HEAT EXCHANGERS
Day 2
Heat Exchanger Types
The Overall Heat Transfer Coefficient
1

UA
1

U c Ac

1
1
 o hA c
U h Ah
"

R f ,c
 o A c
"
 Rw 
R f ,h
 o A h

1
 o hA h
Rf” are fouling factor. They are very important in process maintenance.
Heat Exchanger Analysis
UA-LMTD METHOD
 When LMTD can be fixed
 Operational data are taken from PFD (Process Flow
Diagram)
 Known as ‘Sizing Problem’
e-NTU METHOD
 When LMTD can not be fixed
 Geometrical data are taken from heat exchanger
detail
 Know as ‘Rating Problem’
UA-LMTD METHOD
LMTD  DTLM
Parallel-Flow and Counter-flow Heat Exchanger
UA-LMTD METHOD
q h  m h c p , h T h , i  T h , o 
q c  m c c p , c T c , o  T c , i 
q  UA D T LM
qh  qc  q
UA-LMTD METHOD
Special Operating Conditions
UA-LMTD METHOD
Multipass and Cross-flow Heat Exchangers
D T LM  F  D T LM
F is a correction factor that
compensate ‘non ideal’ flow
Which should be either parallel
or counter-flow
, CF
Example
A counterflow, concentric tube heat exchanger is
used to cool the lubricating oil for a large industrial
gas turbine engine. The flow rate of cooling water
through the inner tube (Di = 25 mm) is 0,2 kg/s,
while the flow rate of oil through the outer annulus
(Do = 45 mm) is 0,1 kg/s. The oil and water enter at
temperatures of 100 and 30oC, respectively. How
long must the tube be made if the outlet
temperature of the oil is to be 60oC.
e-NTU METHOD
q max  C min  T h , i  T c , i 
e 
q
q max
q  e  q max
C  m  c p
C max  max C c , C h 
C min  min C c , C h 
Cr 
C min
C max
e  f  NTU , C r 
NTU 
UA
C min
ε is effectiveness of an heat exchanger.
NTU = Number of Transfer Unit
Example 11.3
Hot exhaust gases, which enter a finned-tube,
cross-flow heat exchanger at 300oC and leave at
100oC, are used to heat pressurized water at a
flow rate of 1 kg/s from 35 to 125oC. The
exhaust gas specific heat is approximately 1000
J/kg.K, and the overall heat transfer coefficient
based on the gas-side surface area is Uh = 100
W/m2.K. Determine the required gas-side
surface area Ah using the NTU method.
Compact Heat Exchangers
• Both methods can be applied to analyze compact
heat exchanger
• Special care must be taken when evaluating fin-side
convection heat transfer coefficient
• Internal side convection heat transfer coefficient is
calculated in the same way as the others heat
exchanger type
• Convection heat transfer coefficient is usually
presented in graphical form for Colburn j Factor (jh)
versus Reynolds Numer (Re)
Compact Heat Exchangers
j H  St  Pr
St 
2/3
h
Gc p
Re 
GD h

G   V max 
 VA fr
A ff
G vi 
Dp 
 1
2 
2


m
A ff
2

m
 A fr
 vo

A vm 

 f

1

v

A ff v i 
 i


Example 11.4
• Consider the heat exchanger design of Example 11.3, that
is, a finned-tube, cross-flow heat exchanger with a gasside overall heat transfer coefficient and area of 100
W/m2.K and 40 m2, respectively. The water flow rate and
inlet temperature remain at 1 kg/s and 35oC. However, a
change in operation conditions for the hot gas generator
causes the gases to now enter the heat exchanger with a
flow rate of 1.5 kg/s and a temperature of 250oC. What is
the rate of heat transfer by the heat exchanger, and what
are the gas and water outlet temperature?
Another Example
Steam in the condenser of a power plant is to be
condensed at a temperature of 30°C with
cooling water from a nearby lake, which enters
the tubes of the condenser at 14°C and leaves at
22°C. The surface area of the tubes is 45 m2, and
the overall heat transfer coefficient is 2100
W/m2·°C. Determine the mass flow rate of the
cooling water needed and the rate of
condensation of the steam in the condenser.
Assumption: 1 Steady operating conditions exist. 2 The heat exchanger is
well insulated so that heat loss to the surroundings is negligible and thus
heat trans-fer from the hot fluid is equal to the heat transfer to the cold fluid. 3
Changesin the kinetic and potential energies of fluid streams are negligible. 4
There isno fouling. 5 Fluid properties are constant.
Properties: The heat of vaporization of water at 30°C is hfg = 2431 kJ/kg
and the specific heat of cold water at the average temperature of 18°C is Cp
= 4184J/kg · °C (Table A–9).
Analysis: The schematic of the condenser is given in Figure 13–19. The
condenser can be treated as a counter-flow heat exchanger since the
temperature of one of the fluids (the steam) remains constant.
The temperature difference between the steam and the cooling water at
the two ends of the condenser is:
Then the heat transfer rate in the condenser is determined from,
Q = UAsΔTlm = 2100 W/m2· °C)(45 m2)(11.5°C) = 1.087 x106 W = 1087 kW
Therefore, the steam will lose heat at a rate of 1,087 kW as it flows through
the condenser, and the cooling water will gain practically all of it, since the
con-denser is well insulated.
The mass flow rate of the cooling water and the rate of the condensation
of thesteam are determined from Q·= [m·Cp(ToutTin)]cooling water =
(m·hfg)steam to be