An adiabatic system operates at steady state 4 kgs of steam

An adiabatic system operates at steady state. 4 kg/s of steam at p = 4 bar and T = 280 C enters the system and is expanded isothermally so that it exits as saturated steam. Find the work.

Solution

>> Now, as Temprature remains constant, and final state is saturated steam

So, Using steam table at temperature T = 280 C = 553 K, and at saturation,

>> P2 = 6.402 MPa = 64.02 bar

and, h2 = 2779.9958 KJ/Kg

>> Now, Initially, T1 = 280 C = 553 K and P1 = 4 bar

So, using steam table at these two conditions,

=> h1 = 3025.7919 KJ/Kg

As, process is isothermal. So, There is no internal energy change

>> So, Work done = Heat Transfer = m*(h2 - h1) = 4*(2779.9958 - 3025.7919) = - 983.18 KJ ....ANSWER.......