2500 kg of water at ambient condition of 1 atm and 25 degrees celsius (u1= 105 k

Question

2500 kg of water at ambient condition of 1 atm and 25 degrees celsius (u1= 105 kJ/kg, h1=105 kJ/kg, s1=0.367 kJ/kg-K) is raised to a temperature of 90 degrees celsius (u2=377 kJ/kg, h2=377 kJ/kg, s2=1.193 kJ/kg-K). The process occurs at a constant pressure and is accomplished using an electric resistance coil in a well-insulated container. Define the water and resistance coil as system, and assume the electric coil has negligible thermal capacity. Determine: a) the heat flow to the system in MJ, b) the shaft work done by the system in MJ, c) the increase in entropy of the system in MJ/K, d) the entropy generated in the process in MJ/K e) the lost work in MJ. Give full details of work to receive good rating.

Explanation / Answer

a) heat flow = m(h2 - h1) = 2500*(377-105) = 680 MJ

b) shaft work = 0

c) increase in entropy = m(s2-s1) = 2500*(1.193-0.367) = 2.065 MJ/K

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