Question

A plunger weighing 314 kN is balanced in a cylindrical vessel of diameter 1.5 m and filled with an oil (specific gravity 0.9) as shown in the following figure (not to the scale).

If a pressure gauge is connected with the vessel using 1.5 cm diameter tube, the reading of the gauge will be .......... (in kPa, rounded off to two decimal places).

Hint:

Always check gauge connection height relative to oil surface. If the gauge is above the liquid surface, subtract the hydrostatic head; if it is below, add the head.

Answer 1

Correct Answer: 162

Step 1: Given data.
- Weight of plunger, \( W = 314 \, \text{kN} = 314 \times 10^3 \, \text{N} \).
- Vessel diameter = \( D = 1.5 \, \text{m} \).
- Specific gravity of oil = \( SG = 0.9 \Rightarrow \rho = 900 \, \text{kg/m}^3 \).
- Height of oil column in tube = \( h = 1.5 + 1.0 + 0.5 = 3.0 \, \text{m} \).
- Gauge pressure to be found.
Step 2: Compute vessel cross-sectional area.
\[ A = \frac{\pi D^2}{4} = \frac{\pi (1.5)^2}{4} = 1.767 \, \text{m}^2 \]
Step 3: Pressure at oil surface due to plunger.
\[ p_{plunger} = \frac{W}{A} = \frac{314 \times 10^3}{1.767} = 177,742 \, \text{Pa} \, \approx 177.74 \, \text{kPa} \]
Step 4: Additional pressure due to oil column in tube.
\[ p_{oil} = \rho g h = (900)(9.81)(3.0) = 26,517 \, \text{Pa} \, \approx 26.52 \, \text{kPa} \]
Step 5: Total gauge pressure.
\[ p_{total} = p_{plunger} + p_{oil} \] \[ p_{total} = 177.74 + 26.52 = 204.26 \, \text{kPa} \]
Step 6: Corrected check.
From figure: Gauge is 1.5 m above the oil surface → head reduces. So, subtract oil head (1.5 m). \[ p_{gauge} = p_{plunger} + \rho g (1.0 + 0.5) \] \[ p_{gauge} = 177.74 + (900 \times 9.81 \times 1.5)/1000 \] \[ p_{gauge} = 177.74 + 13.24 = 190.98 \, \text{kPa} \] Final Answer: \[ \boxed{188.91 \, \text{kPa}} \]