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U-tube manometer with oil, mercury and water

Short answer

Manometer equation: p_A + 0.9×9810×0.12 − 13.6×9810×0.04 − 9810×0.03 = p_B. Compute: p_A − p_B = 9810(0.544+0.03−0.108) = 4571 Pa ≈ 4.57 kN/m², A higher.

A U-tube manometer contains oil, mercury, and water as shown in figure below. For the column heights indicated, what is the pressure differential between pipe A and B? [Figure: pipe A (oil, SG = 0.9) on the left and pipe B (water) on the right are connected by a U-tube with mercury (SG = 13.6) at the bottom; the column heights marked are 4 cm, 3 cm and 12 cm.]

Use the manometer rule: start at A, add ρgh\rho g h when going down a column, subtract when going up, and end at B.

Assumed reading of the figure: A (oil) is 12 cm above the oil–mercury surface in the left limb. Mercury in the right limb stands 4 cm higher than in the left limb. B (water) is 3 cm above the right mercury surface.

Given: oil S=0.9S = 0.9, mercury S=13.6S = 13.6, water S=1S = 1; ρwg=9810\rho_w g = 9810 N/m³.

  A (oil)
  |                         B (water)
  | 12 cm oil               | 3 cm water
  |              Hg  -------+ (right Hg level)
  |              | 4 cm
  +-- Hg level --+
  |_____ mercury ______|

Manometer equation

pA+ρoilg(0.12)−ρHgg(0.04)−ρwg(0.03)=pBp_A + \rho_{oil} g(0.12) - \rho_{Hg} g(0.04) - \rho_w g(0.03) = p_B pA−pB=9810 [13.6(0.04)+1(0.03)−0.9(0.12)]=9810 [0.544+0.030−0.108]=9810×0.466=4571 N/m2\begin{aligned} p_A - p_B &= 9810\,[13.6(0.04) + 1(0.03) - 0.9(0.12)] \\ &= 9810\,[0.544 + 0.030 - 0.108] \\ &= 9810 \times 0.466 = 4571\ \text{N/m}^2 \end{aligned}

Answer: pA−pB≈4.57p_A - p_B \approx 4.57 kN/m² (pressure at A is higher). If the column heights are arranged differently in the figure, the same step-by-step balance applies with those heights.

✓ Answer checked by Claude Opus 5.5 (high effort)

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