Answer:
a. 2257.7 kW b. 6.057kw/K
Step-by-step explanation:
Steam existing at 100KPa with t = 20C, we can obtain values of enthalpy and entropy from table
h1 = hf = 417.4 kJ/kg
s1 = 1.302 KJ/kg.K
h2 = hg = 2675.1 kJ/kg
s2 = 7.359 kJ/K
a. Electrical Power consumption of system is given by Wel = mass x Change in Enthalpy
Wel = 1 x (h2 - h1) = 1 x (2675.1 - 417.4) = 2257.7 kw
b . Entropy Generation is given by Change in Entropy = Entropy generation - heat dissipated/temperature, since we have a well insulated system with no losses, q =0, hence
ΔS = Sgen + 0/T = Sgen
m(S2 - S1) = Sgen
Sgen = 1 x (7.359 - 1.302) = 6.057 kW/K