Question #273278

An ideal Stirling engine with air as the working fluid operates between specified pressure limits. The heat added to the cycle and the network produced by the cycle is to be determined.

Assumptions: Air is an ideal gas with constant specific heats. Properties: The properties of air at room temperature are R = 0.287 kPa·m3 /kg.K, cp = 1.005 kJ/kg·K, cv = 0.718 kJ/kg·K, and k = 1.4 


1
Expert's answer
2021-12-03T12:45:59-0500

Properties:


The properties of air at room temperature are 

R=0.287kPam3kgKR = 0.287 \large\frac{kPa·m^3}{kg·K}

cp=1.005KJkgKc_p = 1.005 \large\frac{KJ}{kg·K}

cv=0.718kJkgKc_v = 0.718 \large\frac{kJ}{kg·K}

k=1.4k = 1.4


Analysis: Applying the ideal gas equation to the isothermal process -4 gives


P4=P3V3V4=(50kPa)(12)=600kPaP_4=P_3\frac{V_3}{V_4}=(50kPa)(12)=600kPa


Since process 4-1 is one of constant volume


T1=T4(P1P4)=(298K)(3600kPa600kPa)=1788KT_1=T_4(\frac{P_1}{P_4})=(298K)(\frac{3600kPa}{600kPa})=1788K


Adapting the first and work integral to the heat addition process gives


qin=W12=RT1lnV2V1=(0.287KJkgK)(1788K)ln(12)=1275KJkgq_{in}=W_{1-2}=RT_1ln\frac{V_2}{V_1}=(0.287\frac{KJ}{kg*K})(1788K)ln(12)=1275\frac{KJ}{kg}


Similarly


qout=W34=RT3lnV4V3=(0.287KJkgK)(298K)ln(112)=212.5KJkgq_{out}=W_{3-4}=RT_3ln\frac{V_4}{V_3}=(0.287\frac{KJ}{kg*K})(298K)ln(\frac{1}{12})=212.5\frac{KJ}{kg}


The net work is then 


Wnet=qinqout=1275212.5=1063KJkgW_{net}=q_{in}-q_{out}=1275-212.5=1063\frac{KJ}{kg}



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