Question 12.CSGP.97: Consider an ideal air-standard Stirling cycle with an ideal ......
Consider an ideal air-standard Stirling cycle with an ideal regenerator. The minimum pressure and temperature in the cycle are 100 kPa, 25°C, the compression ratio is 10, and the maximum temperature in the cycle is 1000°C. Analyze each of the four processes in this cycle for work and heat transfer, and determine the overall performance of the engine.
Ideal Stirling cycle diagram as in Fig. 12.17, with
P _1=100 \,kPa , \quad T _1= T _2=25^{\circ} C , \quad v _1 / v _2=10, \quad T _3= T _4=1000^{\circ} C
From 1-2 at const T: { }_1 w _2={ }_1 q _2= T _1\left( s _2- s _1\right)
=- RT _1 \ln \left( v _1 / v _2\right)=-0.287 \times 298.2 \times \ln (10)=-197.1 \,kJ / kg
From 2-3 at const V: { }_2 w _3=\mathbf{\emptyset}
q _{23}= C _{ V 0}\left( T _3- T _2\right)=0.717(1000-25)= 6 9 9\,k J / k g
From 3-4 at const T; { }_3 w _4={ }_3 q _4= T _3\left( s _4- s _3\right)
=+ RT _3 \times \ln \frac{ V _4}{ V _3}=0.287 \times 1237.2 \times \ln (10)= 8 4 1 . 4 ~ k J / k g
From 4-1 at const V; { }_4 w _1=\mathbf{\emptyset}
Since q _{23} \text { is supplied by }- q _{41} (regenerator)
q _{ H }= q _{34}=841.4 \,kJ / kg , \quad \eta_{ TH }=\frac{ w _{ NET }}{ q _{ H }}=\frac{644.3}{841.4}= 0 . 7 6 6
NOTE: q _{ H }= q _{34}= RT _3 \times \ln (10), \quad q _{ L }=-{ }_1 q _2= RT _1 \times \ln (10)
\eta_{ TH }=\frac{ q _{ H }- q _{ L }}{ q _{ H }}=\frac{ T _3- T _1}{ T _3}=\frac{975}{1273.2}=0.766=\text { Carnot efficiency }