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


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�



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 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.



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.



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.



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 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



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



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 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



( )

( )

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 ��



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.



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 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 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



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.



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.

The E-NTU Method, 250313

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Transcript of The E-NTU Method, 250313