Steam undergoes an adiabatic expansion in a piston–cylinder assembly from 100 bar, 360°c to 1 bar, 160°c. what is work in kj per kg of steam for the process? calculate the amount of entropy produced, in kj/k per kg of steam. what is the magnitude of the maximum theoretical work that could be obtained from the given initial state to the same final pressure?

work is 130.5 kJ/kg

entropy change is 1.655 kJ/kg-k

maximum  theoretical work is 689.4 kJ/kg

Explanation:

piston cylinder assembly

100 bar, 360°C to 1 bar, 160°C

to find out

work  and amount of entropy  and magnitude

solution

first we calculate work i.e heat transfer - work =   specific internal energy @1 bar, 160°C  - specific internal energy @ 100 bar, 360°C    1

so first we get some value from steam table with the help of 100 bar @360°C and  1 bar @ 160°C

specific volume = 0.0233 m³/kg

specific enthalpy = 2961 kJ/kg

specific internal energy = 2728 kJ/kg

specific entropy = 6.004 kJ/kg-k

and respectively

specific volume = 1.9838 m³/kg

specific enthalpy = 2795.8 kJ/kg

specific internal energy = 2597.5 kJ/kg

specific entropy = 7.659 kJ/kg-k

now from equation 1 we know heat transfer q = 0

so - w =   specific internal energy @1 bar, 160°C  - specific internal energy @ 100 bar, 360°C

work = 2728 - 2597.5

work is 130.5 kJ/kg

and entropy change formula is i.e.

entropy change =  specific entropy ( 100 bar @360°C)  - specific entropy ( 1 bar @160°C )

put these value we get

entropy change =  7.659 - 6.004

entropy change is 1.655 kJ/kg-k

and we know maximum  theoretical work = isentropic work

from steam table we know specific internal energy is 2038.3 kJ/kg

maximum  theoretical work = specific internal energy - 2038.3

maximum  theoretical work = 2728 - 2038.3

maximum  theoretical work is 689.4 kJ/kg

the answer is b

explanation: