Answer to Question #277701 in Molecular Physics | Thermodynamics for yuvi

Question

Steam at 2 MPa and 250°C in a rigid cylinder is cooled until the quality is 30%. Find the heat rejected from the cylinder.

@ 2 Mpa and 250°C: υ = 0.11144 m3 /kg

                   u = 2679.6 kJ/kg

 @ 2 Mpa, (saturated):υf = 0.0011767 m3 /kg

                    υg = 0.09963 m3 /kg 

                    uf = 906.44kJ/kg 

                    ug = 1693.8 kJ/kg

Answer 1

As the cylinder is rigid, therefore it's volume remains constant at initial and final states. The calculation part is given in the next step:

Internal energy at state 1:

$u_1 = 2679.6 \;kJ/kg$

Internal energy at state 2:

$u_2 = u_f + x \times u_{fg} \\ = 906.44 + 0.3 \times 1693.8 \\ = 1414.58 \; kJ/kg$

Heat rejected:

$Q = u_2 -u_1 \\ = 1414.58 -2679.6 \\ = -1265.02 \;kJ/kg$