Heat Exchanger Effectiveness

Maximum and Minimum Heat Capacity Rates

Number of Transfer Units

Maximum Temperature Difference

LMTD application is easy when the temperatures are known

When the outlet temperatures are unknown tedious iterations are required to solve problems

Kays & London developed the Effectiveness-NTU

Method for simplifying heat exchanger analysis: RateTransferHeatPossibleMaximum

rateTransferHeatActual

Qactual

Q

max

.

Qmax = the heat transfer rate achievable were it possible to heat the lower heat

capacity stream to the inlet temperature of the other.

Qactual (NRG bal) = Cc(Tc,out – Tc,in) = Ch(Th,in – Th,out)

Cc = mcCpc and Ch = mhCph The heat capacity rates of the cold and hot streams, respectively

ΔTmax=Th,in –Tc,in ( max temp. diff. is hot fluid-cold fluid inlet temp. diff.)

Maximum heat transfer Cold fluid heated to inlet temp of hot fluid Hot fluid cooled to inlet temp of cold fluid Fluid with smaller heat capacity rate has larger temp. change.

Qmax = Cmin(Th,in – Tc,in)

(1)A flow of 1.0 kg/s of an organic liquid of heat capacity 2.0 kJ/kg K is cooled from 350.0 K to 330.0 K by a stream of water flowing counter currently through a double pipe heat exchanger. Estimate the effectiveness of the unit if the water enters the exchanger at 290.0 K.

(2) Cold water enters a counter flow heat exchanger at 10.0 oC at a rate of 8.0 kg/s where it is heated by a hot water stream that enters the heat exchanger at 70.0 oC at a rate of 2.0 kg/s. Assuming the specific heat capacity of water to be constant at Cp=4.18 kJ/kg oC, determine the maximum heat transfer rate and the outlet temperatures of the cold and hot water streams.

Since

So, with the effectiveness of the heat exchanger we can determine the heat transfer rate without knowing the outlet temperatures.

Effectiveness is a function of heat exchanger geometry and flow arrangement

1,1,minmaxmax

ch TTCQQQQ

Recall from alternate method of deriving LMTD

For a parallel flow H/E. Also, from Qactual:

Substituting this into equation at the top and adding & subtracting Tc,1 gives:

h

C

C

T

CPCPhh

Tincinh

outcouth

C

C

C

UA

CmCmUA

TT

TT1

11ln

,

..,,

,,

1,2,1,2, CCh

Chh TT

C

CTT

h

c

c

T

ch

cch

cccch

C

C

C

UA

TT

TTC

CTTTT

1ln1,1,

1,2,2,1,1,1,

Which simplifies to:

h

c

c

T

ch

cc

h

c

C

C

C

UA

TT

TT

C

C111ln

1,1,

1,2,

Manipulating the effectiveness defining equation gives:

cch

cc

ch

ccc

C

C

TT

TT

TTC

TTC

Q

Q min

1,1,

1,2,

1,1,min

1,2,

max

Substituting this result into the equation at the top and solving for ε gives the following relation for the effectiveness of a parallel flow heat exchanger:

ch

c

h

c

c

s

Parallel

C

C

C

C

C

C

C

UA

min1

1exp1

Taking either Cc or Ch to be Cmin gives:

maz

s

Parallel

C

C

C

C

C

UA

min

max

min

min

1

1exp1

UAs/Cmin is a dimensionless group called Number of Transfer Units or NTU, where:

min

.min

P

ss

Cm

UA

C

UANTU

max

min

C

Cc

Heat exchanger effectiveness is a function of both NTU and capacity ratio

Kays and London investigated and developed effectiveness and capacity ratios for many heat exchanger arrangements and made the results available in chart form

Eg.

Repeat the last problem where the oil flow rate was halved, using the effectiveness method.

For greater precision the charts not very userfriendly

For computer based analyses expressions of the curves would be more applicable

Expressions were developed for both NTU and ε relations

1

1exp1

1exp1

1

1exp1

NTU

NTU

cNTUc

cNTU

c

cNTU

Double pipe

parallel flow

counter flow

counter flow, c=1

1ln

11/2

11/2ln1

1ln1ln1

1ln1

1ln

21

2

21

221

2

NTU

cc

cccNTU

ccc

NTU

cc

NTU

Cross flow:

cmax mixed, cmin unmixed

cmax unmixed, cmin mixed

Shell & tube

one shell pass 2,4,6, tube passes

All exchangers, c=0

A process requires a flow of 4.0 kg/s of purified water at 340.0 K to be heated from 320.0 K by 8.0 kg/s of untreated water which can be available at 380.0, 370.0, 360.0, or 350.0 K. Estimate the heat transfer surfaces of one shell pass two tube pass heat exchangers that are suitable for these duties. In all cases the mean heat capacity of the water is 4.18 kJ/kg K. bn

If surface area is the major concern for H/E selection can two half sized H/E suffice for one large one?

Example#4 : A counter flow concentric tube H/E is used to heat 1.25 kg/s of water from 35.0 K to 80.0 K by cooling an oil with cP=2.0 kJ/kg K from 150.0 K to 85.0 K. The overall U= 850.0 kW/m2 K. A similar arrangement is to be built at another plant but for comparison, two smaller counter flow H/Es connected in series on the water side and in parallel on the oil side are contemplated. If the flow is split equally between the two H/Es and the overall U is the same for all the H/Es but the smaller H/Es cost more per unit surface area, which arrangement would be the more economical; the larger or the two equally sized H/Es ?

H/Exchanger 1 H/Exchanger 2

Toil 1=150K

Toil 2=85K

Toil 2,1

Tw3=80KTw2

Toil 22

Tw1=35 K