One dim, steady-state, heat conduction_with_heat_generation

Lectures on Heat Transfer --One-Dimensional, Steady-State

Heat Conduction with Heat Generation

by

Dr. M. ThirumaleshwarDr. M. Thirumaleshwarformerly:

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

Mangalore

Preface

• This file contains slides on One-dimensional, steady-state heat conduction with heat generation.

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

August, 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.

• ��

August, 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

August, 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


One-Dimensional, Steady-State Heat Conduction with Heat

Generation

• Examples - Plane slab – different BC’s -Cylindrical Systems – Solid cylinder – hollow cylinder – different BC’s – Sphere with heat


cylinder – different BC’s – Sphere with heat generation- k varying linearly with T

Examples of situations with internal heat generation are:

• Joule heating in an electrical conductor due to the flow of current in it

• Energy generation in a nuclear fuel rod


due to absorption of neutrons• Exothermic chemical reaction within a

system (e.g. combustion), liberating heat at a given rate throughout the system

Examples (contd.)

• Heat liberated in ‘shielding’ used in Nuclear reactors due to absorption of electromagnetic radiation such as gamma rays

• Curing of concrete


• Curing of concrete• Magnetization of iron• Ripening of fruits and in biological decay

processes

Plane slab with uniform internal heat generation:

• We shall consider three cases of boundary conditions:

• 1. Both the sides of the slab are at the same temperature


same temperature• 2. Two sides of the slab are at different

temperatures, and• 3. One of the sides is insulated

Plane slab with uniform internal heat generation- both the sides at the same

temperature:• Assumptions:• One dimensional conduction i.e. thickness L is small

compared to the dimensions in the y and z directions• Steady state conduction i.e. temperature at any point

within the slab does not change with time; of course, temperatures at different points within the slab will be


within the slab does not change with time; of course, temperatures at different points within the slab will be different.

• Uniform internal heat generation rate, qg (W/m3)• Material of the slab is homogeneous (i.e. constant

density) and isotropic (i.e. value of k is same in all directions).

Plane slab with uniform internal heat generation- both the sides at the same

temperature:k, qg

Tw

To = Tmax

Tw

Temp. distribution-(parabolic)


X

Fig.5.1 Plane slab with internal heat generation - both sides at the same temp.

LL

• For the above mentioned stipulations, governing eqn. in cartesian coordinates reduces to:

d2 T

dx2

q g

k0 ....(5.1)

Solution of eqn. (5.1) gives the temperature profile and then, byusing Fourier’s equation we get the heat flux at any point.

B.C’s:


B.C’s:

(i) at x = 0, dT/dx = 0, since temperature is maximum at the centre line.(ii) at x = ± L, T = Tw

Integrating eqn. (5.1) once,

dT

dx

q g x.

kC1 .....(a)

• Integrating again,

Tq g x2.

2 k.C1 x. C2 ....(5.2)

Applying B.C (i) to eqn.(a): C1 = 0

Applying B.C. (ii) to eqn. (5.2): T wq g L2.

2 k.C2

2.


C2 T wq g L2.

2 k.i.e.

Substituting for C1 and C2 in eqn. (5.2):

T x( )q g x2.

2 k.T w

q g L2.

2 k.

i.e. T x( ) T wq g

2 k.L2 x2. ....(5.3) where L is half thickness of the

slab. (Remember this.)

• Also, by observation, T = Tmax at x = 0. (You can show this easily by differentiating eqn. (5.3) w.r.t x and equating to zero.)

• Then, putting x = 0 in eqn. (5.3):

T max T wq g L2.

2 k.......(5.4)

Then, from eqns. (5.3) and (5.4), we get:


T T w

T max T w

L2 x2

L21

x

L

2....(5.5)

Eqn. (5.5) gives the non-dimensional temperaturedistribution in a slab of half-thickness L, with heatgeneration.Note that the temperature distribution is parabolic, as shown inFig. 5.1.

Convection boundary condition:

• In many practical applications, heat is carried away at the boundaries by a fluid at a temperature Tf flowing on the surface with a convective heat transfer coefficient, h. (e.g. current carrying conductor cooled by ambient air or nuclear fuel rod cooled by a liquid metal


or nuclear fuel rod cooled by a liquid metal coolant).

• Then, by an energy balance at the surface:• heat conducted from within the body to the

surface = the heat convected away by the fluid at the surface.


• If A is the surface area of the slab (normal to the direction of heat flow), we have, from energy balance at the surface:

q g A. L. h A. T w T f.


q g A. L. h A. T w T f.

i.e. T w T fq g L.

h.....(5.6)

• Substituting eqn. (5.6) in eqn. (5.3),

T x( ) T fq g L.

h

q g

2 k.L2 x2. ......(5.7)

Eqn. (5.7) gives temperature distribution in a slab with heat generation, in terms of the fluid temperature, Tf . Remember, again, that L is half-thickness of the slab.

Heat transfer:


Heat transfer:

By observation, we know that the heat transfer rate from either of the surfaces must be equal to half of the total heat generated within the slab, for the B.C. of Tw being the same at both the surfaces.

i.e. Q = qg A L……..(5.8)

Plane slab with uniform internal heat generation – two sides at different

temperatures:• Let T1 > T2. Now, Tmax

must occur somewhere within the slab since heat is being generated in the slab and is flowing from inside to outside, both to

k, qg

Tmax

T1

Temp. distribution


inside to outside, both to the left and right faces. Let Tmax occur at a distance xmax from the origin, as shown in the fig.

X

Fig. 5.3 Plane slab with internal heat generation - two sides at different temp.

T2

L

xmax

• As shown earlier, the general solution for temperature distribution is given by eqn. (5.2), i.e.

Tq g x2.

2 k.C1 x. C2 ....(5.2)

C1 and C2 are obtained by applying the boundary conditions.For the present case, B. C’s are:B.C.(i): at x = 0, T = T1


B.C.(i): at x = 0, T = T1B.C.(ii): at x = L, T = T2

Then, from B.C.(i) and eqn. (5.2), we get: C2 = T1and, from B.C.(ii) and eqn. (5.2), we get:

T2q g L2.

2 k.C1 L. T1

i.e. C1T2 T1

L

q g L.

2 k.

• Substituting for C1 and C2 in eqn. (5.2),

T x( )q g x2.

2 k.T2 T1

L

q g L.

2 k.x. T1

i.e. T x( ) T1 L x( )q g

2 k.. T2 T1( )

Lx. .....(5.12)

Eqn. (5.12) gives the temperature distribution in the slab of thickness L, with heat generation and the two sides maintained at different temperatures of T1 and T2.


at different temperatures of T1 and T2.

Location and value of max. temperature:

Differentiate eqn. (5.12) w.r.t. x and equate to zero; Solving, let the value of x obtained be xmax ;Substitute the obtained value of xmax back in eqn. (5.12) to get the value of Tmax.

• Heat transfer to the two sides:• Total heat generated within the slab is equal to :

Qtot = qg A L• Part of this heat moves to the left and gets dissipated at

the left face; remaining portion of the heat generated moves to the right and gets dissipated from the right face.

• Applying Fourier’s Law:


• Applying Fourier’s Law:Qright = - k A (dT/dx)x=LQleft = - k A (dT/dx)x=0 …this will be -ve since heat flows from right to left i.e. in –ve x- direction

• Of course, sum of Qright and Qleft must be equal to Qtot.

Convection boundary condition:• Let heat be carried away at the left face by a

fluid at a temperature Ta flowing on the surface with a convective heat transfer coefficient, ha, and on the right face, by a fluid at a temperature Tb flowing on the surface with a convective heat transfer coefficient, hb.


• Note that heat generated in the slab in the volume between x = 0 and x = xmax has to move to the left face and the heat generated in the volume between x = xmax and x = L has to move to the right face, since no heat can cross the plane of max. temperature.


• Then, we have, from energy balance at the two surfaces:

On the left face:

q g A. xmax. h a A. T1 T a

. ........(a)


q g A xmax h a A T1 T a ........(a)

On the right face:

q g A. L xmax. h b A. T2 T b

. .........(b)

• From eqns.(a) and (b), we get T1 and T2 in terms of known fluid temperatures Ta and Tb respectively.

• After thus obtaining T1 and T2, substitute them in eqn. (5.12) to get the temperature distribution in terms of fluid temperatures T and T .


fluid temperatures Ta and Tb .

Plane slab with uniform internal heat generation – one face perfectly insulated:

k, qg

Tw

Tmax

Temp. distributionInsulated

h


X

Fig. 5.4 Plane slab with internal heat generation - one side insulated

Tw

L

Ta

ha

• For this case, the general solution for temperature distribution is given by eqn. (5.2), i.e.

Tq g x2.

2 k.C1 x. C2 ....(5.2)

B. C’s are:

B.C.(i): at x = 0, dT/dx = 0, since perfectly insulated.

B.C.(ii): at x = L, T = Tw


B.C.(ii): at x = L, T = Tw

From eqn. (5.2): dT

dx

q g x.

kC1

Then applying B.C.(i), we get: C1 = 0

From B.C.(ii) and eqn. (5.2):

dx k

C2 T wq g L2.

2 k.

• Substituting for C1 and C2 in eqn. (5.2):

T x( ) T wq g

2 k.L2 x2. .......(5.13)

Eqn. (5.13) gives the temperature distribution in a slab of thickness L, with heat generation when one side is perfectly insulated.


Fig. 5.4 sketches the temperature distribution in the slab. Note that now, L is the thickness of the slab andnot half-thickness.

In case of convection boundary condition:

• Since the left face is insulated, all the heat generated in the slab travels to the surface on the right and gets convected away to the fluid.

• Heat generated in the slab: Q gen q g A. L.


Heat convected at surface: Q conv h a A. T w T a.

Equating the heat generated and heat convected, we get:

T w T aq g L.

h a....(a)

• Substituting from (a) in eqn. (5.13),

T x( ) T aq g L.

h a

q g

2 k.L2 x2. .......(5.14)

Maximum temperature:

Obviously, max. temperature occurs at the insulated surface.

Putting x = 0 in eqn. (5.13):


Putting x = 0 in eqn. (5.13):

T max T wq g L2.

2 k......(5.15)

Eq. (5.15) gives Tmax in terms of wall temperature, Tw.

• Substituting for Tw from eqn. (a) in eqn. (5.15):

T max T aq g L.

h a

q g L2.

2 k. .....(5.16)

Eq. (5.16) gives Tmax in terms of fluid temperature, Ta.

From eqn. (5.13) and (5.15), we can write:

T x( ) T w L2 x21

x 2....(5.17)


T x( ) T w

T max T w

L x

L21

x

L....(5.17)

Eqn. (5.17) gives non-dimensional temperature distribution for a slab with heat generation, and one face insulated.

Note that now L is the thickness of the slab.

Solid cylinder with internal heat generation:• Consider a solid

cylinder of radius, R and length, L. There is uniform heat generation within its volume at a

Q

LTi

To

k, qg

Fig. 5.5 (a) Cylindrical system with heat generation

L

R


within its volume at a rate of qg (W/m3). Let the thermal conductivity, k be constant.

• See Fig. 5.5.

Temp. Profile,parabolicTo

Fig. 5.5 (b) Variation of temperaturealong the radius

R

Tw

• Now, the general differential eqn. in cylindrical coordinates reduces to:

d2 T

dr2

1

r

dT

dr.

q g

k0 ....(a)

Multiplying by r: rd2 T

dr2. dT

dr

q g r.

k0

i.e. dr

dT.q g r.


i.e.dr

rdr

.k

Integrating: rdT

dr.

q g r2.

2 k.C1

i.e. dT

dr

q g r.

2 k.C1

r......(b)

Integrating again: T r( )q g r2.

4 k.C1 ln r( ). C2 .....(5.18)

• Eqn. (5.18) is the general relation for temperature distribution along the radius, for a cylindrical system, with uniform heat generation.

• C1 and C2, the constants of integration are obtained by applying the boundary conditions.

• B.C’s are:• B.C. (i): at r = 0, dT/dr = 0 i.e. at the center of the


• B.C. (i): at r = 0, dT/dr = 0 i.e. at the center of the cylinder, temperature is finite and maximum (i.e. To = Tmax) because of symmetry.

• B.C. (ii): at r = R, i.e. at the surface , T = Tw

• From B.C. (i) and eqn. (b), we get: C1 = 0• From B.C. (ii) and eqn. (5.18), we get:

T wq g R2.

4 k.C2


w 4 k.

i.e. C2 T wq g R2.

4 k.

• Eqn. (5.19) is the relation for temperature distribution

Substituting C1 and C2 in eqn. (5.18):

T r( )q g r2.

4 k.T w

q g R2.

4 k.

i.e. T r( ) T wq g

4 k.R2 r2. .......(5.19)


• Eqn. (5.19) is the relation for temperature distribution in terms of the surface temperature, Tw.

• Note that this is a parabolic temperature profile, as shown in Fig. 5.5 (b).

• Maximum temperature:• Max. temperature occurs at the centre, because of

symmetry considerations.

• Therefore, putting r = 0 in eqn. (5.19):

T max T wq g R2.

4 k........(5.20)

From eqns. (5.19) and (5.20),

T T w

T max T w1

r

R

2.......(5.21)

Eqn. (5.21) is the non-dimensional temperature


Eqn. (5.21) is the non-dimensional temperature distribution for the solid cylinder with heat generation.


By an energy balance at the surface:heat generated and conducted from within the bodyto the surface = the heat convected away by the fluidat the surface.

i.e. π R2. L. q g. h 2 π. R. L.( ). T w T a

.

i.e. T w T aq g R.

2 h........(c)

Substituting(c) in eqn. (5.19):

T r( ) T aq g R.

2 h.

q g

4 k.R2 r2. ......(5.22)


Again, for max. temp. put r = 0 in eqn. (5.22):

T max T aq g R.

2 h.

q g R2.

4 k........(5.23)

Eqn. (5.23) gives maximum temperature in the solidcylinder in terms of the fluid temperature, Ta.

Current carrying conductor:• Consider a conductor of cross-sectional area, Ac and

length, L. Let the current carried be I (Amp.). Let the electrical resistivity of the material be ρ (Ohm.m).

• Then, heat generated per unit volume = Qg / Vol. of conductor,

where Qg is the total heat generated (W).

Q g I2 R. where R = electrical resistance of wire, (Ohms)


g

But, Rρ L.

A c

Therefore, q gI2 R.

A c L.

I2 ρ L.

A c

.

A c L.I

A c

2ρ. ...W/m 3

i = I/Ac , is known as the ‘current density’. Note its Units: A/m2

• Therefore, temperature distribution in a current carrying wire (of solid, cylindrical shape) is given by eq. (5.19), viz.

i.e. q g i2 ρ. i2

k ewhere k e

1

ρ= electrical conductivity, (Ohm.m) -1

T r( ) T wq g

4 k.R2 r2. .......(5.19)

Substituting for q , we get:


Substituting for qg, we get:

T r( ) T wi2 ρ.

4 k.R2 r2. .......(5.19 (a))

Max. temperature, which occurs at the centre, is obtained byputting r = 0 in eqn. (5.19, a). i.e.

T max T wi2 ρ. R2.

4 k.........(5.20(a))

• And, from eqns. (5.19 a) and (5.20 a), we get:

T T w

T max T w1

r

R

2

Note that the above eqn. for non-dimensional temperature distribution in a current carrying wire is the same as eqn. (5.21).


Hollow cylinder with heat generation:

• Hollow cylinder with the inside surface insulated:

• We have:

k, qgQ

To Ti To

Insulated

T r( )q g r2.

4 k.C1 ln r( ). C2 .....(5.18)


Fig. 5.7 Hollow cylinder with heat generation, inside surface insulated

ri

ro

4 k.

B.C.’s are:B.C.(i): at r = ri T = Ti and dT/dx = 0 (since inner surface is insulated), andB.C.(ii): at r = ro T= To Get C1 and C2 from these B.C.’s and substitute back in eqn. (5.18) to get the temperature distribution.

Hollow cylinder with the inside surface insulated:

• ALTERNATIVE METHOD: k, qg

Q

To Ti To

Insulated

dr

r

We shall derive the expression fortemperature distribution by a simplermethod of physical consideration


Fig. 5.8 Hollow cylinder with heat generation, inside surface insulated

To Ti To

ri

ro

Since the inside surface is insulated,heat generated within the volumebetween r = ri and r = r, must travelonly outward; and, this heat must beequal to the heat conducted awayfrom the surface at radius r.

and heat balance:

• Writing this heat balance,

q g π. r2 r i2. L. k 2. π. r. L. dT

dr. where dT/dr is the temp. gradient at radius r.

i.e. dTq g r i

2.

2 k.dr

r.

q g

2 k.r. dr.

Integrating, T r( )q g r i

2.ln r( ).

q g r2.C .....(b)


Integrating, T r( )2 k.

ln r( ).4 k.

C .....(b)

The integration constant C is obtained from the B.C.:

At r = ro, T = ToApplying this B.C. to eqn. (b): C T o

q g r o2.

4 k.

q g r i2.

2 k.ln r o

.

• Substituting value of C back in eqn. (b), we get,

T r( )q g r i

2.

2 k.ln r( ).

q g r2.

4 k.T o

q g r o2.

4 k.

q g r i2.

2 k.ln r o

.

i.e. T r( ) T oq g r i

2.

4 k.

r o

r i

2

2 lnr o

r. r

r i

2. .........(5.27)

Putting r = ri and T = Ti in eqn. (5.27), we get,


T i T oq g r i

2.

4 k.

r o

r i

2

2 lnr o

r i

. 1.

i.e. T i T oq g r i

2.

4 k.

r o

r i

2

2 lnr o

r i

. 1. ............(5.28)

• Eqn. (5.28) is important, since it gives the max. temperature drop in the cylindrical shell, when there is internal heat generation and the inside surface is insulated.

• If either of To or Ti is given in a problem, then the other temperature can be calculated using eqn. (5.28).



By an energy balance at the surface:heat generated within the body and conducted to the outersurface is equal to the heat convected away by the fluid at thesurface.

i.e. q g π. r o2 r i

2. L. h a 2. π. r o. L. T o T a

.

• Substituting the value of To from eqn. (c) in eqn. (5.27), we get:

i.e. T o T aq g r o

2 r i2.

2 h a. r o

.......(c)

T r( ) T aq g r o

2 r i2.

2 h a. r o

.

q g r i2.

4 k.

r o

r i

2

2 lnr o

r i

. r

r i

2. .........(5.29)


Eqn. (5.29) gives the temperature distribution in the cylindrical shell with heat generation, inside surface insulated, when the heat generated is carried away by a fluid flowing on the outer surface.

Hollow cylinder with the outside surface insulated:

• Start with:T r( )

q g r2.

4 k.C1 ln r( ). C2 .....(5.18)

Proceeding as earlier, apply the BC’s, get C1 and C2, substitute in (5.18) to get the temp. distribution.

ALTERNATIVELY:


ALTERNATIVELY:

Since the outside surface is insulated, heat generated within thevolume between r = ro and r = r, must travel only inward; and, thisheat must be equal to the heat conducted from the surface atradius r.

We shall derive the expression for temperature distributionby a simpler method of physical consideration and heatbalance:

• Writing this heat balance,

k, qg

Q

Ti

To

Insulateddr

Note that the term on theRHS has +ve sign, since,now, the heat transfer is fromoutside to inside, i.e. in the


Fig. 5.10 Hollow cylinder with heat generation, outside surface insulated

ri

ro

outside to inside, i.e. in the–ve r-direction (because theoutside surface is insulated).

• Integrating:

T r( )q g r o

2.

2 k.ln r( ).

q g r2.

4 k.C .......(b)

Eqn. (b) is the general solution for temperature distribution.The integration constant C is obtained by the B.C.:

At r = ri , T = Ti

Applying this B.C. to eqn. (b):


Applying this B.C. to eqn. (b):

C T iq g r o

2.

2 k.ln r i

.q g r i

2.

4 k.

Substituting value of C back in eqn. (b):

T r( )q g r o

2.

2 k.ln r( ).

q g r2.

4 k.T i

q g r o2.

2 k.ln r i

.q g r i

2.

4 k.

• Putting r = ro and T = To in eqn. (5.31), we get,

i.e. T r( ) T iq g r o

2.

4 k.2 ln

r

r i

.r i

r o

2r

r o

2. ..........(5.31)

T o T iq g r o

2.

4 k.2 ln

r o

r i

.r i

r o

2

1. .......(5.32)


Eqn. (5.32) is important, since it gives the max. temperaturedrop in the cylindrical shell, when there is internal heatgeneration and the outside surface is insulated.

If either of To or Ti is given in a problem, then the other temperature can be calculated using eqn. (5.32).

• Convection boundary condition:• We relate the surface temperature and fluid temperature

by an energy balance at the surface: • heat generated within the body and conducted to the

inner surface is equal to the heat convected away by the fluid at the surface.

i.e. T Tq g r o

2 r i2.

......(c)


i.e. T i T ag o i

2 h a. r i

.......(c)

Using eqn. (c) in eqn. (5.31), we get:

T r( ) T aq g r o

2 r i2.

2 h a. r i

.

q g r o2.

4 k.2 ln

r

r i

.r i

r o

2r

r o

2. .......(5.33)

Hollow cylinder with both the surfaces maintained at constant temperatures:

• We have for temp. distribution:

k, qg

To Ti To

rm

Tm

T r( )q g r2.

4 k.C1 ln r( ). C2 .....(5.18)


Fig. 5.11 Hollow cylinder with heat generation, losing heat from both surfaces

ri

ro

C1 and C2, the constants of integration are obtained by applying the boundary conditions.

• B.C.’s are:• B.C.(i): at r = ri T = Ti , and• B.C.(ii): at r = ro T= To

• Get C1 and C2 from these B.C.’s and substitute back in eqn. (5.18) to get the temperature distribution.

• After lengthy algebraic manipulations, we get, for temp. distribution:


T r( ) T i

T o T i

lnr

r i

lnr o

r i

q g

4 k.

r o2 r i

2

T o T i

.

lnr

r i

lnr o

r i

r

r i

21

r o

r i

2

1

. ......(5.35)

Hollow cylinder with both the surfaces maintained at constant temperatures:

• ALTERNATIVE METHOD:

• Let max. temp. occur at a radius rm . Obviously, rmlies in between ri and ro.

k, qg

To Ti To

rm

Tm


i o

• Therefore, dT/dr at r = rmwill be zero; i.e. surface at rm may be considered as representing an insulated boundary condition.

Fig. 5.11 Hollow cylinder with heat generation, losing heat from both surfaces

ri

ro

• So, the cylindrical shell may be thought of as being made up of two shells; the inner shell, between r = ri and r = rm , insulated on its ‘outer periphery’ and, an outer shell, between r = rm and r = ro, insulated at its ‘inner periphery’.

• Then, max. temperature difference for the inner shell and outer shell can be written from eqn.(5.32) and (5.28) respectively. So, we write:


• For the ‘inner shell’ (insulated on the ‘outer’’ surface):

T m T iq g r m

2.

4 k.2 ln

r m

r i

.r i

r m

2

1. .......(a)

Eqn. (a) is obtained by replacing ro by rm and To by Tm in eqn. (5.32).

•For the ‘outer shell’ (insulated on the ‘inner’’ surface):

T m T oq g r m

2.

4 k.

r o

r m

2

2 lnr o

r m

. 1. ............(b)

Eqn. (b) is obtained by replacing r by r and T by T in eqn. (5.28).


Eqn. (b) is obtained by replacing ri by rm and Ti by Tm in eqn. (5.28).

•Subtracting eqn. (a) from (b):

T i T oq g r m

2.

4 k.

r o

r m

2

2 lnr o

r m

. 1 2 lnr m

r i

.r i

r m

2

1.

i.e. T i T oq g r m

2.

4 k.

r o

r m

2 r i

r m

2

2 lnr m

r o

. 2 lnr m

r i

.. ....(c)

Eqn. (c) must be solved for rm.After some manipulation, we get:


i.e. r mq g r o

2 r i2. 4 k. T i T o

.

q g 2. lnr o

r i

.

........(5.36)

• Substituting the value of rm from eqn. (5.36) in either of eqns. (a) or (b), we get the max. temperature in the shell.

• Then, temperature distribution in the inner shell is determined from eqn. (5.32) and that in the outer shell is determined from eqn. (5.28).

• When Ti and To are equal:• it is seen from eqn. (5.36) that, position of max.

temperature in the shell is given by:


temperature in the shell is given by:

r mr o

2 r i2

2 lnr o

r i

.

i.e. rm depends only on the physical dimensions of the cylindricalshell and not on the thermal conditions.

Solid sphere with internal heat generation:

• Assumptions:• Steady state conduction• One dimensional

conduction, in the r direction only

Q

LTi

To

k, qg

Fig. 5.12 (a) Spherical system with heat generation

R

Solid sphere


direction only• Homogeneous, isotropic

material with constant k• Uniform internal heat

generation rate, qg(W/m3)

Temp. Profile,parabolic

To

Fig. 5.12 (b) Variation of temperature along the radius

R

Tw

• With the above stipulations, the general differential eqn. in spherical coordinates reduces to:

d2 T

dr2

2

r

dT

dr.

q g

k0 ....(a)

We have to solve eqn. (a) to get the temperature profile; then, byapplying Fourier’s Law, we can get the heat flux at any point.

Multiplying eqn. (a) by r2: r2 d2 T. 2 r. dT.q g r2.

0


Multiplying eqn. (a) by r2: rdr2

. 2 r.dr

.k

0

i.e. d

drr2 dT

dr.

q g r2.

k

Integrating: r2 dT

dr.

q g r3.

3 k.C1

i.e. dT

dr

q g r.

3 k.C1

r2......(b)

Integrating again: T r( )q g r2.

6 k.C1

rC2 .....(5.37)

C1 and C2, the constants of integration are obtained by applying the boundary conditions.


by applying the boundary conditions.B.C’s are:B.C. (i): at r = 0, dT/dr = 0 i.e. at the centre of the sphere, temperature is finite and maximum (i.e. To = Tmax) because of symmetry.

B.C. (ii): at r = R, i.e. at the surface , T = Tw

From B.C. (i) and eqn. (b), we get: C1 = 0

• From B.C. (ii) and eqn. (5.37), we get:

T wq g R2.

6 k.C2

i.e. C2 T wq g R2.

6 k.

Substituting C1 and C2 in eqn. (5.37):


T r( )q g r2.

6 k.T w

q g R2.

6 k.

i.e. T r( ) T wq g

6 k.R2 r2. .......(5.38)

Note that this is a parabolic temperature profile, as shownin Fig. 5.12 (b).

• Maximum temperature:• Max. temperature occurs at the centre, because of

symmetry considerations.• Therefore, putting r = 0 in eqn. (5.38):

T max T wq g R2.

6 k........(5.39)

From eqns. (5.38) and (5.39),


From eqns. (5.38) and (5.39),

T r( ) T w

T max T w1

r

R

2.....(5.40)

Eqn. (5.40) is the non-dimensional temperaturedistribution for the solid sphere with heatgeneration.

• Heat flow at the surface:• Heat transfer by conduction at the outer surface of

sphere is given by Fourier’s Law: i.e. Qg = - k A (dT/dr)at r = R

i.e. Q g k 4. π. R2.q g R.

3 k.. ...using eqn. (5.38) for T(r)


i.e. Q g4

3π. R3. q g

. ....(5.41)

Of course, in steady state, heat transfer rate at thesurface must be equal to the heat generation rate inthe sphere, i.e.

Qg = (4/3) ππππ R3 qg

• Convection boundary condition:• Now, heat generated and conducted from within the

body to the surface = heat convected away by the fluid at the surface.

i.e. 4

3π. R3. q g

. h a 4 π. R2.. T w T a.

q g R.


i.e. T w T aq g R.

3 h a.

.......(d)

Substituting(d) in eqn. (5.38):

T r( ) T aq g R.

3 h a.

q g

6 k.R2 r2. ......(5.42)

• Eqn. (5.43) gives the centre temperature of the sphere with heat generation, in terms of the fluid temperature,

Again, for max. temp. put r = 0 in eqn. (5.42):

T max T aq g R.

3 h a.

q g R2.

6 k........(5.43)


with heat generation, in terms of the fluid temperature, when the heat generated is carried away at the surface by a fluid.

• Relations for steady state, one dimensional conduction with internal heat generation, and constant k:


Plane slab, k varying linearly with T


Relations for steady state, one dimensional conduction with internal heat generation, and constant k


Hollow cylinder, k varying linearly with T


Sphere, k varying linearly with T


Ex: Hollow sphere with heat gen.


Ex: Slab with heat gen., convection on both sides

By observation, Max. temp. occurs at the centre of the slab.


One dim, steady-state, heat conduction_with_heat_generation

Engineering

Transcript of One dim, steady-state, heat conduction_with_heat_generation