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Heat Transfer/Heat Exchanger• How is the heat transfer? • Mechanism of Convection• Applications . • Mean fluid Velocity and Boundary and their effect on the rate of heat
transfer.• Fundamental equation of heat transfer• Logarithmic-mean temperature difference.• Heat transfer Coefficients.• Heat flux and Nusselt correlation • Simulation program for Heat Exchanger
How is the heat transfer?
• Heat can transfer between the surface of a solid conductor and the surrounding medium whenever temperature gradient exists.ConductionConvection
Natural convection Forced Convection
Natural and forced Convection
Natural convection occurs whenever heat flows between a solid and fluid, or between fluid layers.
As a result of heat exchange
Change in density of effective fluid layers taken place, which causes upward flow of heated fluid.
If this motion is associated with heat transfer mechanism only, then it is called Natural Convection
Forced Convection
If this motion is associated by mechanical means such as pumps, gravity or fans, the movement of the fluid is enforced.
And in this case, we then speak of Forced convection.
Heat Exchangers• A device whose primary purpose is the transfer of energy
between two fluids is named a Heat Exchanger.
Applications of Heat Exchangers
Heat Exchangers prevent car engine
overheating and increase efficiency
Heat exchangers are used in Industry for
heat transfer
Heat exchangers are used in AC and
furnaces
• The closed-type exchanger is the most popular one.
• One example of this type is the Double pipe exchanger.
• In this type, the hot and cold fluid streams do not come into direct contact with each other. They are separated by a tube wall or flat plate.
Principle of Heat Exchanger• First Law of Thermodynamic: “Energy is conserved.”
generatedsin out
outin ewqhmhmdt
dE
ˆ.ˆ.
outin
hmhm ˆ.ˆ. h
hphh TCmQ ..
ccpcc TCmQ ..
0 0 0 0
•Control Volume
Qh
Cross Section Area
HOT
COLD
Thermal Boundary Layer
Q hot Q cold
Th Ti,wall
To,wall
Tc
Region I : Hot Liquid-Solid Convection
NEWTON’S LAW OF CCOLING
dqx hh . Th Tiw .dA Region II : Conduction Across Copper Wall
FOURIER’S LAW
dqx k.dT
dr
Region III: Solid – Cold Liquid Convection
NEWTON’S LAW OF CCOLING
dqx hc . Tow Tc .dA
THERMAL
BOUNDARY LAYER
Energy moves from hot fluid to a surface by convection, through the wall by conduction, and then by convection from the surface to the cold fluid.
• Velocity distribution and boundary layer
When fluid flow through a circular tube of uniform cross-suction and fully developed,
The velocity distribution depend on the type of the flow.
In laminar flow the volumetric flowrate is a function of the radius.
V u2rdrr0
rD / 2
V = volumetric flowrate
In turbulent flow, there is no such distribution.
• The molecule of the flowing fluid which adjacent to the surface have zero velocity because of mass-attractive forces. Other fluid particles in the vicinity of this layer, when attempting to slid over it, are slow down by viscous forces.
r
Boundary layer
• Accordingly the temperature gradient is larger at the wall and through the viscous sub-layer, and small in the turbulent core.
• The reason for this is 1) Heat must transfer through the boundary layer by conduction.2) Most of the fluid have a low thermal conductivity (k)3) While in the turbulent core there are a rapid moving eddies, which they are equalizing the temperature.
heating
cooling
Tube wall
Twh
Twc
Tc
Metalwall
Warm fluid
cold fluid
qx hAT
qx hA(Tw T)
qx k
A(Tw T)
h
Region I : Hot Liquid – Solid Convection
Th Tiw qx
hh .Ai
qx hhot . Th Tiw .A
Region II : Conduction Across Copper Wall
qx kcopper .2L
lnro
ri
To,wall Ti,wall qx .ln
ro
ri
kcopper .2L
Region III : Solid – Cold Liquid Convection
To,wall Tc qx
hc .Ao
qx hc To,wall Tc Ao
+
Th Tc qx
1
hh .Ai
ln
ro
ri
kcopper .2L
1
hc .Ao
qx U.A. Th Tc 1
1
.
ln.
.
coldicopper
i
oo
ihot
o
hrk
rr
r
rh
rU
U = The Overall Heat Transfer Coefficient [W/m.K]
Th Tc qx
R1 R2 R3
U 1
A.R
r
o
r
i
Calculating U using Log Mean Temperature
coldhot dqdqdq
ch TTT ch dTdTTd )(
hhphh dTCmdq ..
ccpcc dTCmdq ..
Hot Stream :
Cold Stream:
cpc
chph
h
Cm
dq
Cm
dqTd
..)(
dATUdq ..
cpc
hph CmCm
dATUTd.
1
.
1...)(
2
1
2
1
..
1
.
1.
)( A
Acpc
hph
T
TdA
CmCmU
T
Td
outc
inc
outh
inhch TTTT
q
AUTT
q
AU
T
T
...
ln1
2
2
1
2
1
..)( A
Ac
c
h
hT
TdA
q
T
q
TU
T
Td
1
2
12
ln
.
TT
TTAUq
Log Mean Temperature
16
Heat Exchanger (HX) Design MethodsHX designers usually use two well-known methods for calculating the heat transfer rate between fluid streams—the UA-LMTD and the effectiveness-NTU (number of heat transfer units) methods.
Both methods can be equally employed for designing HXs. However, the -NTU method is preferred for rating problems where at least one exit temperature is unknown. If all inlet and outlet temperatures are known, the UA-LMTD method does not require an iterative procedure and is the preferred method.
Energy Balance
Overall Energy Balance
• Assume negligible heat transfer between the exchanger and its surroundings and negligible potential and kinetic energy changes for each fluid.
, ,h i h ohq m i i
, ,c c o c iq m i i
fluid enthalpyi
• Assuming no l/v phase change and constant specific heats, , , ,p h h i h ohq m c T T
, ,h h i h oC T T
, , ,c p c c o c iq m c T T
, ,c c o c iC T T
, Heat capacity r s ateh cC C
• Application to the hot (h) and cold (c) fluids:
18
LMTD (Log Mean Temperature Difference)The most commonly used type of heat exchanger is the recuperative heat exchanger. In this type the two fluids can flow in counter-flow, in parallel-flow, or in a combination of these, and cross-flow.
The true mean temperature difference is the Logarithmic mean Temperature difference (LMTD), is defined as
2
1
21
tt
ln
ttLMTD
CON CURRENT FLOW
1
2
12
lnT
T
TTTLn
731 TTTTT inc
inh
1062 TTTTT outc
outh
COUNTER CURRENT FLOW
1062 TTTTT inc
outh
731 TTTTT outc
inh
U Ý m h . Ý C p
h . T3 T6 A.TLn
Ý m c . Ý C p
c . T7 T10 A.TLn
T1T2
T4 T5
T3
T7 T8 T9
T10
T6
Counter - Current Flow
T1 T2T4 T5
T6T3
T7T8 T9
T10
Parallel Flow
Log Mean Temperature evaluation
T1
A
1 2
T2
T3
T6
T4 T6
T7 T8
T9
T10
Wall∆ T1
∆ T2
∆ A
A
1 2
T1
A
1 2
T2
T3
T6
T4 T6
T7 T8
T9
T10
Wall
q hh Ai Tlm
Tlm (T3 T1) (T6 T2)
ln(T3 T1)
(T6 T2)
q hc Ao Tlm
Tlm (T1 T7) (T2 T10)
ln(T1 T7)
(T2 T10)
Exchanger Fouling
Electron microscope image showing fibers, dust, and other deposited material on aresidential air conditioner coil and a fouled water line in a water heater.
Exchanger Fouling
Overall Coefficient
Overall Heat Transfer Coefficient
• An essential requirement for heat exchanger design or performance calculations.
• Contributing factors include convection and conduction associated with the two fluids and the intermediate solid, as well as the potential use of fins on both sides and the effects of time-dependent surface fouling.
• With subscripts c and h used to designate the hot and cold fluids, respectively, the most general expression for the overall coefficient is:
, ,
1 1 1
1 1
c h
f c f hw
o o o oc c h h
UA UA UA
R RR
hA A A hA
Overall Coefficient
o,
Overall surface efficiency of fin array (Section 3.6.5)
1 1
o
fc or h f
c or h
A
A
total surface area (fins and exposed base) surface area of fins only
t
f
A AA
Assuming an adiabatic tip, the fin efficiency is
,
tanhf c or h
c or h
mL
mL
2 /c or h p w c or hm U k t
, partial overall coe1
fficientp c or hf c or h
hUhR
2 for a unit surfFouling fact ace area (m W)or K/fR
Table 11.1
conduction resistan Wall (K/Wce )wR
25
Heat Exchanger UA-LMTD Design Method
Where U is the overall heat transfer coefficient (and is assumed to be constant over the whole surface area of the heat exchanger).
Lhr
RLk
r/rlnR
LhrR
UA
R
LMTDLMTDUAQ
ooo,f
p
ioi,f
ii
n
i
i
n
i
i
2
1
22
11
1
1
Heat Transfer Duty Overall Heat TransferCoefficient (W/m2.K)
Water to condensingR-12
440-830
Steam to water 960-1650
Water to water 825-1510
Steam to gases 25-2750
26
Calculation of heat transfer——heat transfer coefficient
Table - data of heat resistance from scales
fluidRs
m2· /W℃fluid
Rs
m2· /W℃
water (u<1m/s, t<50 )℃ vapors
sea 0.0001 organic 0.0002
river 0.0006 steam (no oil) 0.0001
well 0.00058 waste (with oil) 0.0002
distilled 0.0001 freezing (with oil) 0.0004
boiler supply 0.00026 gases
for cooling tower
(untreated)0.00058 air 0.0003
for cooling tower
(treated)0.00026 compressed air 0.0004
with solid particles 0.0006 nature gas 0.002
brine 0.0004 coke oven gas 0.002
For reducing this resistance, ( 1 ) cleaning the heat exchanger; ( 2 ) treat the fluid before use.
27
2. Using empirical data or equation to calculate U
Calculation of heat transfer——heat transfer coefficient
Fluid type U W/(m2·K)water—gas 12~60
water—water 800~1800
water—kerosene around 350
water—organic solvent 280~850
gas—gas 12~35
saturated steam—water 1400~4700
saturated steam—gas 30~300
saturated steam—oil 60~350
saturated steam—boiling oil 290~870
Table - empirical data of U for shell - pipe heat exchangers
Empirical equations are given by manufacturers.
28
Effect of h to U
Calculation of heat transfer——heat transfer coefficient
e. g.: using saturated steam with ( h2 ) to heat air (h1) : the pipe size is 25×2.5, air is in the pipe. The conductive heat resistance can be ignored. We have h1
(W/m2.k)
h2
(W/m2.k)
UO
(W/m2.k)comparison
40 5000 31.8 UO 2close to h1;
40 10000 31.9 h2 doubles , UO almost no change
80 5000 63.2 h1 doubles , UO doubles
—— the relatively importance of h to U
the results show that the increase of relative smaller h is key to increase U—— the key resistance is important.
If the value of the two h are not quite different , the increase of any one h is effective to raising the K .
211
2 11
hdh
d
UO
LMTD Method
A Methodology for Heat ExchangerDesign Calculations
- The Log Mean Temperature Difference (LMTD) Method -• A form of Newton’s Law of Cooling may be applied to heat exchangers by using a log-mean value of the temperature difference between the two fluids:
1mq U A T
1 2
11 21n /m
T TT
T T
Evaluation of depends on the heat exchanger type.1 2 and T T
• Counter-Flow Heat Exchanger:
1 ,1 ,1
, ,
h c
h i c o
T T TT T
2 ,2 ,2
, ,
h c
h o c i
T T TT T
LMTD Method (cont.)
• Parallel-Flow Heat Exchanger:
1 ,1 ,1
, ,
h c
h i c i
T T TT T
2 ,2 ,2
, ,
h c
h o c o
T T TT T
Note that Tc,o can not exceed Th,o for a PF HX, but can do so for a CF HX.
For equivalent values of UA and inlet temperatures, 1 , 1 ,m CF m PFT T
• Shell-and-Tube and Cross-Flow Heat Exchangers:1 1 ,m m CFT F T
Figures 11.10 - 11.13F
Special Conditions
Special Operating Conditions
Case (a): Ch>>Cc or h is a condensing vapor .hC
– Negligible or no change in , , .h h o h iT T T
Case (b): Cc>>Ch or c is an evaporating liquid .cC
– Negligible or no change in , , .c c o c iT T T
Case (c): Ch=Cc.1 2 1mT T T –
32
comparison of flow directions
Calculation of heat transfer——temperature difference
21
21
/ln tt
tttm
Given : the inlet and outlet temperatures of two fluids Find : logarithmic average T differences in different situations Solution : cocurrent counter currenthot fluid 240 → 160 ℃ ℃
cold fluid - ) 100 → 140 ℃ ℃
t1=140 20= t2
240 → 160 ℃ ℃
- ) 140 ℃ ← 100 ℃
t1=100 60= t2
140 20
61.7ln 140 / 20mt C
,co
100 60
78.3 Cln 100 / 60mt
,counter
<It isclear thatt m,co < tm,counter
33
Comparison 1——heat transfer area needed
At same flow rate and same T1, T2, t1 and t2 , for same heat
transfer task
Q = w1Cp1(T1 - T2) = w2Cp2(t2 - t1) = UA t△ m
m co counter
co counter
* , for same Q and K from the comparison of Δt A >A ;
*for a given heat exchanger,A and K are constant then Q <Qm
QA
K t
, ,
,
task = capacity
Calculation of heat transfer——temperature difference
34
For a given heating task , Q = W2Cp2(t2 - t1)
consumption of heating medium : W1 = W2Cp2(t2 - t1) / Cp1(T1-T2 )
t
A
t1
t2
T1
T2
cocurrent
counter current : T2min is limited by t1
T2t1
t
A
t2
T1
counter current
cocurrent : T2min is limited by t2
It is obvious that t2 >t1 , then T2co > T2counter , W1co > W1counter
Therefore, for a given task, a counter current heat exchanger consumes little heating medium than a cocurent one.
1 2W T
Calculation of heat transfer——temperature difference
Comparison 2——consumption of heating medium
For a given T1
Problem: Overall Heat Transfer Coefficient
Determination of heat transfer per unit length for heat recoverydevice involving hot flue gases and water.
KNOWN: Geometry of finned, annular heat exchanger. Gas-side temperature and convection coefficient. Water-side flowrate and temperature.
FIND: Heat rate per unit length.
SCHEMATIC:
Do = 60 mm Di,1 = 24 mm Di,2 = 30 mm t = 3 mm = 0.003m L = (60-30)/2 mm = 0.015m
Problem: Overall Heat Transfer Coefficient (cont.)
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) One-dimensional conduction in strut, (4) Adiabatic outer surface conditions, (5) Negligible gas-side radiation, (6) Fully-developed internal flow, (7) Negligible fouling.
PROPERTIES: Table A-6, Water (300 K): k = 0.613 W/mK, Pr = 5.83, = 855 10-6 Ns/m2.
ANALYSIS: The heat rate is
where
m,h m,ccq UA T T
w oc c h1/ UA 1/ hA R 1/ hA
i,2 i,1 4w
ln D / D ln 30 / 24R 7.10 10 K / W.
2 kL 2 50 W / m K lm
Problem: Overall Heat Transfer Coefficient (cont.)
the internal flow is turbulent and the Dittus-Boelter correlation gives
4 / 5 0.44 / 5 0.4 2c i,1 D
0.613 W / m Kh k / D 0.023Re Pr 0.023 9990 5.83 1883 W / m K
0.024m
11 2 3chA 1883 W / m K 0.024m 1m 7.043 10 K / W.
The overall fin efficiency is o f f1 A / A 1
2fA 8 2 L w 8 2 0.015m 1m 0.24m
2 2f i,2A A D 8t w 0.24m 0.03m 8 0.003m 0.31m .
From Eq. 11.4,
ftanh mL
mL
With
D 6 2i,1
4m 4 0.161 kg / sRe 9990
D 0.024m 855 10 N s / m
Problem: Overall Heat Transfer Coefficient (cont.)
Hence f 0.800 /1.10 0.907 o f f1 A / A 1 1 0.24 / 0.31 1 0.907 0.928
11 2 2o hhA 0.928 100 W / m K 0.31m 0.0347 K / W.
It follows that
1 3 4cUA 7.043 10 7.1 10 0.0347 K / W
cUA 23.6 W / K
and q 23.6 W / K 800 300 K 11,800 W < for a 1m long section.
where
1/ 2 1/ 22 1m 2h / kt 2 100 W / m K / 50 W / m K 0.003m 36.5m
1/ 2 1mL 2h / kt L 36.5m 0.015m 0.55
1/ 2tanh 2h / kt L 0.499.
Problem: Overall Heat Transfer Coefficient (cont.)
COMMENTS: (1) The gas-side resistance is substantially decreased by using the fins
,2f iA D and q is increased.
(2) Heat transfer enhancement by the fins could be increased further by using a material of larger k, but material selection would be limited by the large value of Tm,h.
III. Rate equation and TTM
The rate equation for a shell-and-tube heat exchanger is the same as for a concentric pipe exchanger:
However, Ui and TTM are evaluated somewhat differently for shell-and-tube exchangers. We will first discuss how to evaluate TTM and then a little later in the notes we will discuss how to evaluate Ui for shell-and-tube exchangers.In a shell-and-tube exchanger, the flow can be single or multipass. As a result, the temperature profiles for the two fluids in a shell-and-tube heat exchanger are more complex, as shown below.
TMTUAq ii
TMTUAq ii
Computation of TTM:
For the concentric pipe heat exchanger, we showed the following (parallel and countercurrent flow):
TTM =
When a fluid flows perpendicular to a heated or cooled tube bank, and if both of the fluid temperatures are varying, then the temperature conditions do not correspond to either parallel or countercurrent. Instead, this is called crossflow.
Tlm
F o r c ro s s f lo w a n d m u l t ip a s s h e a t e x c h a n g e d e s ig n s , w e m u s t in t ro d u c e a c o r re c t io n f o r th e lo g m e a n te m p e ra tu r e d i f f e re n c e ( L M T D ) :
T T M =
Z T ha T hb
T cb T ca
H Tcb Tca
T ha T ca
FG * TLM
The factor Z is the ratio of the fall in temperature of the hot fluid to the rise in temperature of the cold fluid.
The factor H is the heating effectiveness, or the ratio of the actual temperature rise of the cold fluid to the maximum possible temperature rise obtainable (if the warm-end approach were zero, based on countercurrent flow).
From the given values of H and Z, the factor FG can be read from the text book figures:
Therefore, as with the concentric pipe heat exchanger, the true mean temperature difference for the 1-1 exchanger is equal to the log mean temperature difference (TLM).
For multiple pass shell-and-tube designs, the flow is complex and the TLM is less than that for a pure countercurrent design.
We must account for the smaller temperature driving force using a correction factor, FG, which is less than 1 and typically greater than 0.8.
The rate of heat transfer in multiple pass heat exchangers is written as:
LMTUAFq Gwhere TLM is the log mean temperature difference for pure countercurrent flow
Textbook Figures 15.6 a, b
1-2 exchangers
2-4 exchangers
Ten Minute Problem -- FG for multiple pass HE
For a 2-4 heat exchanger with the cold fluid inside the tubes and the following temperatures:
Tca = 85°F Tha = 200°FTcb = 125°F Thb = 100°F
(a) What is the true mean temperature difference?(answer TTM = 31.7°F)
(b) What exchanger area is required to cool 50,000 lbm/hr of product (shell-side fluid) if the overall heat transfer coefficient is 100 Btu/hr-ft2-°F and Cp for the product is 0.45 Btu/lbm-°F?(answer A = 710 ft2)