1. Objectives
Continue with heat exchangers (ch.11)

2. Coil Extended Surfaces Compact Heat Exchangers
Fins added to refrigerant tubes
Important parameters for heat exchange?

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

4. Heat Exchangers Parallel flow Counterflow Crossflow
Ref: Incropera & Dewitt (2002)

5. Heat Exchanger Analysis - Δtm

6. Heat Exchanger Analysis - Δtm
Counterflow
For parallel flow is the same or

7. Counterflow Heat Exchangers
Important parameters:
Q = U0A0Δtm

8. Heat exchanger effectiveness
Generally for all exchanger
Losses to surrounding  0
Then: Q cold fluid = Q hot fluid
mccp_c(tc,o-tc,i)=mhcp_h(th,i-th,o)
Effectiveness
 = Q exchanged / Q maximum = Q cold or hot fluid / Q maximum

9. Heat Exchanger Effectiveness (ε) (notation in the book)
C=mcp
Mass flow rate
Specific capacity of fluid
THin
TCout
THout
TCin
Location B
Location A

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

11. 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

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

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

16. 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

17. 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

18. 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

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

20. 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