Transcript Q = U 0 A 0 Δt m

Objectives
• Continue with heat exchangers (ch.11)
Coil Extended Surfaces
Compact Heat Exchangers
• Fins added to refrigerant tubes
• Important parameters for heat exchange?
Overall Heat Transfer
Q = U0A0Δtm
Overall Heat
Transfer Coefficient
Mean temperature
difference
Heat Exchangers
• Parallel flow
• Counterflow
• Crossflow
Ref: Incropera & Dewitt (2002)
Heat Exchanger Analysis - Δtm
Heat Exchanger Analysis - Δtm
Counterflow
tm 
or
t m
t
h ,o
 tc ,i   th,i  tc ,o 
 th,o  tc ,i 
ln 



t

t
 h ,i c , o 
t B  t A 

 t B 

ln 
 t A 
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 
m
 1  P  
ln 



1

RP


Important parameters:

th,i  th,o 
R
tc,o  tc,i 

tc, o  tc,i 
P
th,i  tc,i 
Q = U0A0Δtm
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
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
Heat exchangers
Air-liquid
Tube heat exchanger
Air-air
Plate heat exchanger
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:
………
Overall Heat Transfer
Q = U0A0Δtm
Need to find this
AF
AP,o
1
1
U0 

Ro RInternal  Rcond  Pipe  RExternal
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  0.002 sfF/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
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