Chapter 11: Heat Exchangers

Dr Ali Jawarneh
Department of Mechanical Engineering
Hashemite University
 Objectives
When you finish studying this chapter, you should be able to:
• Recognize numerous types of heat exchangers, and classify
them,
• Develop an awareness of fouling on surfaces, and determine
the overall heat transfer coefficient for a heat exchanger,
• Perform a general energy analysis on heat exchangers,
• Obtain a relation for the logarithmic mean temperature
difference for use in the LMTD method, and modify it for
different types of heat exchangers using the correction factor,
• Develop relations for effectiveness, and analyze heat
exchangers when outlet temperatures are not known using the
effectiveness-NTU method,
• Know the primary considerations in the selection of heat
exchangers.
 Types of Heat Exchangers
• Different heat transfer applications
require different types of hardware
and different configurations of heat
transfer equipment.

What should we choose?

 (1) Double-Pipe Heat Exchangers
• The simplest type of heat exchanger is called
the double-pipe heat exchanger.
• One fluid flows through the smaller pipe while
the other fluid flows through the annular space
between the two
pipes.
• Two types of flow
arrangement
– parallel flow,
– counter flow.
 (2) Compact Heat Exchanger
• Large heat transfer surface area per unit volume.
• Area density β ─ heat transfer surface area of a heat
exchanger to its volume ratio.
• Compact heat exchanger β >700 m2/m3.
• Examples:
– car radiators (β ≈1000 m2/m3),
– glass-ceramic gas turbine heat
exchangers (β ≈6000 m2/m3),
– the regenerator of a Stirling
engine (β ≈15,000 m2/m3), and
– the human lung (β ≈20,000 m2/m3).
• Compact heat exchangers are commonly used
in
– gas-to-gas and
– gas-to liquid (or liquid-to-gas) heat exchangers.
• Typically cross-flow configuration ─ the two
fluids move perpendicular to each other.
• The cross-flow is further classified as
– unmixed flow
and
– mixed flow.
(3) Shell-and-Tube Heat Exchanger
• The most common type of heat exchanger in industrial
applications.
• Large number of tubes are packed in a shell with their axes
parallel to that of the shell.
• The other fluid flows outside the tubes through the shell.
• Baffles are commonly placed in the shell.
• Shell-and-tube heat exchangers are relatively large size and
weight.
• Shell-and-tube heat
exchangers are further
classified according to
the number of shell and
tube passes involved.
 (4) Plate and Frame Heat Exchanger
• Consists of a series of plates with corrugated flat flow
passages.
• The hot and cold fluids flow in alternate passages
• Well suited for liquid-to-liquid heat exchange
applications, provided that
the hot and cold fluid
streams are at about the
same pressure.
The Overall Heat Transfer Coefficient
• A heat exchanger typically involves two flowing
fluids separated by a solid wall.
• Heat is transferred
– from the hot fluid to the wall by
convection,
– through the wall by conduction, and
– from the wall to the cold fluid by
convection.
• The thermal resistance network
– two convection and
– one conduction resistances.
• For a double-pipe heat exchanger, the thermal
resistance of the tube wall is
ln ( D0 Di )
Rwall = (11-1)
2π kL
• The total thermal resistance
1 ln ( D0 Di ) 1
Rtotal = Ri + Rwall + Ro = + + (11-2)
hi Ai 2π kL ho Ao
• When one fluid flows inside a circular
tube and the other outside of it, we have
Ai = π Di L ; Ao = π Do L
• It is convenient to combine all the thermal
resistances in the path of heat flow from the
hot fluid to the cold one into a single resistance
R
 ΔT
Q= = UAΔT = U i Ai ΔT = U o Ao ΔT (11-3)
R
U is the overall heat transfer coefficient,
whose unit is W/m2ºC.
• Canceling T, Eq. 11–3 reduces to
1 1 1 1 1
= = =R= + Rwall + (11-4)
UAs U i Ai U o Ao hi Ai ho Ao
• When the wall thickness of the tube is small
and the thermal conductivity of the tube
material is high (Rwall=0) and the inner and
outer surfaces of the tube are almost identical
(Ai≈Ao≈As), Eq. 11–4 simplifies to
1 1 1
≈ + (11-5)
U hi ho
1 1
• When hi>>ho ≈
U ho
1 1
• When hi<<ho ≈
U hi
• The overall heat transfer coefficient U in Eq. 11–5 is dominated
by the smaller convection coefficient { hwater>>hoil>>hgas }

• EXAMPLE: If (hi<< ho), we have 1/hi >> 1/ho, and thus U ≈ hi.
Therefore, the smaller heat transfer coefficient creates a bottleneck
on the path of heat transfer and seriously impedes heat transfer.

• This situation arises frequently when one of the fluids is a gas and
the other is a liquid. In such cases, fins are commonly used on the
gas side to enhance the product UA and thus the heat transfer on
that side

• Usually gases have very low thermal conductivities

• When the tube is finned on one side to enhance heat transfer, the
total heat transfer surface area on the finned side becomes
 Fouling Factor
• The performance of heat exchangers usually
deteriorates with time as a result of accumulation of
deposits on heat transfer surfaces.
• The layer of deposits represents additional resistance
to heat transfer and causes the rate of heat transfer in a
heat exchanger to decrease.
• The fouling factor Rf ─ The net effect of these
accumulations on heat transfer.
• Two common type of fouling:
– precipitation of solid deposits in a
fluid on the heat transfer surfaces.
– corrosion and other chemical fouling.
• The overall heat transfer coefficient needs to
be modified to account for the effects of
fouling on both the inner and the outer surfaces
of the tube.
• For an unfinned shell-and-tube heat exchanger,
it can be expressed as
1 R f ,i ln ( D0 Di ) R f ,o 1
R= + + + + (11-8)
hi Ai Ai 2π kL Ao ho Ao
Rf,i and Rf,o are the fouling factors at those
surfaces.
• Rf,i and Rf,o Table 11-2
EXAMPLE: Water at an average temperature of 107 °C and
an average velocity of 3.5 m/s flows through a 5-m-long
stainless steel tube (k = 14.2 W/m·°C) in a boiler. The inner and
outer diameters of the tube are Di = 1.0 cm and Do = 1.4 cm,
respectively. If the convection heat transfer coefficient at the
outer surface of the tube where boiling is taking place is
ho=8400 W/m2·°C, determine the overall heat transfer
coefficient Ui of this boiler based on the inner surface area of
the tube.

Outer surface
D0, A0, h0, U0 , Rf0
Inner surface
Di, Ai, hi, Ui ,
Rfi
Properties The properties water at 107°C ≈ 110°C are (Table A-9)

υ = μ / ρ = 0.268 × 10 −6 m 2 /s
k = 0.682 W/m 2 .K
Pr = 1.58
Analysis:
Vm Dh (3.5 m/s)(0.01 m)
Re = = = 130,600
υ 0.268 × 10 −6 2
m /s

which is greater than 10,000. Therefore, the flow is turbulent.

Assuming fully developed flow,
hDh
Nu = = 0.023 Re 0.8 Pr 0.4 = 0.023(130,600) 0.8 (1.58) 0.4 = 342
k

k 0.682 W/m.°C
h= Nu = (342) = 23,324 W/m 2 .°C
Dh 0.01 m
 1 ln( Do / Di ) 1
R = Rtotal = Ri + R wall + Ro = + +
hi Ai 2πkL ho Ao
1 ln(1.4 / 1)
= +
( 23,324 W/m 2 .°C)[π (0.01 m)(5 m)] [2π (14.2 W/m.°C)(5 m)]
1
+
(8400 W/m 2 .°C)[π (0.014 m)(5 m)]
= 0.00157°C/W

1 1 1
R= ⎯
⎯→ U i = = = 4055 W/m 2 .°C
U i Ai RAi (0.00157°C/W)[π (0.01 m)(5 m)]
 Analysis of Heat Exchangers
• Two different design tasks:
1) Specified:
- the temperature change in a fluid stream, and
- the mass flow rate.
Required:
- the designer needs to select a heat exchanger.
2) Specified:
- the heat exchanger type and size,
- fluid mass flow rate,
- inlet temperatures.
Required:
- the designer needs to predict the outlet temperatures and heat transfer
rate.
• Two methods used in the analysis of heat exchangers:
– the log mean temperature difference (or LMTD)
• best suited for the #1,
– the effectiveness–NTU method NTU: Number of Transfer Units
• best suited for task #2.
• The analysis of heat exchangers can be greatly
simplify by making the following assumptions,
which are closely approximated in practice:
– steady-flow,
– kinetic and potential energy changes are
negligible,
– the specific heat of a fluid is constant,
– axial heat conduction along the tube is negligible,
– the outer surface of the heat exchanger is perfectly
insulated.
• The first law of thermodynamics requires that
the rate of heat transfer from the hot fluid be
equal to the rate of heat transfer to the cold
one.
 Heat Capacity Rate
• The transfer rate to the cold fluid: {Cc, Ch}

Q = m c (T − T ) = C (T − T
c pc c ,out c ,in c c , out c ,in );C c = m c c pc
(11-9) (11-12) (11-11)
• The transfer rate to the hot fluid:
Q = m h c ph (Th ,in − Th ,out ) = Ch (Th ,in − Th ,out ) ; Ch = m h c ph
(11-10) (11-13) (11-11)
• Two special types of heat exchangers commonly used
in practice are condensers and boilers.
• One of the fluids in a condenser or a boiler undergoes
a phase-change process, and the rate of heat transfer
is expressed as
Q = mh
 fg (11-14)
The Log Mean Temperature Difference
Method
• The temperature difference between the hot and cold
fluids varies along the heat exchanger.
Æ it is convenient to have a mean temperature
difference Tm for use in the relation
Q = UAs ΔTm (11-15)

• Consider the parallel-flow

double-pipe heat exchanger.
• An energy balance on each fluid in a differential
section of the heat exchanger
⎧⎪δ Q = − m h c ph dTh (11-16)
⎨ 
⎪⎩δ Q = m c c pc dTc (11-17)
⎧ δ Q
⎪dTh = −  (11-18)
⎪ mh c ph
⎨ 
⎪dT = δ Q
⎪ c m c c pc (11-19)
⎩
• Taking their difference, we get
⎛ 1 1 ⎞
dTh − dTc = d (Th − Tc ) = −δ Q ⎜ +
⎜ m h c ph m c c pc ⎟⎟
(11-20)
⎝ ⎠
• The rate of heat transfer in the differential section
of the heat exchanger can also be expressed as
δ Q = U (Th − Tc ) dAs (11-21)

• Substituting this equation into Eq. 11–20 and

rearranging give
d (Th − Tc ) ⎛ 1 1 ⎞
= −UdAs ⎜ +
⎜ m h c ph m c c pc ⎟⎟
(11-22)
Th − Tc ⎝ ⎠
• Integrating from the inlet of the heat exchanger to
its outlet, we obtain
Th ,out − Tc ,out ⎛ 1 1 ⎞
= −UAs ⎜ +
⎜ m h c ph m c c pc ⎟⎟
ln (11-23)
Th ,in − Tc ,in ⎝ ⎠
• Solving Eqs. 11–9 and 11–10 for mccpc and mhcph and
substituting into Eq. 11–23 give

Q = UA ΔT (11-24)
Applicable for:
s lm 1- parallel and counter-flow double pipe
ΔT1 − ΔT2
ΔTlm = 2- Condenser

ln ( ΔT1 ΔT2 )
(11-25)
3- Boiler

• ΔTlm is the log mean temperature difference.

• ΔT1 and ΔT2 are the temperature difference between
the two fluids at the two ends
(inlet and outlet).
• It makes no difference which
end of the heat exchanger is
designated as the inlet or the
outlet.
 Counter-Flow Heat Exchangers
• The relation already given for the log
mean temperature difference for
parallel-flow heat exchanger can be used
for a counter-flow heat exchanger.
• ΔT1 and ΔT2 are expressed as
shown in the Fig. 11–15.
• ΔTlm, CF > ΔTlm, PF
• A smaller surface area (a smaller heat
exchanger) is needed to achieve a specified
heat transfer rate in a counter-flow heat
exchanger.
 Multipass and Cross-Flow Heat
Exchangers: Use of a Correction Factor
Shell & Tube H.E Compact H.E

• The log mean temperature difference relation

developed earlier is limited to parallel-flow and
counter-flow heat exchangers only.
• To simplify the analysis of cross-flow and multipass
shell-and-tube heat exchangers, it is convenient to
express the log mean temperature difference relation
as
ΔTlm = F ΔTlm ,CF (11-26) Q = UA s ΔT lm

• F is the correction factor, and ΔTlm, CF is the log

mean temperature for counter-flow case.
 F Charts for Common Shell-and-Tube
and Cross-Flow Heat Exchangers.
Shell and Tube Heat Exchanger Cross Flow Heat Exchanger

Tubes
EXAMPLE: A double-pipe counter-flow heat exchanger is to
cool ethylene glycol (Cp = 2560 J/kg · °C) flowing at a rate of
3.5 kg/s from 80°C to 40°C by water (Cp = 4180 J/kg · °C) that
enters at 20°C and leaves at 55°C. The overall heat transfer
coefficient based on the inner surface area of the tube is
250 W/m2 · °C. Determine (a) the rate of heat transfer, (b) the
mass flow rate of water, and (c) the heat transfer surface area
on the inner side of the tube.
 Analysis (a) The rate of heat transfer is Cold
Water
20°C

Q = [m C p (Tin − Tout )] glycol

Hot Glycol

80°C 40°C
= (3.5 kg/s)(2.56 kJ/kg.°C)(80°C − 40°C) 3.5 kg/s
= 358.4 kW
55°C

(b) The rate of heat transfer from water must be equal to

the rate of heat transfer to the glycol. Then,

Q
Q = [m C p (Tout − Tin )] water ⎯
⎯→ m water =
C p (Tout − Tin )
358.4 kJ/s
= = 2.45 kg/s
(4.18 kJ/kg.°C)(55°C − 20°C)
(c) The temperature differences at the two ends of the heat exchanger are

ΔT1 = Th ,in − Tc,out = 80° C − 55° C = 25° C

ΔT2 = Th ,out − Tc,in = 40° C − 20° C = 20° C

ΔT1 − ΔT2 25 − 20
ΔTlm = = = 22.4° C
ln( ΔT1 / ΔT2 ) ln(25 / 20)

Q 358.4 kW
Q = U i Ai ΔTlm ⎯
⎯→ Ai = = = 64.0 m 2
U i ΔTlm (0.25 kW/m 2 .°C)( 22.4°C)
The Heat Exchanger Design Procedure
using the LMTD
• 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:
1. Select the type of heat exchanger suitable for the application.
2. Determine any unknown inlet or outlet temperature and the
heat transfer rate using an energy balance.
3. Calculate the log mean temperature difference ΔTlm and the
correction factor F, if necessary.
4. Obtain (select or calculate) the value of the overall heat
transfer co-efficient U.
5. Calculate the heat transfer surface area As needed to meet
requirements.
 The Effectiveness – NTU Method
• This method is based on a dimensionless parameter
called the heat transfer effectiveness ε
Q Actual heat transfer rate
ε=  = (11-29)
Qmax Maximum possible heat transfer rate

• The actual heat transfer rate in a heat exchanger

Q = Cc (Tc ,out − Tc ,in ) = Ch (Th ,in − Th ,out ) (11-30)

• The maximum temperature difference

ΔTmax = Th ,in − Tc ,in (11-31)
• The fluid with the smaller heat capacity rate will experience a larger
temperature change, and thus it will be the first to experience the
maximum temperature, at which point the heat transfer will come to a
halt. Therefore, the maximum possible heat transfer rate in a heat
exchanger is

• The maximum possible heat transfer rate in a heat exchanger

Q max = Cmin ΔTmax = Cmin (Th ,in − Tc ,in ) (11-32)

where Cmin:
is the smaller of Ch and Cc
• Once the effectiveness of the heat exchanger is
known, the actual heat transfer rate can be
determined from
Q = ε Q max = ε Cmin (Th ,in − Tc ,in ) (11-33)

• The effectiveness of a heat exchanger depends on:

– the geometry of the heat exchanger, and
– the flow arrangement.
• It can be shown that the effectiveness of double-
pipe parallel-flow heat exchanger is
⎡ UAs ⎛ Cmin ⎞⎤
1 − exp ⎢ − ⎜1 + ⎟⎥
⎣ Cmin ⎝ Cmax ⎠⎦
ε parallel flow =
Cmin (11-38)
1+
Cmax
• Effectiveness relations of the heat exchangers typically
involve a dimensionless group called the number of
transfer units NTU
UA s UA s
NTU= =
Cmin (  p
mc ) (11-39)
min
• For specified values of U and Cmin, the value of NTU is
a measure of the heat transfer surface area As.
• The larger the NTU, the larger the heat exchanger.
• It is also convenient to define a capacity ratio c
Cmin
c= (11-40)
Cmax
• The effectiveness of a heat exchanger is a function of
the number of transfer units NTU and the capacity ratio
c.
Effectiveness for Several Heat Exchangers
Effectiveness Heat Exchangers Plots
{Fig. 13-26}
EXAMPLE: A cross-flow heat exchanger consists of 80 thin
walled tubes of 3-cm diameter located in a duct of 1 m x 1 m
cross-section. There are no fins attached to the tubes. Cold
water (Cp= 4180 J/kg · °C) enters the tubes at 18°C with an
average velocity of 3 m/s, while hot air (Cp= 1010 J/kg · °C)
enters the channel at 130°C and 105 kPa at an average
velocity of 12 m/s. If the overall heat transfer coefficient is 130
W/m2 · °C, determine the outlet temperatures of both fluids
and the rate of heat transfer.
Assumptions: The thickness of the tube is negligible

Analysis: The mass flow rates of the hot and the cold fluids are

 c = ρVAc = (1000 kg/m3 )(3 m/s)[80π(0.03 m)2 /4] = 169.6 kg/s

m

P 105 kPa 3
ρ air = = = 0.908 kg / m
RT (0.287 kPa.m3 / kg.K) × (130 + 273 K)

m h = ρVAc = (0.908 kg / m3 )(12 m / s)(1 m) 2 = 10.90 kg / s

As = nπDL = 80π(0.03 m)(1 m) = 7.54 m 2

Cc = m
 c Cpc = (169.6 kg/s)(4.18 kJ/kg.°C) = 708.9 kW/ °C
Ch = m
 h Cph = (10.9 kg/s)(1.010 kJ/kg.°C) = 11.01 kW/°C
 Cmin = Ch = 11.01 kW/ °C
Cmin 11.01
c= = = 0.01553
Cmax 708.9


Q max = Cmin (Th,in − Tc,in ) = (11.01 kW/ °C)(130°C − 18°C) = 1233 kW

UA s (130 W/m 2 .°C) (7.54 m 2 )

NTU = = = 0.08903
C min 11, 010 W/ °C
Noting that this heat exchanger involves mixed cross-flow, the fluid with Cmin is
mixed, Cmax unmixed, effectiveness of this heat exchanger corresponding to c =
0.01553 and NTU =0.08903 is determined using the proper relation in Table 13-4 to
be
⎡ 1 − c NTU ⎤ ⎡ 1 ⎤
ε = 1 − exp ⎢ − (1 − e ) ⎥ = 1 − exp ⎢ − (1 − e −0.01553×0.08903 ) ⎥ = 0.08513
⎣ c ⎦ ⎣ 0.01553 ⎦
Note: You may use Fig. 13-26, but with large error in this particular example
  =ε Q

Q max = (0.08513)(1233 kW) = 105 kW

Q 105 kW

Q = Cc (Tc,out − Tc,in ) ⎯⎯
→ Tc,out = Tc,in + = 18°C+ = 18.15°C
Cc 708.9 kW / °C
Q 105 kW

Q = Ch (Th,in − Th,out ) ⎯⎯
→ Th,out = Th,in − = 130°C − = 120.5°C
Ch 11.01 kW/ °C
 Selection of Heat Exchangers
• An engineer going through catalogs of heat exchanger
manufacturers will be overwhelmed by the type and
number of readily available off-the-shelf heat
exchangers.
• The proper selection depends on several factors:
– heat transfer rate
– cost
• procurement, maintenance, and power.
– pumping power,
– size and weight,
– Type,
– Materials,
– miscellaneous (leak-tight, safety and reliability, Quietness).