Heat Exchanger Analysis by NTU Method

1. Effectiveness NTU method
Parag Chaware
Department of Mechanical Engineering
Cummins College of Engineering, Pune
Effectiveness NTU method Parag Chaware 1 / 13

2. LMTD I
LMTD method is very suitable for determining the size of a heat exchanger
to realize prescribed outlet temperatures when the mass flow rates and the
inlet and outlet temperatures of the hot and cold fluids are specified.
With the LMTD method, the task is to select a heat exchanger that will meet
the prescribed heat transfer requirements. The procedure to be followed by
the selection process is:
Select Heat Exchanger Determine temperatures
Calculate ∆Tlm
Calculate U
Calculate As
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3. Alternative of LMTD
The heat transfer surface area A and the inlet temperatures are
defined of the heat exchanger but the outlet temperatures are not.
Task is to determine the heat transfer performance
The LMTD method could be used for but the procedure require
tedious iterations
Kays and London came up with a method in 1955 called the
effectiveness NTU method, which greatly simplified heat exchanger
analysis
Effectiveness NTU method Parag Chaware 3 / 13

4. NTU
Effectiveness relations of the heat exchangers typically involve the
dimensionless group UAs/Cmin.
U is the overall heat transfer coefficient and As is the heat transfer
surface area of the heat exchanger.
NTU is a measure of the heat transfer surface area As . Thus, the
larger the NTU, the larger the heat exchanger.
We define another dimensionless parameter Capacity Ratio
Capacity Ratio
C =
Cmin
Cmax
(1)
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5. Heat Capacity
The product of the mass flow rate and the specific heat of a fluid into a
single quantity is called the heat capacity rate.
Ch = ṁhcph and Cc = ṁccpc (2)
The rate of heat transfer needed to change the temperature of the
fluid stream by 1 ◦C as it flows through a heat exchanger.
The fluid with a large heat capacity rate will experience a small
temperature change, and the fluid with a small heat capacity rate
will experience a large temperature change.
Effectiveness NTU method Parag Chaware 5 / 13

6. Effectiveness-NTU method
Effectiveness
ε = Actual Heat transfer
Maximum possible heat transfer
Actual heat transfer rateQ̇ = Ch(Th,i − Th,o) or Cc(Tc,o − Tc,i )
(3)
Maximum temperature difference = ∆Tmax = Th,i − Tc,i (4)
Maximum possible heat transfer = Cmin(Th,i − Tc,i ) (5)
(6)
where,
Cmin = ṁhCh or ṁcCc
Effectiveness NTU method Parag Chaware 6 / 13

7. Parallel Flow I
From energy balance
Q̇ = Ch(Th,i − Th,o) = Cc(Tc,o − Tc,i ) (7)
Th,0 = Th,i −
Cc
Ch
(Tc,o − Tc,i ) (8)
(9)
Th,0 = Th,i −
Cc
Ch
(Tc,o − Tc,i ) (10)
LMTD From elementary analysis in LMTD method
ln
(Th,o − Tc,o)
(Th,i − Tc,i )
= −UAs
1
ṁhCh
+
1
ṁcCc
(11)
Effectiveness NTU method Parag Chaware 7 / 13

8. Parallel Flow II
=
−UAs
Cc
1 +
Cc
Ch
(12)
Add and subtract Tc,i on left hand side and substitute Th,o from eq. (10)
ln




Th,i −
Cc
Ch
(Tc,o − Tc,i ) − Tc,o − Tc,i + Tc,i
(Th,i − Tc,i )



 =
−UAs
Cc
1 +
Cc
Ch
(13)
ln




(Th,i − Tc,i ) −
Cc
Ch
(Tc,o − Tc,i ) − (Tc,o − Tc,i )
(Th,i − Tc,i )



 =
−UAs
Cc
1 +
Cc
Ch
(14)
ln
1 −
1 +
Cc
Ch
(Tc,o − Tc,i )
(Th,i − Tc,i )
=
−UAs
Cc
1 +
Cc
Ch
(15)
Effectiveness NTU method Parag Chaware 8 / 13

9. Parallel Flow...
ε =
Cc(Tc,o − Tc,i )
CminTh,i − Tc,i
(16)
(Tc,o − Tc,i )
(Th,i − Tc,i )
= ε
Cmin
Cc
(17)
(18)
ε =
1 − exp
−
UAs
Cc
1 +
Cc
Ch
Cmin
Cc
1 +
Cc
Ch
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10. NTU
NTU (Number of Transfer Units) =
UAs
Cmin
Taking capacity ratio C =
Cmin
Cmax
Taking Cc or Ch as Cmin
Effectiveness for Parallel flow
εParallel =
1 − exp(−NTU(1 + C))
(1 + C)
Effectiveness for Counter flow
εCounter =
1 − exp[−NTU(1 − C)]
1 − C exp[−NTU(1 − C)]
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11. NTU
NTU
NTU =
UA
Cmin
(19)
=
Heat Capacity rate of heat exchanger W/K
Heat Capacity rate of Fluid(W/K)
(20)
U and Cmin is constant for given flow ;
∴ NTU ∝ Area of heat exchanger
Effectiveness NTU method Parag Chaware 11 / 13

12. LMTD
dQ = −ṁchdTh = −ṁccdTc (21)
dth =
dQ
Ch
(22)
dtc =
dQ
Cc
(23)
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13. dTh − dTc = −dQ
1
Ch
−
1
Cc
(24)
= −UdA(th − tc)
1
Ch
−
1
Cc
(25)
dθ = −UdAθ
1
Ch
−
1
Cc
(26)
Z
dθ
θ
= −UA
1
Ch
−
1
Cc
(27)
ln
θ2
θ1
= −UA
1
Ch
−
1
Cc
(28)
ln
(Th,o − Tc,o)
(Th,i − Tc,i )
= −UAs
1
ṁhCh
+
1
ṁcCc
(29)
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