Objective Heat Exchangers Learn about different types Define Heat Exchanger Effectivness (ε) Analyze how geometry affects ε Solve examples

Systems: residential Outdoor Air Indoor Air

Large building system Chiller

Large building system Chiller Outdoor air 95oF 53oF Water from building Water to building 43oF

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

Some Heat Exchanger Facts All of the energy that leaves the hot fluid enters the cold fluid If a heat exchanger surface is not below the dew point of the air, you will not get any dehumidification Water takes time to drain off of the coil Heat exchanger effectivness varies greatly

Heat Exchanger Effectivness (ε) C=mcp Mass flow rate Specific capacity of fluid THin TCout THout TCin Location B Location A

Example: What is the saving with the residential heat recovery system? Outdoor Air 32ºF 72ºF 72ºF Combustion products 52ºF Exhaust Furnace Fresh Air Gas For ε=0.5 and if mass flow rate for outdoor and exhaust air are the same 50% of heating energy for ventilation is recovered! For ε=1 → free ventilation! (or maybe not)

Air-Liquid Heat Exchangers Extended surfaces (fins) from air side Fins added to refrigerant tubes

Analysis of Compact Heat Exchangers Geometry is very complex Assume flat circular-plate fin

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

Δtm for Heat Exchangers Depends on flow direction: Parallel flow Counterflow Crossflow Ref: Incropera & Dewitt (2002)

Heat Exchanger Analysis - Δtm Counterflow Logarithmic mean temperature difference For parallel flow is the same or

Counterflow Heat Exchangers Important parameters:

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

Reading Assignment Chapter 11 - From 11.1-11.7

Final project topics Beside 3 introduced in last class: Duct design DOAS design VAV design