Question

The gas expanding in the combustion space of a reciprocating engine has an initial pressure of 5 MPa and an initial temperature of 1623°C. The initial volume is 0.05 m^3 and the gas expands through a volume ratio of 20 according to the law PV^1.25 = constant. Calculate: a) Work transfer b) Heat transfer

Answer 1

Answer:

a). Work transfer = 527.2 kJ

b). Heat Transfer = 197.7 kJ

Explanation:

Given:

$P_{1}$ = 5 Mpa

$T_{1}$ = 1623°C

                       = 1896 K

$V_{1}$ = 0.05 $m^{3}$

Also given $\frac{V_{2}}{V_{1}} = 20$

Therefore, $V_{2}$ = 1  $m^{3}$

R = 0.27 kJ / kg-K

$C_{V}$ = 0.8 kJ / kg-K

Also given : $P_{1}V_{1}^{1.25}=C$

   Therefore, $P_{1}V_{1}^{1.25}$ = $P_{2}V_{2}^{1.25}$

                     $5\times 0.05^{1.25}=P_{2}\times 1^{1.25}$

                     $P_{2}$ = 0.1182 MPa

a). Work transfer, δW = $\frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}$

                                  $\left [\frac{5\times 0.05-0.1182\times 1}{1.25-1} \right ]\times 10^{6}$

                              = 527200 J

                             = 527.200 kJ

b). From 1st law of thermodynamics,

Heat transfer, δQ = ΔU+δW

   = $\frac{mR(T_{2}-T_{1})}{\gamma -1}+ \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}$

  =$\left [ \frac{\gamma -n}{\gamma -1} \right ]\times \delta W$

  =$\left [ \frac{1.4 -1.25}{1.4 -1} \right ]\times 527.200$

  = 197.7 kJ