1. 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)

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

3. 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

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

6. 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

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

8. 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:

9. 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

10. 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) = 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

11. 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

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

13. 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

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

18. 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

19. 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

20. Reading Assignment
Chapter 11
- From

21. 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)

22. 1. Redo Fin Theory
Same result

23. 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)

25. 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)

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

27. 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

28. Results

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

30. 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.

31. 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?