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99-05-21
HEAT TRANSFER - LAB LESSON NO. 3
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TRANSIENT HEAT CONDUCTION , ANALOGY
METHODS AND HEAT FLOW MEASURING
Before the start of the lab you should be able to answer the following questions:
1. Explain the Schmidt’s method. How does it work?
2. Describe (at least) two different analogy methods which may be used in solving heat
transfer problems.
3. What are the benefits using analogy methods?
4. In these analogies, what are the entities corresponding to temperature, heat transfer
resistance, heat capacity and heat flow?
5. Explain how a two-dimensional, steady state, heat conducting problem can be solved
using finite difference equation approach.
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1 OBJECTIVE
In this lab, Schmidt´s method for solving a transient heat conducting problem through a
wall will be used. Experiment will be conducted to verify the calculated solution. A simple
computer program will also be used for comparison.
Further more, two analogy methods, Hydraulic and Electrical, will be demonstrated. Two
different electrical approaches will be investigated, one continuous and one discrete.
Agreement between them is investigated. The hydraulic analogy will be demonstrated on a
wall with changing temperature on one side.
A MS-Excel program is written for numerical study of the problem at hand. Also a
commercial FEM software is demonstrated.
Some different methods of measuring the heat flow through a wall is demonstrated, one
old and two more modern methods.
2 AIM
Ø Get hands on experience of some heat flow measuring methods.
Ø Experience of analogy method in solving heat transfer problems.
Ø Introduction in solving heat transfer problem numerically, understand the approach of
FDE (Finite Difference Equations).
KEEP IN MIND - KEEP IT SIMPLE
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3 THEORY
3.1 Introduction
The general heat transfer problem has four dimensions, i.e. three dimension in space in one
in time. However, it is often possible to simplify to a certain limit. Often, the steady-state
solution is wanted. Many problem can be treated with only two dimensions in space and
sometimes even with only one dimension in space.
The general governing equation for heat transfer in a solid, assuming constant material
properties and no internal heat generation, is:
∂
∂+
∂α
τ∂
2
222
z
TTT(1)
T = Temperature
τ = time
x, y, z = distance
( )pck ⋅ρ=α
Eq. (1) can be solved numerically which will be showed in this lab for two different cases,
transient heat transfer in an one dimensional wall (Schmidt´s method) and steady-state heat
transfer in a two dimensional wall.
Another way to investigate the influence of different parameters in heat transfer problems
is to use analogy methods. Electrical currents, hydraulic flows and heat flows are governed by
the same type of differential equations, and thus it is possible to use electric or hydraulic
analogies in the study of heat transfer.
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3.2 Analogy methods
As stated earlier there are analogy between heat transfer and electricity and hydraulic. A
former professor of our department, professor Bo Pierre, investigated the influence of the ribs
on the insulating capability of the hull of refrigerated freight ships by using the electric
analogy. A couple of article were published on the subject (in Swedish, “Spantens inverkan på
isoleringen vid kyllastfartyg, belyst med elektriska analogiförsök”), see appendix.
The following equations are valid for the thermal and electric flows:
q = U · A · ∆T = ∆T / Rth (2)
I = ∆V / R (3)
Thus, the heat flow q correspond to the current I, the temperature difference ∆T correspond
to the difference in electric potential ∆V, and the heat transfer resistance Rth correspond to the
electric resistance.
In transient problems, there is an additional correspondence between heat capacity and
electric capacitance.
A hydraulic model may be used for visualizing the heat flow and temperature distribution
through a wall. The heat flow correspond to the liquid flow, the heat transfer resistance in
different layers in the wall correspond to the flow resistance of capillary tubes of different
length and diameter. The temperature in these layers correspond to the pressure inside the
tube, in our case visualized as the water level in the vertical tubes.
In transient problems, there is an additional correspondence between the amount of heat
stored in each layer and the amount of water stored in the vertical tubes. A large tube diameter
is able to store more water at a given water level, which then correspond to a large amount of
heat stored at a given temperature.
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To evaluate the results of the analogy test, a number of scale factors are needed. These are
shown in table 1.
Table 1: Analogy models, analogous entities and scale factors.
Thermal Hydraulic Electric
Heat
Q
Liquid volume
v
Electric charge
∫ τ⋅= elel dIQ
Analogous quantities Temperature
T
Water level
hv
Electric potential
V
Time
τ
Time
τh
Time
τel
Relations:
Transport function
(corresponding to heat
transfer)
Q / τ = K · ∆t
Conduction:
K = A · k / ∆x
Surface heat transfer:
K = A · h
v / τh = Ch · hv
Ch = Capillary tube
constant
Rh = 1 / Ch
Qel / τh = I = ∆V / R
Conservation functions
(corresponding to heat
balance)
Q = C · T
C = m · cp = A · ∆x · ρ · cp
v = Ar · hv
Ar = tube area
Qel = Cel · V
Cel = capacitance
Scales: Q = = vnQh ⋅ = = el
Qel Qn ⋅
T = = vTh hn ⋅ = = VnT
el ⋅
τ = = hhn τ⋅τ = = eleln τ⋅τ
Relations between scale
factors
r
hh
Th
r
Qh
hTh
h
Qh
A
C
K
Cn
nA
Cn
nnC
Kn
⋅=
⋅=
⋅⋅=
τ
τ
elelel
Tel
el
Qel
elTelel
Qel
CRK
Cn
nC
Cn
nnRKn
⋅⋅=
⋅=
⋅⋅⋅=
τ
τ
1
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3.3 Finite Difference Equations
An analytical solution for eq.(1) can be attained for some simple geometries, however for
the general case no such solution has been reported. To be able solve the problem at hand,
some other method must be used. With the fast and cheap computers of today it is now
possible for almost anyone to solve eq.(1) by using numerical methods.
T
xi-1 i+1ii-1/2 i+1/2
Figure 1: Temperature as a function of space
From eq.(1) it is apparent that we must find a way of expressing the derivatives in an
simply way. We are interested in solving the temperature in position i,j,k. If we only study the
temperature dependency of the x coordinate, realizing that the dependency in y- and z-
dimension is treated in an analogous manner, the temperature gradients can be expressed as:
( ) ( ) ( )
( ) ( ) ( )
x
TT
x
T
x
TT
x
T
mkji
mkji
m
kji
mkji
mkji
m
kji
∆
−=
∂∂
∆
−=
∂∂
−
−
+
+
,,1,,
,,21
,,,,1
,,21(4)
where i, j, k denotes points in x, y, z dimension, respectively, and (m) denotes time step.
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The second derivative of temperature in point i,j,k is
( )
( ) ( )
x
x
T
x
T
x
T
m
kji
m
kji
m
kji∆
∂∂
−∂∂
=∂∂ −+ ,,21,,21
,,
2
2
(5)
Inserting eq.(4) into eq.(5) yields
( )( ) ( ) ( ) ( )
( ) ( ) ( )
2
,,,,1,,1
,,1,,,,,,1
,,
2
2 2
x
TTT
xx
TT
x
TT
x
Tm
kjim
kjim
kji
mkji
mkji
mkji
mkji
m
kji∆
⋅−+=
∆∆
−−
∆
−
=∂∂ −+
−+
(6)
and for y- and z-dimension
( )
( ) ( ) ( ) ( )
( ) ( ) ( )
( )( ) ( ) ( ) ( )
( ) ( ) ( )
2
,,1,,1,,
1,,,,,,1,,
,,
2
2
2
,,,1,,1,
,1,,,,,,1,
,,
2
2
2
2
z
TTT
zz
TT
z
TT
z
T
y
TTT
y
y
TT
y
TT
y
T
mkji
mkji
mkji
mkji
mkji
mkji
mkji
m
kji
mkji
mkji
mkji
mkji
mkji
mkji
mkji
m
kji
∆
⋅−+=
∆∆
−−
∆
−
=∂∂
∆
⋅−+=
∆∆
−−
∆
−
=∂∂
−+
−+
−+
−+
(7)
The time derivative can be expressed as
( ) ( )
τ∆
−=
τ∂∂ + m
kjim
kjim
kji
TTT ,,1
,,
,,
(8)
Inserting eqs.(6) - (8) into eq.(1) yields
( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
∆
⋅−++
+∆
⋅−++
∆
⋅−+
α=τ∆
−
−+
−+−++
2
,,1,,1,,
2
,,,1,,1,
2
,,,,1,,1
,,1
,,
2
22
z
TTT
y
TTT
x
TTT
TTm
kjim
kjim
kji
mkji
mkji
mkji
mkji
mkji
mkji
mkji
mkji (9)
which is the finite difference equation of eq.(1).
3.4 Steady-state, two dimensional heat transfer
In a steady state analysis the temperature at any given point is constant with respect to
time, 0=τ∂
∂T. In eq.(9) the left hand side is then zero. For a two dimensional analysis (x,y)
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the temperature in z-direction is constant, 0=∂∂
z
T. Eq.(9) can then be simplified, assuming
that the computational domain is spaced equal in both directions (∆y = ∆x), as:
41,1,,1,1
,−+−+ +++
= jijijijiji
TTTTT (10)
which easily can be solved.
3.5 Transient, one dimensional heat transfer, Schmidt´s method
Again, we use eq.(9) as a starting point. Now we have a transient problems, which means
that the left hand side is not equal to zero. We are only interested in one space dimension, say
x. Eq.(9) then simplifies to:
( ) ( ) ( ) ( ) ( )
∆
⋅−+α=
τ∆− −+
+
211
1 2
x
TTTTT mi
mi
mi
mi
mi (11)
Collecting terms with temperature of the new time step on the left hand side and the old
time step on the right hand side.
( ) ( ) ( ) ( )( ) ( )mi
mi
mi
mi
mi TTTT
xT +⋅−+
∆τ∆⋅α
= −++ 21121 (12)
which can be rearranged even more
( ) ( ) ( )( ) ( )
∆τ∆⋅α
⋅−++∆
τ∆⋅α= −+
+2112
1 21x
TTTx
T mi
mi
mi
mi (13)
Introducing M defined as
τ∆⋅α∆
=2x
M (14)
and inserting in eq.(13) yields
( )( ) ( )
( )
−+
+= −++
MT
M
TTT m
i
mi
mim
i
21111 (15)
By making the clever observation and setting M = 2, i.e. the relation between step in space
and step in time, Schmidt attained an equation which is easy to calculate. It can be seen that
the second term on the right hand side vanishes, and eq.(15) simplifies to
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( )( ) ( )
2111m
im
imi
TTT −++ +
= (16)
which says that the temperature at node i and time m+1 is the arithmetic mean of the
surrounding nodes, i+1 and i-1, at the previous time step, m. Because of it’s simplicity,
eq.(16) can be solved graphically, without any calculations required. At the pre-computer era
this was a necessity.
By setting M=2 we have linked the increment in time together with the increment in space
and the properties of the material. We must fulfill this relation if Schmidt’s method is used,
i.e. eq.(16). However, for the modern engineer, calculation capability is no problem and the
requirement of setting M=2 is somewhat obsolete. But, for numerical stability reasons it can
be shown that M should be equal to or greater than 2.
4 Experiment
4.1 Analogy method 1: The influence of the ribs on the insulating
capability of the hull of refrigerating freight ships
The apparatus consists of a plastic tray with aluminum rails at two sides and a number of
loose pieces of rail. On the bottom of the tray there is a plastic sheet suitable for drawing lines
with a pencil. The tray is filled with slightly salt water. With the loose rails, a profile model of
the ship’s ribs is built, see figure 2.
To calculate the factor, y, by which the heat transfer in to the refrigerated room is increased
with the ribs compared to without the ribs is calculated as:
zhz
lzky
−δ+δ⋅
= (17)
where d, k and l is defined as
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040.0015.099.0
90.0
11
int
+⋅−⋅−=+=
λδ
+α
+α
λ+δ=δ ∑
bhzhl
zhkb
b
extii
(18)
bδiδ
z
b
h
Figure 2: Schematic model of the ship’s ribs.
An electrical potential is applied between the ribs and a parallel rail, and this potential
corresponds to the temperature difference between the outside wall of the hull and the inside
wall of the refrigerated room. A volt-meter is supplied for reading the potential at different
positions in the water, and a Ampére-meter is connected in the circuit so that the current
through the model can be read.
TEST PROCEDURE
A. Use the volt meter to find five points of equal potential in between the ribs and the inside
wall. Connect the points to a curve. Draw such lines for three different potentials.
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B. Read the current through the model with the ribs connected. Then disconnect the ribs and
remove them from the tray. Read the new current without the ribs. By what factor, y, is
the heat flow (current) increased with the ribs compared to without the ribs?
C. Compare the results from B with the equation of Pierre, eq.(17), using the following data:
αext = αext = ∞, δb = 0, li = 0.04 W/(m·K). z, b, h and δi are found by measuring the model.
D. What kinds of simplifications have we done in the model? How do you think that these
simplifications affect the accuracy of the results?
4.2 Steady state, two-dimensional heat transfer
A similar model as above is built by using a mesh of electric resistances. In this case, the
potential may only be found in a finite number of discrete points. Note the similarity between
the mesh model and the numerical model later used in the computer simulation.
TEST PROCEDURE
A. Measure the electrical potential in two nodal points in the mesh. Note which nodes and the
results in supplied figure, figure 3.
B. Measure the potential of the four surrounding nodes for each of the two nodes measured in
A.
C. Compare the average of the four surrounding nodes with the potential of the node itself.
D. Measure the current through the model with and without the connections representing the
rib. Compare the result to that from the model in section 4.1.
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A B C D E F G H I J K L M N O P Q
1
2
3
4
5
6
7
8
9
10
11
12
13
14
T1= °C
T2= °C
Figure 3: Electrical resistance mesh.
USING MS-EXCEL
The nodal mesh of the electric model may also be constructed in a spread sheet, e.g. MS-
Excel. To do this, let the cells of the spread sheet represent the nodal points, specify the
boundary temperatures in the appropriate cells and let the program calculate the temperatures
in the rest of the cells as the average of the four surrounding cells. An iterative procedure is
needed. The calculations should be continued until no further change in the temperatures is
found. The model is then said to be relaxed. A model template similar to figure 3 is supplied.
Compare the result from the excel model to those from the mesh!
An example of solution is given in figure 4.
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A B C D E F G H I J K L M N O P Q
1 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0
2 1.3 1.3 1.3 1.3 1.4 1.4 1.5 1.6 1.7 1.8 1.9 2.0 2.2 2.3 2.3 2.4 2.4
3 2.6 2.6 2.6 2.7 2.7 2.8 3.0 3.1 3.3 3.5 3.8 4.1 4.3 4.5 4.7 4.8 4.8
4 3.9 3.9 3.9 4.0 4.1 4.2 4.4 4.6 4.9 5.3 5.7 6.1 6.6 6.9 7.1 7.2 7.2
5 5.1 5.1 5.2 5.3 5.4 5.6 5.8 6.1 6.4 6.9 7.5 8.2 8.9 9.4 9.6 9.7 9.8
6 6.4 6.4 6.4 6.5 6.6 6.8 7.1 7.4 7.9 8.5 9.3 10.3 11.6 12.1 12.3 12.3 12.4
7 7.5 7.6 7.6 7.7 7.9 8.1 8.3 8.7 9.2 9.8 10.7 12.2 15.0 15.0 15.0 15.0 15.0
8 8.7 8.7 8.8 8.9 9.0 9.2 9.5 9.8 10.3 10.9 11.7 12.8 14.0 14.5 14.8 14.9 15.0
9 9.8 9.8 9.9 10.0 10.1 10.3 10.5 10.8 11.2 11.7 12.4 13.1 13.9 14.4 14.6 14.8 15.0
10 10.9 10.9 11.0 11.0 11.1 11.3 11.5 11.8 12.1 12.5 13.0 13.5 14.0 14.3 14.6 14.8 15.0
11 11.9 12.0 12.0 12.1 12.1 12.3 12.4 12.6 12.9 13.2 13.5 13.8 14.2 14.4 14.7 14.8 15.0
12 13.0 13.0 13.0 13.1 13.1 13.2 13.3 13.4 13.6 13.8 14.0 14.2 14.4 14.6 14.8 14.9 15.0
13 14.0 14.0 14.0 14.0 14.1 14.1 14.2 14.2 14.3 14.4 14.5 14.6 14.7 14.8 14.9 14.9 15.0
14 15.0 15.0 15.0 15.0 15.0 15.0 15.0 15.0 15.0 15.0 15.0 15.0 15.0 15.0 15.0 15.0 15.0
T1= 15 °C
T2= 0 °C
Figure 4: A solution of the two dimensional wall.
It is of course also possible to write a computer program in any programming language to
find the relaxed temperatures of the nodal mesh. A separate program has the advantage of
letting you steer the iteration process. For example, you could change only one temperature in
each step, the temperature in the point where the difference between the nodal value and the
average of the surrounding values is the largest. This procedure is said to ensure a convergent
solution. It is, however, considerably slower than just calculating new averages for every
point.
The lab assistant will demonstrate a commercial software (ANSYS) by simulating the
above problem.
4.3 Heat flow metering
The lab assistance will talking about methods of measuring the heat flow. Some small
experiment will be conducted to let the student with both modern and historical equipment.
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4.4 Electrical and Hydraulic analogy for transient, one dimensional heat
transfer through a wall
Two models of transient heat transfer through a wall will be demonstrated by the lab
assistant, one electrical and one hydraulic, see also table 1.
For both models:
A. Try to figure out what each part of the model corresponds to in a real wall.
B. Point out which part of the wall that has the lowest thermal conductivity and which has
the highest heat capacity.
C. In which part of the wall is the phase delay the largest? Why?
4.5 Transient, one dimensional heat transfer through a wall, Schmidt’s
method
The apparatus in this test consists of a pile of five plates made of an insulating material.
The temperatures in between the plates and on both sides of the pile are measured by
thermocouples. The pile is placed on top of a copper plate, kept at room temperature by water
cooling. An electrically heated copper plate, kept at approximately +40°C, can be placed on
top of the pile. the temperatures are read by a data acquisition system and transferred to a
computer, where the temperature can be read. At any instance the students can store the
temperature in MS-Excel. The insulating material has the following properties: ρ=25 kg/m³,
k=0.041 W/(m·K), cp=1.340 kJ/(kg·K) and plate thickness ∆x=10 mm.
TEST PROCEDURE
A. Using Schmidt’s method, calculate the time step, ∆τ, and the absolute time corresponding
to six time steps. ∆τ = sec.
B. Place the heated plate on top of the pile, and start the timer in the computer program.
C. Save the temperatures in MS-Excel for time step 0, 2, 4 and 6. Then note them in the table
below.
D. Calculate the temperature for six time steps using Schmidt’s method.
E. Compare your calculated result with experiment. Is the agreement good? If not, why?
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F. Use Schmidt’s graphical method to estimate the temperature distribution for six time
steps.
G. Compare your graphical solution with your numerical solution.
H. Together with the lab assistant compare your result with a computer program, which uses
Schmidt’s method. See how the choice of plate thickness influence the accuracy of the
result and also increase the number of time step needed to reach the same ending time.
Measured temperatures in experiment
Position: 0 1 2 3 4 5 6Time step
0 _____ _____ _____ _____ _____ _____ _____
2 _____ _____ _____ _____ _____ _____ _____
4 _____ _____ _____ _____ _____ _____ _____
6 _____ _____ _____ _____ _____ _____ _____
Calculated temperature using Schmidt’s method
Position: 0 1 2 3 4 5Time step
0 _____ _____ _____ _____ _____ _____
1 _____ _____ _____ _____ _____ _____
2 _____ _____ _____ _____ _____ _____
3 _____ _____ _____ _____ _____ _____
4 _____ _____ _____ _____ _____ _____
5 _____ _____ _____ _____ _____ _____
6 _____ _____ _____ _____ _____ _____
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Schmidt’s method, graphically (See Holman):
T (°C)
Position x
1817
22212019
26252423
30292827
34333231
38373635
42414039
4 5 0 1 2 3
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