Answer:
A)Back work ratio = 33.51%
Thermal Efficiency = 36.79%
B) Back work ratio = 33.51%
Thermal efficiency = 55.32%
Step-by-step explanation:
To solve this question, we have to work under the following conditions;
1. Standard air conditions
2. KE = PE = 0
3. Air is an ideal gas with variable specific heats
A) From ideal gas properties of air table A-17 I attached, we see that;
At 300K temperature,
h1 = 300.19 kJ/kg and Pr1 = 1.386
From the formula, P2/P1 = Pr2/Pr1
Let's make Pr2 the subject;
Pr2 =(P2 /P1) Pr1
From the question, the pressure ratio is 3 and so;
Pr2 = 3 x (1.386) = 4.158
Also from the table A-17, at Pr2 = 4.158, we see that, At a value of h is approximately 411.26 kj/kg and thus;
h2 = h4 = 411.26 kJ/kg
Still from the same table, at T5 = T7 = 1200 K
h5 = h7 = 1277.79 kJ/kg
and Pr5 = 238.0
So using the pressure ratio formula,
Pr6/Pr5 = P6/P5
So, Pr6 = (P6/P5) Pr5
Ratio of P5 to P6 is 3, thus, ratio of P6/P5 is 1/3 and ;
Pr6 = (1/3)238 = 79.33
At Pr of 79.33,on the same table, we have h approximately as 946.36
Thus; h6 = h8 = 946.36 kJ/kg
Now, the work into the comperssor is given by ;
Wc,in = 2(h2 − h1)
Plugging in the relevant values, we obtain;
Wc,in = 2(411.26 − 300.19) = 2(111.07)
= 222.14 kJ/kg
The work out of the turbine is given by;
wt,out = 2(h5 − h6)
= 2(1277.79 − 944.36) = 2(333.43)
662.86 kJ/kg
Now, let's find the back work ratio. It is given by a formula;
rbw = wc,in/wt,out
So rbw = 222.14/662.86 = 0.3351 or simply, 33.51%
Now,for the thermal efficiency ;
The energy into the cycle is expressed as;
qin = (h5 − h4)+(h7 − h6)
And thus;
qin = (1277.79 − 411.26) + (1277.79 − 946.36) = 866.53 + 331.43 = 1197.96 kJ/kg
Efficiency is given by the formula ;
η = wnet/qin
Where wnet = wt,out - wc,in
So, wnet = 662.86 - 222.14 = 440.72 Kj/kg
Thus ;
η = 440.72/1197.96 = 0.3679 or 36.79%
B) In this second part,since a regenerator is in place, the work of the turbine stages and the compressor stages remains the same, and thus the back work ratio is also the same while the energy input will be different.
So, back ratio ;
rbw = wc,in/wt,out
= 222.14/662.86 = 0.3351 or simply, 33.51%
And energy input ;
qregen =ε(h8 − h4) where ε is the effectiveness which is 0.75 from the question
Plugging in the relevant values to obtain;
=0.75(946.36 − 411.26) kJ/kg = 0.75 x 535.1 = 401.325 kJ/kg
Now, new qin is given by the formula;
qin = qin,old − qregen
Thus, plugging in values,
qin = (1197.96 − 401.325) kJ/kg = 796.635 kJ/kg
So now to calculate efficiency ;
η = wnet/qin
Where, wnet = wt,out - wc,in
So, wnet = 662.86 - 222.14 = 440.72 Kj/kg
η = 440.72/796.635 = 0.5532 or 55.32%