1. Objectives Finish with Heat Exchangers
Duct Design AND Diffuser Selection

2. 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

3. 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

4. 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

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

6. 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

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

10. 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

11. Reading Assignment
Chapter 11
- From

12. 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)

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

15. 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

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

17. Relationship Between Static and Total Pressure

18. System Characteristic

19. Electrical Resistance Analogy

20. 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

21. Frictional Losses

22. Ductulator

23. Non-circular Ducts
Parallel concept to wetted perimeter

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

25. Loss Coefficients
ΔPt = CoPv,0