Design for a Storage Room for Frozen Food

1. Samir Abusetta
Shehryar Niazi
MEEN 4360 – Thermal System Design | Dr. Lea Der Chen | 17th December, 2013

2. Outline
 Introduction
 Objectives & Background
 Proposal
 System Description
 Analysis
 Results
 Conclusion

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
•
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
•
CP = Specific heat capacity
•
m = Mass (amount of beef)
•
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

13. Sizing the Cascade System
 R410a System:
•
Compressor = 60kJ/kg*0.3 kg/s = 18 kW
 R134a System:
•
Compressor = 35.4kJ/kg * 0.1464 kg/s = 5.18 kW

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

15. Questions?