GATE 2026 ME – Question 57
A rigid closed vertical cylindrical vessel of 15 cm diameter contains 5 kg water at 80 °C with 10% quality. The water is heated till its temperature reaches 130 °C. Considering only a horizontal separated interface between liquid and vapor, the dip in the liquid level after the heating process is ________ cm (*rounded off to 2 decimal places*).
Properties of water at various saturation temperatures are given in the table below.
| Temperature $T$ (°C) | $v_f$ (m$^3$/kg) | $v_g$ (m$^3$/kg) | $u_f$ (kJ/kg) | $u_g$ (kJ/kg) |
|---|---|---|---|---|
| 80 | 0.001029 | 3.4053 | 334.97 | 2481.60 |
| 130 | 0.001070 | 0.66808 | 546.10 | 2539.50 |
$T$, $v$, and $u$ are temperature, specific volume, and specific internal energy, respectively. Subscripts $f$ and $g$ represent saturated liquid and saturated vapor, respectively.
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Correct answer: 11.32 to 11.44
Explanation
The vessel is rigid and closed, so the specific volume stays constant: $v = 0.001029 + 0.1(3.4053 - 0.001029) = 0.34146$ m$^3$/kg. At 130 °C the quality is $x = \frac{0.34146 - 0.001070}{0.66808 - 0.001070} = 0.5103$. The liquid volume at the start is $4.5 \times 0.001029 = 0.0046305$ m$^3$. At the end it is $5(1 - 0.5103) \times 0.001070 = 0.0026198$ m$^3$. The loss of liquid volume is $0.0020107$ m$^3$. The cross-section area is $\frac{\pi}{4}(0.15)^2 = 0.017671$ m$^2$, so the level dips by $\frac{0.0020107}{0.017671} = 0.1138$ m, which is 11.38 cm.