41.4p_A - p_B \approx 41.4 kN/m² (A at higher pressure), for the heights read as above. The method is the same for any other placement of the marked heights.

✓ Answer checked by Claude Opus 5.5 (high effort)

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