Multi-fluid manometer — gage pressure at A: 20-Bio-A7 Bioinstrumentation, December 2019, Q2

Question 2: Multi-fluid manometer — gage pressure at A (25 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 verified solution is reproduced below.

Given.

Quantity	Value
γ_water	9790 N/m³
γ_mercury	133,100 N/m³
Oil specific gravity	SG = 0.85
Water column (A → air–water interface)	45 − 30 = 15 cm
Oil column (top of middle tube → mercury surface)	40 cm

Find. The gage pressure at A, and whether it is higher or lower than atmospheric.

Compound water–air–oil–mercury manometer connecting point A to the atmosphere.

Approach. Trace a hydrostatic path from A to the open (atmospheric) end, applying the manometer rule — pressure increases by $\gamma h$ descending through a fluid, decreases by $\gamma h$ ascending — through each fluid layer in turn; air's weight is negligible over these heights.

Convert the oil's specific gravity to a unit weight. $$\gamma_{oil} = SG \times \gamma_{water} = 0.85 \times 9790 = 8321.5\ \text{N/m}^3$$
Traverse from A to the top of the left tube (up through water, then air). Moving up through the 15 cm of water loses $\gamma_{water}(0.15)$; the following 30 cm of air is weightless in comparison, so the pressure at the top of the bend (and, via the connected air space, at the top of the middle tube where oil begins) is $$p_{top} = p_A - \gamma_{water}(0.15)$$
Traverse down through the oil to the mercury surface. Descending 40 cm of oil gains $\gamma_{oil}(0.40)$: $$p_{Hg,\,mid} = p_{top} + \gamma_{oil}(0.40) = p_A - \gamma_{water}(0.15) + \gamma_{oil}(0.40)$$
Traverse through the mercury pool and up the open tube to atmosphere. The mercury leg descends 15 cm and the connected leg on the open-tube side rises the same 15 cm back to the same elevation, so mercury's net contribution is zero; the remaining open-tube air column is likewise weightless. Setting the resulting pressure equal to atmospheric (gage $=0$): $$0 = p_A - \gamma_{water}(0.15) + \gamma_{oil}(0.40)$$ $$\boxed{p_A = \gamma_{water}(0.15) - \gamma_{oil}(0.40) = (9790)(0.15) - (8321.5)(0.40) = 1468.5 - 3328.6 = -1860\ \text{Pa}}$$

The gage pressure at A is negative, so A is below atmospheric pressure by about 1.86 kPa — the heavier oil column (40 cm at $\gamma_{oil}=8321.5$ N/m$^3$) outweighs the shallower water column (15 cm at $\gamma_{water}=9790$ N/m$^3$) that supports it, even though water is the denser fluid, because the oil path is more than twice as long.

Quantity	Value
γ_oil	8321.5 N/m³
Gage pressure at A	−1860 Pa
Relative to atmosphere	Lower, by ≈ 1.86 kPa

Topic. Multi-fluid (compound) manometry.

Key relations. Hydrostatic/manometer equation $p_2 = p_1 + \gamma \Delta h$ (descending) or $p_2 = p_1 - \gamma\Delta h$ (ascending); $\gamma = SG\times\gamma_{water}$.

Why this works. Within any single connected static fluid, pressure depends only on vertical elevation, not on horizontal path or tube shape, so a manometer can be solved by walking from a known point to an unknown one, layer by layer, adding or subtracting $\gamma h$ as elevation falls or rises. A trapped gas (air here) is treated as a single uniform pressure throughout because its weight is negligible compared with the liquids.

Common pitfall. Losing track of sign when crossing several fluids in sequence, or forgetting that descending and re-ascending through the same fluid by equal depths (the mercury pool here) contributes nothing net to the pressure difference — only the elevation change of the interfaces at the top and bottom of each fluid layer matters, not the pool's own depth.

Source. White, Fluid Mechanics, 8th ed., §2.4 (hydrostatic pressure, multiple fluids, manometry).

Question 2 of 4: Multi-fluid manometer — gage pressure at A

Question 2: Multi-fluid manometer — gage pressure at A (25 marks)