Question #277701

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



1
Expert's answer
2021-12-09T15:48:21-0500

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:

u1=2679.6  kJ/kgu_1 = 2679.6 \;kJ/kg

Internal energy at state 2:

u2=uf+x×ufg=906.44+0.3×1693.8=1414.58  kJ/kgu_2 = u_f + x \times u_{fg} \\ = 906.44 + 0.3 \times 1693.8 \\ = 1414.58 \; kJ/kg

Heat rejected:

Q=u2u1=1414.582679.6=1265.02  kJ/kgQ = u_2 -u_1 \\ = 1414.58 -2679.6 \\ = -1265.02 \;kJ/kg


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