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3. Heat Exchanger Design - Wikis · PDF file3. Heat Exchanger Design John Richard Thome 1er...
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3. Heat Exchanger Design
John Richard Thome
1er mars 2008
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 1 / 41
3.1 Function and configuration of heat exchangers
Heat exchanger goal : get energy from one fluid mass to another.Simple or composite wall of some kind divides the two flows and provides anelement of thermal resistance between them.
Figure 3.1 Heat exchange.
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 2 / 41
3.1 Function and configuration of heat exchangers
Exception : Direct-contact form of heat exchanger
Steam is bubbled into water. It condenses and the water is heated at the sametime.
Figure 3.2 A direct-contact heat exchanger.
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 3 / 41
3.1 Function and configuration of heat exchangers
Three basic types of heat exchangerThe simple parallel or counterflow configuration
Figure 3.3 Parallel or counterflow heat exchangers.
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 4 / 41
3.1 Function and configuration of heat exchangers
Figure 3.4 Heliflow compact counterflow heat exchanger. (Photograph coutesy ofGraham Manufacturing Co., Inc., Batavia, New York.)
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 5 / 41
3.1 Function and configuration of heat exchangers
The shell-and-tube configuration
Figure 3.3 Two kinds of shell-and-tube heat exchangers.
Figure 3.5 Typical commercial one-shell-pass, two-tube-pass heat exchangers.
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 6 / 41
3.1 Function and configuration of heat exchangersThe cross-flow configuration
Figure 3.6 c The basic 1 ft/1 ft/2 ft module for a waste heat recuperator. It is aplate-fin, gas-to-air cross-flow heat exchanger with neither flow mixed.
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 7 / 41
3.1 Function and configuration of heat exchangersFour typical single-shell-pass heat exchangers (Nomenclature on page 106)
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 8 / 41
3.1 Function and configuration of heat exchangers
Figure 3.7 Four typical heat exchanger configurations. Drawings courtesy of theTubular Exchanger Manufacturers’ Association (TEMA).
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 9 / 41
3.1 Function and configuration of heat exchangersAnother variation on the single-pass configuration
Figure 3.9 The temperature distribution through a condenser.John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 10 / 41
3.2 Evaluation of the mean temperature difference in aheat exchangerOverall heat transfer (LMTD)
Q = UA ∆Tmean with a constant U (3.1)
Figure 3.8 The temperature variation through single-pass heat exchangers.John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 11 / 41
3.2 Evaluation of the mean temperature difference in aheat exchangerHeat transfer area
dQ = U ∆T dA (3.2)
where ∆T = Th − Tc
dQ = −(mcp)h dTh = −Ch dTh (3.2 − 3.3)
dQ = (mcp)c dTc = Cc dTc (3.2 − 3.3)
Where Ch and Cc are the hot and the cold fluid heat capacity rates
This equation can be integrated from the lefthand side :Parallel flow :
Th = Th, in Tc = Tc, in
Counterflow :Th = Th, in Tc = Tc, out
"in" is for inlet, "out" is for outlet, "h" is for the hot fluid and "c" is for the cold
fluid.John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 12 / 41
3.2 Evaluation of the mean temperature difference in aheat exchanger
The temperatures inside areParallel flow :
Th = Th, in −CcCh
(Tc − Tc, in) = Th, in −QCh
(3.4)
Counterflow :
Th = Th, in −CcCh
(Tc, out − Tc) = Th, in −QCh
(3.4)
Equations (3.4) can be solved for the local temperature differences
∆Tparallel = Th − Tc = Th, in −(
1 +CcCh
)Tc +
CcCh
Tc, in (3.5)
∆Tcounter = Th − Tc = Th, in −(
1 − CcCh
)Tc +
CcCh
Tc, out (3.5)
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 13 / 41
3.2 Evaluation of the mean temperature difference in aheat exchanger
Substitution of these in equation (3.2)Parallel flow :
U dACc
=dTc[
−(
1 + CcCh
)Tc + Cc
ChTc, in + Th, in
] (3.6)
Counterflow :U dACc
=dTc[
−(
1 − CcCh
)Tc − Cc
ChTc, out + Th, in
] (3.6)
Equations (3.6) can be integrated across the exchanger∫ A
0
UCc
dA =
∫ Tc, out
Tc, in
dTc[−−−]
(3.7)
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 14 / 41
3.2 Evaluation of the mean temperature difference in aheat exchanger
If U and Cc can be treated as constantParallel flow :
ln
−(
1 + CcCh
)Tc, out + Cc
ChTc, in + Th, in
−(
1 + CcCh
)Tc, in + Cc
ChTc, in + Th, in
= −UACc
(1 +
CcCh
)(3.8)
Counterflow :
ln
−(
1 − CcCh
)Tc, out − Cc
ChTc, out + Th, in
−(
1 − CcCh
)Tc, in − Cc
ChTc, out + Th, in
= −UACc
(1 − Cc
Ch
)(3.8)
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 15 / 41
3.2 Evaluation of the mean temperature difference in aheat exchanger
with the help of the definitions of ∆Ta and ∆Tb, given in Fig. 3.8Parallel flow :
ln
(
1 + CcCh
)(Tc, in − Tc, out) + ∆Tb
∆Tb
= −UA(
1Cc
+1
Ch
)(3.9)
Counterflow :
ln ∆Ta(−1 + Cc
Ch
)(Tc, in − Tc, out) + ∆Ta
= −UA(
1Cc
− 1Ch
)(3.9)
Conservation of energy (Qc = Qh) requires that
CcCh
= −Th, out − Th, inTc, out − Tc, in
(3.10)
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 16 / 41
3.2 Evaluation of the mean temperature difference in aheat exchanger
Then equation (3.9) and equation (3.10) giveParallel flow :
ln
∆Ta−∆Tb︷ ︸︸ ︷
(Tc, in − Tc, out) + (Th, out − Th, in) +∆Tb∆Tb
= ln(
∆Ta∆Tb
)= −UA
(1
Cc+
1Ch
)
Counterflow :
ln(
∆Ta∆Tb − ∆Ta + ∆Ta
)= ln
(∆Ta∆Tb
)= −UA
(1
Cc− 1
Ch
)(3.11)
Finally, we write
1Cc
=Tc, out − Tc, in
Q and 1Ch
=Th, in − Th, out
Q
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 17 / 41
3.2 Evaluation of the mean temperature difference in aheat exchanger
in equation (3.11) and we get for either parallel or counterflow
Q = UA(
∆Ta − ∆Tbln(∆Ta/∆Tb)
)(3.12)
Logarithmic mean temperature difference (LMTD)
∆Tmean = LMTD =∆Ta − ∆Tb
ln(∆Ta/∆Tb)(3.13)
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 18 / 41
3.2 Evaluation of the mean temperature difference in aheat exchanger
Example 3.1
Suppose that we had asked, "What mean radius of pipe would have allowed us tocompute the conduction through the wall of a pipe as though it were a slab ofthickness L = r0 − ri ?" (see Fig. 3.10). To answer this, we compare
Q = kA ∆TL = 2πkl∆T
(rmean
r0 − ri
)with equation (2.21)
Q = 2πkl∆T 1ln(r0 − ri)
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 19 / 41
3.2 Evaluation of the mean temperature difference in aheat exchangerExample 3.1 (bis)
It follows that
rmean =r0 − ri
ln(r0/ri)= logarithmic mean radius
Figure 3.10 Calculation of the mean radius for heat conduction through a pipe.
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 20 / 41
3.2 Evaluation of the mean temperature difference in aheat exchanger
Example 3.2
Suppose that the temperature difference on either end of a heat exchanger, ∆Ta,and ∆Tb, are equal. Clearly, the effective ∆T must equal ∆Ta and ∆Tb in thiscase. Does the LMTD reduce to this value ?
SOLUTION. If we substitute ∆Ta = ∆Tb in equation (3.13), we get
LMTD = 0
Therefore it is necessary to use L’Hospital’s rule
LMTD = ∆Ta = ∆Tb
It follows that the LMTD reduces to the intuitively obvious result in the limit.
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 21 / 41
3.2 Evaluation of the mean temperature difference in aheat exchanger
Extended use of the LMTD
Limitations :
LMTD is restricted to the single-pass parallel and counterflow configurations(can be overcome by adjusting the LMTD for other configurations)
Value of U must be negligibly dependent on T to complete the integration ofequation (3.7)
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 22 / 41
3.2 Evaluation of the mean temperature difference in aheat exchangerExtended use of the LMTD
Figure 3.11 A typical case of a heat exchanger in which U varies dramatically.John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 23 / 41
3.2 Evaluation of the mean temperature difference in aheat exchanger
Extended use of the LMTD
Correction factor, F : is derived analytically from the temperature differencevariation with respect to the log mean temperature difference
Q = UA(LMTD) · F
Tt, out − Tt, inTs, in − Tt, in︸ ︷︷ ︸
P
,Ts, in − Ts, outTt, out − Tt, in︸ ︷︷ ︸
R
(3.14)
where Tt and Ts are temperatures of tube and shell flows, respectively
P is the relative influence of the overall temperature difference(Ts, in − Tt, in) on the tube flow temperature.R, according to eqn. (3.10), equals the heat capacity ratio Ct/Cs
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 24 / 41
3.2 Evaluation of the mean temperature difference in aheat exchangerExtended use of the LMTD
Figure 3.13 The basis of the LMTD in a multipass exchanger, prior to correction.
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 25 / 41
3.2 Evaluation of the mean temperature difference in aheat exchanger
Figure 3.14 LMTD correction factors, F, for multipass shelland- tube heatexchangers and one-pass cross-flow exchangers.
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 26 / 41
3.2 Evaluation of the mean temperature difference in aheat exchanger
Figure 3.14 LMTD correction factors, F, for multipass shelland- tube heatexchangers and one-pass cross-flow exchangers.
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 27 / 41
3.2 Evaluation of the mean temperature difference in aheat exchanger
Figure 3.14 LMTD correction factors, F, for multipass shelland- tube heatexchangers and one-pass cross-flow exchangers.
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 28 / 41
3.2 Evaluation of the mean temperature difference in aheat exchanger
Figure 3.14 LMTD correction factors, F, for multipass shelland- tube heatexchangers and one-pass cross-flow exchangers.
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 29 / 41
3.2 Evaluation of the mean temperature difference in aheat exchanger
Example 3.4
5.795 kg/s of oil flows through the shell side of a two-shell pass, four tube-pass oilcooler. The oil enters at 1810C and leaves at 380C. Water flows in the tubes,entering at 32.C and leaving at 490C. In addition, cp poil = 2282 J/kg·K and U =416 W/m2K. Find how much area the heat exchanger must have. Solution
LMTD = 40.76 K
R = 8.412 P = 0.114
with figure (3.14) F = 0.92 and
Q = UA(LMTD)F
we find the areaA = 121.2 m2
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 30 / 41
3.3 Heat exchanger effectiveness
LMTD method can only be used if all 4 temperatures are knownNTU method can be used if only the 2 intlet temperatures are known
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 31 / 41
3.3 Heat exchanger effectivenessHeat exchanger effectiveness
ε =Ch (Th, in − Th, out)
Cmin (Th, in − Tc, in)=
Cc (Tc, out − Tc, in)
Cmin (Th, in − Tc, in)(3.16)
where Cmin is smaller of Ch and Cc
ifCh < Cc , then Qmax = Ch (Th, in − Tc, in)
Ch > Cc , then Qmax = Cc (Th, in − Tc, in)
ε is actual heat transferred divided by the maximum heat that could possibly betransferred from one stream to the other
It follows that
Q = ε Cmin (Th, in − Tc, in) (3.17)
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 32 / 41
3.3 Heat exchanger effectivenessNumber of transfer units (NTU)
NTU =UACmin
(3.18)
can be viewed as a comparison of the heat capacity of the heat exchanger withthe heat capacity of the flow.
Reduce the parallel-flow result from equation (3.9) based on these definitions
−(
CminCc
+CminCh
)NTU = ln
[−
(1 +
CcCh
)εCminCc
+ 1]
(3.19)
For the parallel single-pass heat exchanger
ε =1 − exp
{−NTU
[1 +
(CminCmax
)]}1 +
(CminCmax
) = f(
NTU,CminCmax
)(3.20)
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 33 / 41
3.3 Heat exchanger effectiveness
The corresponding expression for the counterflow case
ε =1 − exp
{−NTU
[1 −
(CminCmax
)]}1 −
(CminCmax
)exp
{−NTU
[1 −
(CminCmax
)]} (3.21)
Similar calculations give the effectiveness for the other heat exchangerconfigurations.
Example 3.5
Consider the following parallel-flow heat exchanger specification :cold flow enters at 40oC : Cc = 20,000 W/Khot flow enters at 150oC : Ch = 10,000 W/KA = 30 m2 U = 500 W/m2K.
Determine the heat transfer and the exit temperatures.
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 34 / 41
3.3 Heat exchanger effectivenessSolution
In this case we do not know the exit temperatures, so it is not possible tocalculate the LMTD. Instead, we can go either to the parallel-flow effectivenesschart in Figure (3.16) or to equation (3.20), using
NTU =UACmin
= 1.5
CminCmax
= 0.5
and we obtain ε = 0.596. Now from equation (3.17), we find that
Q = ε Cmin (Th, in − Tc, in) = 655.6 kW
From energy balances such as are expressed in equation (3.4), we get
Th ,in = 84.44oC
Th ,out = 72.78oCJohn Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 35 / 41
3.3 Heat exchanger effectivenessFor a single fluid stream flowing through an isothermal pipe, the equation for theeffectiveness in any configuration must reduce to the same common expression asCmax approaches infinity.
In this case ε becomelim
Cmax→∞ε = 1 − e−NTU (3.22)
Figure 3.16 The effectiveness of parallel and counterflow heat exchangers.John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 36 / 41
3.3 Heat exchanger effectiveness
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 37 / 41
3.3 Heat exchanger effectiveness
Figure 3.17 The effectiveness of some other heat exchanger configurations.
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 38 / 41
3.4 Heat exchanger design
Determination of h in a baffled shell remains a problem that cannot be solvedanalytically.
Apart from predicting heat transfer, a host of additional considerations must beaddressed in designing heat exchangers. The primary ones are :
Minimization of pumping power
pumping power =m∆p
ρ(3.23)
wherem is the mass flow rate of the stream∆p is the pressure drop of the streamρ is the fluid density
Determining the pressure drop can be relatively difficult.Minimization of fixed costs
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 39 / 41
3.4 Heat exchanger design
Optimizing the design of an exchanger is not just a matter of making ∆p as smallas possible.
Augmentation of heat exchange by employing fins or roughening elements :will invariably increase the pressure dropcan also reduce the fixed cost of an exchanger by increasing U and byreducing the required areacan reduce the required flow rate by increasing the effectiveness and thusbalance the increase of ∆p
Taborek’s list of design considerations for a large shell-and-tube exchanger :Choise of fluidsCost of the calculationRough estimate of the size of the heat exchangerEvaluate the heat transfer, pressure drop, and cost of various exchangerconfigurations
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 40 / 41
3.4 Heat exchanger design
Once the exchanger configuration is set :U will be approximately setarea becomes the basic design variablproceed along the lines of Section 3.2 or 3.3.If it is possible to begin with a complete specification of inlet and outlettemperatures,
Q︸︷︷︸C ∆T
= U︸︷︷︸know
AF (LMTD)︸ ︷︷ ︸calculable
A can be calculated and the design completed
Usually, a reevaluation of U and some iteration of the calculation is needed.
More often, we begin without full knowledge of the outlet temperatures :invent an appropriate trial-anderror method to get the areamore complicated sequence of trials for pressure drop and cost optimization
John Richard Thome (LTCM - SGM - EPFL) Heat transfer - Heat Exchanger Design 1er mars 2008 41 / 41