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Objectives

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

  3. 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

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

  5. 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

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

  7. 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:

  8. 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

  9. 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

  10. 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

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

  12. 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

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

  14. ExampleHW4 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

  15. 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

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

  17. 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)

  18. 1. Redo Fin Theory • Same result

  19. 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)

  20. Lewis Number • Lewis number, Le = α/Dc = hc/hD/cP • Ratio of heat transfer to mass transfer • Table 9.1 (for forced convection c = 2/3)

  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. Approximate Expression for Mean Enthalpy Difference h1 enthalpy of entering air stream h2 enthalpy of leaving air stream hs,R,1 fictitious enthalpy of saturated air at entering refrigerant temp. hs,R,2 fictitious enthalpy of saturated air at leaving refrigerant temp.

  26. 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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