Question

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.

Answer 1

Answer:

7.15

Explanation:

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

              $COP_{HP}=\frac{T_H}{T_H-T_L}$

Therefore, the temperature relationship, $T_H=1.15\;T_L$

Then, we should apply the values in the COP.

                           $=\frac{1.15\;T_L}{1.15-1}$

                           $=7.67$

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

                   $Q_H=COP_{HP}\times W_{nst}$

                          $=7.67\times5=38.35$

We must then calculate the refrigerant mass flow rate.

                   $m=0.264\;kg/s$

                   $q_H=\frac{Q_H}{m}$

                         $=\frac{38.35}{0.264}=145.27$

The $h_g$ 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.

                   $T_L=\frac{T_H}{1.15}$

                        $=\frac{64+273}{1.15}=293.04$

                        $=19.89\°C$

the lowest reservoir temperature = 258.703  kpa                    

So, the pressure ratio should be = 7.15