Answer:
Step-by-step explanation:
For the given stirling circle with air as the working fluid, temperature, pressure, and volume i^th stage are
and
respectively
Obtain the properties of air at room temperature from molar mass, gas constant and critical point properties table
Gas constant R = 0.3704 psia ft³ / ibm.R
Express the thermal efficiency of an ideal stirling engine
![n_(th)=(W_(net))/(Q_(in)) =1-(T_L)/(T_H)](https://img.qammunity.org/2021/formulas/engineering/college/1wvp1yrwn7zwcbuy9ybt50ygu5ys8zicis.png)
Here, network done by the stirling engine is
heat input to the engine is
Temperature of the heat source is
![T_H](https://img.qammunity.org/2021/formulas/physics/college/vb8v5g7bbbu4qx3mrh8k0fwovn9ivpclx7.png)
Temperature of the sink is
![T_L](https://img.qammunity.org/2021/formulas/physics/high-school/ek5ij0vwuq9k8u7amkhjfmkw1i4pspg8th.png)
substitute 535 for
, 2Btu for
, 5Btu for
to find
![T_H](https://img.qammunity.org/2021/formulas/physics/college/vb8v5g7bbbu4qx3mrh8k0fwovn9ivpclx7.png)
![(2Btu)/(5Btu)=1-(535R)/(T_H) \\\\T_H=(535)/(0.6)\\\\=891.7R](https://img.qammunity.org/2021/formulas/engineering/college/p8maqoiuamq3fpavcf7717pxs7e698x2m2.png)
Hence , the temperature of the source of energy reservoir is 891.7R
Express the idea gas equation for air at state 3 of the cycle
![m=(P_3V_3)/(RT_3)](https://img.qammunity.org/2021/formulas/engineering/college/sso0r8bd05hia93rdtz2xfihxmesa41yad.png)
substitute
![15psia \ \ for \ \ p_3 \\\\0.5ft^3 \ \ for \ \ V_3 \\\\ 0.3704 psia.ft^/ibm.R \ \ for \ \ R \\\\535R \ \ for \ \ R](https://img.qammunity.org/2021/formulas/engineering/college/qe8vgw66nrtuqfuyalqhkrcceg4fa1otd2.png)
![m=((15)(0.5))/((0.3785)(535)) \\\\=0.3785Ibm](https://img.qammunity.org/2021/formulas/engineering/college/41liz49u5pia632y6aetzszi6swawwp2vd.png)
Hence ,the amount of air conditioned in the engine is 0.3785 Ibm
Express the ideal gas equation for air at state 1 of the cycle
![P_1=(mRT_1)/(V_1)](https://img.qammunity.org/2021/formulas/engineering/college/310esf3zl88p15dqu13u2jv8b1edryzhvt.png)
substitute 0.03785 for m,
![0.3704 psia ft^3/Ibm.R for R,\\\\ 891.7R for T_1 \ \ and \ \ 0.06ft^3 for V_1](https://img.qammunity.org/2021/formulas/engineering/college/f17k6298cbuutgdboxsicx1gvu8b6idxs7.png)
![P_1=((0.03785)(0.3704)(891.7))/((0.05)) \\\\=250 \texttt{psia}](https://img.qammunity.org/2021/formulas/engineering/college/fkhfoc99tsxqv9tupoaciyhyjd5fjuhsz7.png)
Hence, the maximum air pressure during the cycle is 250 psia