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Objectives

Objectives. Heat exchanger Geometry and effectivness (through examples) Fin theory Dry HX vs. Vet HX (if we have time). Reading . 11.1-11.6 (including 11.6). Heat exchangers. Air-liquid . Tube heat exchanger. Air-air . Plate heat exchanger. Example.

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Objectives

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  1. Objectives • Heat exchanger • Geometry and effectivness (through examples) • Fin theory • Dry HX vs. Vet HX (if we have time)

  2. Reading • 11.1-11.6 (including 11.6)

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

  4. 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

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

  6. 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

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

  8. 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:

  9. 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

  10. 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

  11. 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

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

  13. 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

  14. Heat exchanger performance (11.3) • NTU – absolute sizing (# of transfer units) • ε – relative sizing (effectiveness)

  15. Example problem AHU M For the problem 9 HW assignment # 2 (process in AHU) calculate: a) Effectiveness of the cooling coil b) UoAo value for the CC Inlet water temperature into CC is coil is 45ºF OA CC CC (mcp)w steam RA tc,in=45ºF Qcc=195600Btu/h tM=81ºF tCC=55ºF

  16. Summary • Calculate efficiency of extended surface • Add thermal resistances in series • If you know temperatures • Calculate R and P to get F, ε, NTU • Might be iterative • If you know ε, NTU • Calculate R,P and get F, temps

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

  18. Analysis of Moist Coils • Redo fin theory • Energy balance on fin surface, water film, air Introduce Lewis Number • Digression – approximate enthalpy • Redo fin analysis for cooling/ dehumidification (t → h)

  19. 1. Redo Fin Theory • Same result

  20. 2. Energy and mass balances • Steady-state energy equation on air • Energy balance on water • Mass balance on water • Lewis number • Rewrite energy balance on water surface • Reintroduce hg0 (enthalpy of sat. water vapor at 0 °C or °F)

  21. 4. Fin analysis for wet fins Heat conduction only occurs in y-direction through water film

  22. Overview of Procedure • Same approach as for dry fin with addition of conduction through water film • Define “fictitous moist air enthalpy” define at water surface temperature • Define heat-transfer coefficient • Develop new governing equation

  23. Results

  24. Overall Heat Transfer Coefficients • Very parallel procedure to dry coil problem • U-values now influenced by condensation • See Example 11.6 for details

  25. Wet Surface Heat Transfer • If you know dry surface heat transfer • Reynolds number changes – empirical relationships • Approximate wet-surface • Does a wet or a dry coil have higher or lower heat exchange? • Does a wet or a dry coil have higher or lower pressure drop?

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