Consider an ideal gas-turbine cycle with two stages of compression and two stages of expansion. The pressure ratio across each stage of the compressor and turbine is 3. The air enters each stage of compressor at 300 K and each stage of the turbine at 1200 K. Determine the back-work ratio and the thermal efficiency of the cycle, assuming a) no regenerator is used, and b) a regenerator with 75% effectiveness is used. Use variable specific heats.

A)Back work ratio = 33.51%

Thermal Efficiency = 36.79%

B) Back work ratio = 33.51%

Thermal efficiency = 55.32%

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%

whichever one it is, it's definitely not one that you included on the list you provided.