Two-Fluid Manometer Between Pipes A and B: 04-BS-7, December 2017, Q1

Question 1 of 13: Two-Fluid Manometer Between Pipes A and B

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

04-BS-7 Mechanics of Fluids — National Examination, 2017-Dec. Three (3) hours duration, closed book. Section A (Calculative, 9 questions, do 7) and Section B (Graphical & Analytical, 4 questions, do 3); every question is answered below regardless of the exam's "do N of M" instruction, so the set is a complete study resource.

Reference texts: White, F.M., Fluid Mechanics (8th ed.) — fluid statics and manometry (Ch. 2), Bernoulli and flow measurement (Ch. 3), viscous flow in ducts and the Moody chart (Ch. 6), flow past immersed bodies and drag (Ch. 7), turbomachinery and wind turbines (Ch. 11).

Check — assumptions used across this paper:

Q1's manometer is read from the printed figure as a benzene(hatched)–mercury(black)–carbon-tetrachloride(clear) chain: benzene fills pipe A up and over the first bend and down the U-tube's left arm to the benzene/mercury interface at the UPPER dimension line (2.0 m + 400 mm = 2.4 m above A); the mercury stands 400 mm lower in the right arm, at the mercury/CCl₄ interface on the LOWER line (2.0 m above A); CCl₄ then fills the rest of the run over the second bend down to B, 3.0 m below A. (The mercury is higher on the A side, so p_A < p_B.)
Q6's spillway/gate width is read from the drawing as the 8.76 m dimension (the question's own hint: "width of each gate is slightly greater than its height" – 8.76 m > 8.23 m); the closed gate's wetted height at F.S.L. is F.S.L. − Crest = 7.92 m (the gate's own 8.23 m height extends slightly above F.S.L., matching the paper's note that "the top of the gate is higher than F.S.L.").
Q7 and Q9's friction/drag coefficients come from the Colebrook–White equation and the plotted drag curve — the same relations the attached Moody and drag charts plot.
Q9's Reynolds number (≈5.9×10⁷) is beyond the attached drag chart's plotted range (10⁻¹ to 10⁶); C_D = 0.3 is taken from the right-hand end of the cylinder curve (minimum ≈0.3 past the drag crisis, ≈0.33 at 10⁶) — the best available reading — and flagged here as an extrapolation.
Q8(c)'s fuel density (needed to convert a fuel mass into litres) is not stated in Q8 itself; the paper lists no fuel density on its Constants page, so the gasoline SG = 0.75 given in Q4 of this same paper is adopted (Q2's 0.72 would give 3.50 L/100 km instead of 3.36).

Question 1: Two-Fluid Manometer Between Pipes A and B (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, $p_B$	200 kPa
Elevation, mercury/CCl₄ interface (right arm, lower line), above A	2.0 m
Elevation, benzene/mercury interface (left arm, upper line), above A	2.4 m (2.0 m + 400 mm)
Elevation of B relative to A	−3.0 m (B below A)
Specific gravities (Constants, p.11)	benzene 0.90, mercury 13.56, CCl₄ 1.59

Elevation ladder of the manometer chain: benzene from A up to its mercury interface at +2.4 m (left arm of the U-tube), mercury between +2.4 m and its CCl₄ interface at +2.0 m (right arm, 400 mm lower), carbon tetrachloride from +2.0 m down to B at −3.0 m.

Find. The pressure $p_A$ in pipe A.

Approach. Walk the hydrostatic law $p_{\text{down}} = p_{\text{up}} + \rho g\,\Delta z$ along the connected fluid chain from B to A, switching density at each interface.

CCl₄ leg: B (−3.0 m) up to the Hg/CCl₄ interface (+2.0 m, right arm). Rising through carbon tetrachloride reduces pressure: $$p_{i2}=p_B-\rho_{CCl_4}\,g\,(z_{i2}-z_B)=200{,}000-(1590)(9.81)(2.0-(-3.0))=200{,}000-77{,}990=122{,}010\ \text{Pa}$$
Mercury leg: interface (+2.0 m) up to the C₆H₆/Hg interface (+2.4 m, left arm). Rising through mercury reduces pressure: $$p_{i1}=p_{i2}-\rho_{Hg}\,g\,(z_{i1}-z_{i2})=122{,}010-(13{,}560)(9.81)(0.4)=122{,}010-53{,}209=68{,}801\ \text{Pa}$$
Benzene leg: interface (+2.4 m) down to A (0 m). Descending through benzene raises pressure: $$p_A=p_{i1}+\rho_{benzene}\,g\,(z_{i1}-z_A)=68{,}801+(900)(9.81)(2.4)=68{,}801+21{,}190=\boxed{89{,}991\ \text{Pa} \approx 90.0\ \text{kPa}}$$

Quantity	Value
Pressure at Hg/CCl₄ interface	122.0 kPa
Pressure at C₆H₆/Hg interface	68.8 kPa
Pressure in pipe A, $p_A$	90.0 kPa

Concept & Theory

Topic. Multi-fluid U-tube manometry.

Key relations. Hydrostatic law $p_2=p_1+\rho g(z_1-z_2)$, applied leg-by-leg along a chain of immiscible, stationary fluids.

Why this works. Pressure is continuous across a fluid interface at rest, so the hydrostatic law can be re-applied with a new density each time the tube crosses into a different liquid; the total change from B to A is just the sum of the individual leg contributions, independent of the tube's actual bends — only the endpoint elevations of each leg matter.

Common pitfall. Using the wrong density on a leg, or adding instead of subtracting when moving upward through a fluid — always re-derive the sign from "pressure increases going down." Also read which arm of the U-tube carries the higher mercury level: here the mercury stands 400 mm HIGHER on the benzene (A) side, so the benzene/mercury interface is at +2.4 m, not +2.0 m; swapping the two interfaces gives 186.6 kPa instead of 90.0 kPa.

Source. White, Fluid Mechanics, Ch. 2 (Manometry).