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Objective

Objective . Heat Exchangers Learn about different types Define Heat Exchanger Effectivness ( ε ) Analyze how geometry affects ε Solve examples. Systems: residential. Outdoor Air. Indoor Air. Large building system Chiller . Large building system Chiller . Outdoor air . 95 o F.

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Objective

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  1. Objective • Heat Exchangers • Learn about different types • Define Heat Exchanger Effectivness (ε) • Analyze how geometry affects ε • Solve examples

  2. Systems: residential Outdoor Air Indoor Air

  3. Large building system Chiller

  4. Large building system Chiller Outdoor air 95oF 53oF Water from building Water to building 43oF

  5. Heat exchangers Air-liquid Tube heat exchanger Air-air Plate heat exchanger

  6. Some Heat Exchanger Facts • All of the energy that leaves the hot fluid enters the cold fluid • If a heat exchanger surface is not below the dew point of the air, you will not get any dehumidification • Water takes time to drain off of the coil • Heat exchanger effectivness varies greatly

  7. Heat Exchanger Effectivness (ε) C=mcp Mass flow rate Specific capacity of fluid THin TCout THout TCin Location B Location A

  8. Example: What is the saving with the residential heat recovery system? Outdoor Air 32ºF 72ºF 72ºF Combustion products 52ºF Furnace Exhaust Fresh Air Gas For ε=0.5 and if mass flow rate for outdoor and exhaust air are the same 50% of heating energy for ventilation is recovered! For ε=1 → free ventilation! (or maybe not)

  9. Air-Liquid Heat Exchangers Extended surfaces (fins) from air side • Fins added to refrigerant tubes

  10. Analysis of Compact Heat Exchangers • Geometry is very complex • Assume flat circular-plate fin

  11. Overall Heat Transfer Q = U0A0Δtm Overall Heat Transfer Coefficient Mean temperature difference

  12. Δtm for Heat Exchangers Depends on flow direction: • Parallel flow • Counterflow • Crossflow Ref: Incropera & Dewitt (2002)

  13. Heat Exchanger Analysis - Δtm Counterflow Logarithmic mean temperature difference For parallel flow is the same or

  14. Counterflow Heat Exchangers Important parameters:

  15. Example Assume that the residential heat recovery system is counterflow heat exchanger with ε=0.5. Calculate Δtm for the residential heat recovery system if : mcp,hot= 0.8· mc p,cold Outdoor Air 32ºF 72ºF mcp,hot= 0.8· mc p,cold mc p,cold 0.2· mc p,cold 72ºF Combustion products Furnace Exhaust Fresh Air th,i=72 ºF, tc,i=32 ºF For ε = 0.5 → th,o=52 ºF, tc,o=48 ºF Δtm,cf=(20-16)/ln(20/16)=17.9 ºF

  16. What about crossflow heat exchangers? Δtm= F·Δtm,cf Correction factor Δt for counterflow Derivation of F is in the text book: ………

  17. Example: Calculate the real Δtm for the residential heat recovery cross flow system (both fluids unmixed): For: th,i=72 ºF, tc,i=32 ºF , th,o=52 ºF, tc,o=48 ºF R=1.25, P=0.4 → From diagram → F=0.92 Δtm=Δtm,cf · F =17.9 ·0.92=16.5 ºF

  18. Overall Heat Transfer Q = U0A0Δtm Need to find this AP,o AF

  19. Heat Transfer Heat transfer from fin and pipe to air (External): tP,o t tF,m where is fin efficiency Heat transfer from hot fluid to pipe (Internal ): Heat transfer through the wall:

  20. Resistance model Q = U0A0Δtm From eq. 1, 2, and 3: • We can often neglect conduction through pipe walls • Sometime more important to add fouling coefficients R Internal R cond-Pipe R External

  21. Example The air to air heat exchanger in the heat recovery system from previous example has flow rate of fresh air of 200 cfm. With given: Calculate the needed area of heat exchanger A0=? Solution: Q = mcp,coldΔtcold = mcp,hotΔthot = U0A0Δtm From heat exchanger side: Q = U0A0Δtm→ A0 = Q/ U0Δtm U0 = 1/(RInternal+RCond+RFin+RExternal) = (1/10+0.002+0+1/10) = 4.95 Btu/hsfF Δtm = 16.5 F From air side: Q = mcp,coldΔtcold = = 200cfm·60min/h·0.075lb/cf·0.24Btu/lbF·16 = 3456 Btu/h Then: A0 = 3456 / (4.95·16.5) = 42 sf

  22. For Air-Liquid Heat Exchanger we need Fin Efficiency • Assume entire fin is at fin base temperature • Maximum possible heat transfer • Perfect fin • Efficiency is ratio of actual heat transfer to perfect case • Non-dimensional parameter tF,m

  23. Fin Theory k – conductivity of material hc,o – convection coefficient pL=L(hc,o /ky)0.5

  24. Reading Assignment • Chapter 11 - From 11.1-11.7

  25. Final project topics • Beside 3 introduced in last class: • Duct design • DOAS design • VAV design

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