Question

Refrigerant-134a enters the condenser of a residential heat pump at 800 kPa and 50°C at a rate of 0.02 kg/s and leaves at 750 kPa subcooled by 3°C. The refrigerant enters the compressor at 200 kPa superheated by 4°C. Determine the isentropic efficiency of the compressor, the rate of heat supplied to the room, COP of the Heat Pump and the rate of heat supplied to this room if the heat pump operated on an ideal vapor compression cycle between pressure limits of 200 and 800 kpa

Answer 1

Step-by-step explanation:

The value of enthalpy and entropy at state 1 will be determined according to the given pressure and temperature as follows using interpolation from A-13 is as follows.

`h_(1)` = 247.88 kJ/kg,
`S_(1)` = 0.9579 kJ/kg K

At state 2, isentropic enthalpy will be determined from the condition
`S_(2) = S_(1)` and given pressure at 2 with data from A-13 using interpolation is:

`h_(2s)` = 279.45 kJ/kg

We will calculate actual enthalpy at state 2 using given pressure and temperature from A-13 as follows.

`h_(2)` = 286.71 kJ/kg

Hence, isentropic compressor efficiency will be calculated using standard relation as:

`\eta_(c) = (h_(2s) - h_(1))/(h_(2) - h_(1))`

=
`(279.45 - 247.88)/(286.71 - 247.88)`

= 0.813

Now, at state 3 enthaply is determined by temperature at state 3, that is,
`26^(o)C` for given pressure as per saturated liquid approximation and data from A-11.

`h_(3)` = 87.83 kJ/Kg

Using energy balance in 2-3, the rate of heat supplied to the heated room is as follows.

`Q_(H) = m(h_(2) - h_(3))`

= 0.022 (286.71 - 87.83) kW

= 4.38 kW

Now, COP will be calculated using power that is expressed through energy balance in 1-2 as follows.

COP =
`(Q_(H))/(W)`

=
`(Q_(H))/(m(h_(2) - h_(1)))`

=
`(4.38)/(0.022 (286.71 - 246.88))`

= 5.13

In an ideal vapour-compression cycle, the enthalpy and entropy at state 1 will be obtained from given pressure and state with data from A-12:

`h_(1)` = 244.5 kJ/kg

`S_(1)` = 0.93788 kJ/kg K

`h_(2)` = 273.71 kJ/kg

At state 3, enthalpy will be determined from given pressure and state with data from A-12 as follows.

`h_(3)` = 95.48 kJ/kg

Hence, using energy balance in 2-3 the rate of heat supplied will be calculated as follows.

`Q_(H) = m(h_(2) - h_(3))`

= 0.022 (273.31 - 95.48) kW

= 3.91 kW

The power input which is expressed through energy balance in 1-2 will be used to determine COP as follows.

COP =
`(Q_(H))/(W)`

=
`(Q_(H))/(m (h_(2) - h_(1)))`

=
`(3.91)/(0.022(273.31 - 244.5))`

= 6.17