Three-Fluid Differential Manometer: 04-BS-7, May 2015, Q1

Question 1: Three-Fluid Differential Manometer (5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Pressure in pipe B	200 kPa
SG benzene (pipe A fluid)	0.90
SG carbon tetrachloride (pipe B fluid)	1.59
SG mercury (U-tube fluid)	13.56
Benzene column, A up to the upper Hg surface	2.0 m + 0.40 m = 2.40 m
Mercury differential (upper surface above lower)	0.40 m
CCl₄ column, lower Hg surface down to B	3.0 m

Find. The gauge pressure in pipe A.

Fig. Q1 — benzene rises from A over the left bend to the upper mercury surface; mercury drops 40 cm to the lower (right) surface; carbon tetrachloride continues over the right bend down to B.

Approach. Traverse the tube from A to B one fluid at a time, adding ρgΔz going down and subtracting it going up, and set the result equal to the known p_B.

Write the manometer traverse from A to B. Starting at A, rise 2.40 m through benzene to the upper mercury surface (pressure falls), fall 0.40 m through mercury to the lower surface (pressure rises), then fall 3.0 m through CCl₄ to B (pressure rises): $$p_A - \rho_{benz}g(2.40) + \rho_{Hg}g(0.40) + \rho_{CCl_4}g(3.0) = p_B$$
Substitute the fluid densities (ρ = SG×1000 kg/m³, g = 9.81 m/s²): $$\rho_{benz}=900,\ \rho_{Hg}=13\,560,\ \rho_{CCl_4}=1590\ \text{kg/m}^3$$
Solve for p_A. Rearranging the traverse of step 1 for the starting pressure p_A: $$p_A = p_B + \rho_{benz}g(2.40) - \rho_{Hg}g(0.40) - \rho_{CCl_4}g(3.0)$$ Each term (benzene 21.19 kPa lost rising to the upper Hg surface, mercury 53.20 kPa gained descending 0.40 m, CCl₄ 46.79 kPa gained descending to B) combines with the known p_B = 200 kPa: $$p_A = 200 + 21.19 - 53.20 - 46.79 = \boxed{121.19\ \text{kPa}}$$

Quantity	Result
Pressure in pipe A	121.19 kPa (gauge)

Topic: Multi-fluid differential manometry.

Key relations: hydrostatic traverse p + ρgΔz (down) / p − ρgΔz (up); SG×1000 kg/m³ for density.

Why this works. Pressure in a connected static fluid depends only on elevation, never on the path taken between two points or on the tube's horizontal runs and bends. Walking the tube fluid-by-fluid from the known pressure to the unknown one, adding or subtracting each column's ρgΔz, is therefore exact regardless of how many bends the tube makes.

Common pitfall. Mixing up which mercury surface is higher reverses the sign of the 53.2 kPa term and produces a badly wrong answer (224 kPa instead of 121 kPa here) — always identify the higher interface directly from the figure before writing the traverse.

Source: White, Fluid Mechanics, Ch. 2 (manometry).

Question 1 of 13: Three-Fluid Differential Manometer

Question 1: Three-Fluid Differential Manometer (5 marks)