Numerical design (Mathematical modelling) of a compressor for refrigeration and air conditioning(cylinder volume calculation)

Compressor is the heart of any refrigerating equipment as it pump the refrigerant in to the all components like our heart supply the blood throughout the body parts. Compressor decide the cooling capacity and power consumption for refrigerating machine.
Its main work is to take the refrigerant from suction pipe at very low pressure(temperature as well) and compress it to a very high pressure (discharge pressure).
As we all know that refrigerant condense in condenser at constant temperature (for zeotropic blends it vary), and that temperature is corresponding to the discharge pressure of refrigerant.
Discharge pressure of refrigerant is such that its corresponding temperature should be higher than the ambient temperature only than refrigerant will condense in to liquid if the discharge saturation temperature(corresponding to discharge pressure) is lower than the ambient temperature, than refrigerant will not condense.

All compressors can be described on their application(use) as per below:

(i) Low Back Pressure (LBP) (Evaporator temperature (-23.3 deg C)
(ii) Medium or commercial back pressure (MBP or CBP) (Evaporator temperature -6.7 deg C)
(iii) High back pressure (HBP) (Evaporator temperature 7.2 deg C)

Let us design the compressor on the basis of medium back pressure (MBP):

Assumption: (i) Assume pressure loss is negligible in connecting tubes and components.
                     (ii) Assume the compression process is an isentropic process. Entropy at suction and discharge are equal.
           

Test conditions for ASHRAE MBP are as per below:

Evaporator temperature: -6.7 deg C
Condensing temperature: 54.4 deg C
Liquid sub cool temperature: 32.3 deg C
Suction temperature: 32.2 deg C
Cooling capacity required: 300 W (We can assume any value, it is just for example)
Compressor RPM = 3000 (Any value can be selected)
Refrigerant = R600a


We have to calculate the compressor cylinder volume.

The refrigeration cycle for above test conditions is shown as per below:

Refrigeration cycle is 1-2-3-4-5-5a-6-1.

Step1: Make a PH chart using above test conditions.

Step2: Calculate the enthalpies at 6 and 2.

enthalpy of refrigerant entering in to evaporator = h6
enthalpy of refrigerant leaving the evaporator = h2

we can also calculate the amount of vapor present in to the refrigerant at the entry of evaporator as below:

enthalpy at condenser outlet = h5 = h6 (adibatic process in capillary, 5-6 expansion in capillary)

enthalpy of liquid refrigerant at evaporator temperature = hle
enthalpy of vapor refrigerant at evaporator temperature = h1

dryness fraction(x) = (h6-hle)/(h1-hle)------------------------------(i)

in current example, hle = 184.7 kJ/kg
                                h1 = 546.1 kJ/kg
                                h6 = 277.4 kJ/kg

so using equation 1,

x= (277.4-184.7)/(546.1-184.7) = 0.256 or 25.6 %

Step3: calculate the mass flow rate of refrigerant required to obtain the cooling capacity.

m_ref= mass flow rate of refrigerant

m_ref = cooling capacity/((h2-h6)*1000)---------------------------(ii)

where:  mass flow rate of refrigerant is in kg/s
              h2 and h6 are in kJ/kg
             cooling capacity is in W

for current example, cooling capacity = 300 W
                                                       h2  = 610.5 kJ/kg
                                                       h6 = 277.4 KJ/kg

so using equation 2,

m_ref = 300/((610.5-277.4)*1000) = 0.0009 kg/s = 3.24 kg/hr

Step4: Calculate specific volume of refrigerant at suction temperature using PH chart.

for current example it is, v_s = 0.3456 m3/kg

Compressor speed = N = 3000 rpm
Compressor speed in revolution per second = N/60 = 3000/60 = 50 rps

Step5: Calculate the volume required for 300 W cooling capacity at Medium back pressure.

m_ref = V*N/v_s----------------------------------------------(iii)

Here,  m_ref = mass flow of refrigerant in kg/s
           V= compressor cylinder volume
           N = compressor motor speed in revolution per second,
           v_s = specific volume of refrigerant at suction

so by reshuffling equation (iii)

V = m_ref*v_s/N--------------------------------------------------------(iv)

By putting the values of m_ref, N & v_s in equation (iv), we can calculate the compressor cylinder required as per below:

V = 0.0009*0.3456/50 = 0000062208 m3 = 6.2208 cc (because 1m3 = 1000000 cc)

We can calculate the power consumed by compressor as per below:

P_c = m_ref*(h3-h2)*1000--------------------------------------------(v)

where, P_c = power consumed by compressor in W
            m_ref = mass flow rate of refrigerant in kg/s
              h3 = enthalpy of refrigerant at discharge in kJ/kg (691.3 kJ/kg in current example)
              h2 = enthalpy of refrigerant at suction in kJ/kg

Putting the values in equation (v), we get

P_c = 0.0009*(691.3-610.5)*1000 = 72.72 W

Difference from actual: Above power consumption is calculated by assuming isentropic compression of refrigerant (entropy at suction and discharge of compressor is constant), But in actual there is always an increase in entropy for a successful completion of any process as per second law of Thermodynamics. Isentropic efficiency of compressor which can be calculated as per below:

Isentropic efficiency = 0.874-0.0135*(P_d/P_s)-------------------------(vi)

Where, P_d = discharge pressure(761.4 kPa in current example)
              P_s = suction pressure (122.6 kPa assuming no pressure loss in suction pipe)

So, Isentropic efficiency using equation (vi) = 79%

So actual discharge enthalpy = h2+(h3-h2)/Isentropic efficiency------(vii)

Actual discharge enthalpy = 610.5+(691.3-610.5)/0.79 = 712.88 kJ/kg

So actual power consumption is replacing ideal discharge enthalpy in equation no. (v), by actual discharge enthalpy

P_c = 0.0009*(712.88-610.5)*1000 = 92.142 W

But actual power consumption is high, because motor itself convert electricity in to mechanical (rotary motion). So it has also an efficiency (winding and slip losses). 

so Actual power consumption depend upon the winding temperature. at around 90deg C of winding temperature motor has an eficiency of 40%.

Actual power consumption is = P_c/0.4

P_act = 92.142/0.4 = 230 W

So COP of compressor is = Cooling capacity/P_c-----------------------(viii)

COP = 300/230 = 1.3 W/W

EER is calculated as per below;

Cooling capacity in btu/hr = 300*12000/3500 = 1028.57 btu/hr

EER = btu/hr/P_c---------------------------------------------------(ix)

EER = 1028.57/92.142 = 11.16 Btu/Whr

If we assumed the phase factor (phai) to be unity and rated voltage is 230 V, than current consumed by compressor is given as per below:

P_c = V*I-------------------------------------------------------------(x)

where, P_c = Power consumed by compressor in W
            V = Voltage in V
             I = Current in Ampere

By reshuffling equation number (x),

I = P_c/V

I = 92.142/230 = 0.4 A