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Double U-tube manometer, oil and brine pipes

Short answer

Stepwise manometer (add γh downward, subtract upward) using 20 cm oil, 60 cm Hg, 50 cm air, 10 cm brine gives p₂ ≈ 120.5 kPa. From p₂ through 50 cm brine, 10+30 cm Hg, 30 cm water yields p₃ ≈ 69.6 kPa (gauge). Assumed brine SG = 1.1.

The two pipes are connected by a double U-tube manometer as shown in figure where the brine pipe is connected to a tank filled with different fluids. Oil and brine are flowing in parallel horizontal pipes. The pressure at the centre of oil pipe is 200 kPa. Calculate pressures at point 2 & 3. [Figure: oil pipe (S.G. = 0.85) on the left connected through a mercury (S.G. = 13.6) U-tube and air (S.G. = 0.001) segments to the brine pipe (S.G. [?]) at point 2; the brine pipe is connected by a second U-tube to a closed tank with air (S.G. = 0.001) at the top (point 3) and layers of water and mercury (S.G. = 13.6) below, each layer 30 cm; marked heights include 20 cm, 50 cm, 60 cm, 50 cm, 10 cm, 30 cm, 10 cm, 30 cm]

The figure does not give every number legibly, so the reading used is stated below. Pressures are in gauge kPa, g=9.81 m/s2g=9.81\ \text{m/s}^2, and moving down a fluid adds γh\gamma h, moving up subtracts γh\gamma h.

Specific weights: oil γo=0.85×9.81=8.34\gamma_o=0.85\times9.81=8.34; mercury γHg=13.6×9.81=133.4\gamma_{Hg}=13.6\times9.81=133.4; brine (assumed S.G. 1.1) γb=10.79\gamma_b=10.79; air γa=0.00981\gamma_a=0.00981; water γw=9.81\gamma_w=9.81 kN/m³.

Assumed reading (heights from the figure):

  • pipe 1 (oil) →\to down 20 cm in oil →\to up 60 cm in mercury →\to down 50 cm in air →\to up 10 cm in brine to pipe 2 (net rise = 0, since both pipe centres are at the same level)
  • pipe 2 →\to down 50 cm in brine →\to up 10 cm in mercury →\to up 30 cm in mercury and 30 cm in water inside the tank, to the air at point 3
 (1) oil pipe          (2) brine pipe
   |                      |
   oil 20cm  Hg 60cm  air 50cm  brine 10cm

Pressure at point 2

p2=p1+γo(0.20)−γHg(0.60)+γa(0.50)−γb(0.10)=200+1.668−80.05+0.005−1.079=120.54 kPa\begin{aligned} p_2&=p_1+\gamma_o(0.20)-\gamma_{Hg}(0.60)+\gamma_a(0.50)-\gamma_b(0.10)\\ &=200+1.668-80.05+0.005-1.079\\ &=120.54\ \text{kPa} \end{aligned}

Pressure at point 3

p3=p2+γb(0.50)−γHg(0.10)−γHg(0.30)−γw(0.30)=120.54+5.395−13.34−40.02−2.943=69.63 kPa\begin{aligned} p_3&=p_2+\gamma_b(0.50)-\gamma_{Hg}(0.10)-\gamma_{Hg}(0.30)-\gamma_w(0.30)\\ &=120.54+5.395-13.34-40.02-2.943\\ &=69.63\ \text{kPa} \end{aligned}

Answer: p2≈120.5p_2\approx120.5 kPa and p3≈69.6p_3\approx69.6 kPa (gauge). For any other reading of the figure, use the same step-by-step rule with the correct heights and brine S.G.

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