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

2. Systems: residential
Outdoor Air
Indoor Air

3. Large building system Chiller

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

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

6. 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

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

8. 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)

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

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

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

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

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

14. Counterflow Heat Exchangers
Important parameters:

15. 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

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

18. 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

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

20. 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:

21. 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

22. 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

23. 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

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

27. Reading Assignment
Chapter 11
- From

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