Project ForgeProcess Calculations
Laws of Thermodynamics
GATE 2019
Hard
A vessel contains steam at 180°C, initial quality 0.5, initial volume 1 m³. The vessel loses heat at constant rate q̇ under isobaric conditions so quality reduces to 0.1 after 10 hours. At 180°C: vg=0.19405, vf=0.001127 m³/kg; hg=2778.2, hf=763.21 kJ/kg. The rate of heat loss q̇ (kJ/hour) is (rounded off to nearest integer)
826 kJ/hour(± 15 accepted)
Explanation
v1=vf+0.5(vg−vf)=0.09759 m³/kg, mass m=1/0.09759=10.247kg (fixed, closed system). h1=763.21+0.5×2014.99=1770.7 kJ/kg. h2=763.21+0.1×2014.99=964.71 kJ/kg. At constant P, Q=ΔH=m(h2−h1)=10.247×(−806.0)=−8259 kJ over 10h. q̇=8259/10=826 kJ/hour.
Source: GATE CH 2019 (2019)
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