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Filed Trip is Tomorrow at 9 am ,1095m/data=!3m1!1e3!4m2!3m1!1s0x8644b492ae61201b:0x1142c282cbe51336.

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Presentation on theme: "Filed Trip is Tomorrow at 9 am ,1095m/data=!3m1!1e3!4m2!3m1!1s0x8644b492ae61201b:0x1142c282cbe51336."— Presentation transcript:

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 = U 0 A 0 Δt m Overall Heat Transfer Coefficient Mean temperature difference

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

6 Heat Exchanger Analysis - Δt m

7 Counterflow For parallel flow is the same or

8 Counterflow Heat Exchangers Important parameters: Q = U 0 A 0 Δt m

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

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12 Overall Heat Transfer Q = U 0 A 0 Δt m Need to find this A P,o AFAF

13 Heat Transfer From hot fluid to pipe Through the wall From the pipe and fins

14 Resistance model Q = U 0 A 0 Δt m 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 With given: Calculate the needed area of heat exchanger A 0 =? The air to air heat exchanger in the heat recovery system from previous example has flow rate of fresh air of 200 cfm. Solution: Q = mc p,cold Δt cold = mc p,hot Δt hot = U 0 A 0 Δt m From heat exchanger side: Q = U 0 A 0 Δt m → A 0 = Q/ U 0 Δt m U 0 = 1/(R Internal +R Cond +R Fin +R External ) = (1/10+0.002+0+1/10) = 4.95 Btu/hsfF Δt m = 16.5 F From air side: Q = mc p,cold Δt cold = = 200cfm·60min/h·0.075lb/cf·0.24Btu/lbF·16 = 3456 Btu/h Then: A 0 = 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 t F,m

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

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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20 Heat exchanger performance (Book Section 11.3) NTU – absolute sizing (# of transfer units) ε – relative sizing (effectiveness) Criteria NTU εPRP crcr

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22 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

23 Reading Assignment Chapter 11 - From 11.1-11.7

24 Analysis of Moist Coils 1.Redo fin theory 2.Energy balance on fin surface, water film, air I ntroduce Lewis Number 3.Digression – approximate enthalpy 4.Redo fin analysis for cooling/ dehumidification (t → h)

25 Energy and mass balances Steady-state energy equation on air Energy balance on water Mass balance on water

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