ENERGY CONVERSION
MME 9617A
Eric Savory
www.eng.uwo.ca/people/esavory/mme9617a.htm
Lecture 8 – Basics of heat exchangers
Department of Mechanical and Material Engineering
University of Western Ontario
Heat Exchangers
The most common types of energy conversion
systems (e.g. internal combustion engines,
gas/steam turbines, boilers) consist of three parts:
1. a combustion process generating heat and
kinetic energy (K.E.)
2. a device for converting K.E. to mechanical
(useful) energy
3. heat exchangers to recuperate the heat
either for heating purposes or to increase
efficiency.
The different applications of heat exchangers
require different designs (geometries):
Heat Exchangers are classified according to
their function and geometry:
Function:
1. Recuperative: two fluids separated by a solid
wall (this is the most common type)
2. Evaporative: enthalpy of evaporation of one
fluid is used to heat or cool the other fluid
(condensers/evaporators and boilers)
3. Regenerative: use a third material which
stores/releases heat
Geometry: 1. Double Tube
2. Shell and Tube
3. Cross-flow Heat Exchangers
4. Compact Heat Exchangers
Underlying calculation approach
The heat transfer rate for most heat exchangers
can be calculated using the LMTD-method (Log
Mean Temperature Difference), if the inlet (T1) and
outlet (T2) temperatures are known:
Q  U A T
T2  T1
T 
F
ln T2 / T1 
U = Overall heat transfer coefficient [ W/m2-oC ]
A = Effective heat transfer surface area [ m2 ]
F = Geometry correction factor
T = Log mean temperature difference
Otherwise, the Effectiveness () – Number of
Transfer Units (NTU) method may be used:
Q

Q max
UA
NTU 
 Cmin
m
General Formulation for Heat Exchanger
Analysis (LMTD-method)
Most heat exchangers are characterized relative to a
double-pipe heat exchanger (H = Hot, C = Cold):
T1
T2
We now want to derive the expression for LMTD
for a counter-flow double-pipe heat exchanger.
This will be done by considering the first law (for
counter flow):
First globally:
 CH THi  THo   m
 CC TCo  TCi 
Q  U A T  m
Then locally: Apply the first law between points 1
and 2 (for counter-flow)
Heat lost by hot side = Heat gained by cold side
For counter-flow:
QUA
THo  THi   TCo  TCi 
 THo  TCi 

ln 
 THi  TCo 
UA
THo  TCi   THi  TCo 
 THo  TCi 

ln 
 THi  TCo 
T2  T1
QUA
 T2 

ln 
 T1 
By using the notation 1 and 2, as shown on the
graphs, this definition is valid for both Countercurrent and Co-flow (parallel) double-pipe heat
exchangers.
ξ-NTU (Effectiveness – Number of Transfer
Units) Method
If the inlet or outlet temperatures are not given,
the LMTD-method becomes cumbersome to use.
It is thus advisable to use the Effectiveness-NTU
method. The method can be formulated from
the following definitions:
Effectiveness:
Actual heat transfer
Q


Theoretical max . heat transfer Q max
UA
NTU 
 Cmin
m
Minimum thermal capacity
 max. temp. difference
In general:
Actual heat transfer is given by
 CH THi  THo   m
 CC TCo  TCi 
Q  m
Theoretical maximum heat transfer by:
 Cmin THi  TCi 
Q  m
Hence, we obtain the effectiveness as:
 CH THi  THo  m
 CC TCo  TCi 
m


 Cmin THi  TCi  m
 Cmin THi  TCi 
m
For a counter-flow heat exchanger:
Let
and
 CH  m
 Cmin
m
 Cmin m
 CH
m
R

 Cmax m
 CC
m
Which, on using the definition for LMTD, leads to
an expression for the effectiveness as:
 NTU 1  R 
1 e

 NTU 1  R 
1 R e
If, instead
 CC  m
 Cmin
m
then
 Cmin m
 CC
m
R

 Cmax m
 CH
m
We end up with the same effectiveness:
 NTU 1  R 
1 e

 NTU 1  R 
1 R e
Counter flow
 CC  m
 CH
m
 CC  m
 CH
m
Parallel flow
Similar expressions are used for other types of
geometry.
For example, for a parallel double-pipe heat
exchanger, the effectiveness is:
 NTU 1  R 
1 e

1 R
Next we shall look at some applications of these
concepts.
Typical thermal design problems
• Problem #1
– Given the entrance temperature of the two
streams, given one exit temperature;
– Find heat transfer area, A.
• Problem #2
– Given entrance temperature of the two
streams, given the heat transfer area, A;
– Find the exit temperatures of the two
streams.
Objective: Calculation procedure and advantages
/ disadvantages of:
Double pipe
Shell and tube
Cross flow heat exchangers
1. Double Pipe Heat Exchangers:
Double Pipe Heat Exchangers
Arrangements:
Advantages:
- low pressure loss
- small applications (simple, cheap to build)
- counter flow: high effectiveness; parallel flow:
quick (short) fetches.
Disadvantage:
- requires large surface area (footprint on floor) if
large heat transfer rates are needed.
2. Shell-and-Tube Heat Exchangers:
Advantages:
- ideal for large scale applications
- commonly used in petrochemical industry where
dangerous substances are present (protective
shell)
- compact design or double tube heat exchanger.
Disadvantages:
- very bulky (heavy construction), baffles are used
to increase mixing
- subject to water hammer and corrosion (behind
baffles)
- high pressure loses (recirculation behind baffles)
Heat transfer calculations:
Q  U A T
T2  T1
T 
F
ln T2 / T1 
Using counter flow, double pipe heat exchanger
definition for the temperatures
T2  THo  TCi
T1  THi  TCo
Heat exchanger correction factor plot for one
shell pass and an even number of tube passes
=
+
Heat exchanger correction factor plot for two shell
passes and twice an even number of tube passes
For n-shell passes with an even number of tubes:
Again, for boiling or evaporation R  0
so that  = 1 – e-NTU
Cross flow and compact heat exchangers
Overview: Cross-flow and compact heat
exchangers are used where space is limited. These
aim to maximize the heat transfer surface area.
Cross-flow Heat Exchangers:
Commonly used in gas (air) heating applications.
The heat transfer is influenced by whether the
fluids are unmixed (i.e. confined in a channel) or
mixed (i.e. not confined, hence free to contact
several different heat transfer surfaces).
e.g.: both fluids unmixed: air-conditioning devices
e.g.: both fluids mixed: boilers
Advantage: large
surface area-good
for transferring
heat to gases
Cross-Flows
may be mixed
or unmixed
Disadvantages:
heavy, high
pressure losses
In a cross-flow heat exchanger the direction of fluids are
perpendicular to each other. The required surface area,
Across for this heat exchanger is usually calculated by
using tables. It is between the required surface area for
counter-flow (Acounter) and parallel-flow (Aparallel) i.e.
Acounter< Across <Aparallel
Both fluids unmixed
One fluid unmixed
Both fluids unmixed
Cross-flow heat exchangers have the same
analysis equations as before:
Q  U A T
T2  T1
T 
F
ln T2 / T1 
with F as the correction factor (see graphs). The
-NTU method may also be used
Heat exchanger correction factor plot for single
pass, cross-flow with one fluid mixed
Heat exchanger correction factor plot for single
pass, cross-flow with both fluids unmixed
Compact heat exchangers: These are cross-flow
heat exchangers characterized by very large heat
transfer area per unit volume. In fact, the contact
area is so large that much of the flow behaves as
duct or channel flow.
For this reason, the heat-transfer is dominated by
wall effects and the characteristics cannot be
evaluated as for the other types.
For these heat exchangers, the heat transfer rate is
directly related to pressure loss.
Advantages:
- very small
- ideal for transferring heat to / from fluids with
very low conductivity or where the
heat transfer must be done in very small
spaces (e.g. electronic component cooling,
cryogenic cooling, domestic furnaces).
Disadvantages:
- high manufacturing costs
- very heavy
- extremely high pressure losses.
Examples of compact heat exchangers
To solve problems involving design and selection
(sizing) of compact heat exchangers it is first
required to find the effective pressure (static)
loss. This loss can be shown, based on
fundamental heat transfer principles, to be
directly related to the heat transfer rate based on
Colburn’s analogy:
2
f
jH   St Pr 3
8
f – friction factor, St – Stanton number,
Pr – Prandtl number and jH = Colburn factor
These calculations can be quite involved and so
most design or sizing applications use data in
tables and graphs.
All material properties are calculated at the bulk
average temperature, i.e. at (T1+T2)/2, if T1 = inlet,
T2 = exit
Pr andtl number
Friction factor


Pr  
  Cp
dP

DH
dx
f
2
 Umax
Reynolds number at the smallest diameter:
DH G
Re 


m
G
Ac
DH = hydraulic diameter at smallest cross-section
= 4 Ac / P
Ac = smallest cross-sectional area
P = perimeter (circumference) of tube
µ = dynamic viscosity
 = thermal diffusivity
G = maximum mass flow rate flux
 = mass flow rate
m
Ac

A
h
St 
G Cp
 = ratio of open area to total frontal area (A)
h = heat transfer coefficient
Cp = specific heat capacity

V1 G2 
A Vm 
2  V2
1      1  f
p 


2 
A c V1 
 V1

p = pressure loss through heat exchanger
Vm = (V2 + V1) / 2
Overall heat transfer coefficient UA is computed
from:
1
1
1


U A h A c h A h
(h A)h = hot fluid
(h A)c = cold fluid
A = effective heat transfer area
Then the heat transfer Q is:
Q  U A T
Heat transfer and friction factor for a finned flat tube
heat exchanger
Heat transfer and friction factor for a finned circulator-tube
heat exchanger (details on next slide)
Summary
Summary of effectiveness equations
Heat exchanger
type:
Effectiveness:
Heat exchanger
type:
Effectiveness:
=
+
Heat exchanger
type:
Effectiveness:
Example questions
Example 1 – Finned flat tube heat exchanger
Air at 1 atm and 300 K enters a finned flat tube
heat exchanger (as in graph in an earlier slide)
with a velocity of 15 m/s. Calculate the heat
transfer coefficient (h).
Note at this temperature the air properties
(found from tables) are:
 = 1.1774 kg/m3
 = 1.983 x 10-5 kg/ms
Cp = 1.0057 kJ/KgoC
Pr = 0.708
Example 2 – Shell and tube heat exchanger
Hot oil at 100oC is used to heat air in a shell and
tube heat exchanger. The oil makes 6 tube passes
and the air makes 1 shell pass. 2.0 kg/s of air
(specific heat of 1009 J/kgoC) is to be heated from
20 to 80oC. The specific heat of the oil is 2100
J/kgoC and its flow rate is 3.0 kg/s. Calculate the
area required for the heat exchanger for U = 200
W/m2oC.
Example 3 – Finned-tube (both fluids unmixed)
cross-flow heat exchanger
A finned-tube exchanger is used to heat 2.36 m3/s
of air (specific heat of 1006 J/kgoC) at 1 atm from
15.55 to 29.44oC. Hot water enters the tubes at
82.22oC and the air flows across the tubes,
producing an average overall heat transfer
coefficient of 227 W/m2oC. The total surface area of
the exchanger is 9.29m2. Calculate the heat transfer
rate (kW) and the exit water temperature.
Note: We don’t know whether the air or the water is
the minimum thermal capacity fluid. So try with the
air as the minimum fluid first and see if the -NTU
equations give a possible solution. If not then we
have to use water as the minimum and iterate to a
solution.