Answer:
A)W'/m = 311 KJ/kg
B)σ'_gen/m = 0.9113 KJ/kg.k
Explanation:
a).The energy rate balance equation in the control volume is given by the formula;
Q' - W' + m(h1 - h2) = 0
Dividing through by m, we have;
(Q'/m) - (W'/m) + (h1 - h2) = 0
Rearranging, we have;
W'/m = (Q'/m) + (h1 - h2)
Normally, this transforms to another equation;
W'/m = (Q'/m) + c_p(T1 - T2)
Where;
W'/m is the rate at which power is developed
Q'/m is the rate at which heat is flowing
c_p is specific heat at constant pressure which from tables at a temperature of 980k = 1.1 KJ/kg.k
T1 is initial temperature
T2 is exit temperature
We are given;
Q'/m = -30 kj/kg (negative because it leaves the turbine)
T1 = 980 k
T2 = 670 k
Plugging in the relevant values;
W'/m = -30 + 1.1(980 - 670)
W'/m = 311 KJ/kg
B) The Entropy produced from the entropy balance equation in a control volume is given by the formula;
(Q'/T_boundary) + m(s1 - s2) + σ'_gen = 0
Dividing through by m gives;
((Q'/m)/T_boundary) + (s1 - s2) + σ'_gen/m = 0
Rearranging, we have;
σ'_gen/m = -((Q'/m)/T_boundary) + (s2 - s1)
Under the conditions given in the question, this transforms normally to;
σ'_gen/m = -((Q'/m)/T_boundary) - c_p•In(T2/T1) - R•In(p2/p1)
σ'_gen/m is the rate of entropy production in kj/kg
We are given;
p2 = 100 kpa
p1 = 400 kpa
T_boundary = 315 K
For an ideal gas, R = 0.287 KJ/kg.K
Plugging in the relevant values including the ones initially written in answer a above, we have;
σ'_gen/m = -(-30/315) - 1.1(In(670/980)) - 0.287(In(100/400))
σ'_gen/m = 0.0952 + 0.4183 + 0.3979
σ'_gen/m = 0.9113 KJ/kg.k
