Respuesta :
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.
[tex]h_{1}[/tex] = 247.88 kJ/kg, [tex]S_{1}[/tex] = 0.9579 kJ/kg K
At state 2, isentropic enthalpy will be determined from the condition [tex]S_{2} = S_{1}[/tex] and given pressure at 2 with data from A-13 using interpolation is:
[tex]h_{2s}[/tex] = 279.45 kJ/kg
We will calculate actual enthalpy at state 2 using given pressure and temperature from A-13 as follows.
[tex]h_{2}[/tex] = 286.71 kJ/kg
Hence, isentropic compressor efficiency will be calculated using standard relation as:
[tex]\eta_{c} = \frac{h_{2s} - h_{1}}{h_{2} - h_{1}}[/tex]
= [tex]\frac{279.45 - 247.88}{286.71 - 247.88}[/tex]
= 0.813
Now, at state 3 enthaply is determined by temperature at state 3, that is, [tex]26^{o}C[/tex] for given pressure as per saturated liquid approximation and data from A-11.
[tex]h_{3}[/tex] = 87.83 kJ/Kg
Using energy balance in 2-3, the rate of heat supplied to the heated room is as follows.
[tex]Q_{H} = m(h_{2} - h_{3})[/tex]
= 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 = [tex]\frac{Q_{H}}{W}[/tex]
= [tex]\frac{Q_{H}}{m(h_{2} - h_{1})}[/tex]
= [tex]\frac{4.38}{0.022 (286.71 - 246.88)}[/tex]
= 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:
[tex]h_{1}[/tex] = 244.5 kJ/kg
[tex]S_{1}[/tex] = 0.93788 kJ/kg K
[tex]h_{2}[/tex] = 273.71 kJ/kg
At state 3, enthalpy will be determined from given pressure and state with data from A-12 as follows.
[tex]h_{3}[/tex] = 95.48 kJ/kg
Hence, using energy balance in 2-3 the rate of heat supplied will be calculated as follows.
[tex]Q_{H} = m(h_{2} - h_{3})[/tex]
= 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 = [tex]\frac{Q_{H}}{W}[/tex]
= [tex]\frac{Q_{H}}{m (h_{2} - h_{1})}[/tex]
= [tex]\frac{3.91}{0.022(273.31 - 244.5)}[/tex]
= 6.17