Thermal system design 3

MEEN 4360 – Thermal System Design | Dr. Lea Der Chen | 17th December, 2013

Samir AbusettaShehryar Niazi

Outline

Introduction

Objectives & Background

Proposal

System Description

Analysis

Results

Conclusion

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.

Proposal

Cascade Refrigeration System with one refrigeration cycle operating with R-410a and the other cycle with R-134a systems

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

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

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

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)

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.

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

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.

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

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

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

Questions?