3. Introduction
Freeze food for consumption at a later time
ABC food company in need of a 40 ton storage room to freeze food
Consider multi stage refrigeration cycle
Keep it at -35 F in 20ft. x 20ft. x 22ft.
4. Proposal
Cascade Refrigeration System with one refrigeration cycle operating
with R-410a and the other cycle with R-134a systems
5. Background of Cascade System
Like a reverse Rankin cycle
Used in industrial applications where quite low temperatures are
required – high efficiency
The large temp difference requires a large pressure difference
Refrigeration cycle is performed in stages
The refrigerant in the two stages doesn’t mix
Four basic thermodynamics principals: compression, evaporator,
expansion valve, condenser
6. Analysis:
Heat Leakage Load
Q = U * A * ΔT
0.11(Btu/ft.2 F. h)* (2560 ft^2)*(97-(-37)) F= 37734.4 9(Btu/h)
•
Q = Total heat transfer
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U = The rate of heat flow through the walls, floor, and ceiling of the
refrigerated space
•
A = Total Surface Area Outside of the refrigerated space
•
ΔT= Temperature Difference
7. Analysis:
Product Load
Q = (m* CP*ΔT)entering temp to freezing + (m*Hfg)freezing + (m*CP*ΔT)sub freezing
Q = (17600*0.77 *85) + (17600*100) + (17600*0.41*67)
Q = 3395392 Btu/day or 141474.5 Btu/hour
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CP = Specific heat capacity
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m = Mass (amount of beef)
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Hfg = Latent heat
•
ΔT= Temperature Difference
8. Analysis:
Misc. Load & Service Load
Misc. load is heat introduced by lights, motors, and other heat producing
devices located in the refrigerated area
Service load is the heat that enters the refrigerated area when doors or other
access means are opened
250Btu/hour (more feasible to assume)
Q total = 37734 + 141474 + 250 = 179458 Btu/hour
= 189328.2 (kJ/hour)
9. Analysis:
R-410a Cycle
Evaporator: Cooling Capacity = 262.8 - 87=175.8 kJ/kg
Compressor: Work = 322.4 – 262.8 = 59.6 kJ/kg
Condenser: Heat Loss from Condenser = 322.4 – 87= 235.4 kJ/kg
•
This heat must be removed by the 134a refrigerant cycle.
Expansion Valve: Pressure drops from 1400kPa to 175kPa and the
temperature from 18.5C at P=1400kPa to -40C.
10. Analysis:
R-134a Cycle
Evaporator: Cooling Capacity = 398.6 – 284.4=114.2 KJ/kg
Compressor: Work = 433.9 – 398.6 = 35.3 KJ/kg
Condenser: Heat loss from condenser to the environment = 284.4 – 87.4 = 197 KJ/kg
Expansion valve: Pressure will drop from 1600kPa to 293kPa and temperature
from 57.9 C (Tsat of R-134a at1600kpa), to 0 C
11. Analysis:
Mass Flow Rate Ratio of Cycles
Mass Flow Ratio:
Ratio of heat gained by the evaporator of the R-134a cycle to the heat
lost by the R410 cycle in the condenser
•
mR134a/mR410a = (398.6 - 284.1) / (322.4 - 87.4) = 0.4872
p.s: Thermodynamics first law.
12. Analysis:
Mass Flow Rates (R410a & R134a)
Mass Flow Rate (R410a) = Total Heat / Cooling Capacity
= (189328.2 kJ/hr) / (175.8kJ/kg) = 1082kg/hr
= 0.3kg/s
Mass Flow Rate (R134a) = mass ratio * 0.3kg/s = 0.1464kg/s
14. Conclusion
The over all system must provide cooling capacity of
179485 Btu/hr or 179485/12000 = 15 ton of refrigeration
Install 18 ton of refrigeration system with margin of safety of 3 ton or 20 %
overcapacity.
p.s: 1 ton refrigeration = 12000Btu