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Consider a Carnot heat pump cycle executed in a steady-flow system in the saturated mixture region using R-134a flowing at a rate of 0.264 kg/s. The maximum absolute temperature in the cycle is 1.15 times the minimum absolute temperature, and the net power input to the cycle is 5 kW. If the refrigerant changes from saturated vapor to saturated liquid during the heat rejection process, determine the ratio of the maximum to minimum pressures in the cycle.

Respuesta :

Limosa

Answer:

7.15

Explanation:

Firstly, the COP of such heat pump must be measured that is,

              [tex]COP_{HP}=\frac{T_H}{T_H-T_L}[/tex]

Therefore, the temperature relationship, [tex]T_H=1.15\;T_L[/tex]

Then, we should apply the values in the COP.

                           [tex]=\frac{1.15\;T_L}{1.15-1}[/tex]

                           [tex]=7.67[/tex]

The number of heat rejected by the heat pump must then be calculated.

                   [tex]Q_H=COP_{HP}\times W_{nst}[/tex]

                          [tex]=7.67\times5=38.35[/tex]

We must then calculate the refrigerant mass flow rate.

                   [tex]m=0.264\;kg/s[/tex]

                   [tex]q_H=\frac{Q_H}{m}[/tex]

                         [tex]=\frac{38.35}{0.264}=145.27[/tex]

The [tex]h_g[/tex] value is 145.27 and therefore the hot reservoir temperature is 64° C.

The pressure at 64 ° C is thus 1849.36 kPa by interpolation.

And, the lowest reservoir temperature must be calculated.

                   [tex]T_L=\frac{T_H}{1.15}[/tex]

                        [tex]=\frac{64+273}{1.15}=293.04[/tex]

                        [tex]=19.89\°C[/tex]

the lowest reservoir temperature = 258.703  kpa                    

So, the pressure ratio should be = 7.15

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