In: Mechanical Engineering
A 4-cylinder ammonia compressor with a bore 0.04 m & stroke
0.03 m running 2500. The system operates in a simple VCC with
evaporating & condensing temperatures zero C & 40 C,
respectively generating cooling capacity 5 tons.
Determine:
a. Draw the PH diagram indicating all enthalpies & sp.
vol.
b. Piston displacement, m3/sec
c. Suction Volume, m3/sec
d. Volumetric efficiency
e. Mass of refrigerant, kg/sec
f. Work of compressor, kW
We will assume that there is no superheating or subcooling. Using this, let us draw the cycle on the Ph diagram. Please note that due to legal issues, I am not allowed to post an actual diagram from the internet. However, you can easily cross-verify from any random chart available.
From the pH chart for ammonia, the properties at the 4 points are obtained as follows
Point | Enthalpy (kJ/kg) | Specific Volume (m3/kg) |
1 | 1462 | 0.289 |
2 | 1630 | 0.11 |
3 | 390.6 | 0.001725 |
4 | 390.6 | 0.05 |
From here refrigeration capacity per unit mass of refrigerant = h1 - h4 = 1462 - 390.6 = 1071.4 kJ/kg
Thus mass flow rate of refrigerant m = 5*3.5 / 1071.4 = 0.016 kg/s
Thus Compressor work is m*(h2-h1) = 0.016*(1630-1462) = 0.016 * 168 = 2.69 kW
Now for the compressor it is given
The piston displacement in each stroke is the swept volume. Thus swept volume is obtained as
Suction volume is the volume of the refrigerant at intake. Thus suction volume is m*v1 = 0.016*0.289 = 0.00462 m3/s
The volumetric efficiency of the compressor is the ratio of the volume it is actually taking in, and the maximum volume it can intake. The volume it is actually intaking is the suction volume. The maximum volume it can intake is the swept volume, or piston displacement volume. Thus volumetric efficieny of the compressor is 0.00462/0.00628 = 0.736 = 73.6 %
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