Download - The E-NTU Method, 250313
Dr Grant Campbell University of Manchester February 2013
Heat Transfer & Process Integration
The εεεε-NTU Method of Heat Exchanger Design
1. Introducing the εεεε-NTU Method, illustrating its use, and contrasting it with the LMTD
method
You have learnt previously, by way of introduction, that heat exchanger design revolves
around the equation
mTUAQ ∆=�
There are two main types of heat exchanger problem:
• the sizing problem, in which we know the required rate of heat transfer, and need to
decide what size of heat exchanger (of a particular configuration) is required to deliver
that rate of heat transfer, i.e. we know Q� , and we need to calculate A;
• the rating problem (sometimes called the performance problem), in which we have an
existing heat exchanger, of known configuration and size, and we want to calculate what
rate of heat transfer it will deliver, i.e. we know A, and we need to calculate Q� .
In either case, we need to be able to calculate U, the overall heat transfer coefficient, and
∆Tm, the effective average temperature driving force within the heat exchanger.
The basis of the LMTD method of heat exchanger design is the equation:
lmT TUAFQ ∆=�
where ∆Tlm is the log-mean temperature difference. The factor FT indicates how far the flow
of the two fluids in the heat exchanger deviates from counterflow behaviour, and hence how
much less the effective temperature driving force is than that operating under true
counterflow behaviour. The problem is that to calculate ∆Tlm requires us to know all four
temperatures, i.e. the inlet and outlet temperatures for both the hot and the cold stream.
Frequently we only know the inlet temperatures, and want to know what outlet temperatures
will be achieved, without having to specify them in advance. This is particularly the case for
rating problems, in which we have an existing heat exchanger, and want to know how it will
perform if we use it for two fluids of known initial temperatures. In this situation, using the
LMTD method would require an iterative solution in order to find combinations of outlet
temperatures that give equal heat transfer between the two fluids and across the heat
exchanger. The ε-NTU method offers the advantage that only the two inlet temperatures
need to be known, and avoids the need for iterative solutions.
The starting point for the ε-NTU method, by contrast, is the equation:
( )( )inoutp TTcmQ −= ��
For a heat exchanger, the heat lost by the hot stream is equal to the heat gained by the cold
stream, which is equal to the heat transfer rate:
( ) ( ) ( ) ( )1221 cccphhhp TTcmTTcmQ −=−= ���
The ε-NTU Method of Heat Exchanger Design 2
Dr Grant Campbell University of Manchester February 2013
Considering the cold fluid, entering at a temperature Tc1, the maximum possible rate of heat
transfer would correspond to the maximum possible temperature rise of the cold fluid. Under
countercurrent flow, the theoretical maximum temperature that the cold fluid could reach is
the inlet temperature of the hot fluid, i.e. if Tc2 = Th1. (This is a theoretical maximum, as it
would require an infinitely large heat exchanger to achieve.) Thus, the maximum theoretical
rate of heat transfer is given by
( ) ( )11max chcp TTcmQ −= ��
T
Th1
Th2
Tc2
Tc1
�H|∆ �H | = �Q
Slope = hPcm )(
1
�
Slope = cPcm )(
1
�
T
Th1
Th2
Tc2
Tc1
�H�Q
maxQ�
The ε-NTU Method of Heat Exchanger Design 3
Dr Grant Campbell University of Manchester February 2013
Equally, the maximum heat transfer rate from the hot fluid would occur if the hot fluid exited
at the temperature of the inlet cold fluid:
( ) ( )11max chhotp TTcmQ −= ��
In practice, one of these will be limiting. The temperature change is greatest for the fluid that
has the smallest capacity to hold heat, i.e. the one with the smaller ( )pcm� . The other fluid
will have a smaller temperature change – it could therefore never reach the inlet temperature
of the first fluid, no matter how large the heat exchanger. Therefore, the maximum
theoretical heat transfer is given by
( ) ( )11minmax chp TTcmQ −= ��
Now, the actual rate of heat transfer is smaller than this theoretical maximum, because the
heat exchanger is not infinitely large. We can define an effectiveness, ε, such that the actual
heat transfer is less than the maximum theoretical heat transfer:
( )( )
( )( )11
12
11
21
max
ferheat trans al theoreticMaximum
achievedfer heat trans Actual
ch
cc
ch
hh
TT
TTor
TT
TT
Q
Q
−
−
−
−==
=
�
ε
depending which stream has the larger temperature change.
Then, the actual heat transfer achieved is given by
( ) ( )11minmax chp TTcmQQ −== ��� εε
Now, we generally know the inlet temperatures of our two fluids – Th1 and Tc1. We also
generally know their respective mass flowrates and specific heat capacities. So, if we can
find a way of calculating the effectiveness, ε, then we can calculate the actual heat transfer.
(Note that ε looks suspiciously like P, one of the parameters used to calculate FT in the
LMTD method. In fact, P is similarly defined as the thermal effectiveness:
( )
( )
( ) ( )
( ) ( )max11
12
11
12
Q
Q
TTcm
TTcm
TT
TTP
chcoldp
cccoldp
ch
cc
�
�
�
�
=−
−=
−
−=
The subtle difference is that P is defined relative to the cold fluid, irrespective of whether it is
the one that could theoretically achieve the greatest temperature change. By contrast, ε is
defined specifically with respect to the fluid that could in theory achieve the greater
temperature change, i.e. the one with the smaller ( )pcm� .)
For different heat exchanger configurations, the effectiveness, ε, can be shown to be a
function of two further dimensionless parameters, C* and NTU:
( )ionconfiguratexchanger heat NTU, ,*Cf=ε
The ε-NTU Method of Heat Exchanger Design 4
Dr Grant Campbell University of Manchester February 2013
The first of these two dimensionless parameters is C*:
max
min*
C
CC =
where ( )minmin pmcC = , the lower of the heat capacity flowrates of the two fluids, and Cmax is
the higher of the two heat capacity flowrates; C* is thus the ratio of these heat capacity
flowrates. Again, this looks spookily familiar; as R in the LMTD method is also defined as
the ratio of the heat capacity flowrates. The subtle difference here is that in calculating R, the
hot fluid heat capacity flowrate forms the denominator, and the cold fluid heat capacity
flowrate is the numerator, irrespective of which is larger, such that R can be greater than 1 –
in fact, R can vary from 0 to ∞ . By contrast, C* is defined such that the smaller of the two
heat capacity flowrates is on the numerator, such that C* can only vary between 0 and 1.
So, for two fluids,
( ) ( )cpchph cmCcmC �� == ;
One of these will be smaller, giving Cmin, while the other one gives Cmax. Note the units of
heat capacity flowrates: K
W
Ks
J
Kkg
J
s
kg==× . This means that C indicates the rate of heat
uptake corresponding to a temperature rise of 1 K.
Now, the other dimensionless parameter that determines the value of ε is NTU, the Number
of Transfer Units. This is defined as
min
NTUC
UA=
Cmin is, as just noted, the smaller of the heat capacity rates of the two streams. The units,
remember, were W/K – the rate at which this stream can take up or lose heat per degree
change in temperature. The product UA also has the units W/K, indicating in this case the
rate of heat transfer per degree of temperature difference – same units, hence NTU is
dimensionless. It represents in dimensionless form the amount of heat transfer available
(which depends on both U and A) to heat a particular heat capacity rate of fluid (where the
heat capacity rate depends on both the flowrate and the heat capacity). So it clearly relates
the size and performance of the heat exchanger to the required amount of heating. A large
value of NTU would imply a lot of heat available to heat a given fluid at a given flowrate.
(Alternatively, NTU can be viewed as the ratio of where the heat has come from, to where the
heat goes. It comes from a temperature driving force between the two fluids, ∆T. It goes into
a temperature rise of the fluid with the smaller heat capacity. NTU therefore represents the
ratio of the temperature rise achieved for every degree difference in temperature driving
force. A high value of NTU will therefore deliver a large temperature rise for a given
temperature driving force, indicating the heat is being transferred effectively – because of a
high heat transfer coefficient and/or a large area.)
Similarly to FT, charts are available that relate ε, NTU and C*. Let’s do an example to
introduce such a chart and see how it’s used. Then we’ll explore some examples that
explicitly derive the relationships between ε, NTU and C*, so that you can see where these
charts some from.
The ε-NTU Method of Heat Exchanger Design 5
Dr Grant Campbell University of Manchester February 2013
Example: Ethyl alcohol (specific heat capacity = 3840 J kg–1
K–1
) has an initial temperature
of 78°C and a flowrate of 30,000 kg h–1
(8.333 kg s–1
). A one shell pass, two tube pass heat
exchanger is available, consisting of 35 tubes in each pass. The tubes are made of stainless
steel (thermal conductivity = 19 W m–1
K–1
), each of 33 mm external diameter, 26 mm
internal diameter, with a total length over the two passes of 15 m. Cooling water (specific
heat capacity = 4179 J kg–1
K–1
) is available at 9°C and a flowrate of 24665 kg h–1
(6.851 kg
s–1
). Find the heat transfer rate that this heat exchanger would deliver, and the outlet
temperatures of the two streams that would result. A graph of ε versus NTU for a one shell
pass, two tube pass heat exchanger is provided.
Answer
Note that this is the same question asked previously in the tutorial questions, in which you
were asked to determine, using the LMTD method, whether this exchanger was large enough
to cool the ethyl alcohol to 44°C. It is an example of a rating problem – the heat exchanger
already exists, we know what size it is, and we want to know what rate of heat transfer could
be achieved using it. Knowing the rate of heat transfer, we could then calculate the outlet
temperatures for the two streams. However, to do this using the LMTD, we need to know
those outlet temperatures in advance. In practice, we’d need to iterate – guess an outlet
temperature for, say, the hot stream, calculate the heat transfer using ( ) ( )21 hhhp TTcmQ −= �� ,
calculate the cold stream outlet temperature using ( ) ( )12 cccp TTcmQ −= �� , calculate the LMTD,
calculate P and R, calculate FT (or read it from a graph), insert into lmT TUAFQ ∆=� , and see if
this value of Q� equalled that given by the first equation – altering the initial guess until the
two values of Q� were equal. The ε-NTU method avoids this need for iteration.
The ε-NTU Method of Heat Exchanger Design 6
Dr Grant Campbell University of Manchester February 2013
From the tutorial question, we have already calculated that U is around 400 W m–2
K–1
, and
that the outside area of the tubes is around 55 m2. We need to determine Cmin and Cmax:
( )
( ) W/K2863041796.851
W/K319993840333.8
=×==
=×==
cpc
hph
cmC
cmC
�
�
So Cc is Cmin, and:
768.028630
55400NTU
min
=×
==C
UA
8947.031999
28630
max
min*===
C
CC
From the chart, ε = 0.42.
( )
( )
kW830
W829697
9782863042.0
11minmax
=
=
−××=
−== ch TTCQQ εε ��
So, 830 kW is the heat transfer that would actually be achieved in this heat exchanger using
these two fluids at these two flowrates. This would result in outlet temperatures for the two
fluids of 52°C and 38°C, for the hot and cold fluids, respectively. Comparing this with the
original question, which asked whether the heat exchanger was adequate to cool the ethyl
alcohol to 44°C, the answer is once again No – but now we know the temperature we could
actually achieve, something we could only have worked out by iteration using the LMTD
method.
The ε-NTU method avoids the need for iteration, particularly in relation to rating problems.
Note that there is still a problem – that calculating U requires knowledge of the bulk mean
temperature of the two fluids in order to work out their thermophysical properties in order to
calculate convective heat transfer coefficients – and until we know the outlet temperatures,
we can’t know the bulk mean temperatures and hence U accurately. So we should perhaps do
one iteration, using the outlet temperatures we have just calculated in order to calculate U
more accurately. In practice, we can make a good guess for the appropriate bulk mean
temperatures to use, as U won’t change much with temperature, so we can get away with a
direct calculation. If we did decide that one iteration was worth doing, this is still less than
the numerous iterations that would be required using the LMTD method.
This example has illustrated how the ε-NTU method is applied, and has demonstrated its
advantages over the LMTD method. But the ε-NTU method has the further advantage that,
with a little practice, it is easier to understand what the various parameters and their relations
to each other mean – it is not such a ‘black box’ as the LMTD method.
The ε-NTU Method of Heat Exchanger Design 7
Dr Grant Campbell University of Manchester February 2013
2. Understanding where the εεεε-NTU method comes from
Let’s consider a heat exchanger operating under countercurrent flow. The heat transfer is
described by:
( ) ( )1221 ccchhh TTCTTCQ −=−=�
Assuming heat losses from the heat exchanger are negligible, then over a small area, dA,
within the heat exchanger, a small amount of heat transfer, Qd � , will occur.
The size of the temperature change always depends on both the mass flowrate of a stream and
its specific heat capacity, which is why these two are always grouped together. The analysis
will be easier to follow if we replace these groupings with the heat capacity flowrate, C:
( ) ( )cpchph cmCcmC �� == ;
Now, the change in heat content of the two streams, in moving a distance dA in the positive A
direction (i.e. from one end of the heat exchanger to the other), is
cchh dTCdTCQd ==�
which is also equal to the heat transfer rate at that point. Clearly, then
c
h
ch dT
C
CdT =
T
Th1
Th2
Tc2
Tc1
Q�
Slope = hPcm )(
1
�
Slope = cPcm )(
1
�
dA
dA
Th
Tc
+ve A direction
( )ch TTUdAQd −=�
ccdTCQd =�
hhdTCQd =�
The ε-NTU Method of Heat Exchanger Design 8
Dr Grant Campbell University of Manchester February 2013
The change in temperature difference is
( )
−=−=
−=−
1h
cccc
h
c
chch
C
CdTdTdT
C
C
dTdTTTd
But
c
cC
QddT
�
=
so
( )
−=
−=−
ch
h
c
c
ch
CCQd
C
C
C
QdTTd
11
1
�
�
( )
−
−=
ch
ch
CC
TTdQd
11
�
Now, from TUAQ ∆=� , at that point in the heat exchanger:
( )( )
−
−=−=
ch
chch
CC
TTddATTUQd
11
�
( )( )
ch
ch
ch TT
TTddA
CCU
−
−=
−
11
Assuming U, Ch and Cc are constant over the whole length of the heat exchanger, integrating
gives:
( )( )
( )( )
( )( )21
12
12
21
0
lnln11
11 21
12
ch
ch
ch
ch
ch
TT
TT ch
ch
A
ch
TT
TT
TT
TT
CCUA
TT
TTddA
CCU
ch
ch
−
−−=
−
−=
−
−
−=
− ∫∫
−
−
−−=
−
−
chch
ch
CCUA
TT
TT 11ln
21
12
The ε-NTU Method of Heat Exchanger Design 9
Dr Grant Campbell University of Manchester February 2013
or
−−=
−
−
chch
ch
CCUA
TT
TT 11exp
21
12
Now, at this point we are clearly quite close to deriving the log-mean temperature difference,
and could do so if we wished (by substituting ( )inout TTQC −= for the two streams).
However, having defined the effectiveness, ε, let’s make use of that and see where it takes us:
( )
( ) ( )1221
11minmax
ccchhh
ch
TTCTTC
TTCQQ
−=−=
−== εε ��
Rearranging, we can write explicit equations for Th2 and Tc2:
( )
h
chhh
C
TTCTT 11min
12
−−=
ε
( )
c
chcc
C
TTCTT 11min
12
−+=
ε
We can then use these to eliminate them from the above equation:
( )
( )
( )
( )
−
−
=
−−
−−
=
−+−
−
−−
=
−
−=
−−
c
h
c
ch
h
ch
c
chch
c
h
chh
ch
ch
ch
C
C
C
C
C
CTT
C
CTT
C
TTCTT
TC
TTCT
TT
TT
CCUA
min
min
min11
min11
11min11
111min
1
21
12
1
1
1
1
11exp
ε
ε
ε
ε
ε
ε
from which:
−=
−−
−
hchc C
C
CCUA
C
C minmin 111
exp1εε
The ε-NTU Method of Heat Exchanger Design 10
Dr Grant Campbell University of Manchester February 2013
Now, let’s suppose, for the sake of example, that Ch is the smaller of the two, i.e. Ch = Cmin.
( )εε
−=
−−
− 1
11exp1
maxminmax
min
CCUA
C
C
Rearranging to give an explicit expression for ε gives:
−−
−
−−−
=
−−−=
−−
−
=
−−
−
−−+−
=
−−
−+−
maxminmax
min
maxmin
maxminmaxminmax
min
maxminmax
min
maxmin
maxminmax
min
11exp1
11exp1
11exp1
11exp1
011
exp11
exp1
011
exp11
CCUA
C
C
CCUA
CCUA
CCUA
C
C
CCUA
C
C
CCUA
CCUA
C
C
ε
ε
εε
εε
Remembering once again that we defined
max
min*
C
CC =
and
min
NTUC
UA=
we can simplify our expression to give:
−−
−
−−−
=
max
min
min
min
minmax
min
max
min
min
min
min
exp1
exp1
C
C
C
C
C
UA
C
C
C
C
C
C
C
UA
ε
( )[ ]
( )[ ]( )**
*
1NTUexp1
1NTUexp1
CC
C
−−−
−−−=ε
The ε-NTU Method of Heat Exchanger Design 11
Dr Grant Campbell University of Manchester February 2013
The equivalent analysis for parallel flow would give
( )[ ]*
*
1
1NTUexp1
C
C
+
+−−=ε
Both of these expressions remain the same whether it is the hot or the cold fluid that has Cmin!
Having got these expressions, we don’t need a chart of ε versus NTU – we can calculate ε
directly, for a counterflow or parallel flow heat exchanger. Equivalent expressions can be
developed for other heat exchanger configurations.
Example: A 30 kg s–1
stream of toluene, initially at a temperature of 15°C, is to be heated to
50°C, using hot water initially at 95°C and available at a flowrate of 15 kg s–1
. A heat
exchanger is available with a surface area for heat transfer of 30 m2, in which the overall heat
transfer coefficient for these two fluids is estimated to be 1500 W m–2
K–1
. Calculate whether
the heat exchanger would be able to perform the required duty under (a) counterflow and (b)
parallel flow.
Specific heat capacity of toluene at the mean temperature of 32.5°C = 1800 J kg–1
K–1
.
Specific heat capacity of water = 4180 J kg–1
K–1
Answer
W/K54000180030
W/K62700418015
=×=
=×=
c
h
C
C
So Cc = Cmin.
8333.054000
301500NTU
min
=×
==C
UA
8612.062700
54000
max
min*===
C
CC
For countercurrent flow,
( )[ ]( )[ ]( )
( )[ ]( ) ( )[ ]
469.0
2328.0
1092.0
8612.013338.0exp8612.01
8612.013338.0exp1
1NTUexp1
1NTUexp1**
*
=
=
−−−
−−−=
−−−
−−−=
CC
Cε
The ε-NTU Method of Heat Exchanger Design 12
Dr Grant Campbell University of Manchester February 2013
( )
( )
W4320000
159554000
11minmax
=
−×=
−= ch TTCQ�
W20260004320000469.0max =×== QQ ε
For the cold (toluene) stream:
( ) ( )
( )
C5.52
15540002026000
2
2
12
°=
−=
−=
c
c
cccp
T
T
TTcmQ ��
For the hot (water) stream:
( ) ( )
( )
C7.62
95627002026000
2
2
21
°=
−=
−=
h
h
hhhp
T
T
TTcmQ ��
The heat exchanger would (probably) just be adequate to heat the toluene to 50°C – only
probably, because design procedures are not perfect or precise. Under the conditions
described, the heat exchanger is not particular effective – the effectiveness value is under
47%. This is because the number of transfer units is small – less than one.
For parallel flow,
( )[ ]
( )[ ]
423.0
8612.1
7880.0
8612.01
8612.018333.0exp1
1
1NTUexp1*
*
=
=
+
+−−=
+
+−−=
C
Cε
W18270004320000423.0max =×== QQ �� ε
For the cold (toluene) stream:
( ) ( )
( )
C8.48
15540001827000
2
2
12
°=
−=
−=
c
c
cccp
T
T
TTcmQ ��
The ε-NTU Method of Heat Exchanger Design 13
Dr Grant Campbell University of Manchester February 2013
For the hot (water) stream:
( ) ( )
( )
C9.65
95627001827000
2
2
21
°=
−=
−=
h
h
hhhp
T
T
TTcmQ ��
Under parallel flow, the exchanger is less efficient than under counterflow (its effectiveness
drops to 42%), and would not be adequate to heat the toluene to 50°C.
Now, the above example raises some interesting questions. What if Cmin and Cmax were
equal? (We could achieve this by altering the water flowrate so that the two were equal.)
Would this be more efficient? It would imply that the temperature change for both streams
was the same and, in a countercurrent heat exchanger, that the temperature driving force was
therefore uniform throughout the heat exchanger. This seems to offer an efficient
arrangement.
The problem is that when we set Cmin equal to Cmax in the countercurrent flow equation, the
equation becomes indeterminate. However, by applying l’Hopital’s rule, it can be shown that
for Cmin = Cmax, the equation simplifies to:
NTU1
NTU
+=ε
The equation for parallel flow does not give this problem of an indeterminate form, and
simplifies straightforwardly to:
( )NTU212
1 −−= eε
For the example above, for NTU = 0.8333, these equations yield effectiveness values of
0.455 for counterflow, and 0.406 for parallel flow – less than was obtained for different
values of Cmin and Cmax. So in fact, having Cmin = Cmax is the least efficient arrangement.
How about the other extreme – Cmin/Cmax = 0? This would occur when one of the streams
undergoes phase change at constant temperature, which implies a Cmax of ∞ . In this
situation, the equations for parallel flow and counterflow both simplify to:
NTU1 −
−= eε
For the example just given, if we were to use low pressure steam condensing at 95°C as our
hot stream, instead of hot water, then the effectiveness becomes 565.01 8333.0=−=
−eε . So
having Cmin/Cmax = 0 is the most efficient arrangement. Also, in this situation in which one of
the streams has a constant temperature, there is no difference between parallel flow and
counterflow – their heat transfer behaviour is the same. In fact, all heat exchanger
configurations default to this same behaviour if one of the streams has a constant
temperature. For the same values of U and A, and hence for the same value of NTU, all heat
exchanger designs give the same effectiveness when operating as either condensers or
boilers.
The ε-NTU Method of Heat Exchanger Design 14
Dr Grant Campbell University of Manchester February 2013
3. Understanding what the various parameters of the εεεε-NTU method and their relations to
each other mean
Having derived ε-NTU relationships for counterflow and parallel flow configurations, and
explored their behaviour, we have begun to get an appreciation of what these various
parameters mean and how they relate to each other. Considering once again the graph of ε
versus NTU for a one shell pass, two tube pass heat exchanger, we can begin to see why the
curves fall as they do.
The graph shows curves of ε as a function of NTU for different values of C*. As we have just
discovered for counterflow and parallel flow, we can see that for a 1-2 HE, C* = 1 gives the
least efficient heat transfer, and C* = 0 gives the most efficient heat transfer. For any given
value of C*, increasing NTU increases ε – this makes sense, as increasing the overall heat
transfer coefficient or the area ought to give more heat transfer. For a given value of NTU,
we could modify C* to achieve more effective heat transfer by altering the flowrates of our
two fluids in order to alter Cmin or Cmax.
The graph also shows two other curves. The dotted curve shows the point at which we are
achieving 95% of the maximum effectiveness for a given C*. Increasing the effectiveness by
increasing NTU beyond this point is not good economics – a large increase in size will be
needed for a small increase in heat transfer, and the increased cost would not be justified. We
should aim to design our heat exchanger to deliver less than 95% of εmax. The solid curve
that crosses through the curves for different values of C* indicates the temperature meet
condition, i.e. the point at which the two outlet temperatures are equal. To the right of this
curve, temperature cross occurs, which can result in heat transfer in the wrong direction in
parts of the heat exchanger.
The ε-NTU Method of Heat Exchanger Design 15
Dr Grant Campbell University of Manchester February 2013
The graphs below show the equivalent charts for countercurrent and co-current heat
exchangers; these show the same relationships, and give the same results, as the equations
just derived. Note some interesting features: e.g. because εmax is always 1 for a countercurrent
heat exchanger, 95% of εmax is a horizontal line, in contrast to co-current heat exchangers.
The ε-NTU Method of Heat Exchanger Design 16
Dr Grant Campbell University of Manchester February 2013
The tables below lists the explicit expressions for ε as a function of NTU and C* for
counterflow and parallel flow, and for several other heat exchanger configurations. It also
shows the inverse expressions for NTU = f(ε, C*), which can be used for sizing problems, i.e.
to find the area (given by NTU) when a known heat transfer rate is required.
Heat exchanger configuration ε = f(NTU, C*)
Counterflow ( )[ ]( )[ ]( )**
*
1NTUexp1
1NTUexp1
CC
C
−−−
−−−=ε
Parallel flow ( )[ ]*
*
1
1NTUexp1
C
C
+
+−−=ε
Shell and tube:
1 shell pass, even no. tube
passes ( )
( )
( )
+−−
+−+
+++
=
212*
212*
212**
1
1NTUexp1
1NTUexp1
11
2
C
C
CC
ε
Shell and tube: n shell passes,
2n, 4n etc. tube passes
−
−
−
−
−
−
=
*
1
*
1
1
*
1
1
11
1
1
1
CCC
nn
ε
ε
ε
ε
ε
Heat exchanger configuration NTU = f(ε, C*)
Counterflow
1Cfor1
NTU
1Cfor1
1ln
1
1NTU
*
**
*
=−
=
<
−
−
−=
ε
ε
ε
εC
C
Parallel flow ( )[ ]*
*
1
11lnNTU
C
C
+
+−=
ε
Shell and tube: 1 shell pass,
even no. tube passes
( )
( )
( )
+++−
+−+−
+
=212**
212**
212*112
112
ln
1
1NTU
CC
CC
C ε
ε
Shell and tube: n shell passes,
2n, 4n etc. tube passes
Too complicated – look it up in a textbook
if you ever need it.
Remember, in all cases, for C* = 0, these equations simplify to
( )NTUexp1 −−=ε
The ε-NTU Method of Heat Exchanger Design 17
Dr Grant Campbell University of Manchester February 2013
A further interesting question is: What is the maximum possible effectiveness, when NTU →
∞ ? The table below summarises this. Putting C* = 1, which represents the worst case,
shows the relative efficiencies of the various configurations, from which we can see that
parallel flow is the worst configuration.
Heat exchanger configuration Value of ε when NTU→ ∞ Value when NTU→ ∞ and
C* = 1
Counterflow 1=ε , for all values of C* 1
Parallel flow *1
1
C+=ε
0.5
Shell and tube: 1 shell pass,
even no. tube passes ( ) 212** 11
2
CC +++=ε 5858.0
22
2=
+
One final benefit of the ε-NTU method. Returning to our earlier example, we can ask the
question: What value of cold water mass flowrate would we need, in order to achieve the
required cooling of ethyl alcohol?
The required cooling is given by
( )
( )
kW1088
W1087966
447831999
21
=
=
−×=
−= hhh TTCQ�
If we increase the cooling water flowrate sufficiently, then Ch will become Cmin. The
required effectiveness can then be calculated from
( )
( )
493.0
978319991088000
11minmax
=
−××=
−==
ε
ε
εε ch TTCQQ ��
688.031999
55400NTU
min
=×
==C
UA
From the chart, a value of ε of 0.5 and a value of NTU of 0.7 corresponds to a value of C* of
0! This implies that we could only achieve the required cooling in this heat exchanger if we
had an infinite value of Cc. We could achieve this by using some suitably low boiling liquid
to provide the cooling, as boiling (and condensation) give C* = 0. Alternatively, we could
use cooler cooling water, i.e. lower its inlet temperature. Again, the ε-NTU chart lets us
easily calculate whether this would work.
These examples demonstrate, in addition to avoiding iterative solutions, the greater power
and understanding delivered by the ε-NTU method compared with the LMTD method.
The ε-NTU Method of Heat Exchanger Design 18
Dr Grant Campbell University of Manchester February 2013
4. Summary
It is often helpful, when attempting problems using this method, to write down the
fundamental equations on which this method is based:
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( )11min2112
11min2112
chhhhccc
chphhhpcccp
TTCTTCTTC
TTcmTTcmTTcmQ
−=−=−=
−=−=−=
ε
ε ����
From this, we are reminded that there are various ways of calculating ε, depending on the
information given in the question:
( ) ( ) ( )11min11min chchp TTC
Q
TTcm
Q
−=
−=
�
�
�
ε
or
( )( )
( )( )11
12
11
21
ch
cc
ch
hh
TT
TTor
TT
TT
−
−=
−
−= εε
depending on whether ( )21 hh TT − or ( )12 cc TT − is larger.
It is worth remembering that C* can also be calculated directly from the temperatures:
max
min
max
min*
T
T
C
CC
∆
∆==
Understanding that the ε-NTU method employs ratios, and understanding the physical
significance of those ratios, will help in deploying the equations appropriately and correctly.