Question

In a vapor compression refrigeration system the following observations were noted. Condenser pressure 2 bar. Evaporator pressure 0.02 MPa. Sub-cooling 10°C. Super heating 20°C. Estimate the power consumed by compressor, heat absorbed in evaporator, heat rejected in condenser and coefficient of performance. Mark all given data in PH and TS diagram.

R134a refrigerant

Answer 1

Before we start, a few words. It is possible that you won't find some data used here on a random R134a thermodynamic property saturation/ superheated table. That is natural because the practicality of this system is very bad (I get that this might be just a theoretical problem, but even by that standard also). No refrigeration cycle functions completely in sub-zero temperature and with almost vaccum evaporator pressure. The superheated property table I use starts from 1 bar. As such, I had to scour the internet to get a table that provided the required data. If you are in a hurry, I suggest cross-validating the data from a pH chart, it will be relatively easier.

SUBCOOL SUBWOOL 10 2 bath FNT.........0-2 tot SUPERMEAT 20°c SUPERHEAT I 2000 1-2 Evaporator with superheat 2-3 → Compressor 3-4 → Condenser with subcool 4-1 → Empanalon device We have to calculate enthalpies at the 4 vertices. This can be easily done with R134a pl chart, but we will use R13a properties table and calculations The evaporator is at 0.02MP = 0.2 bar... From table, at P = 0.2 bar, saturation temperature is - 56.54c There is 20 superheat and 1 So at point 2, P = 0.2 bar, temperature = -56.54 + 20 = -36.54% from superheated table data, we can calculate enthalpy of this point as hy = 377.5 k I lkg Also we get So= 1.8549 kJ/AK

Now asuming the compressor.is_doing is entropie.work, we.com write Sz = = 1.8549 kJ /kok. & The compressor outlet is at 2 bar = 0.2MPa. from superheated tables for R-134a, we see that at p = 0.2MPa, T=30°c, s=1&549 kJ /kgK. Thus for p = 0.2 MPa, T = 30°c, enthalpy (hg = 426.87 kJ/kg Now from saturation tables for R-134a, at condensor pressure 2 bar, the saturation temperature is - 10.08°c So at point '4', temperature = -10-08-10 = - 20-08°C From saturated data for P = 0.2MPa and T = - 20.08°c, we get enthalpy 1h4 = 173.54 killeg Expansion is an isenthalpic process. So hath = 17 3 54 kJ / kg. © Power consumed by compressen-hy-hy = 426-87-377.5 = 49.37 kJ/kg LP.S. - This is not the power, it is the energy consumed per unit mass of R-134 ... You will get the power by multiplying this with the mass flow rate, which is not provided in question ]

& Heat absorbed in evaporator = hah = 377.5-173.54 = 203.96 kJ/kg 2 den 3 Heat rejected in condenser = hz - ha = 426-87 - 173.54 = 253.33 kJ lkg A COP of syskm = h, h_ 203-96 = 4.13 hz-h 49.37

ANSWERS

• The energy consumed per unit mass of refrigerant in the compressor is 49.37 kJ/kg
• The heat absorbed in the evaporator is 203.96 kJ/kg
• The heat rejected in the condenser is 253.33 kJ/kg
• The COP of the system is 4.13