Heat exchangers
Aims and objectives
Identify the relationship between heat transfer
and fluid capacity rate
Develop the log mean temperature difference
for heat transfer in heat exchangers
Develop the concept of effectiveness of a
heat exchanger
Calculations of heat exchange surface area
and fluid temperature changes
Heat exchangers – two main types:
type 1
Regenerator
Air
side
Seal
Q
Gas
side
Howden Buffalo Utility - Air Heaters & Replacement Baskets
This type of rotary regenerator is only suitable when the hot
and cold stream pressures are not very different.
http://www.pias-usa.com/products/utility/airheaters.html
https://www.youtube.com/watch?v=roImOiIxrjo
Regenerative heat exchanger – just to be aware
https://www.youtube.com/watch?v=XuLfASz0YkU
Heat exchangers – two main types:
type 2, the focus of this topic
Recuperator
(Most common type
of heat exchanger) Fluid B, hot
Fluid A,
cool
As Fluid B travels the length of the tube it is in, its temperature decreases because
of the temperature difference between A and B, as the Fluid A travels in the reverse
direction warming up as it goes.
The temperature difference between the fluids at any point causes heat transfer at a
position, 1, Q=UATA-B,1, and the moving fliud has mcp capacity of thermal ‘storage’
which changes temperature between position 1 and 2 by T1-2.
So there is a local heat transfer rate at any position on the heat exchanger at steady
state, but there is a positional change of temperature of the fluids as they pass
through the heat exchanger
Recuperative heat exchanger – direct heat transfer, subject of
this topic.
https://www.youtube.com/watch?v=Jv5p7o-7Pms
Looking directly at the
combustion gas/water
heat exchanger
Domestic gas fired
water heater
Burner array
Looking up at the
combustion gas/water
heat exchanger
Example: Calculate the rate of heat lost from a domestic boiler producing combustion gases at
140C with a specific heat capacity cp =1.2 kJ/kgK and at a rate of 3 g/s. The ambient
temperature is 15C.
Shell and tube heat exchanger:
Single pass, counter flow
See spreadsheet for a
calculation of the
temperature of each
fluid
Temperature distribution in double pipe
(simplest) exchanger
B(hot)
B(hot)
Counter
Parallel
flow
flow
A(cold)
T
B
T
Wall
temperature
A
x
B
A
Wall
temperature
x
160
Counter flow
140
temperature, C
120
oil, 1
100
water, 2
80
60
40
20
0
0
500
160
1000
length, mm
1500
2000
Parallel
temperature, C
140
120
oil, 1
100
water, 2
80
60
40
20
0
0
200
400
600
length, mm
800
1000
Calculation in spreadsheet is slightly more complicated than thermo 2, but is
of interest, therefore included here for information
𝑇𝑤𝑛+1 = 𝑇𝑤𝑛 −
and
𝑇𝑜𝑛+1 = 𝑇𝑜𝑛 −
and
𝑇𝑜𝑛 − 𝑇𝑤𝑛
𝑅𝑡ℎ
𝑇𝑜𝑛 − 𝑇𝑤𝑛
𝑅𝑡ℎ
𝑇𝑜𝑛 − 𝑇𝑤𝑛
,𝑊
𝑅𝑡ℎ
and
1000𝐶𝑜 , 𝑊/𝐾
1000𝐶𝑤
1000𝐶𝑜
Heat transfer calculation
Formulae applicable:
For an element:
Hot
stream,
mh, cp,h
q =
 dq
TB
TA
dA dq
over all elements on length
Cold
stream,
mc, cp,c
Heat transfer analysis for double pipe Heat exchanger
Energy balance
Heat transfer from the hot fluid = Heat transfer to
cold fluid
Q = mcp Tin − Tout
= mcp Tout − Tin
hot
cold
What is needed to solve this equation is Q or
inlet temperatures and at least one outlet
temperature.
Example of heat exchanger overall heat transfer
coefficient
A tube and shell heat exchanger has tube internal radius 6 mm and a wall thickness of
1 mm. The inner heat transfer coefficient is 1500 Wm-2K-1 and that for the outer
surface is 2000 Wm-2K-1. The thermal conductivity of the pipe wall is 100 Wm-1K-1.
Calculate thermal resistance and hence overall heat transfer coefficient.
Calculation check – neat version:
Rth,I = 1/hA = 1/1500*2*0.006*L = 1/18L
Rth,o = 1/hA = 1/2000*2*0.007*L = 1/28L
Rth,wall =ln(7/6)/2*100*L = 1/1297 L
Overall heat transfer coefficient is per m2 of heat exchanger surface. Choose the
surface area as the outer radius surface of the tube. The length per 1 m2 is:
Area = 2rL
r=0.007m, A = 1m2
L = 22.7m
RTH = (1/22.7)*(1/18+1/28+1/1297)
= 0.014*0.092=0.0013
U = 1/0.0013 = 769 W/m2K
Note: Even if U is constant, (TB-TA) varies along the
heat exchanger. This must be taken into account.
There are two approaches available, namely LMTD –
and ε-NTU.
LMTD method for sizing heat exchangers
In a parallel and counter flow single pass tubular heat
exchanger, the temperature difference between the
two fluid streams varies as the flow proceeds from
inlet to outlet.
The LMTD (Log Mean Temperature Difference)
analysis gives the equivalent constant temperature
difference such that the heat transfer is the same.
Derivation
Heat transferred between hot, h, and cold, c is:
𝑑𝑞 = 𝑇ℎ − 𝑇𝑐 𝑈𝐴 𝑑𝐴
From capacity of each fluid, heat absorbed (or lost) is:
𝑑𝑞 𝑑𝑞
𝑑 𝑇ℎ − 𝑇𝑐 = 𝑑𝑇ℎ − 𝑑𝑇𝑐 = −
−
𝐶ℎ 𝐶𝑐
Combine equations to eliminate q:
Shuffle and integrate: 𝑙𝑛
𝑇ℎ2 −𝑇𝑐2
𝑇ℎ1 −𝑇𝑐1
𝑑 𝑇ℎ −𝑇𝑐
−
= 𝑇ℎ − 𝑇𝑐 𝑈𝐴 𝑑𝐴
1
1
+
𝐶ℎ 𝐶𝑐
=−
1
𝐶ℎ
+
1
𝐶𝑐
𝑈𝐴 𝐴
𝑇ℎ2 − 𝑇𝑐2 = 𝛥𝑇2
𝑇ℎ1 − 𝑇𝑐1 = 𝛥𝑇1
Also,
1
𝐶ℎ
+
1
𝐶𝑐
=
1
𝑞
𝛥𝑇1 − 𝛥𝑇2 , so 𝑞 =
𝛥𝑇2 −𝛥𝑇1
𝑙𝑛
𝛥𝑇2
𝛥𝑇1
𝑈𝐴 == 𝛥𝑇𝑚 𝑈𝐴
LMTD method
Δ Ta - Δ T b
LMTD = ΔTm =
ln (Δ Ta /Δ Tb)
where a refers to inlet and b to outlet conditions or
any two far apart points along the heat exchange
surface.
F1
Δ Ta = ( T F2 - T F1 )a
F2
Δ T b = ( T F2 - T F1 )b
a
b
These are temperature differences between streams.
These relationships apply to both parallel and counter
flow arrangements.
Example
Calculate the LMTD for (  ) and ( ) cases which
have the same inlet and outlet stream temperatures.
Hot in = 100oC, Hot out = 50oC
Cold in = 20oC, Cold out = 40oC
T
100
80
T 60
Parallel Flow
50
10
40
20
x
LMTD= (80-10)/[ln(80/10)]
LMTD= 33.7oC
100
Counter Flow
40
50
20 30 x
LMTD= (60-30)/[ln(60/30)]
LMTD= 43.3oC
For sizing heat exchangers (i.e. working out
required size for flow rate and temp change), Log
Mean Temperature Difference (LMTD) approach is
used to give the area required from:
Q
A=
U ΔTm
Rating of heat exchangers (i.e. working out temp
change in a given heat exchanger) can be done by
iteration using the LMTD approach.
Example: for the counter flow and parallel flow heat exchangers above, determine the
surface area required to exchange heat at the rate of 1 kW; the overall heat transfer
coefficient, U, is 120 W/m2K.
Advantage of parallel and counter flow
The counter flow type is the most compact for a
given heat transfer rate.
The parallel flow type gives a lower Tmax for the tube
wall temperature.
The LMTD analysis requires the inlet and outlet
temperatures of the fluid streams to be known. This
may necessitate an iterative solution involving
guessed outlet values. This is the main disadvantage of
the approach.
Procedure
1: Guess outlet temperatures
2: Calculate LMTD
3: Calculate q from step 2
4: Calculate outlet temperature for step 1
i.e. if temperatures are not known, iterative
calculation is required
Example: calculate the outlet temperatures for a heat exchanger with area 1.3 m2,
given the overall heat transfer coefficient, U = 700 W/m2K and the capacity rates are
0.12 kW/K and 0.42 kW/K for the hot and cold fluids respectively. The hot fluid
enters at 140C and the cold fluid at 20C.
More complex exchangers
In more complex designs of similar type, the "tube"
flow passes through the heat exchanger more than
once, and the "shell" side flow can be directed
across the tube to increase turbulent heat transfer.
Generally, these are called Shell and tube heat
exchangers.
Shell and Tube heat exchanger
Tube side inlet
Shell side outlet
Coates building
boiler room
Tube side
outlet
Shell side
inlet
Crossflow and
multipass.
The stream
temperature
differences are more
complicated in
crossflow or multipass
configurations.
The effects of crossflow and multipass are accounted
for by introducing a factor F such that: Q = U A F ΔTm
The factor F is derived from calculations similar to the
derivation of LMTD or read from graphs.
A simpler and more direct approach for finding the
temperatures is the ε-NTU method.
Correction factor - LMTD
The correction factor allows us to use the LMTD
type of calculation for more complex geometries. Two
Subscripts:
new variables are introduced:
1: inflow
t 2  t1
P
T1  t1
T1  T2
R
t 2  t1
2: outflow
Temperatures:
T – fluid 1, t – fluid 2
R is the ratio of the temperature changes in each
fluid and hence of the thermal capacities
(e. g. 𝐶1 =
𝑄
).
∆𝑇1
P is the ratio of temp change in one
fluid to the maximum temp change available.
Correction factor - LMTD
Read the correction factor from charts, knowing P on
the horizontal axis, a selection of curves for various
values of R, and corresponding correction factor, F
from the vertical axis.
Three cases shown from a book by Bowman, Mueller
and Nagle in Heat Transfer, Bejan, 1993.
Example: calculate the geometrical correction factor for for a heat exchanger with required
flow temperatures: Thi = 100C, Tho = 50C, Tci = 20C, Tco = 60C) for the 3 heat exchangers
shown in the charts above. Compare with the case for Thi = 200C, Tho = 100C, Tci = 30C, Tco
= 90C.
If P0, the stream having temperatures t1 and t2 has
change of phase, i.e. if pressure drop is not too great
t1 t2.
If R0, the stream having temperatures T1 and T2 has
a change of phase.
If a stream has a phase change, the capacity rate is
effectively infinite, because the fluid will absorb heat
without changing temperature.
These are the two limits on the charts.
Cross flow example– Heat Transfer, Bejan
Care using LMTD
It is important to understand the ‘capacity rate’ of the
fluids in the heat exchanger, i.e.
 hot cp,hot and
Chot  m
 coldcp,cold
Ccold  m
Must always use the formula for LMTD:
q  UA ΔTm
Together with the heat capacity formulae:
 cp Tin  Tout hot  m
 cp Tout  Tin cold
q  m
Using the former on its own suggests that the heat
transfer does not depend on the capacity rate of the
fluids.