Here are the solutions to your problems:
2b) Determine chemical formula of the following refrigerants
- (i) R12: CCl2F2 (Dichlorodifluoromethane)
- (ii) R22: CHClF2 (Chlorodifluoromethane)
- (iii) CO2: CO2 (Carbon dioxide)
- (iv) NH3: NH3 (Ammonia)
- (v) H2O: H2O (Water)
3) A Freon 12 vapour compression system operating at a condenser temperature of 40∘C and evaporator temperature of −5∘C develops 15 tons of refrigeration.
To solve this problem, a P-h diagram for Freon 12 (R12) is required to determine the specific enthalpy (h) and specific volume (v) values at different points in the cycle. Since the diagram is not provided, I will outline the steps and formulas.
Given Data:
- Refrigerant: Freon 12 (R12)
- Condenser temperature, Tc=40∘C
- Evaporator temperature, Te=−5∘C
- Refrigeration capacity, Qe=15 tons of refrigeration (TR)
Conversion:
1 ton of refrigeration (TR) = 3.517 kW
Qe=15TR×3.517TRkW=52.755 kW
Cycle Points (from P-h diagram):
- State 1 (Evaporator outlet, compressor inlet): Saturated vapor at Te=−5∘C.
- Read h1 (enthalpy), v1 (specific volume), s1 (entropy) from the diagram.
- State 2 (Compressor outlet, condenser inlet): Superheated vapor after isentropic compression from State 1 to the condenser pressure (corresponding to Tc=40∘C).
- Follow an isentropic line (s2=s1) from State 1 to the condenser pressure. Read h2.
- State 3 (Condenser outlet, expansion valve inlet): Saturated liquid at Tc=40∘C.
- Read h3 from the diagram.
- State 4 (Expansion valve outlet, evaporator inlet): Liquid-vapor mixture after isenthalpic expansion from State 3.
Calculations:
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i) Mass flow rate of refrigerant circulated (m˙):
The refrigeration capacity is given by:
Qe=m˙(h1−h4)
m˙=h1−h4Qe
Numerical value cannot be determined without h1 and h4 from the P-h diagram.
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ii) Theoretical piston displacement of the compressor (V˙) and piston displacement per ton of refrigeration:
The theoretical piston displacement is:
V˙=m˙⋅v1
Piston displacement per ton of refrigeration:
TRV˙=Qe/TRm˙⋅v1
Numerical values cannot be determined without m˙ and v1 from the P-h diagram.
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iii) Theoretical horse power of the compressor (Wc) and horse power per ton of refrigeration:
The theoretical compressor work is:
Wc=m˙(h2−h1)
To convert to horsepower (HP), use 1HP=0.7457 kW:
Wc(HP)=0.7457m˙(h2−h1)
Horse power per ton of refrigeration:
TRWc(HP)=Qe/TRWc(HP)
Numerical values cannot be determined without m˙, h1, and h2 from the P-h diagram.
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iv) Heat rejected in the condenser (Qc):
The heat rejected in the condenser is:
Qc=m˙(h2−h3)
Numerical value cannot be determined without m˙, h2, and h3 from the P-h diagram.
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v) Carnot COP and actual COP of the cycle:
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Carnot COP (COPCarnot):
Convert temperatures to Kelvin:
Te=−5\circC+273.15=268.15 K
Tc=40\circC+273.15=313.15 K
COPCarnot=Tc−TeTe=313.15K−268.15K268.15K=45268.15≈5.959
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Actual COP (COPactual):
COPactual=WcQe=m˙(h2−h1)m˙(h1−h4)=h2−h1h1−h4
Numerical value cannot be determined without h1, h2, and h4 from the P-h diagram.