Absolute Pressure in a Pipe from a Multi-Fluid Manometer: 04-BS-7, May 2016, Q1

Question 1: Absolute Pressure in a Pipe from a Multi-Fluid 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
Water column below the pipe (to the mercury surface)	180 mm
Net mercury column (datum-to-datum, first bend)	180 mm
Glycerine column (mercury surface up to the air gap)	360 mm
Air gap below the open top (weight negligible)	100 mm
SG glycerine / SG mercury	1.26 / 13.56
Atmospheric head (as instructed)	10 m of water = 98.1 kPa

Find. The absolute pressure P in the pipe, in kPa.

Fig. Q1 — traced manometer path used to solve for P: water (180 mm) and mercury (net 180 mm) below the pipe, glycerine (360 mm) up to the open, atmospheric top.

Check: the printed figure has two U-bends and two atmospheric vents; the first (left-hand) mercury U-tube already forms a complete, self-contained path from the pipe to an open end (the vent at the top of the glycerine leg), so it alone is sufficient to solve for P — consistent with the paper's own Note 6 allowing a stated assumption where a figure needs interpretation. The datum used below is the pipe centreline; the second U-tube's readings (323 mm, 30 mm) are additional figure detail not needed once this loop is closed.

Approach. Trace the pressure from the pipe, through each fluid layer of the first (left) U-tube, to the open (atmospheric) top of its right-hand leg — adding ρgh going down, subtracting going up — then add the given atmospheric head to convert the resulting gauge pressure to absolute.

Set up the manometer equation (gauge, at the pipe). Starting at P and working down through water, down-then-up through the connected mercury (a net drop of 180 mm from the left leg's mercury surface to the right leg's, since both are measured 180 mm and 360 mm below the same datum), then up through glycerine to the open vent (gauge = 0, air weight negligible): $$P + \rho_w g(0.180) + \rho_w g(0.180)\,\text{SG}_{Hg} - \rho_w g(0.360)\,\text{SG}_{gly} = 0$$
Solve for the gauge pressure. $$P = \rho_w g\Big[(0.360)(1.26) - (0.180)(1) - (0.180)(13.56)\Big]$$ $$P = 9810\big[0.4536 - 0.180 - 2.4408\big] = 9810(-2.1672) = \boxed{-21.26\ \text{kPa (gauge)}}$$
Convert to absolute using the given atmospheric head. $$P_{atm} = \rho_w g(10) = 1000(9.81)(10) = 98.10\ \text{kPa}$$ $$P_{abs} = P_{gauge} + P_{atm} = -21.26 + 98.10 = \boxed{76.84\ \text{kPa (absolute)}}$$

Quantity	Result
Gauge pressure in the pipe	−21.26 kPa
Absolute pressure in the pipe P	76.84 kPa

Topic: Multi-fluid (compound) manometry.

Key relations: p = ρgh per fluid layer; manometer rule — add going down, subtract going up, summed layer by layer between two known points.

Why this works. Pressure in a connected static fluid depends only on vertical elevation change within each fluid, so a manometer can be "walked" fluid-by-fluid from an unknown point to a known one (here, atmosphere); using several fluids of different density (water, mercury, glycerine) lets the instrument be tuned for sensitivity while keeping the tube a manageable length.

Common pitfall. Forgetting that a net height difference in mercury between two legs (not simply "the mercury column height" in one leg) is what matters when the mercury pool connects two different reference points; here that net difference happens to equal the smaller of the two given readings (180 mm), taken from the shared datum.

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

Question 1 of 13: Absolute Pressure in a Pipe from a Multi-Fluid Manometer

Question 1: Absolute Pressure in a Pipe from a Multi-Fluid Manometer (5 marks)