Saturated liquid water flows steadily into a well-insulated electrical water heater (see Anim. 4-1-1) with a mass flow rate (m⋅) of 1 kg/s at 100 kPa. Determine (a) the electrical power consumption (W⋅el), and (b) the rate of entropy generation (S⋅gen) in the water heater's universe if the heater turns water into saturated vapor at the exit. Assume no pressure loss, neglect changes in ke and pe, and use the PC model. The ambient atmospheric conditions are 100 kPa and 20oC.

Question

Cristofer

Chemistry

31 October, 14:37

Saturated liquid water flows steadily into a well-insulated electrical water heater (see Anim. 4-1-1) with a mass flow rate (m⋅) of 1 kg/s at 100 kPa. Determine (a) the electrical power consumption (W⋅el), and (b) the rate of entropy generation (S⋅gen) in the water heater's universe if the heater turns water into saturated vapor at the exit. Assume no pressure loss, neglect changes in ke and pe, and use the PC model. The ambient atmospheric conditions are 100 kPa and 20oC.

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Answers (1)

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Hazel Nunez · Answer 1

a. 2257.7 kW b. 6.057kw/K

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