Question

Liquid water at 325 K and 8000 kPa flows into a boiler at a rate of 10 kg⋅s− 1 and is vaporized, producing saturated vapor at 8000 kPa. What is the maximum fraction of the heat added to the water in the boiler that can be converted into work in a process whose product is water at the initial conditions, if Tσ = 300k? what happens to the rest of the heat?

What is the rate of entropy change in the surroundings as a resuit ot the work- producing process? in the system?Total?

Answer 1

0.4058

Step-by-step explanation:

From the steam table

At 325 K, saturated liquid

Enthalpy, Hf = 217 kJ/kg

Entropy Sf = 0.7274 kJ/kg-K

Specific volume Vf = 1.013 cm3/gm

Saturated pressure Psat = 12.87 kPa

For compressed liquid

P1 = 8000 kPa

Temperature T = 325 K

Thermal expansion coefficient B = 460 x 10^{-6} K^{-1}

Enthalpy at initial conditions

H1 = Hf + Vf (1-BT)(P1 - P_sat)

= 217 + 1.013*10^-3 (1 - 460*10^{-6})(8000 - 12.87)

= 223.881 kJ/kg

Entropy at initial conditions

S1 = Sf - BVf (P1 - Psat)

= 0.7274 - 460×10^{-6}*1.013*10^-3 (8000 - 12.87)

= 0.724 kJ/kg-K

At 8000 kPa, saturated vapor

H2 = 2759.9 kJ/kg

S2 = 5.7471 kJ/kg-K

T = 300 K

Heat added Q = H_2 - H_1

= 2759.9 - 223.881

= 2536 kJ/kg

Maximum work

W = (H1 - H2) - T (S1 - S2)

= 223.881 - 2759.9 - 300(0.724 - 5.7471)

= - 1029 kJ/kg

Fraction of heat added = W/Q

= 1029/2536

= 0.4058