Question 1.28: A 1-m^3 rigid tank contains hydrogen at 250 kPa and 420 K. T...

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.

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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
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