Compound Manometer — Absolute Pressure in a Pipe: 04-BS-7, May 2013, Q1

Question 1: Compound Manometer — Absolute Pressure in a Pipe (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 (read from the figure). The pipe carries water; the water fills the connecting tube down the left leg of the first U-tube to a mercury surface 100 mm below the pipe centreline. The mercury (SG 13.56) fills the bottom of that U and stands in its right leg at a level 180 mm lower than the left-leg mercury surface. Above the right-leg mercury sits a 360 mm column of glycerine (SG 1.26), topped by the air in the crossover bend, which is vented to atmosphere at the top (opening 1). The second U-tube (323 mm of glycerine over mercury, and mercury standing 30 mm higher in the open right-hand tube, opening 2) runs from that vented air to the open tube, so both of its ends are at atmospheric pressure. Atmospheric pressure at both openings is equivalent to a 10 m head of water.

[Figure not reproduced: Source figure, Question 1 (exam page 2): compound manometer. See the official exam paper or the cited reference text.]

Exam figure (page 2). Water (stippled) reaches the mercury 100 mm below P. The mercury (dark) stands 180 mm lower in the right leg, under 360 mm of glycerine (hatched). The air bend above it is vented, so the second U-tube only links two atmospheric points.

Find. Absolute pressure P at the pipe centreline, in kPa.

Approach. Walk the hydrostatic path from P to the open (atmospheric) end, adding ρgh descending and subtracting it ascending, fluid by fluid, then set the result equal to the atmospheric-equivalent 10 m of water.

Reading the figure: the air in the crossover bend is vented through the top opening, so it is at atmospheric pressure, and the first U-tube alone closes the path from P to atmosphere. The second U-tube is a self-check on the specific gravities, not part of the path to P. Both of its ends are atmospheric, so its two columns must balance: $1.26\times323 = 407.0$ mm of water against $13.56\times30 = 406.8$ mm of water, equal to within 0.05%.

Atmospheric reference. The open tubes see a pressure equal to 10 m of water: $P_{atm}' = \rho_w g (10) = 1000 \times 9.81 \times 10 = 98{,}100\ \text{Pa} = 98.1\ \text{kPa}$.
Water column, P down to the left-leg mercury surface. Descending 100 mm of water: $\Delta p_w = 1000\times9.81\times0.100 = 981.0\ \text{Pa}$.
Mercury, down to the level of the right-leg mercury surface. Descending a further 180 mm, now in mercury: $\Delta p_{Hg} = 13560\times9.81\times0.180 = 23{,}944.3\ \text{Pa}$.
Glycerine, up to the vented air. Ascending 360 mm of glycerine to the atmospheric air in the bend: $\Delta p_{gly} = 1260\times9.81\times0.360 = 4449.8\ \text{Pa}$.
Assemble and solve for P. Tracing P → atmosphere: $P + \Delta p_w + \Delta p_{Hg} - \Delta p_{gly} = P_{atm}'$, so $$P = 98{,}100 - 981.0 - 23{,}944.3 + 4449.8 = \boxed{77{,}624\ \text{Pa} = 77.62\ \text{kPa (absolute)}}$$ In gauge terms, $P = -20.48\ \text{kPa}$. The pipe is under suction, which is why the mercury is drawn up the left leg toward it.

Quantity	Value
Absolute pressure P in pipe	77.62 kPa (−20.48 kPa gauge)

Topic: Compound (multi-fluid) manometry.

Key relations: hydrostatic pressure $p = \rho g h$ (descending adds, ascending subtracts); pressure is uniform along a connected body of a single low-density fluid (here, the vented air) regardless of small elevation changes.

Why this works: a manometer is solved by summing $\rho g h$ along any continuous fluid path between two points of known or sought pressure — the path taken through the tubing does not matter, only the net elevation change through each fluid. The key step in a compound manometer is finding the shortest closed path to a known pressure. Here the vented air bend is atmospheric, so the second U-tube never enters the answer.

Common pitfall: misreading which fluid fills each leg. The 180 mm below the water is mercury, not water or glycerine, and it carries most of the pressure difference. Another mistake is chaining through every labelled dimension, including a second U-tube whose ends are both at atmospheric pressure.

Source: Crowe, Elger & Roberson, Engineering Fluid Mechanics, Ch. 3 (Fluid Statics/manometry).

Question 1 of 13: Compound Manometer — Absolute Pressure in a Pipe

Question 1: Compound Manometer — Absolute Pressure in a Pipe (5 marks)