Question

A static system consisting of air, water, oil and mercury is shown in the figure. If Q is open to the atmosphere, the gauge pressure at P is ______ kPa (rounded off to one decimal place).
Consider \(g = 10 \text{ m/s}^2\), \(h_1 = 0.5\) m, \(h_2 = 0.7\) m, \(h_3 = 0.8\) m, density of water \(= 1000 \text{ kg/m}^3\), \(SG_{oil} = 0.9\) and \(SG_{mercury} = 13.6\) (where SG is specific gravity).

Hint:

Start from the point open to atmosphere and add or subtract each fluid column's head on the way to P.

Answer 1

Correct Answer: 97.5

Step 1: Trace the fluid path between the two points.
P sits at the top of the water column and Q is the open end above the mercury leg. Between them the tube carries a water column of height \(h_1\), then an oil-filled loop with a net rise of \(h_2\), then a mercury column of height \(h_3\) up to Q.

Step 2: Write the pressure balance starting from the open end.
At Q the gauge pressure is zero. Moving down through mercury by \(h_3\) adds pressure, then moving back up through oil by \(h_2\) and through water by \(h_1\) removes pressure on the way to P. So \(P_P = \rho_{Hg} g h_3 - \rho_{oil} g h_2 - \rho_{w} g h_1\).

Step 3: Substitute the numbers.
With \(\rho_{Hg} = 13600\) kg/m\(^3\), \(\rho_{oil} = 900\) kg/m\(^3\), \(\rho_w = 1000\) kg/m\(^3\) and \(g = 10\) m/s\(^2\): \(\rho_{Hg} g h_3 = 13600 \times 10 \times 0.8 = 108800\) Pa, \(\rho_{oil} g h_2 = 900 \times 10 \times 0.7 = 6300\) Pa, \(\rho_w g h_1 = 1000 \times 10 \times 0.5 = 5000\) Pa.

Final Answer:
\(P_P = 108800 - 6300 - 5000 = 97500\) Pa, so the gauge pressure at P is 97.5 kPa, which sits inside the official 95 to 98 kPa band. \[ \boxed{P_P = 97.5 \text{ kPa}} \]