GATE 2022 ME (ME2) – Question 60
A rigid tank of volume of 8 m$^3$ is being filled up with air from a pipeline connected through a valve. Initially the valve is closed and the tank is assumed to be completely evacuated. The air pressure and temperature inside the pipeline are maintained at 600 kPa and 306 K, respectively. The filling of the tank begins by opening the valve and the process ends when the tank pressure is equal to the pipeline pressure. During the filling process, heat loss to the surrounding is 1000 kJ. The specific heats of air at constant pressure and at constant volume are 1.005 kJ/kg.K and 0.718 kJ/kg.K, respectively. Neglect changes in kinetic energy and potential energy. The final temperature of the tank after the completion of the filling process is ______ K (round off to the nearest integer).
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Correct answer: 395
Explanation
Take the tank as a control volume. It starts empty, so the energy balance for a filling process (no work, kinetic and potential energy neglected) is
$$Q_{in}=m_2u_2-m_ih_i,\qquad Q_{in}=-1000\text{ kJ}\ (\text{heat loss}).$$
The tank is initially evacuated, so the mass in the tank at the end is the mass that came in, $m_i=m_2=m$, and the supply enthalpy is $h_i=c_pT_i$:
$$-1000=mc_vT_2-mc_pT_i .$$
**Step 1: gas constant.**
$$R=c_p-c_v=1.005-0.718=0.287\text{ kJ/kg K}.$$
**Step 2: mass from the ideal gas law.**
$$m=\frac{p_2V}{RT_2}=\frac{600\times8}{0.287\,T_2}=\frac{4800}{0.287\,T_2}.$$
**Step 3: solve for $T_2$.** Substituting $m$:
$$-1000=\frac{4800}{0.287\,T_2}\big(0.718\,T_2-1.005(306)\big)$$
$$-1000=\frac{4800}{0.287}\left(0.718-\frac{307.53}{T_2}\right)$$
$$\frac{307.53}{T_2}=0.718+\frac{1000(0.287)}{4800}=0.718+0.05979=0.77779$$
$$T_2=\frac{307.53}{0.77779}=395.4\text{ K}.$$
The final temperature is **395 K**.