Question 2 of 4: Multi-fluid manometer — gage pressure at A
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, May 2015 — 04-Bio-A7, 3 hours, open book (any non-communicating calculator permitted). Four questions constitute a complete exam paper, each of equal value, and all require calculation. All four are solved here.
Check — subject-identity note: every page of the paper is headed and footed “04-Bio-A7, Fluid Mechanics” and all four questions (centrifugal-pump performance and similarity scaling, a multi-fluid manometer, laminar pipe flow between two reservoirs, and a turbulent parallel-pipe network) are genuine Fluid Mechanics content, with no bioinstrumentation, signal-processing, or physiological material at all. This solution follows the paper's true subject and cites a fluid-mechanics reference accordingly.
Check — figure reading. The dimension lines in the source figure are read as: the left tube's total height from its top bend down to A is 45 cm, of which the top 30 cm (down to the air–water interface) is air and the remaining 15 cm (down to A) is water; the right (open) tube's 40 cm dimension runs from the atmosphere opening down to the mercury surface, matching the oil column length in the middle tube (both tube tops sit at the same elevation, connected by the air-filled bend); the mercury forms a single connected pool whose two legs (in the middle and right tubes) are drawn at equal depth (15 cm), so the descent into mercury and the equal ascent out of it on the other side contribute zero net elevation change to the pressure traverse.
Given.
Quantity
Value
γwater
9790 N/m³
γmercury
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.
