Respuesta :
Explanation:
The given data is as follows.
MW = 29 g/mol, [tex]C_{p}[/tex] = 3.5 R
Formula to calculate minimum amount of work is as follows.
[tex]W_{s} = C_{p}T_{1}[(\frac{P_{2}}{P_{1}})^{\frac{R}{C_{p}}} - 1][/tex]
= [tex]3.5 \times 8.314 J/k mol \times 306 \times [(\frac{550}{125})^{\frac{1}{3.5}} - 1][/tex]
= 4.692 kJ/mol
Therefore, total work done will be calculated as follows.
Total work done = [tex]m \times W_{s}[/tex]
Since, m = [tex]\frac{150 \times 10^{3}g/min}{29}[/tex]. Therefore, putting these values into the above formula as follows.
Total work done = [tex]m \times W_{s}[/tex]
= [tex]\frac{150 \times 10^{3}g/min}{29} \times 4.692 kJ/min[/tex]
= 24268.96 kJ/min
It is known that 1 kJ/min = 0.0166 kW. Hence, convert 24268.96 kJ/min into kW as follows.
[tex]24268.96 kJ/min \times \frac{0.0166 kW}{1 kJ/min}[/tex]
= 402.86 kW
Thus, we can conclude that the minimum work required by an air compressor is 402.86 kW.
