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Differential manometer pressure at point B

Short answer

Pressure at X–X from left: 60 + 9.81×0.2 = 61.962 kN/m². Right: p_B + 0.9×9.81×0.45 + 13.6×9.81×0.1. Equating: p_B = 61.962 − 17.315 = 44.65 kN/m².

A differential manometer is connected to points A and B, as shown in the figure below. Given that the air pressure at A is 60 kN/m², determine the pressure at point B. [Figure: pipe A contains air above water (S₁ = 1) and pipe B contains oil (S₂ = 0.9); they are connected by a U-tube with mercury (S = 13.6) in its bottom; the vertical distances marked on the figure are 200 mm, 550 mm and 100 mm, measured relative to the datum X–X at the mercury surface.]

Write the pressure balance at the datum X–X (mercury surface in the left limb), moving through each fluid column.

Assumed reading of the figure: the air–water surface on the A side is 200 mm above X–X. On the B side, mercury stands 100 mm above X–X, and B is 550 mm above X–X, so the oil column is 550−100=450550 - 100 = 450 mm.

Given: pA=60p_A = 60 kN/m² (air; its weight is neglected), water S1=1S_1 = 1, oil S2=0.9S_2 = 0.9, mercury S=13.6S = 13.6, ρwg=9.81\rho_w g = 9.81 kN/m³.

   A (air)               B (oil)
   |~~~~~| 200 mm water    |  450 mm oil
   |     |                 |------ Hg top (+100)
 X-|-----|------------ X   | 100 mm Hg
   |__ mercury ____________|

Pressure at X–X, left limb

pL=pA+ρwg(0.20)=60+9.81(0.20)=60+1.962=61.962 kN/m2p_L = p_A + \rho_w g(0.20) = 60 + 9.81(0.20) = 60 + 1.962 = 61.962\ \text{kN/m}^2

Pressure at X–X, right limb

pR=pB+0.9(9.81)(0.45)+13.6(9.81)(0.10)=pB+3.973+13.342p_R = p_B + 0.9(9.81)(0.45) + 13.6(9.81)(0.10) = p_B + 3.973 + 13.342

Equate

61.962=pB+17.315pB=44.647 kN/m2\begin{aligned} 61.962 &= p_B + 17.315 \\ p_B &= 44.647\ \text{kN/m}^2 \end{aligned}

Answer: pB≈44.65p_B \approx 44.65 kN/m² (gauge), i.e. about 15.35 kN/m² less than at A. (If the figure's heights are placed differently, use the same column-by-column balance with those heights.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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