Question

Consider a Carnot heat-engine cycle executed in a closed system using 0.025 kg of steam as the working fluid. It is known that the maximum absolute temperature in the cycle is twice the minimum absolute temperature, and the net work output of the cycle is 60 kJ. If the steam changes from saturated vapor to saturated liquid during heat rejection, determine the temperature of the steam during the heat rejection process. The temperature of the steam during the heat rejection process is

Answer 1

Answer:

The temperature of the steam during the heat rejection process is 42.5°C

Explanation:

Given the data in the question;

the maximum temperature T$_H$ in the cycle is twice the minimum absolute temperature T$_L$ in the cycle

T$_H$  = 0.5T$_L$

now, we find the efficiency of the Carnot cycle engine

η$_{th$ = 1 - T$_L$/T$_H$

η$_{th$ = 1 - T$_L$/0.5T$_L$

η$_{th$ = 0.5

the efficiency of the Carnot heat engine can be expressed as;

η$_{th$ = 1 - W$_{net$/Q$_H$

where W$_{net$ is net work done, Q$_H$ is is the heat supplied

we substitute

0.5 = 60 / Q$_H$

Q$_H$ = 60 / 0.5

Q$_H$ = 120 kJ

Now, we apply the first law of thermodynamics to the system

W$_{net$ = Q$_H$ - Q$_L$

60 = 120 -  Q$_L$

Q$_L$ = 60 kJ

now, the amount of heat rejection per kg of steam is;

q$_L$ = Q$_L$/m

we substitute

q$_L$ = 60/0.025

q$_L$ = 2400 kJ/kg

which means for 1 kilogram of conversion of saturated vapor to saturated liquid , it takes 2400 kJ/kg of heat ( enthalpy of vaporization)

q$_L$ = h$_{fg$ = 2400 kJ/kg

now, at  h$_{fg$  = 2400 kJ/kg from saturated water tables;

T$_L$ = 40 + ( 45 - 40 ) ( $\frac{2400-2406.0}{2394.0-2406.0}\\}$ )

T$_L$ = 40 + (5) × (0.5)

T$_L$ = 40 + 2.5

T$_L$ = 42.5°C  

Therefore, The temperature of the steam during the heat rejection process is 42.5°C