Answer:
(a) 0.0064 kg/s
(b) 800 KPa
(c) 2.03
Explanation:
The ideal vapor compression cycle consists of following processes:
Process 1-2 Isentropic compression in a compressor
Process 2-3 Constant-pressure heat rejection in a condenser
Process 3-4 Throttling in an expansion device
Process 4-1 Constant-pressure heat absorption in an evaporator
For state 4 (while entering compressor):
x₄ = 34% = 0.34
P₄ = 120 KPa
from saturated table:
h₄ = hf + x hfg = 22.4 KJ/kg + (0.34)(214.52 KJ/kg)
h₄ = 95.34 KJ/kg
For State 1 (Entering Compressor):
h₁ = hg at 120 KPa
h₁ = 236.99 KJ/kg
s₁ = sg at 120 KPa = 0.94789 KJ/kg.k
For State 3 (Entering Expansion Valve)
Since 3 - 4 is an isenthalpic process.
Therefore,
h₃ = h₄ = 95.34 KJ/kg
Since this state lies at liquid side of saturation line, therefore, h₃ must be hf. Hence from saturation table we find the pressure by interpolation.
P₃ = 800 KPa
For State 2 (Leaving Compressor)
Since, process 2-3 is at constant pressure. Therefore,
P₂ = P₃ = 800 KPa
T₂ = 70°C (given)
Saturation temperature at 800 KPa is 31.31°C, which is less than T₂. Thus, this is super heated state. From super heated property table:
h₂ = 306.9 KJ/kg
(a)
Compressor Power = m(h₂ - h₁)
where,
m = mass flow rate of refrigerant.
m = Compressor Power/(h₂ - h₁)
m = (0.450 KJ/s)/(306.9 KJ/kg - 236.99 KJ/kg)
m = 0.0064 kg/s
(b)
Condenser Pressure = P₂ = P₃ = 800 KPa
(c)
The COP of ideal vapor compression cycle is given as:
COP = (h₁ - h₄)/(h₂ - h₁)
COP = (236.99 - 95.34)/(306.9 - 236.99)
COP = 2.03
The Ph diagram is attached
