A 1-m^{3} rigid tank contains hydrogen at 250 kPa and 420 K. The gas is now cooled until its temperature drops to 300 K. Determine (a) the final pressure in the tank and (b) the amount of heat transfer from the tank.
A 1-m^{3} rigid tank contains hydrogen at 250 kPa and 420 K. The gas is now cooled until its temperature drops to 300 K. Determine (a) the final pressure in the tank and (b) the amount of heat transfer from the tank.
The hydrogen gas in a rigid tank is cooled until its temperature drops to 300 K. The final pressure in the tank and the amount of heat transfer are to be determined.
Assumptions 1 Hydrogen is an ideal gas since it is at a high temperature and low pressure relative to its critical point values of -240°C and 1.30 MPa. 2 The kinetic and potential energy changes are negligible, \Delta ke \cong \Delta pe \cong 0 .
Properties The gas constant of hydrogen is R = 4.124 kPa.m^{3}/kg.K (Table A-1).
Analysis (a) We take the hydrogen in the tank as our system. This is a closed system since no mass enters or leaves. The final pressure of hydrogen can be determined from the ideal gas relation,
\frac{P_{1} V}{T_{1}} = \frac{P_{2} V}{T_{2}} \rightarrow P_{2} = \frac{T_{2}}{T_{1}} P_{1} = \frac{300 K}{420 K}\left ( 250 kPa \right ) = 178.6 kPa
(b) The energy balance for this system can be expressed as
\underbrace{E_{in} – E_{out}}_{Net energy transfer by heat, work, and mass} = \underbrace{\Delta E_{system}}_{Change in internal, kinetic, potential, etc. energies}– Q_{out} = \Delta U
Q_{out} = – \Delta U = – m \left ( u_{2} – u_{1} \right ) \cong m C_{v} \left ( T_{1} – T_{2} \right )
where
m = \frac{P_{1} V}{RT_{1}} = \frac{\left ( 250 kPa \right )\left ( 1.0 m^{3} \right )}{\left ( 4.124 kPa \cdot m^{3}/kg \cdot K \right )\left ( 420 K \right )} = 0.1443 kg
Using the C_{\nu } = \left ( C_{p} – R \right ) = 14.516 – 4.124 = 10.392 kJ/kg.K value at the average temperature of 360 K and substituting, the heat transfer is determined to be
Q_{out} = \left ( 0.1443 kg \right )\left ( 10.392 kJ/kg \cdot K \right )\left ( 420 – 300 \right )K = 180.0 kJ