Absolute Pressure in a Pipe from a Multi-Fluid Manometer: 04-BS-7, December 2019, 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.

the recovered figure shows TWO U-tube manometers in series, but only the FIRST (left) loop, from the pipe down through water and mercury and up through glycerine to its own open vent, is needed to solve for P. The second loop's readings (323 mm, 30 mm) are additional figure detail, confirmed identical to 2016-May's own unused second-loop numbers, which is itself strong evidence the reconstruction is correct.

Given.

Quantity	Value
Water column below the pipe (to the mercury surface)	80 mm
Net mercury column (first loop)	60 mm
Glycerine column (mercury surface up to the open vent)	100 mm
SG glycerine / SG mercury (Constants, p.12)	1.26 / 13.56
Atmospheric head (as instructed)	10 m of water = 98.10 kPa

Fig. Q1 — traced manometer path: water (80 mm) and mercury (60 mm net) below the pipe, glycerine (100 mm) up to the open, atmospheric top.

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

Approach. Trace the pressure from the pipe, through each fluid layer of the first (left) loop, 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 through the net mercury column, then up through glycerine to the open vent (gauge = 0 there): $$P + \rho_w g(0.080) + \rho_w g(0.060)\,\text{SG}_{Hg} - \rho_w g(0.100)\,\text{SG}_{gly} = 0$$
Solve for the gauge pressure. $$P = \rho_w g\Big[(0.100)(1.26) - (0.080) - (0.060)(13.56)\Big]$$ $$P = 9810\big[0.1260 - 0.0800 - 0.8136\big] = 9810(-0.7676) = \boxed{-7.53\ \text{kPa (gauge)}}$$
Convert to absolute using the given atmospheric head. $$P_{atm} = \rho_w g(10) = 1000(9.81)(10) = 98.10\ \text{kPa}$$ $$P_{abs} = P_{gauge} + P_{atm} = -7.53 + 98.10 = \boxed{90.57\ \text{kPa (absolute)}}$$

Quantity	Result
Gauge pressure in the pipe	−7.53 kPa
Absolute pressure in the pipe P	90.57 kPa

Concept & Theory

Topic. Multi-fluid (compound) manometry.

Key relations. $p=\rho gh$ 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); the much denser mercury column dominates the pressure difference even though its column is shorter than the glycerine's, which is exactly why a small mercury height is chosen for a compact instrument.

Common pitfall. Trying to include the SECOND loop's readings in the pressure walk — once one continuous path from P to a known atmospheric point is closed, any other tube in the figure is redundant information for finding P.

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)