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Thermal system design 3

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Design for a Storage Room for Frozen Food

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Thermal system design 3

1. 1. Samir Abusetta Shehryar Niazi MEEN 4360 – Thermal System Design | Dr. Lea Der Chen | 17th December, 2013
2. 2. Outline  Introduction  Objectives & Background  Proposal  System Description  Analysis  Results  Conclusion
3. 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. 4. Proposal  Cascade Refrigeration System with one refrigeration cycle operating with R-410a and the other cycle with R-134a systems
5. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 15. Questions?
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