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Consider an ideal gas refrigeration cycle using helium as the working fluid. Helium enters the compressor at 120 kPa and 21°C and compressed to 440 kPa. Helium is then cooled to 16°C before it enters the turbine. Assume constant heat capacity of helium. For a mass flow rate of 0.24 kg/s, the net power input required is:

Answer :

Answer:[tex]W_{net}=103.72 kJ/s[/tex]

Explanation:

Given

[tex]P_1=120kPa[/tex]

[tex]T_1=21^{\circ}C\approx 294K[/tex]

[tex]P_2=440kPa[/tex]

[tex]T_3=16^{\circ}C\approx 289 K[/tex]

[tex]T_2[/tex] is the compressor exit temperature and is given by

[tex]\gamma [/tex] for helium is [tex]\frac{5}{3}[/tex]

[tex]C_p[/tex] for helium is [tex]5.1926 kJ/kg-K[/tex]

[tex]T_2=T_1\left [ \frac{440}{120}\right ]^\frac{\gamma -1}{\gamma }[/tex]

[tex]T_2=494.37 K[/tex]

Now Turbine Exit Pressure is given By

[tex]T_4=T_3\left [ \frac{120}{440}\right ]^\frac{\gamma -1}{\gamma }[/tex]

[tex]T_4=289\left [ \frac{120}{440}\right ]^\frac{\frac{5}{3} -1}{\frac{5}{3}}[/tex]

[tex]T_4=171.865 K[/tex]

[tex]W_{net}=mC\left ( \left ( T_2-T_1\right )-\left ( T_3-T_4\right )\right )[/tex]

[tex]W_{net}=0.24\times 5.1926\left ( \left ( 494.37-294\right )-\left ( 289-171.865\right )\right )[/tex]

[tex]W_{net}=0.24\times 5.1926\times 83.235[/tex]

[tex]W_{net}=103.72 kJ/s[/tex]

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