Transient Heat-conduction-Part-II

Lectures on Heat Transfer --TRANSIENT HEAT

CONDUCTION: Part-II

by

Dr. M. ThirumaleshwarDr. M. Thirumaleshwarformerly:

Professor, Dept. of Mechanical Engineering,St. Joseph Engg. College, Vamanjoor,

MangaloreIndia

Preface

• This file contains slides on Transient Heat conduction: Part-II

• The slides were prepared while teaching Heat Transfer course to the M.Tech. Heat Transfer course to the M.Tech. students in Mechanical Engineering Dept. of St. Joseph Engineering College, Vamanjoor, Mangalore, India, during Sept. – Dec. 2010.

Aug. 2016 2MT/SJEC/M.Tech.

• It is hoped that these Slides will be useful to teachers, students, researchers and professionals working in this field.

• For students, it should be particularly useful to study, quickly review the subject, useful to study, quickly review the subject, and to prepare for the examinations.

• ��

Aug. 2016 3MT/SJEC/M.Tech.

References• 1. M. Thirumaleshwar: Fundamentals of Heat &

Mass Transfer, Pearson Edu., 2006• https://books.google.co.in/books?id=b2238B-

AsqcC&printsec=frontcover&source=gbs_atb#v=onepage&q&f=false

• 2. Cengel Y. A. Heat Transfer: A Practical Approach, 2nd Ed. McGraw Hill Co., 2003

Aug. 2016 MT/SJEC/M.Tech. 4

Approach, 2nd Ed. McGraw Hill Co., 2003• 3. Cengel, Y. A. and Ghajar, A. J., Heat and

Mass Transfer - Fundamentals and Applications, 5th Ed., McGraw-Hill, New York, NY, 2014.

References… contd.

• 4. Incropera , Dewitt, Bergman, Lavine: Fundamentals of Heat and Mass Transfer, 6th

Ed., Wiley Intl.• 5. M. Thirumaleshwar: Software Solutions to • 5. M. Thirumaleshwar: Software Solutions to

Problems on Heat Transfer – CONDUCTION-Part-II, Bookboon, 2013

• http://bookboon.com/en/software-solutions-problems-on-heat-transfer-cii-ebook


TRANSIENT HEAT CONDUCTION: Part-II ..Outline:

• Semi-infinite solids with different BC’s• Problems • Product solution for multi-dimension

systems


systems• Summary of Basic relations for transient

conduction

One-dimensional transient conduction in Semi-infinite solids:

• A solid which has one exposed surface and extends to infinity in other directions is known as a semi-infinite solid.

• Consider a semi-infinite solid, extending from x = 0 to x = ∞, initially at an uniform temperature, T .


0 to x = ∞, initially at an uniform temperature, Ti. There is no internal energy generation.

• Now, if there is a change in the thermal conditions at the exposed surface at x = 0, transient conditions will be induced in the solid. Fig. 7.10 illustrates three possible boundary conditions at the surface:

• Case (i): Constant surface temperature:• See Fig. 7.10(a). The solid is initially at an uniform

temperature Ti and for times τ > 0, the boundary surface at x = 0 is maintained at temperature To.

• Non-dimensional temperature distribution in the solid is obtained as:

T x τ,( ) T 0

T i T 0erf

x

2 α τ.........(7.29)


T i T 0 2 α τ..

where, erf(z) is the Gaussian error function defined as:

erf ζ( )2

π 0

ζuexp u2 d. .....(7.30)

Values of error function are tabulated in Handbooks.

• Gaussian error function is also shown plotted in Fig. (7.11) below:

erf ζ( )0.5

0.6

0.7

0.8

0.9

1Gaussian error function


ζ0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3

0

0.1

0.2

0.3

0.4

Fig. 7.11. Gaussian error function erf(ζ) vs. ζ

• Then, from eqn. (7.29), we have the temperature distribution as:

T x τ,( ) T 0 T i T 02

π

.0

x

4 α. τ.uexp u2 d. .....(7.31)

Once the temperature distribution is known, heat flux at any point isobtained by applying Fourier’s Law, viz.

dT x τ,


Q i k A.dT x τ,

dx. W....instantaneous heat flow rate at a given x location

Performing the differentiation on T(x,τ) given by eqn. (7.31) by Leibnitz’s rule, we get,

dT

dx

T i T 0

π α. τ.exp

x2

4 α. τ..

• Substituting this in Fourier’s Law, we get:

i.e. Q i k A. T i T 0.

expx2

4 α. τ.

π α. τ.. W......(7.32)

Heat flow rate at the surface (x = 0):Putting x = 0 in eqn. (7.32),

Q k A.T 0 T i. W......(7.33)


Q surface k A.T 0 T i

π α. τ.. W......(7.33)

Total heat flow during ττττ = 0 to ττττ = ττττ:This is obtained by integrating eqn. (7.33) from τ = 0 to τ = τ.

Q totalk A. T 0 T i

.

π α.0

ττ1

τd.

i.e. Q total k A. T 0 T i. 2. τ

π α..

i.e. Q total 1.13 k. A. T 0 T i. τ

α. J......(7.34)

Criterion to apply these relations for a finite slab:For a slab of finite thickness L, above relations for a semi-infinite slab


For a slab of finite thickness L, above relations for a semi-infinite slabcan be applied if:

L

2 α τ..0.5

Penetration depth and penetration time:‘Penetration depth’ is the distance from the surface where theTemperature change is within 1% of the change in the surfacetemperature, i.e.

From the ‘Error function’ Table 7.3, this corresponds to:

x

2 α τ..1.8

i.e. penetration depth, ‘d’ is given by:

T T 0

T i T 00.99


d 3.6 α τ..

‘Penetration time’ is the time taken for the surface perturbation to be felt at that depth. Therefore,

d

2 α τ p..

1.8 or, τ pd2

13 α.

• Case (ii): Constant surface heat flux:• See Fig. 7.10(b). The solid is initially at an uniform

temperature Ti , and for times τ > 0, the boundary surface at x = 0 is subjected to a constant heat flux qo.(W/m2).

• Then, the non-dimensional temperature distribution in the solid is given as:


T x τ,( ) T i

2 q 0. α τ.

π.

kexp

x2

4 α. τ..

q 0 x.

k1 erf

x

2 α τ... .....(7.35)

where erf is the error function defined earlier.

• Case(iii): Convection at the exposed surface:

• See Fig. 7.10(c). The solid is initially at an uniform temperature Ti , and for times τ > 0, the boundary surface at x = 0 is subjected to convection with a fluid at temperature Ta and heat transfer coeff., h.

• Then, an energy balance at the surface gives:


kdT

dx x 0

. h T a T 0 τ,( ).

and, the non-dimensional temperature distribution inthe solid is given as:

T x τ,( ) T i

T a T i1 erf

x

2 α τ..exp

h x.

k

h2 α. τ.

k21 erf

x

2 α τ..

h α τ..

k. .....(7.36)

To represent eqn. (7.36) in graphical form:


Put ζ x

2 α τ..and, η h α τ..

k

θ η ζ,( ) 1 erf ζ( ) exp 2 ζ. η. η 21 erf ζ η( )( ). .....(7.37)

then, h x.

k2 ζ. η. and, eqn. (7.36) can be written as:

We use Mathcad todraw the graph. Remember:

ζ x

2 α τ..and, η h α τ..

k


and,

θ η ζ,( )T x τ,( ) T i

T a T i

• Example 7.11: A thick copper slab (� = 1.1 x 10-4 m2/s, k = 380 W/(m.C)) is initially at an uniform temperature of 250 C. Suddenly, its surface temperature is lowered to 60 C.

• (i) How long will it take the temperature at a depth of 3 cm to reach 100 C?


a depth of 3 cm to reach 100 C? • (ii) What is the heat flux at the surface at

that time?• (iii) What is the total amount of heat

removed from the slab per unit surface area till that time?


• Time reqd. to reach 100 C at a depth of 3 cm from the surface:

• Since this is a very large slab, we will consider it as a semi-infinite medium, with the surface suddenly brought to and maintained at a constant temperature, To. This belongs to case (i), refer to Fig. 7.10, (a).

• So, eqn. (7.29) is applicable, to get temperature variation as function of position and time, i.e.

T x τ,( ) T 0

T i T 0erf

x

2 α τ.........(7.29)

Now, we get:T T 0

T i T 00.211= since all temperatures are given.

From Table 7.3 for values of error function, or from Fig. 7.11, it is seen that


• Heat flux at the surface:


Note: -ve sign indicates that energy is leaving the surface, which is true,since the slab is being cooled.

• Total amount of heat removed, per unit surface area:


• Example 7.12: A large block of steel (� = 1.4 x 10-5

m2/s, k = 45 W/(m.C)) is initially at an uniform temperature of 25 C. Suddenly, its surface is exposed to a constant heat flux of 3 x 105 W/m2. Calculate the temperature at a depth of 3 cm after a period of 1 min.


• Temp. at a depth of 3 cm, after a time period of 60 s:

• This is the case of a semi-infinite slab, with constant heat flux conditions at its exposed surface. So, this is case (ii), refer Fig. 7.10 (b).



T x τ,( ) T i

2 q 0. α τ.

π.

kexp

x2

4 α. τ..

q 0 x.

k1 erf

x

2 α τ... .....(7.35)

Substituting and calculating, we get,

i.e. T x τ,( ) 98.949= C...temp. at a depth of 3 cm, after a time period of 60 s..Ans. .

• Example 7.14: In areas where ambient temperature drops to subzero temperatures and remains so for prolonged periods, freezing of water in underground pipelines is a major concern. It is of interest to know at what depth the water pipes should be buried so that the water does not freeze.


• At a particular location, the soil is initially at an uniform temperature of 15 C and the soil is subjected to a subzero temperature of -20 C continuously for 50 days.

• (i) What is the minimum burial depth required to ensure that the water in the pipes does not freeze? i.e. pipe surface temperature should not fall below 0 C.

• (ii) Plot the temp. distributions in the soil for different times i.e. after 1 day, 1 week, etc.

• Properties of soil may be taken as: � = 0.138 x


• Properties of soil may be taken as: � = 0.138 x 10-6 m2/s, r = 2050 kg/m3, k = 0.52 W/(m.K), C.p = 1840 J/kg.K.


• Depth at which temp. reaches 0 C:• We shall consider earth's surface as a semi-infinite medium, with the

surface suddenly brought to and maintained at a constant temperature, To. This belongs to case (i), refer to Fig. 7.10, (a).


T x τ,( ) T 0

T i T 0erf

x

2 α τ.........(7.29)

Now, we get:T T 0

T i T 00.571= since all temperatures are given.


• To plot the temp. distributions in the soil at a depth of 1 m for different times, :

• Again, we use eqn. (7.29).

T x τ,( ) T 0 T i T 0 erfx

2 α τ... ....eqn. (A)


Transient heat conduction in multi-dimensional systems – Product solution:• There are many practical cases where

assumption of one dimensional conduction may not be valid, i.e. temperature gradients may be significant in more than one dimension.

• For example, in a ‘short cylinder’ whose length is


• For example, in a ‘short cylinder’ whose length is comparable to diameter, and for a long rectangular bar, it is reasonable to say that heat transfer will be significant in both the x and r (or y )directions, and in a parallelpiped, heat transfer will be three dimensional.

Transient heat conduction in multi-dimensional systems – Product solution:

• Now, in such cases, it has been shown that the desired two dimensional solution is given as a product of the one dimensional solutions of the individual systems which form the two dimensional


systems which form the two dimensional body by their intersection. So, in general,

we write:θ

θ i solid

θθ i system1

θθ i system2

. θθ i system3

. .....(7.38)

• Some of the combinations of such one dimensional systems and the resulting two dimensional bodies are shown in Fig. 7.13.

• In Fig.7.13, for convenience, we use the following notations:

θ wall x τ,( )T x τ,( ) T a

T i T a wall

θθ i wall

....(7.39, a)


wall

θ cyl r τ,( )T r τ,( ) T a

T i T a long_cyl

θθ i long_cyl

....(7.39, b)

θ semi_inf x τ,( )T x τ,( ) T a

T i T asemi_inf

θθ i

semi_inf

.....(7.39, c)

• With the notation indicated above, two dimensional solution for a long, rectangular bar is given by:

T x y, τ,( ) T a

T i T a rect_bar

θ wall x τ,( ) θ wall y τ,( ). .....(7.40)

And, two dimensional solution for a short cylinderis given by:


is given by:

T r x, τ,( ) T a

T i T ashort_cyl

θ wall x τ,( ) θ cyl r τ,( ). .....(7.41)

• Important Note:• For a semi-infinite plate, x is measured from the

surface and for an infinite plate, x is measuredfrom the mid-plane.

• If temperature is to be calculated after a given time for the multi-dimensional body, the solution is straight forward;


• however, if the time is to be calculated to attain a given temperature, then, a trial and error solution will be required.

Heat transfer in transient conduction in multi-dimensional systems :

• Heat transfer in a multidimensional body in transient conduction can be obtained by using the Grober charts (given earlier) for Q/Qmax for the one dimensional systems constituting the given multi-dimensional body.


given multi-dimensional body. • For a body formed by the intersection of two

one-dimensional systems 1 and 2, we have:

Q

Q max total

Q

Q max 1

Q

Q max 2

1Q

Q max 1

. .....(7.42)

Heat transfer in transient conduction in multi-dimensional systems :

For a body formed by the intersection of threeone-dimensional systems 1, 2 and 3, we have:


Q

Q max total

Q

Q max 1

Q

Q max 2

1Q

Q max 1

. Q

Q max 3

1Q

Q max 1

. 1Q

Q max 2

......(7.43)

Basic relations for transient conduction:


Basic relations for transient conduction (contd.):


Multi-dimensional transient conduction:


Transient Heat-conduction-Part-II

Engineering

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