Question 2.45: In a Carnot cycle the lowest pressure and temperature are eq...
In a Carnot cycle the lowest pressure and temperature are equal to 1 bar and 20°C and isothermal compression pressure is 4 bar. Pressure after isentropic compression is 10 bar and after isothermal heat addition process is 5 bar. Calculate (a) the highest temperature in the cycle, (b) the change in entropy during isothermal expansion, (c) heat addition to the cycle, (d) heat rejected by the cycle,(e) power developed with 150 cycles/min, (f) thermal efficiency of the cycle, and (g) mean effective pressure. Assume \gamma = 1.4.
The cycle is shown in Fig 2.56
\begin{aligned}&T_2=20+273=293 \ K , p_4=1 \ bar \\&p_1=4 \text { bar, } p_2=10 \ bar , p_3=5 \ bar\end{aligned}
At point 4, ν_4=\frac{R T_4}{p_4}=\frac{0.287 \times 293}{1 \times 10^2}\\ = 0.8409 m³/kg
Isothermal compression 4-1 :
\begin{aligned}ν_1 &=\frac{p_4 ν_4}{p_1}=\frac{1 \times 10^5 \times 0.8409}{4 \times 10^5} \\&=0.2102 \ m ^3 / kg\end{aligned}
Isentropic compression 1-2 :
ν_2=ν_1\left\lgroup \frac{p_1}{p_2}\right\rgroup ^{\frac{1}{\gamma}}=0.2102\left(\frac{4}{10}\right)^{\frac{1}{1.4}}=0.1092 \ m ^3 / kg
Swept volume, ν_s=ν_4-ν_2=0.8409-0.1092=0.7317 \ m ^3 / kg
(a) T_1=T_2 \left\lgroup\frac{ν_1}{ν_2}\right\rgroup ^{\gamma-1}=293\left(\frac{0.2102}{0.1092}\right)^{0.4}=380.7 \ K
(b) Isothermal heat addition 2-3:
\begin{aligned}ν_3 &=\frac{p_2 ν_2}{p_3}=\frac{0.1092 \times 10}{5}=0.2184 \ m ^3 / kg \\s_3-s_2 &=\left\lgroup\frac{p_2 ν_2}{T_2}\right\rgroup \ln \left\lgroup \frac{ν_3}{ν_2}\right\rgroup \\&=\left[\frac{10 \times 10^2 \times 0.1092}{380.7}\right] \ln \left[\frac{0.2184}{0.1092}\right] \\&=0.1988 \ kJ / kg K\end{aligned}
(c) Heat added, q_s=p_2 ν_2 \ln \frac{ν_3}{ν_2}=T_1\left(s_3-s_2\right)=380.7 \times 0.1988=75.68 \ kJ / kg
(e) Work done per cycle,
w=q_s-q_r=75.68-58.25=17.43 \ kJ / kg
\text { Power developed }=\frac{1.743 \times 150}{60}=43.575 \ kW
(f) Thermal efficiency, \eta_{ th }=1-\frac{T_2}{T_1}=1-\frac{293}{380.7}=0.2303 \text { or } 23.03 \%
(g) p_m=\frac{w}{v_s}=\frac{17.43}{0.7317}=23.82 kN / m ^2=0.2382 \ bar
