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Filed Trip is Tomorrow at 9 am

Filed Trip is Tomorrow at 9 am. https://www.google.com/maps/place/Saint+Edwards+University/@30.2273026,-97.7529731,1095m/data=!3m1!1e3!4m2!3m1!1s0x8644b492ae61201b:0x1142c282cbe51336. Objectives. Continue with heat exchangers (ch.11). Coil Extended Surfaces Compact Heat Exchangers.

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Filed Trip is Tomorrow at 9 am

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  1. Filed Trip is Tomorrow at 9 am https://www.google.com/maps/place/Saint+Edwards+University/@30.2273026,-97.7529731,1095m/data=!3m1!1e3!4m2!3m1!1s0x8644b492ae61201b:0x1142c282cbe51336

  2. Objectives • Continue with heat exchangers (ch.11)

  3. Coil Extended Surfaces Compact Heat Exchangers • Fins added to refrigerant tubes • Important parameters for heat exchange?

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

  5. Heat Exchangers • Parallel flow • Counterflow • Crossflow Ref: Incropera & Dewitt (2002)

  6. Heat Exchanger Analysis - Δtm

  7. Heat Exchanger Analysis - Δtm Counterflow For parallel flow is the same or

  8. Counterflow Heat Exchangers Important parameters: Q = U0A0Δtm

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

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

  11. Heat Transfer • From hot fluid to pipe • Through the wall • From the pipe and fins

  12. 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

  13. 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

  14. 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

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

  16. 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

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

  18. 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

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

  20. 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)

  21. Energy and mass balances • Steady-state energy equation on air • Energy balance on water • Mass balance on water

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