CHAPTER-4

TRANSIENT HEAT

CONDUCTION

By. Wubishet Degife

18/19/2020 MEng 3292

Whenever the boundary temperatures change the

temperature at each point of the system change with

time until steady-state temperature distribution is

reached. These are categorized as unsteady, or

transient heat conduction. Cooling of a hot metal

billet with air or water is a typical example.

The simplest situation is where temperature gradients

within the solid are small such that uniform

temperature can be assumed at any time. The

analysis that uses this approach is termed as lumped

capacitance method. Following this, analytical and

numerical methods will also be seen.

28/19/2020 MEng 3292

4.1 THE LUMPED CAPACITANCE METHOD

A typical application in heat treatment is quenching

where a metal or an alloy with initial temperature Ti

is suddenly immersed in a liquid of lower

temperature, T∞ < Ti. This is shown in fig-

chp5\fig5.1.pptx . If quenching begins at time t = 0,

then given sufficient time, the temperature of the

solid will decrease eventually to T∞. The heat

transfer is due to convection on the surface. The

assumption, here, is that the temperature of the solid

is spatially uniform. This assumption, according to

Fourier’s law implies infinite k which is clearly

impossible. But when38/19/2020 MEng 3292

compared to the convective heat transfer resistance,

it can approximately satisfy the above condition.

Energy balance on the surface gives

After substitution

Let

0EEEEEE ginstoutgin ==+=+

dt

dTVc)TT(hAEE sstout =−−=−

−=

−

dt

d

hA

Vc

givewillthisand,TT

s

48/19/2020 MEng 3292

Separating the variables and integration gives

where θi = Ti – T∞ is the initial condition

Evaluating the integrals will give

The above equation gives the

• time required for the temperature to reach T or

• Temperature at a prescribed time t

−=

i

t

0s

dtd

hA

Vc

−=

−

−=

−=

tVc

hAexp

TT

TT

ortlnhA

Vc

s

ii

is

58/19/2020 MEng 3292

The solution also indicates exponential decay as

shown in fig-chp5\fig5.2.pptx .

If we define τt as the thermal time constant, which is

an indicator of how fast the solid will respond to

surrounding temperature change

Where

Rt = convection heat transfer resistance

Ct = thermal capacitance (lumped)

Increase in τt (due to increase in Rt or Ct or both)

tt

s

t CR)Vc(hA

1=

=

68/19/2020 MEng 3292

the solid will respond slowly.

Total heat transfer up to some time t will be

Substitution of the solution gives

For quenching Q is positive and the solid experiences a decrease in energy.

==

=−==

t

0s

t

0

ss

dthAdtqQ

dthAdt)TT(hAdtqdQ

stout

t

i EEt

exp1)Vc(Q −==

−−=

78/19/2020 MEng 3292

5.2 VALIDITY OF THE LUMPED

CAPACITANCE METHOD

The previous method is the simplest and the most

convenient method for the solution of transient

problems. But it comes at a cost

- the assumption made-infinite thermal conductivity

Under what condition will this assumption hold?

Consider the steady-state condition shown in fig-

chp5\fig5.3.pptx . T∞<Ts,2<Ts,1.

Under steady-state condition

)TT(hA)TT(L

kA2,s2,s1,s −=−

88/19/2020 MEng 3292

Rearranging results in

Bi-Biot number-a dimensionless parameter which is

a ratio of the two thermal resistances.

Bi<<1: Rt,conv >> Rt,cond-equivalent to k→∞

Bi>>1: Rt,conv<< Rt,cond – not good for lumped

capacitance approach

For transient processes, consider fig-

chp5\fig5.4.pptx. The block is initially at Ti and

experiences cooling when it is immersed in a fluid

of T∞<Ti.

Bik

hL

R

R

hA/1

kA/L

TT

TT

conv,t

cond,t

2,s

2,s1,s ===−

−

98/19/2020 MEng 3292

For Bi<<1 the temperature gradient in the block is

small and T(x,t)≈T(t).

For moderate to large values of Bi-temperature

gradient is significant.

The lumped capacitance method will hold true for

Formula to be used for different geometries would

require the definition of Lc, a characteristic length,

as

Lc = V/As

1.0k

hLBi c =

108/19/2020 MEng 3292

Plane wall: wall thickness L, Lc=(LhW)/2hW=L/2

Cylinder: radius ro, Lc=(πro2W)/2πroW=ro/2

Sphere: radius ro, [4/3(πro3)]/4πro

2 = ro/3

For conservative approach, Lc values

Plane wall: same cylinder and sphere: ro

Using Lc = V/As, the exponent of the transient

equation may be expressed as (after some

manipulation)

2

c

c

2

c

c

c

s

L

t

k

hL

L

t.

c

k.

k

hL

cL

ht

Vc

thA

===

118/19/2020 MEng 3292

The above is a product of two dimensionless numbers

Fo is called Fourier number-dimensionless time.

Substitution in the solution gives

2

s

L

tFowhereFo.Bi

Vc

thA

=

)Fo.Biexp(TT

TT

ii

−=−

−=

128/19/2020 MEng 3292

Example 5.1

A thermocouple junction, which may be approximated

as a sphere, is to be used for temperature

measurement in a gas stream. The convection

coefficient between the junction surface and the gas

is h = 400 W/m2.K, and the junction thermophysical

properties are k = 20 W/m.K, c = 400 J/kg.K, and

ρ=8500 kg/m3. Determine the junction diameter

needed for the thermocouple to have a time constant

of 1 s. If the junction is at 25oC and is placed in a

gas stream that is at 200oC, how long will it take for

the junction to reach 199oC?

138/19/2020 MEng 3292

Figure for example 5.1148/19/2020 MEng 3292

Solution

1. As = πD2 and V = πD3/6

Substituting numerical values

With Lc = ro/3

Lumped capacitance method is an excellent approxim.

c6

Dx

Dh

1 3

2t

=

m10x06.7400x8500

1x400x6

c

h6D 4t −===

34

o 10x35.220x3

10x53.3x400

k

)3/r(hBi −

−

===

158/19/2020 MEng 3292

2. The time required for the junction to reach 199oC is

t

4

i

2

3

5s2.5t

200199

20025ln

400x6

400x10x06.7x8500

TT

TTln

)D(h

c)6/D(t

=

−

−=

−

−=

−

168/19/2020 MEng 3292

5.3 GENERAL LUMPED CAPACITANCE

ANALYSIS

Transient heat conduction can also be initiated by

radiation; by a heat flux from a sheet of electrical

heater attached to a surface, etc.

fig-chp5\fig5.5.pptx depicts the influence of

convection, radiation, an applied surface heat flux

and internal energy generation.

Applying energy conservation principle

dt

dTVcA)]TT()TT(h[EAq

dt

dTVcA)qq(EAq

)r,c(s

4

sur

4

gh,s

''

s

)r,c(s

''

rad

''

convgh,s

''

s

=−+−−+

=+−+

178/19/2020 MEng 3292

The above is a non-linear, first order, non-

homogeneous differential equation which can not be

integrated to obtain an exact solution

-requires approximate solution by numerical (finite

difference) approach.

Exact solution for simplified equation:

(a) If no imposed heat flux or internal generation and

convection is absent (vacuum) or negligible relative

to radiation, the above equation simplifies to

)(,

44 -= surrs TTAdt

dTVc

188/19/2020 MEng 3292

Separating the variables and integrating

The above will not give T explicitly and no solution

for Tsur = 0 (deep space). However for Tsur = 0 in the

above integral equation, it yields

−

+

−

+−

−

+=

−=

−−

sur

i1

sur

1

isur

isur

sur

sur

3

surr,s

T

T 44

sur

t

0

r,s

T

Ttan

T

Ttan2

TT

TTln

TT

TTln

TA4

Vct

givesTT

dTdt

Vc

A

i

−=

3

i

3

r,s T

1

T

1

A3

Vct

198/19/2020 MEng 3292

(b)For negligible radiation, and for θ = T-T∞ , dθ/dt =

dT/dt the equation reduces to a linear, first-order,

non-homogeneous differential equation

The above can be converted to a homogeneous

equation by using

Separating variables and integrating from 0 to t

gives the

+=

==−+

Vc

EAqb

Vc

hAa0ba

dt

d gh,s

''

c,s

0adt

dinresults

a

b ''

' =+−

20)to( ''

i 8/19/2020 MEng 3292

solution as

For steady state t→∞, the equation reduces to

(T-T∞)=(b/a)

)]atexp(1[TT

a/b)atexp(

TT

TT

)atexp()a/b(TT

)a/b(TT)atexp(

ii

ii'

'

−−−

+−=−

−

−=−−

−−−=

218/19/2020 MEng 3292

Example 5.2

Consider the thermocouple and convection condition

of example 5.1, but now allow for radiation

exchange with the walls of a duct that encloses the

gas stream. If the duct walls are at 400oC and the

emissivity of the thermocouple bead is 0.9, calculate

the steady-state temperature of the junction. Also,

determine the time for the junction temperature to

increase from initial condition of 25oC to a

temperature that is within 1oC of its steady-state

value.

228/19/2020 MEng 3292

Figure for example 5.2238/19/2020 MEng 3292

Solution

1. Energy balance on the thermocouple

Substituting numerical values gives

T = 218.7oC

2. As this involves the transient part, the complete

equation is

Numerical solution for T = 217.7oC gives t = 4.9 s

0EE outin =−

0A)]TT(h)TT([ s

44

sur =−−−

stoutin EEE =−

dt

dTVcA)]TT()TT(h[ s

44sur =−+−−

248/19/2020 MEng 3292

5.4 SPATIAL EFFECTS

For a one dimensional problem, the transient heat

conduction equation can be determined from the

general heat conduction equation as

The solution will require two boundary conditions and

one initial condition given as (fig-chp5\fig5.4.pptx)

IC: T(x,0) = Ti

BC: (1) (∂T/∂x)@x=0= 0

(2) [-k(∂T/∂x)@x=L] = h[T(L,t)-T∞]

t

T1

x

T2

2

=

258/19/2020 MEng 3292

The solution will have a functional form of

T=T(x, t, Ti, T∞, L, k, α, h) (too many variables!)

Non-dimensionalising the dependent variable T will

reduce the number of variables as follows:

Let θ=T-T∞, θi = Ti – T∞, dT = dθ

Define θ*=θ/θi then, dT = dθ =θidθ*

A dimensionless spatial coordinate is defined as

x* =x/L → dx = Ldx*

and dimensionless time

t*= (αt/L2) = Fo (Fourier No.), dt = (L2/α)dt*

268/19/2020 MEng 3292

Substituting the above in the transient conduction

equation will give

And the initial and the boundary conditions become

*

*

2*

*2

*2

*

i

2*2

*2

i

txor

)t()/L(

1

xL

=

=

)t,1(Bix

0x

1)0,x(

**

1x@*

*

0x@*

*

**

*

*

−=

=

=

=

=

278/19/2020 MEng 3292

]where Bi = hL/k

In dimensionless form, the functional dependence

becomes

θ* = f(x*, Fo, Bi)

much simpler than the representation of function T.

5.5 PLANE WALL WITH CONVECTION

This has an exact solution and also an approximate

solution derived from the exact solution.

5.5.1 Exact Solution

No attempt will be made to go through the steps of the

exact solution. Referring to fig-chp5\fig5.6.pptx, a

plane wall of thickness 2L with the assumption that 288/19/2020 MEng 3292

this thickness is much smaller than the width and

height (allows one dimensional justification). For

initial wall temperature, T(x,0) = Ti and suddenly

immersed in a fluid of T∞, the exact solution is in

series form and determined as

)2sin(2

sin4C

)xcos()Foexp(C

nn

nn

1n

*

n

2

nn

*

+=

−=

=

298/19/2020 MEng 3292

and the discrete values ζn, called eigenvalues are

positive roots of the transcendental equation

ζn tan ζn =Bi or tan ζn =Bi/ ζn

The first few solutions for Bi=10 are given in the

accompanying graph (fig-chp5\eigen1.pptx). Roots

for different values of Bi are given in the handout.

5.5.2 Approximate Solutions

For values of Fo>0.2, the infinite series solution can

be approximated by the first term of the series only.

This will give

)xcos()Foexp(C *

1

2

11

* −=308/19/2020 MEng 3292

At midplane (x*=0) fig-chp5\fig5.7.pptx

This will give θ* = θo* cos(ζ1x

*) fig-

chp5\fig5.8.pptx

The above equation shows that the time dependence of

the temperature at any location within the wall is the

same as that of the midplane temperature.

5.5.3 Total Energy Transfer

For a time interval from t=0 to any time t>0, energy

conservation equation can be written as

Ein –Eout = ΔEst

)Foexp(CTT

TT 2

11

i

o*

o −=−

−

318/19/2020 MEng 3292

For heat transfer from the surface

Ein = 0 and this gives Eout = -ΔEst = Q

Or Q = -[E(t) – E(0)] = -∫ρc[T(x,t) – Ti] dV

To nondimensionalise, let

Qo = ρcV(Ti – T∞) Qmax at t→∞

The nondimensional expression will be fig-

chp5\fig5.9.pptx

Use the approximate solution and with V= 2LWH,

dV=(dx)WH=(dx*L)WH to get (x*→0 to 1)

−=−

−−=

dV)1(V

1

V

dV

TT

]T)t,x(T[

Q

Q *

i

i

o

*

o

1

1

o

sin1

Q

Q

−=

328/19/2020 MEng 3292

5.5.4 Additional Considerations

The above solution is applicable to a plane wall,

thickness L and insulated on one side (x* = 0).

The foregoing results may be used to determine the

transient response of a plane wall to a sudden

change in surface temperature →equivalent to h=∞

which gives Bi = ∞, and T∞ replaced by Ts.

5.6 RADIAL SYSTEMS WITH CONVECTION

5.6.1 Infinite Cylinder

For an infinite cylinder, the temperature change is in

the radial direction only. This approximation can

hold true for .10r/L o 338/19/2020 MEng 3292

5.6.1 Exact Solution

In dimensionless form, the solution is given in series

form as

And the eigenvalues of ζn are positive roots of the

transcendental equation

fig-chp5\eigen2.pptx

)(J)(J

)(J2C

andr

tFowhere)r(J)Foexp(C

n

2

1n

2

o

n1

n

n

2

o1n

*

no

2

nn

*

+=

=−=

=

Bi)(J

)(J

no

n1n =

348/19/2020 MEng 3292

where J1 and Jo are Bessel functions of the first kind of

orders one and zero respecively.

5.6.2 Sphere

For a sphere with radius ro, the exact solution is given

by

2

o

*

n

1n*

n

2

nn

*

r

tFowhere)rsin(

r

1)Foexp(C

=−=

=

)2sin(2

)]cos()[sin(4C

nn

nnnn

−

−=

358/19/2020 MEng 3292

where the discreet values of ζn are the positive roots of

the transcendental equation (fig-chp5\eigen3.pptx)

1 – ζncot ζn = Bi

5.6.2 Approximate Solutions

For the infinite cylinder and sphere the series solution

can be approximated by a single term for Fo>0.2

and the time dependence of the temperature at any

location is the same as that of the centerline or

centerpoint.

Infinite Cylinder

Tcenterlineiswhere)r(J

or)r(J)Foexp(C

*

o

*

1o

*

o

*

*

1o

2

11

*

=

−=

368/19/2020 MEng 3292

given by (@r = r* = 0)

fig-chp5\fig5.10.pptx , fig-chp5\fig5.11.pptx

Sphere

Given by

fig-chp5\fig5.13.pptx , fig-chp5\fig5.14.pptx

)Foexp(C 2

11

*

o −=

Tintcenterpoiswhere)rsin(r

1

or)rsin(r

1)Foexp(C

*

o

*

1*

1

*

o

*

*

1*

1

2

11

*

=

−=

)Foexp(C 2

11

*

o −=

378/19/2020 MEng 3292

5.6.3 Total Energy Transfer

Using similar procedure as that of the plane wall, the

heat transfers can be determined as:

Infinite Cylinders

fig-chp5\fig5.12.pptx

Sphere

fig-chp5\fig5.15.pptx

)(J2

1Q

Q11

1

*

o

o

−=

)]cos()[sin(3

1Q

Q1113

1

*

o

o

−−=

388/19/2020 MEng 3292

5.6.4 Additional Considerations

The above also gives the solution when the cylinder or

sphere is subjected to a sudden change in surface

temperature to Ts. Replace T∞ by Ts which results

due to infinite h value, hence infinite Bi.

Example 5.3

A new process for treatment of a special material is to

be evaluated. The material, a sphere of radius ro = 5

mm, is initially in equilibrium at 400oC in a furnace.

It is suddenly removed from the furnace and

subjected to a two-step cooling process.

398/19/2020 MEng 3292

Step 1

Cooling in air at 20oC for a period of time ta until the

center temperature reaches a critical value,

Ta(0,ta)=335oC. For this situation, the convective

heat transfer coefficient is ha = 10 W/m2.K.

After the sphere has reached this critical temperature,

the second step is initiated.

Step 2

Cooling in a well-stirred water bath at 20oC, with a

convective heat transfer coefficient of hw =

6000W/m2.K.

408/19/2020 MEng 3292

ρ = 3000 kg/m2, k=20 W/m.K, c=1000 J/kg.K and

α = 6.66 x 10-6 m2/s

1. Calculate the time ta required for step 1 of the

cooling process to be completed.

2. Calculate the time tw required during step 2 of the

process for the center of the sphere to cool from

335oC to 50oC.

Solution

1. To check if lumped capacitance method can be

used with Lc = ro/3,

41

4oo 10x33.820x3

005.0x10

k3

rhBi −===

8/19/2020 MEng 3292

Figure for example 5.3 428/19/2020 MEng 3292

Indeed the lumped capacitance method can be used.

This will give

2. To check if the lumped capacitance method can be

used

Lumped capacitance method can not be used. A one

term approximation can be used.

43

s9420335

20400ln

20x3

1000x005.0x3000

TT

TTln

h3

crln

Ah

Vct

a

i

a

o

o

i

sa

a

=−

−=

−

−==

1.050.020x3

005.0x6000

k3

rhBi oo ===

8/19/2020 MEng 3292

With the Biot number now defined as

Bi=hwro/k=(6000x0.005)/20 = 1.50

The table gives C1 = 1.376 and ζ1 = 1.8 rad

One term approximation is justifiable.

44

2.082.020335

2050x

376.1

1ln

8.1

1

TT

T)t,0(Tx

C

1ln

1

Cln

1Fo

2

i

w

1

2

11

*

o

2

1

=

−

−−=

−

−−=

−=

s1.310x66.6

005.082.0

rFot

6

22

ow ===

−

8/19/2020 MEng 3292

5.7 THE SEMI-INFINITE SOLID

This is a geometry that is infinite in size in all but one

direction having one surface only. The interior is

well removed from the surface that it is unaffected

by the surface condition.

i.e. T(x→∞,t) = Ti

A one dimensional transient equation can be used for

such situation. Heat transfer near the surface of the

earth or transient response of a thick slab are a few

of the examples that can be mentioned.

Three possible situations can exist on the surface as

shown in fig-chp5\fig5.16.pptx .458/19/2020 MEng 3292

The familiar equation will be used

Interior boundary condition

T(x→∞,t) = Ti

The above equation can be transformed into an ODE

by using a function of the form η = η(x,t) that will

result in T(η) instead of T(x,t).

Transformation is done as follows:

Use the transformation equation

η ≡ x/(4αt)1/2

i2

2

T)0,x(T.C.It

T1

x

T=

=

468/19/2020 MEng 3292

to get the following derivatives

d

dT

t2d

dT

)t4(t2

x

td

dT

t

T

d

Td

t4

1

)t4(

1

d

dT

)t4(

1

d

d

xx

T

d

d

x

T

d

dT

)t4(

1

xd

dT

x

T

2/1

2

2

2/12/12

2

2/1

−=−=

=

=

=

=

=

=

478/19/2020 MEng 3292

Substitution gives

Boundary and initial conditions for case (1):

For x = 0 → η = 0

T(0,t) → T(η=0) = Ts

For x→∞, and t=0 (both corresponding to η→ ∞)

T(η→∞) = Ti

The equation to be solved is

d

dT2

d

Td2

2

−=

d2

ddT

ddTd

ord

dT2

d

ddTd

−=

−=

488/19/2020 MEng 3292

Integration gives

Integrating a second time, we obtain

u is a dummy variable.

Applying the boundary condition T(η=0) = Ts gives

C2 = Ts The resulting equation will be

)exp(Cd

dTorC

d

dTln 2

1

'

1

2

−=+−=

20

2

1 Cdu)uexp(CT +−=

+−=

0s

2

1 Tdu)uexp(CT

498/19/2020 MEng 3292

Applying the second boundary condition

T(η→∞) = Ti results in

Evaluating the definite integral gives

Hence the temperature distribution may be expressed

as

+−=0

s

2

1i Tdu)uexp(CT

2/1

si1

)TT(2C

−=

erfdu)uexp()/2(TT

TT

0

22/1

si

s −=−

−

508/19/2020 MEng 3292

erf η, called Gaussian error function is a standard

mathematical function whose values are given in

tables.

The surface heat flux may be obtained by using

Fourier’s law at x = 0 as follows:

2/1

is''

s

0@2/122/1

si

0@si

0x@

''

s

)t(

)TT(kq

)t4)(exp()/2)(TT(k

dx

d

d

)erf(d)TT(k

x

Tkq

−=

−−−=

−−=

−=

=−

==

518/19/2020 MEng 3292

Analytical solutions can also be determined for Case (2)

and Case (3). The summary of the results is given

below.

Case (1) Constant Surface Temperature: T(0,t)=Ts

Temperature within the medium monotonically approach

Ts with increasing t. Surface temperature gradient, and

hence the surface heat flux decreases as t-1/2.

t

)TT(k)t(q

t2

xerf)(erf

TT

T)t,x(T

is''

s

si

s

−=

==

−

−

528/19/2020 MEng 3292

Case (2) Constant Surface Heat Flux:

Surface temperature T(0,t) = Ts(t) for a constant heat

flux increases monotonically as t1/2.

Case (3) Surface Convection:

''

o

''

s qq =

−

−=−

t2

xerfc

k

xq

t4

xexp

k

)/t(q2T)t,x(T

''

o

22/1''

oi

)]t,0(TT[hx

Tk

0x@

−=

−

=

+

+−

=

−

−

k

th

t2

xerfc

k

th

k

hxexp

t2

xerfc

TT

T)t,x(T2

2

i

i

538/19/2020 MEng 3292

erfc η ≡ 1 - erf η - Complementary error function

For h→∞, the 2nd expression goes to zero, T∞=Ts.

Gives the same result as case 1.

Ts and temperatures within the medium approach the

fluid temperature T∞ with increasing time. As Ts

approaches T∞, there will be a reduction in heat flux.

fig-chp5\fig5.17.pptx shows the temperature

distribution on the surface and in the medium .

An interesting application of case 1 is when two semi-

infinite solids, at temperatures TA,i and TB,i (TA>TB)

are placed in

548/19/2020 MEng 3292

contact at their free surfaces as shown in fig-

chp5\fig5.18.pptx . For negligible contact

resistance, this will dictate a surface temperature Ts

at time of contact on both surfaces.

And solving for Ts gives

2/1

B

i,BsB

2/1

A

i,AsA''

B,S

''

A,S)t(

)TT(k

)t(

)TT(k,qq

−−=

−−=

2/1

B

2/1

A

i,B

2/1

Bi,A

2/1

A

s)ck()ck(

T)ck(T)ck(T

+

+=

558/19/2020 MEng 3292