Pressure in Pipe A from a Two-Fluid Differential Manometer: 04-BS-7, Undated paper, Q1

Question 1: Pressure in Pipe A from a Two-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
Fluid in pipe A / connecting leg	benzene, SG = 0.90
Fluid in U-tube	mercury, SG = 13.56
Fluid in pipe B / connecting leg	carbon tetrachloride, SG = 1.59
Manometer reading (mercury column offset)	400 mm
Elevation, lower mercury meniscus above A	2.0 m
Elevation, A above B	3.0 m

Find. The gauge pressure in pipe A.

Benzene fills pipe A up to the upper mercury meniscus; the mercury column steps down 400 mm through the U-bend to its lower meniscus, above which carbon tetrachloride fills the leg down to pipe B, 2.0 m below the lower meniscus and a further 3.0 m above pipe B.

Approach. Trace the hydrostatic pressure from A to B through each fluid in turn, adding ρgh for each downward step and subtracting it for each upward step; the local up-and-over shape of the tube is irrelevant — only the net elevation change through each single fluid matters.

Rise through benzene, A to the upper mercury meniscus. The benzene leg rises a net 2.0 + 0.4 = 2.4 m from A to the meniscus: $$ P_1 = P_A - \rho_{benzene}\,g\,(2.4) $$
Step down through mercury, 400 mm net. The two mercury menisci differ by the given 400 mm reading, with the far (B-side) meniscus the lower of the two: $$ P_2 = P_1 + \rho_{Hg}\,g\,(0.4) $$
Descend through carbon tetrachloride to B. From the lower mercury meniscus down to B is a net 2.0 + 3.0 = 5.0 m: $$ P_B = P_2 + \rho_{CCl_4}\,g\,(5.0) $$
Combine and solve for P_A. Substituting Steps 1–3 and rearranging for the unknown: $$ P_A = P_B + \rho_{benzene}\,g(2.4) - \rho_{Hg}\,g(0.4) - \rho_{CCl_4}\,g(5.0) $$ $$ P_A = 200\,000 + (900)(9.81)(2.4) - (13\,560)(9.81)(0.4) - (1590)(9.81)(5.0) $$ $$ \boxed{P_A \approx 90.0\ \text{kPa}} $$

Quantity	Result
Pressure in pipe A	90.0 kPa

Topic: Multi-fluid differential manometry.

Key relations: hydrostatic pressure at a point, p = ρgh; superposition along a continuous static fluid path.

Why this works: pressure in a connected static fluid depends only on vertical elevation, never on the horizontal run or the shape of the tube, so a manometer can be traced fluid-by-fluid from one pipe to the other by simple bookkeeping of ups and downs. Each fluid contributes only over the net elevation change it actually spans; going up subtracts ρgh, going down adds it.

Common pitfall: students often add every physical rise-and-fall segment separately instead of collapsing each single-fluid run to its net elevation change, or mix up which fluid occupies which leg once the tube changes direction.

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

Question 1 of 13: Pressure in Pipe A from a Two-Fluid Differential Manometer

Question 1: Pressure in Pipe A from a Two-Fluid Differential Manometer (5 marks)