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).
Show Hint
Start from the point open to atmosphere and add or subtract each fluid column's head on the way to P.
Step 1: Change every fluid column into an equivalent water head.
Using specific gravity, a mercury column of height $h_3$ carries the same pressure as $SG_{mercury} h_3$ metres of water, and an oil column of height $h_2$ carries the same pressure as $SG_{oil} h_2$ metres of water.
Step 2: Add up the heads from P to the open end Q.
Going from P down through water by $h_1$, then up through the oil loop by $h_2$, then up through mercury by $h_3$, brings us to atmospheric pressure at Q. In head terms: $P_{atm} = P_P + \rho_w g h_1 - \rho_w g (SG_{oil} h_2) - \rho_w g (SG_{mercury} h_3)$.
Step 3: Solve for the head at P and convert.
Rearranging, $P_P = \rho_w g (SG_{mercury} h_3 - SG_{oil} h_2 - h_1)$. Numerically $SG_{mercury} h_3 = 13.6 \times 0.8 = 10.88$ m, $SG_{oil} h_2 = 0.9 \times 0.7 = 0.63$ m, so the net water head is $10.88 - 0.63 - 0.5 = 9.75$ m.
Final Answer:
$P_P = 1000 \times 10 \times 9.75 = 97500$ Pa = 97.5 kPa, matching the 95 to 98 kPa key window.
\[ \boxed{P_P = 97.5 \text{ kPa}} \]
