Objectives Heat exchanger Dry HX vs. Vet HX (if we have time) 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 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 mc p,cold mcp,hot= 0.8· mc p,cold 0.2· mc p,cold 72ºF Combustion products Exhaust Furnace 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

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

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

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

Heat Transfer Heat transfer from fin and pipe to air (External): t tP,o tF,m where is fin efficiency Heat transfer from hot fluid to pipe (Internal ): Heat transfer through the wall:

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

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,cold Δtcold = mcp,hot Δthot = 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,cold Δtcold = = 200cfm·60min/h·0.075lb/cf·0.24Btu/lbF·16 = 3456 Btu/h Then: A0 = 3456 / (4.95·16.5) = 42 sf

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

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

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

Heat exchanger performance (11.3) NTU – absolute sizing (# of transfer units) ε – relative sizing (effectiveness) Criteria NTU ε P RP cr

Example HW4 problem AHU M OA CC CC RA tc,in=45ºF Qcc=195600Btu/h 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

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

Reading Assignment Chapter 11 - From 11.1-11.7

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)

1. Redo Fin Theory Same result

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)

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)

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

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

Results

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

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.

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?