Water, initially saturated vapor at 4 bar, fills a closed, rigid container. the water is heated until its temperature is 400°c. for the water, determine the heat transfer, in kj per kg of water. kinetic and potential energy effects can be ignored.

q=471.19kj.kg

Explanation:water constant volume = C

First state

Quantity X1=1

Pressure=1 bar

Second state

T2=400C

Required heat transferred q in kj.kg

Assumption:

System is at equilibrium

Kinetic and Potential energy are neglected

Therefore, from 1st law of thermodynamics q - w=∆u. . . . 1

From steam table A

U1=Ug=2553.6kj.kg

U2=Vg=1625m³.kg

V1=V2=4625m³.kg

From steam Table B

Px= 7 bar

Vx=4667kj.kg

Ux=3026.6kj.kg

Py= 10 bar

Vy= 3066kj.kg

Uy=2957.5kj.kg

Assuming a Linear interpolation

Ux-V2/Ux-Vy=Vx-U2/Ux-Vy

U2 =Vx-{vx-vy}× vx-v2/Ux-Vy

U2 = 3026.6 - (3026.6-2957.3)× 4667-4625/4667-3066

=3024.79kj.kg

Therefore, from q-w=∆u

q=3024.79-253.6

=471.19kj.kg

Where u = internal energy

V= specific volume