heat exchanger effectiveness maximum and minimum heat capacity rates number of transfer units...
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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