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Objectives

Objectives. Continue with heat exchangers (ch.11). Coil Extended Surfaces Compact Heat Exchangers. Fins added to refrigerant tubes Important parameters for heat exchange?. Overall Heat Transfer. Q = U 0 A 0 Δ t m Overall Heat Transfer Coefficient. Mean temperature difference.

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Objectives

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

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

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

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

  5. Heat Exchanger Analysis - Δtm

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

  7. Counterflow Heat Exchangers Important parameters: Q = U0A0Δtm

  8. Heat exchanger effectiveness • Generally for all exchanger • Losses to surrounding  0 • Then: Q cold fluid = Q hot fluid • mccp_c(tc,o-tc,i)=mhcp_h(th,i-th,o) • Effectiveness  = Q exchanged / Q maximum = Q cold or hot fluid / Q maximum

  9. Heat Exchanger Effectiveness (ε)(notation in the book) C=mcp Mass flow rate Specific capacity of fluid THin TCout THout TCin Location B Location A

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

  11. 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

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

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

  14. 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

  15. 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

  16. 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

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

  18. 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

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