Question 2 of 13: Manometer Fluid Identification

Question 2 — Manometer Fluid Identification (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. All pressures absolute; atmospheric p₀ = 100 kPa; SG_water=1.00, SG_glycerine=1.26, SG_mercury=13.56 (General Reference Information, p.12). Column heights and end pressures as marked on the attachment (85, 110, 108, 150, 200, 160 kPa; 1530, 810, 815, 1000, 300, 260 mm).

Find. The fluid (air, water, glycerine or mercury) in every section of all four manometers, written into the label spaces.

Approach. For a static fluid column of height h connecting two points of known pressure, ρ = Δp/(gh); match the computed density to the given specific gravities (using the shading change noted in the instructions to confirm where one fluid ends and the next begins).

1. Column 1 (1530 mm, 85 kPa bulb to the open/atmospheric leg). $$\rho_1 = \frac{{|100,000-85,000|}}{{9.81\times1.530}} = \frac{{15,000}}{{15.01}} = 999\text{{ kg/m}}^3 \Rightarrow \boxed{{\text{{WATER}}}}\ (SG\approx1.00)$$
2. Column 2 (810 mm, atmosphere to 110 kPa bulb). $$\rho_2 = \frac{{|110,000-100,000|}}{{9.81\times0.810}} = \frac{{10,000}}{{7.946}} = 1259\text{{ kg/m}}^3 \Rightarrow \boxed{{\text{{GLYCERINE}}}}\ (SG=1.26,\text{{ exact match}})$$
3. Column 3 (815 mm, atmosphere to 108 kPa bulb). $$\rho_3 = \frac{{|108,000-100,000|}}{{9.81\times0.815}} = \frac{{8,000}}{{7.995}} = 1001\text{{ kg/m}}^3 \Rightarrow \boxed{{\text{{WATER}}}}$$
4. Air spaces in the upper manometers. The unshaded crown of the 810 mm (glycerine) manometer and the unshaded inverted U above the 815 mm column, running across to the flared open end, hold AIR at atmospheric pressure. That is why the free surfaces of columns 2 and 3 sit at p₀ = 100 kPa; air's weight over these heights is negligible.
5. Lower-right manometer (108 kPa bulb → 150 kPa bulb, 260 mm deflection). The 108 kPa bulb is the same water-filled bulb as column 3, so the long leg that drops from it (the 1000 mm level difference to the 150 kPa bulb) is water. If the leg rising to the 150 kPa bulb is also water, the loop closes with mercury in the bottom of the U: $$p_{150} = 108 + \frac{1000\times9.81\times1.000}{1000} + \frac{(13{,}560-1000)\times9.81\times0.260}{1000} = 108 + 9.81 + 32.04 = 149.85\text{ kPa} \approx 150\text{ kPa}$$ Water in both upper legs closes to 0.1%. Glycerine would give 151.7 kPa, and air over the mercury would give 142.6 kPa. So the answer is WATER (both legs and the 150 kPa bulb) over MERCURY (the dark bottom of the U).
6. Lower-left manometer (200 kPa and 160 kPa bulbs at the same level, 300 mm deflection). Take a fluid of density ρ above the mercury on both sides. Because the bulbs are level, the 40 kPa difference must equal (ρ_Hg − ρ)g(0.300): $$\rho = 13{,}560 - \frac{200{,}000-160{,}000}{9.81\times0.300} = 13{,}560 - 13{,}592 \approx 0\ \text{kg/m}^3 \Rightarrow \boxed{\text{AIR}}$$ Check: 300 mm of mercury alone gives 13,560×9.81×0.300 = 39.9 kPa ≈ 40 kPa. Water above the mercury would give only (13,560 − 1000)×9.81×0.300 = 37.0 kPa. So the hatched bulbs and legs here are AIR over MERCURY (dark), even though they are shaded like the water elsewhere. This is exactly the trap the note "similar shading does not necessarily mean the same fluid" warns about.