Of an ideal bravton refrigeration cycle at 1 atm and 270 k...

Of an ideal Brayton refrigeration cycle at 1 atm and 270 K, with air entering the compressor at a volumetric flow rate of 1.5 m³/s. If the compressor pressure ratio is 3 and the turbine inlet temperature is 300 K, the net power input required for the cycle is: assuming the turbine and compressor are 100% efficient. Assume specific heat ratio (γ) = 1.4; R = 0.287 kJ/kgK. 196.98 kW (b) 158.86 kW 36.91 kW 29.18 kW

Transcript

00:01 Here in this question we are given the idea vibrator cycle which is here in this way this cycle is moving in this way let's say this is one this is two this point is three and this point is four and this is let's say four dash and this point is let's say two dash so from here this is moving in this way here wc force is acting in var and and w r is acting downward here this is pressure p 2 this is pressure p 1 kv2 is acting involved in this direction so here we are given the value of p which is 1 a tm we are given the value of t 1 that is 270 kelvin we are given the value of q which is equal to 1 .4 meter cube per second so from here we can say that m dot will be equals to s multiply by a kov dot that will be equals to 1 .2 multiply by a 1 .4 that will comes out to be 1 .68 kilogram per second now from here nita e is equal to nita t that will be equals to 0 .8 so from here gamma p will be equals to p2 divided by the p1 that will come up to be 3 so from here we are given the value of t3 that is equals to 300 kelvin for air and gamma is given which is 1 .4 cp is given which is 1 .005 kilogram per kilo meter kelvin so from here we can say that the value of t2 divided by the t will be equals to p2 t1 p2 divided by the p1 d s 2 divided by the p 1 ds to the power gamma minus 4 divided by the gamma so from here t 2 dash will be equals to t 1 gamma p gamma minus 1 divided by the gamma and from here we get the value of t 2 dash is 369 .56 kelvin so from here now we can set up compression efficiency which is netta c will be equals to ideal work input divided by the actual work input that will be equals to cp multiply by the t2 dash minus t1 divided by the cp t2 minus t1 so putting the values here this will come t2 will come t2 will come up to be 394 .45 kelvin and so this is the value of t2 here and now for the process 3 to 4 which is isoentropic compression process using isotropic relation t3 divided by a t 4 dash will be equals to p2 divided by the p1, raised to the power gamma minus 1 divided by the gamma...

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