Question

Consider a Carnot cycle executed in a closed system with 0.6 kg of air. The temperature limits of the cycle are 300 and 1100 K, and the minimum and maximum pressures that occur during the cycle are 20 and 3000 kPa. Assuming constant specific heats, determine the net work output per cycle.

Answer 1

Answer:

63.8 kJ

Explanation:

The net work output per cycle is the difference in heat input and heat output. The heat input and heat output are expressed as a function of volume ratios, while volume is expressed as a function of pressure and pressure as a function of temperature.

R = 287 J/kg.K, k = 1.4

Hence the net work input (W) is given as:

$W=Q_{in}-Q_{out}\\\\W=mR[T_Hln\frac{V_2}{V_1} -T_Lln\frac{V_3}{V_4}]\\\\=mR[T_Hln\frac{P_1}{P_2} -T_Lln\frac{P_4}{P_3}]\\\\=mR[T_Hln(\frac{P_1}{P_3}(\frac{T_L}{T_H} )^\frac{k}{k-1}) -T_Lln(\frac{P_1}{P_3}(\frac{T_L}{T_H} )^\frac{k}{k-1})]\\\\=mR(T_H-T_L)ln(\frac{P_1}{P_3}(\frac{T_L}{T_H} )^\frac{k}{k-1})\\\\Substituting\ values:\\\\W=mR(T_H-T_L)ln(\frac{P_1}{P_3}(\frac{T_L}{T_H} )^\frac{k}{k-1})=0.6*287(1100-300)ln(\frac{3000*10^3}{2-*10^3}(\frac{300}{1100} )^\frac{1.4}{1.4-1})$

$W=63.8\ kJ$