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
43oF building
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
Heat exchanged
Maximumposible heat exchange
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
Overall Heat
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
tm 
or
th,o  tc,i  th,i  tc,o 
 th,o  tc,i 
ln 



t

t
 h,i c,o 
t m
t B  t A 

 t B 

ln
 t A 
Logarithmic mean temperature difference
For parallel flow is the same
t m

t B  t A 

 t B 

ln
 t A 
Counterflow Heat Exchangers


th,o  tc,i   th,i  tc,o 


tc,o  tc,i R  1
tm 
t 

ln 

th,o  tc,i 

th,i  tc,o 
Important parameters:

t h , i  t h, o 
R
tc,o  tc,i 

tc, o  tc ,i 
P
th,i  tc,i 
m
 1  P  
ln 



1

RP


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
mc p,cold
72ºF
mcp,hot= 0.8· 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
0.2· mc p,cold
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
AF
AP,o
1
1
U0 

Ro RInternal  Rcond  Pipe  RExternal
Heat Transfer
Heat transfer from fin and pipe to air (External):
Q  hc ,o AP ,o (t P,o  t )  hc ,o AF (t F,m  t )
Q  hc ,o AP ,o   AF (t P,o  t )
where
t
(eq.1)
  (tF , m  t ) /(tF , B  t )
is fin efficiency
Heat transfer from hot fluid to pipe (Internal ):
Q  hi AP,i (t f  tP,i ) (eq. 2)
Heat transfer through the wall:
k P AP,m (t P,i  t P,o )

Q
(eq. 3)
xP
tF,m tP,o
Resistance model
Q = U0A0Δtm
From eq. 1, 2, and 3:
U0 
1
Ao
AP ,i hi
R Internal

Ao x p
AP ,m k p

R cond-Pipe
1
AP ,o

hc ,o  A  
 F

 hc1,o
R External
• We can often neglect conduction through pipe walls
• Sometime more important to add fouling coefficients
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:
h Internal  10 Btu/hsfF, R cond  0.002sfF/Btu/h , h External  10 Btu/hsfF
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
  (tF , m  t ) /(tF , B  t )
• Non-dimensional parameter
tF,m
Fin Theory
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