Transient Heat Conduction

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HEAT TRANSFER - LAB LESSON NO. 3

C:\Mina dokument\Institutionell tjänstgöring\VT-labbar\#3\Lab-pek.doc 1 (16)

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

Transient Heat Conduction

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