Objectives Finish with Heat Exchangers Duct Design AND Diffuser Selection

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 (Book Section 11.3) NTU – absolute sizing (# of transfer units) ε – relative sizing (effectiveness) Criteria NTU ε P RP cr

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)

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

Duct Design Total and static pressure drops are proportional to square of velocity Plot of pressure drop vs. volumetric flow rate (or velocity) is called system characteristic

Pressures Static pressure Velocity pressure Total pressure – sum of the two above

Relationship Between Static and Total Pressure

System Characteristic

Electrical Resistance Analogy

Duct Design Static pressure drop (friction losses) in a duct is proportional to square of velocity or flow It is a function of Dynamic pressure Length Pipe diameter Friction coefficient

Frictional Losses

Ductulator

Non-circular Ducts Parallel concept to wetted perimeter

Dynamic losses Losses associated with Two methods Changes in velocity Obstructions Bends Fittings and transitions Two methods Equivalent length and loss coefficients

Loss Coefficients ΔPt = CoPv,0