conduction - rgcetpdy.ac.in year/heat and mass transfe… · the general heat conduction equation...

CONDUCTION

Fourier Law of Conduction

Rate of heat conduction is proportional to the area measured normal to the direction of

heat flow and to the temperature gradient in that direction.

dx

dTkAQ

dx

dTAQ

Where

A – Area in m2

dx

dTTemperature gradient in k/m

k – Thermal conductivity in W/mK

Thermal conductivity is defined as the ability of a substance to conduct heat.

[The negative sign indicates that the heat flows in a direction along which there is a

decrease in temperature]

GENERAL HEAT CONDUCTION EQUATION IN CARTESIAN COORDINATES

Consider a small rectangular element of sides dx, dy and dz as shown in Fig. 1.1.

The energy balance of this rectangular element is obtained from first law of

thermodynamics.

Net heat conducted into Heat generated Heat stored

element form all the + within the = stored in the ..(1.1)

coordinate directions element element

Net heat conducted into element from all the coordinate directions

Let q x be the heat flux in a direction of face ABCD and q x + d x be the heat flux in a

direction of face EFGH.

The rate of heat flow into the element in x direction through the face ABCD is

….. (1.2)

Where k – Thermal conductivity, W/mK

x

TTemperature gradient

The rate of heat flow out of the element in x direction through the face EFGH is

dxdydzx

Tk

xdydz

x

Tk

dxQx

QdQ

xx

xxxx

dydzx

TkdydzqQ xxx

dxdydzx

Tk

xdydz

x

TkdQ xxxx

Subtracting (1.2) – (1.3)

dxdydz

x

Tk

axdydz

x

Tkdydz

x

TkdQ xxxdxxx

x

Tk

xdydz

x

Tkdydz

x

Tk xxx dx dy dz

dxdydzx

Tk

xQQ xdxxx

…. (1.4)

Similarly

dxdydzy

Tk

ydQ ydyyy

…. (1.5)

dxdydzy

Tk

yqQZ ydyy

…. (1.6)

ADDING (1.4) + (1.5) + (1.6)

Net heat conducted

z

Tk

zdxdydz

y

Tk

ydxdydz

x

Tk

xzyx dx dy dz

z

Tk

zy

Tk

yx

Tk

xzyx dx dy dz

Heat Stored in the element

We know that,

Heat stored in the Mass of the Specific heat of Rise in

element = element Heat of the temperature

element of element

Net heat conducted into element from all the coordinate directions

z

Tk

zy

Tk

yx

Tk

xzyx dx dy dz

t

TCpm

dx dy dz t

TC p

[ Mass = Density Volume ]

….. (1.9)

Heat Stored within the element

Heat generated within the element is given by

Q = q dx dy dz ….. (1.9)

Substituting equation (1.7), (1.8) and (1.9) in equation (1.1)

(1.1) ==>

z

Tk

zy

Tk

yx

Tk

xzyx dx dy dz + q dx dy dz

t

TC p

dx dy dz

==> t

TCq

zkz

zy

Tky

yx

Tkx

xp

Considering the material is isotropic. So,

kx = ky = kz = k = constant.

==>

2

2

2

2

2

2

z

T

y

T

x

Tk + q

t

TC p

Divided by k,

t

T

k

C

k

q

z

T

y

T

x

T p

2

2

2

2

2

2

t

T

k

q

z

T

y

T

x

T

12

2

2

2

2

2

….. (1.10)

Heat stored in the element t

TC p

dx dy dz

It is a general three dimensional heat conduction equation in Cartesian coordinates

Where, Thermal diffusivity smC

k

p

/2

Thermal diffusivity is nothing but how fast heat is diffused through a material during

changes of temperature with time.

Case (i): No heat sources

In the absence of internal heat generation, equation (1.10) reduces to

t

T

z

T

y

T

x

T

12

2

2

2

2

2

….. (1.11)

This equation is known as diffusion equation (or) Fourier‟s equation.

Case (ii): Steady state conditions

In steady state condition, the temperature does not change with time. So, 0

t

T. The

heat conduction equation (1.10) reduces to

02

2

2

2

2

2

k

q

z

T

y

T

x

T ….. (1.12)

(or)

02

k

qTV

This equation is known as known as Poisson‟s equation.

In the absence of internal heat generation, equation (1.12) becomes:

02

2

2

2

2

2

z

T

y

T

x

T …… (1.13)

(or)

02

TV

This equation is known as Laplace equation.

Case (iii): One dimensional steady state heat conduction

If the temperature varies only in the x direction, the equation (1.10) reduces to

02

2

k

q

x

T …… (1.14)

In the absence of internal heat generation, equation (1.14) becomes:

02

2

x

T …… (1.15)

Case (iv): Two dimensional steady state heat conduction

If the temperature varies only in the x and y direction, the equation (1.10) becomes:

02

2

2

2

k

q

y

T

x

T ….. (1.16)

In the absence of internal heat generation, equation (1.16) reduces to

02

2

2

2

y

T

x

T … .(1.17)

Case (v): Unsteady state, one dimensional, without internal heat generation

In unsteady state, the temperature changes with time, i.e., 0

t

T. So, the general conduction

equation (1.10) reduces to

t

T

x

T

12

2

…… (1.18)

GENERAL HEAT CONDUCTION EQUATION IN CYLINDRICAL CO-ORDINATES

The general heat conduction equation in Cartesian coordinates derived in the previous

section is used for solids with rectangular boundaries like squares, cubes, slabs etc. But, the

Cartesian coordinate system is not applicable for the solids like cylinders, cones, spheres etc. for

cylindrical solids, a cylindrical coordinate system is used.

Consider a small cylindrical element of sides dr, d and dz as shown in fig. 1.2.

The volume of the element dv = r d dr dz.

Let us assume that thermal conductivity k, Specific heat Cp and density are constant.

The energy balance of this cylindrical element is obtained from first law of thermodynamics.

Net heat conducted into Heat generated Heat stored

element form all the + within the = stored in the ..(1.19)

coordinate directions element element

Net heat conducted into element from all the co-ordinate directions

Heat entering in the element through (r, ) plane in time d

dz

TdrrdkQ z

Heat leaving from the element through (r, ) plane in time d .

dzQz

QQ zzdzz

Net heat conducted into the element through plane in time dӨ.

(r, )

ddzrddrz

Tk

dzdz

Tdrrdk

z

dzQz

QQ

z

dzzz

..2

2

..

….. (1.20)

Heat entering in the element through ( , z) plane in time d .

dr

T

rr

Tdzrddrk

dr

Tr

rdzdrdk

drdr

Tdzrdk

r

drQr

QQ

r

drrr

1..

...

..

2

2

Heat entering in the element through (z, r) plane in time d .

d

r

TdzdrkQ

.

Heat leaving from the element through (z, r) plane in time d .

rdQr

QdQ

Net heat conducted through (r, ) plane = ddzrddrz

T..

2

2

Net heat conducted through ( , z) plane dr

T

rr

Tdzrddrk

1..

2

2

Net heat conducted into the element through (z, r) plane in time d .

ddzdrrdT

rk

ddzdrdT

rk

rddr

Tdrdzk

r

rdQr

dQd

2

2

2

1

1

.


2

2

z

Tk

(dr rd dz) d

+

ddzdrrdT

rk

dr

T

rr

Tdzdrrdk

2

2

2

2

2

1

1

[Adding equation 1.20, 1.21 and 1.22]

2

2

2

2

22

2

2

2

22

2

2

2

11

11

z

TT

rr

T

rr

Tddzdrrdk

T

rr

T

rr

T

z

Tddzdrrdk

….. (1.23)


2

2

2

2

22

211

z

TT

rr

T

rr

Tddzdrrdk

Heat generated within the element

Total heat generated within the element is given by

Q =q (dr rd dz) d …… (1.24)

Heat stored in the element

The increase in internal energy of the element is equal to the net heat stored in the

element.

Increase in internal energy

= Net heat stored in the element

= ( dr rd dz)

dT

C p

……. (1.25)

Substituting equation (1.23). (1.24) and (1.25) in (1.19)

k19.1 (dr rd dz)d qz

TT

rr

T

rr

T

2

2

2

2

22

211

(dr rd dz)

d

TC p

= (dr rd dz)

dT

C p

Divided by (dr rd dz) d

TCq

z

TT

rr

T

rr

Tk p.

112

2

2

2

22

2

==> …… (1.26)

It is a general three dimensional heat conduction equation in cylindrical co-ordinates.

==>

pC

kT

k

q

z

TT

rr

T

rr

T

1112

2

2

2

22

2

If the flow is steady, one dimensional and no heat generation, equation (1.26) becomes:

T

k

C

k

q

z

TT

rr

T

rr

T p

2

2

2

2

22

211

01

2

2

r

T

rr

T ……. (1.27)

(or)

==> 0.1

r

Tr

dr

d

r …….. (1.28)

CONDUCTION OF HEAT THROUGH A SLAB OR PLANE WALL

Consider a slab of uniform thermal conductivity k, thickness L, with inner temperature

T1, and outer temperature T2.

Let us consider a small elemental area of thickness „d x ‟.

From Fourier law of conduction, we know that,

kAdTdxQ

dx

dTkAQ

.

Integrating the above equation between the limits of 0 to L and T1 to T2.

21

12

0

0

0

0

2

1

2

1

2

1

TTkALQ

TTkALQ

AkAxQ

dTkAdxQ

kAdTQdx

T

T

L

L T

T

L T

T

21 TTL

kAQ ….. (1.29)

R

TQ

kA

L

TTQ

overall

21

……. (1.30)

Where

21 TTT

kA

LR Thermal resistance of slab.

CONDUCTION OF HEAT THROUGH HOLLOW CYLINDER

Consider a hollow cylinder inner radius r1, outer radius r2, inner temperature T1, outer

temperature T2 and thermal conductivity k.

Let us consider a small elemental area of thickness “dr”

From Fourier law of conduction, we know that,

dr

dTkAQ

Area of a cylinder is rL2

rLA 2

So,

LdTkr

drQ

dr

dTrLkQ

2

2

Integrating the above equation from r1 to r2 and T1 to T2.

2

1

2

1

2

r

r

T

T

dTLkr

drQ

21

1212

21

2

2

2 2

1

2

1

TTLkr

rQIn

TTLkInrInrQ

RLkInrQT

T

r

r

1

2

212

r

rIn

TTLkQ

…… (1.31)

1

2

21

2

1

r

rIn

Lk

TTQ

R

TQ overall ….. (1.32)

Where

21 TTT

1

2

2

1

r

rIn

LkR

Thermal resistance of the hollow cylinder.

CONDUCTION OF HEAT THROUGH HOLLOW SPHERE

Consider a hollow sphere of inner radius r1, outer radius r2, inner temperature T1, outer

temperature T2 and thermal conductivity k.

Let us consider a small elemental area of thickness „dr‟. From Fourier law of heat conduction, we

know that

dr

dTkAQ

Area of sphere is 24 r

dr

dTrkQ

rA

2

2

4

4

……. (1.33)

dTkr

drQ 4

2

Integrating on both sides

2121

12

12

21

2

2

4

411

41

4

4

2

1

2

1

2

1

2

1

2

1

2

1

TTkrr

rrQ

TTkrr

Q

Tkr

Q

dTkr

drQ

kdTr

drQ

T

T

r

r

T

T

r

r

T

T

r

r

21

12

214

rr

rr

TTkQ

…… (1.34)

21

12

21

4 rrk

rr

TTQ

R

TQ loveral …….. (1.35)

Where

21 TTT

21

12

4 rrk

rrR

Thermal resistance of hollow sphere.

Newton’s Law of Cooling

Heat transfer by convection is given by Newton law of cooling

TThAQ s ……. (1.36)

Where

A - Area exposed to heat transfer in m2

h – Heat transfer co-efficient in W/m2K

Ts – Temperature of the surface in K

T - Temperature of the fluid in K.

6. Heat Transfer Through a Composite Plane Wall with Inside and Outside Convection

Consider a composite wall of thickness L1, L2 and L3 having thermal conductivity k1, k2

and k3 respectively. It is assumed that the interior and exterior surface of the system are

subjected to convection at mean temperatures Ta and Tb with heat transfer co-efficient ha and hb

respectively. Within the composite wall, the slabs are subjected to conduction.

From Newton‟s law of cooling, we know that,

Heat transfer by convection at side A is

1TThaAQ a [From equn. (1.36)] …… (1.37)

Heat transfer by conduction at slab (1) is

1

211

L

TTAkQ

[From equn. (1.29)] …….. (1.38)

Heat transfer by conduction at slab (2) is

2

122

L

TTAkQ

…….. (1.39)

Similarly at slab (3) is

3

433

L

TTAkQ

……… (1.40)

Heat transfer by convection at side B is

bTThbAQ 4 ……… (1.41)

We know that,

AhQTT

a

a

11 [From equn. (1.37)]

Ak

LQTT

1

1

21 [ From equn. (1.38)]

Ak

LQTT

2

2

32 [From equn. (1.39)]

Ak

LQTT

3

3

43 [From equn. (1.40)]

AhQTT

b

b

14 [From equn. (1.410]

Adding both sides of the above equations

AhAk

L

Ak

L

Ak

L

Ah

TTQ

AhAk

L

Ak

L

Ak

L

AhQTT

ba

ba

ba

ba

11

11

3

3

2

2

1

1

3

3

2

2

1

1

R

TQ overall …….. (1.42)

Where

ba TTT

Thermal resistance, ba RRRRRR 321

=AhAk

L

Ak

L

Ak

L

Ah ba

11

3

3

2

2

1

1

We know that,

UA

TTQ

UAR

ba

1

1

ba TTUAQ …… (1.43)

Where

„U‟ is the overall heat transfer co-efficient (W/m2K).

HEAT TRANSFER THROUGH COMPOSITE PIPES OR CYLINDERS WITH INSIDE

AND OUTSIDE CONVECTION

A hot fluid at a temperature Ta, with heat transfer co-efficient ha, flowing through a pipe is

separated by two layers from atmosphere as shown in fig. 1.7. Let the thermal conductivities be

k1 and k2. On the outside surface heat is being transferred to a cold fluid at a temperature Tb with

heat transfer co-efficient hb.

Heat transfer by convection at side A is

1TTAhQ aa [From equn. No. (1.36)]

Here Area, A = Lr12

So,

112 TTLhrQ aa …….. (1.44)

Heat transfer by conduction at section 1 is

1

2

2112

r

rIn

TTLkQ

……… (1.45)

[From equn. (1.310]

Similarly

At section 2

2

3

3222

r

rIn

TTLkQ

……….. (1.46)

Heat transfer by convection at side B is

b

b

TTLhbrQ

TThbAQ

3

3

32 ………… (1.47)

We know that,

a

ahLr

QTT

1

12

[From equn. (1.440]

1

2

1

212 r

rIn

Lk

QTT

[From equn. (1.45)]

2

3

2

322 r

rIn

Lk

QTT

[From equn. (1.46)]

hbLr

QTT b

3

32

[From equn. (1.47)]

Adding both sides of the above equations

32

23

1

12

1

32

23

1

12

1

32

23

1

12

1

1//1

2

1

1//1

2

1//1

21

rhk

rrIn

k

rrIn

rhL

TTQ

rhk

rrIn

k

rrIn

rh

TTLQ

rhk

rrIn

k

rrIn

rhL

QTTa

ba

ba

ba

ba

ba

R

TQ overall

Where

32

23

1

12

1

1//1

2

1

rhk

rrIn

k

rrIn

rhLR

ba

baoverall TTT

We know that,

UA

TTQ

UAR

ba

1

1

ba TTUAQ ……… (1.49)

Where

U = Overall heat transfer co-efficient, W/m2K

A = Area = Lr32

CRITICAL RADIUS OF INSULATION

Addition of insulating material on a surface does not reduce the amount of heat transfer

rate always. In fact under certain circumstances it actually increases the heat loss up to certain

thickness of insulation. The radius of insulation for which the heat transfer is maximum is called

critical radius of insulation and the corresponding thickness is called critical thickness. If the

thickness is further increased, the heat loss will be reduced.

Critical Radius of Insulation For A Cylinder

Consider a cylinder having thermal conductivity k. Let r1 and r0 inner and outer radii of

insulation.

Heat transfer, Q =

kL

r

rIn

TTi

2

1

0

[From equn. No. (1.31)]

Considering h be the outside heat transfer co-efficient.

hAkL

r

r

rIn

TTQ i

0

1

1

0

1

2

Here A0 = Lr02

LhkL

r

rIn

TTQ i

0

1

0

2

1

2

To find the critical radius of insulation, differentiate Q with respect to r0 and equate it to

zero.

01

0

2

00

0

2

1

2

1

2

1

2

10

hLrr

rIn

kL

hLrkLrTT

dr

dQi

Since 0 TTi

02

1

2

12

00

hLrkLr

……. (1.50)

HEAT CONDUCTION WITH HEAT GENERATION

In many practical cases, there is a heat generation within the system. Typical examples

are

Electric coils

Resistance heater

Nuclear reactor

Combustion of fuel in the fuel bed of boiler furnaces.

PLANE WALL WITH INTERNAL HEAT GENERATION

Consider a slab of thickness L, thermal conductivity k, as shown in fig. 1.10.

Consider a small elemental area of thickness d x .

crh

kr 0

From Fourier‟s law of conduction, we know that

Heat transfer at dx

dTkAQxx , …… (1.51)

Heat conducted out at x +d x

dxdx

TdkA

dx

dTkAQ dxx 2

2

…… (1.52)

Heat generated within d x

qAdxQg …….. (1.53)

We know that,

02

2

2

2

2

2

qAdxdx

TdkA

dx

TdkAqAdx

dxdx

TdkA

dx

dTkAqAdx

dx

dTkA

QQQ dxxgx

02

2 dx

k

q

dx

Td ………. (1.54)

Integrating above equation

(1.54)

1

2

2

0

Cxk

q

dx

dT

dxk

q

dx

Td

……….. (1.55)

Integrating

(1.55)

21

2

1

2CxC

x

k

qT

Cxk

q

dx

dT

……… (1.56)

The temperature on the two faces of the slab (Tw) is the same because it loses it loses the

same amount of heat by convection on two sides.

Apply boundary conditions; C1 = 0

(1.56) 2

2

2

1Cx

k

qT

Apply T 2

,L

xTw

2

2

2

2

22

1

22

1

L

k

qTC

CL

k

qT

w

w

Substituting C1 and C2 value in equation (1.56)

k

qLTx

k

qT w

80

2

12

2

224

8xL

k

qTT w ……. (1.57)

The maximum temperature Tmax (at the centre) is obtained by putting 0x in Equation

(1.57).

21

2

2

1CxCx

k

qT

k

qLTC w

8

2

2

…….. (1.58)

Heat flow rate

qALQ2

1

Heat transfer by convection

h

qLTTw

qALhAThAT

hAThATqAL

TThAqALQ

TThAQ

w

W

w

w

2

2

1

2

1

2

1

Surface or Wall temperature

…….. (1.59)

CYLINDER WITH INTERNAL HEAT GENERATION

Consider a cylinder of radius r and thermal conductivity k. Heat is generated (Qg) in the

cylinder due to passage of an electric current.

From Fourier‟s law of conduction, we know that,

02

2

k

q

dr

Tdr r

………. (1.60)

Integrating

k

qLTT w

8

2

max

h

qLTTw

2

r

C

k

qr

dr

dT

Cr

k

q

dr

dTr

k

q

dr

Tdr r

1

1

2

2

2

2

2

0

Integrating

r

Cr

k

q

dr

dT 1

2

2122

2CInrC

k

qrT

……. (1.61)

212

2CInrC

k

qrT

Apply boundary conditions

C1 = 0

(1.61) 2

2

0

2C

k

qrTw [Put T = Tw, r = r0]

k

qrTC w

2

2

0

2

Apply C1 and C2 value in Equation (1.61)

22

04

rrk

qTT w

At centre

r = 0, T = Tmax

==> 2

0max4

rk

qTT w

k

qrTw

k

qrT

40

4

2

0

2

……. (1.62)

We know that,

Heat generated

LqrQ2

0 ……….. (1.63)

Heat transfer due to convection

TAhAQ w

TTLrhQ w02 ………. (1.64)

Equation (1.63) and (1.64)

h

qrTT

hTqrhT

hThTqr

TThqr

TTLrhLqr

w

w

w

w

w

2

22

22

2

2

0

0

0

0

0

2

0

……. (1.65)

Similarly,

For sphere, temperature at the centre

…… (1.66)

Maximum Temperature, 2

0max4

rk

qTT w

Surface temperature, h

qrTTw

2

0

k

qrTT wc

6

2

0

3

2

32

050.03/4

050.04250

3/4

250

4

q

rr

q

Temperature at the centre of the sphere

k

qrTT wc

6

2

[From Equn no. (1.66)]

18.06

050.01555281

2

Result:

Heat generated, q = 15,000 W/m3

Center temperature, Tc = 315.7 K

FINS

It is possible to increase the heat transfer rate by increasing the surface of heat transfer.

The surfaces used for increasing heat transfer are called extended surfaces of fins.

Types of fins

Some common types of fin configuration are shown in fig. 1.11.

3/000,15 mWq

KTc 7.315

(i) Uniform straight fin

(ii) Tapered straight fin

(iii) Splines

(iv) Annular fin

(v) Pin fins

Commonly there are three types of fin

Infinitely long fin

Short fin (end is insulated)

Short fin (end is insulated)

TEMPERATURE DISTRIBUTION AND HEAT DISSIPATION IN FIN

Fig. 1.12 (a) and (b) shows the straight fin or longitudinal fin of rectangular section and

circular section respectively. One end of the fin is enclosed in a heating chamber and the other

end is exposed to atmospheric air.

Heat is transferred across the rectangular fin and circular rod by conduction. From the

surface of the fin, heat is transferred to air by convection. Let us consider a small elemental area

of thickness dx, which is at a distance of x from the base.

A steady state conditions, heat balance equation for that element is as follows.

Heat conducted into the element = Heat conducted out of the element + heat convected to the

surrounding air.

convdxxx QQQ ……. (1.67)

Where,

TTpdxh

TThAQconv

dx

AdkA

dx

dTkAdxQx

dx

dTkAQx

dx

2

2

Substituting Q x , Q dxx and Qconv values in equation (1.67)

(1.67)

0

0

2

2

2

2

2

2

2

2

2

2

2

TTmdx

Td

TTkP

hP

dx

Td

TTkA

hP

dx

Td

TThpdx

TdkA

TTpdxhdxdx

TdkA

dx

dTkA

dx

dTkA

Where, kA

hPm

2

02

2

2

mdx

Td

kA

hPm

……. (1.68)

TT

Equation (1.68) shows that the temperature is a function of x and m. It is a second order,

linear differential equation. Its general solution is,

mxmxeCeC 21

……… (1.69)

The temperature distribution and heat dissipation depends upon the following fin

conditions.

Case (i): Infinitely long fin

If a fin is infinitely long, the temperature at its end is equal to that of the surrounding

fluid.

TbTAtx ;0 and TTAtX ;

bT Base temperature of fin

From equation (1.69), we know that,

mx

mx

mxme

eCeCTT

eCeC

21

21

…… (1.70)

TT

Substituting

0

2

0

170.1

;0

eCeCTT

TTAtX

b

b

==> ……. (1.71)

Substituting

21 CCTTb

TTAtX ; in equation (1.70)

0

0

2

21

21

m

mm

mmx

eC

eCeC

eCeCTT

,,0 Soem

Substituting C2 = 0 value in equation (1.71)

070.1 1 CTTb


mx

b

mx

b

eTT

TT

eTTTT

070.1

…….. (1.72)

Where,

Tb – Base temperature, K

T - Surrounding temperature, K

T – Intermediate temperature, K

x - Distance, m

kA

hPm

h – heat transfer co-efficient, W/m2K

P – Perimeter, m

k – Thermal conductivity, W/mK

C2 = 0

1CTTb

Temperature distribution of fin, mx

b

eTT

TT

A – Area, m2

Heat dissipation through the fin is obtained by integrating the heat lost by convection

over the entire fin surface.

We know that,

Heat lost by convection, TThAQconv

mx

b

mx

b

b

mx

eTTTTeTT

TT

dxTTehPQ

dxTThPQ

TThPdxQ

0

0

TThPkAQ

kA

hPmTThP

kA

hP

TThPm

mTThP

mTThP

mxem

TThP

dxeTTbhPQ

b

b

b

b

b

b

mx

1

1

1

11

10

0

…. ( 1.73)

Case (ii): Fin with insulated end (Short fin)

The fin has a finite length and the tip of fin is insulated.

Heat transferred, TThPkAQ b

,

bTTAtx

dx

dTLAtx

;0

;0;


meCmeCdx

dT

eCeCTT

mxmx

mxme

21

21

Applying the first boundary condition, i.e., at x = L, 0dx

dT

mLmL

mLmL

mLmL

eCeC

emCemC

meCmeC

21

21

210

……. (1.74)


mxmxeeCCTT 21

Applying the Second boundary condition, i.e., at x = 0, T = Tb

21

0

2

0

1

CCTT

eCeCTT

b

b

2

2

2 CeCTTmL

b mLeCC

2

21

12

2 mL

b eCTT

… (1.75)

Substituting C2 value in equation (1.74)

mLmLeCeC 21

12

2

mL

b

e

TTC

mLmLmL

mL

b

mL

mL

b

eee

TTbC

emLe

TTC

ee

TTC

222

21

2

21

1

1

12

1

==> …. (1.76)


mL

mx

mL

mx

b

mL

mx

mL

mx

b

mx

mL

bmx

mL

b

e

e

e

e

TT

TT

e

e

e

eTTTT

ee

TTe

e

TTTT

22

22

22

11

11

11

Multiplying the numerator and denominator by emL

and e-mL

mLmL

xLm

mLmL

Lxm

mL

mL

mL

mx

mL

mL

mL

mx

b

ee

e

ee

e

e

e

e

e

e

e

e

e

TT

TT

2211

mLmL

xLmxLm

b ee

ee

TT

TT

…… (1.77)

In terms of hyperbolic function it can be written as,

mL

xLm

TT

TT

b cosh

cosh

Temperature distribution of fin with insulated end

……… (1.78)

mL

b

e

TTC

21

1

mL

xLm

TT

TT

b cosh

cosh

mL

xLmmTTb

dx

dT

mL

xLmTTTT b

cosh

sinh

cosh

cosh

We know that,

Heat transferred, dx

dTkAQ

mLTTkA

hPkA

mLTTbkAm

mL

mLTTkAmQ

atx

mL

xLmTTkAmQ

mL

xLmmTTkA

b

b

b

b

tanh

tanh

cosh

sinh

0

cosh

sinh

cosh

sinh

kA

hPm

mLTThPkA b tanh

…..(1.79)

Applications

The main application of fins are

Cooling of electronic components

Cooling of motor cycle engines.

Cooling of small capacity compressors

Cooling of transformers

Cooling of radiators and refrigerators etc.

Heat transferred for insulated fin mLTThPkAQ b tanh

Fin efficiency

The efficiency of a fin is defined as the ratio of actual heat transferred fin to the

maximum possible heat transferred by the fin

maxQ

Q fin

fin

For insulated end

mL

mLfin

tanh

Fin effectiveness

It is defined as the ratio of heat transfer with fin to heat transfer without fin

Fin effectiveness, withoutfin

withfin

Q

QE

For insulated end

Fin effectiveness,

kP

hA

mLE

tanh

TRANSIENT HEAT CONDUCTION (OR) UNSTEADY STATE CONDUCTION

If the temperature of a body does not vary with time, it is said to be in a steady state. But

if there is an abrupt change in its surface temperature, it attains a steady state after some period.

During this period the temperature varies with time and the body is said to be in an unsteady or

transient state.

Transient heat conduction occur in cooling of IC engines, automobile engines, boiler

tubes, heating and cooling of metal billets, rocket nozzles, electric irons etc.

Transient heat conduction can be divided in to periodic heat flow and non periodic heat

flow.

(i) Periodic heat flow

In periodic heat flow, the temperature varies on a regular basis.

Examples: Cylinder of an IC engine,

Surface of earth during a period of 24 hours.

(ii) Non periodic heat flow:

In non periodic heat flow, the temperature at any point within the system varies non-

linearly with time.

Examples: heating of an ingot in a furnace, cooling of bars.

Biot Number

The ratio of internal conduction resistance to the surface convection resistance is known

as Biot number.

Biot Number ceisvectionresSurfacecon

cesisnductionreInternalco

tan

tan

k

hLcBi

Where

K – Thermal conductivity, W/mK

H – Heat transfer co-efficient, W/m2K

Lc – Characteristic length or significant length

Characteristic length, A

V

aSurfaceAre

VolumeLc

For slab:


LA

A

VLc

2

Where

L – Thickness of the slab

2

LLc

For cylinder:

Characteristic length, RL

LRRLc

22

2

Where

R – Radius of cylinder

For sphere:


VLc

2

3

4

3

4

R

R

Where

R – Radius of the sphere.

For Cube:


VLc

2

3

6L

L

Where

L – Thickness of the cube.

2

RLc

3

RLc

6

LLc

LUMPED HEAT ANALYSIS [NEGLIGIBLE INTERNAL RESISTANCE]

The process in which the internal resistance is assumed as negligible in comparison with

its surface resistance is known as Newtonian heating or cooling process.

In a Newtonian heating or cooling process the temperature is considered to be uniform at

a given time. Such an analysis is called lumped parameter analysis.

Let us consider a solid whose initial temperature is T0 and it is placed suddenly in

ambient air or any liquid at a constant temperature T . The transient response of the body can

be determined by relating its rate of change of internal energy with convective exchange at the

surface.

Convective heat loss from the body = Rate of change of internal energy

TThA = dt

dTVC p

TT

dT = dt

VC

hA

p

Integrating

TT

dT =

dt

VC

hA

p

TTIn = 1CtVC

hA

p

…. (1.80)

Apply boundary conditions.

At t = 0, T = T0

10 080.1 CTTIn

TTInC 01

Substituting C1 value in equation (1.80)

tVC

hA

TT

TTIn

tVC

hATTInTTIn

TTIntVC

hATTIn

p

p

p

0

0

0

…… (1.81)

Where

T0 – Initial temperature of the solid, K

T – Intermediate temperature of the solid, K

T

- Surface temperature of the solid (or) Finial temperature of the solid, K

h – Heat transfer co-efficient, W/m2K

A – Surface area of body, m2

P – Density of the body, kg/m3

V – Volume of the body, m3

Cp – Specific heat of the body, J/kg K.

t – Time, s.

Note

1. In lumped parameter system, Biot number value is less than 0.1.

i.e., 1.01.0 k

hLB c

i

VC

hAt

pe

TT

TT

0

2. T0 – Initial temperature, K

T – Intermediate temperature, K

T

- Surface temperature or Final temperature, K

HEAT FLOW IN SEMI-INFINITE SOLIDS

A solid which extends itself infinitely in all directions of space is known as infinite solid.

If an infinite solid is split in the middle by a plane, each half is known as semi infinite solid.

In a semi infinite solid, at any instant of time, there is always a point where the effect of

heating (or cooling) at one of its boundaries is not felt at all. At this point the temperature

remains unchanged.

Semi Infinite Plate

Consider a semi infinite body and it extends to infinity in the +ve x direction. The entire body is

initially at uniform temperature Ti including the surface at x = 0 is suddenly raised to T0.

The governing equation is

dt

dT

dx

Td

12

2

The boundary conditions are

T iTx 0,

T 0,0 0 fortTt

T 0, fortTt i

The analytical solution for this case is given by

t

xerf

TT

TT

i

x

20

0 ….. (1.82)

Where erf indicates “error function of” and the definition of error function is generally

available in mathematical texts. Usually tabulation of error values are available in data books.

- Thermal diffusivity, m2/s

t – Time, s

X – Distance, m

Ti – Initial temperature, K

T0 – Surface temperature (or) Final temperature, K

T x - Intermediate temperature, K

Note

1. In semi infinite solid, head transfer co-efficient or biot number value is .

i.e., h

or

iB

2. Ti – Initial temperature, K

T0 – Surface temperature (or) Final temperature, K

T x - Intermediate temperature, K

TRANSIENT HEAT FLOW IN AN INFINITE PLATE

A solid which extends itself infinitely in all directions of space is known as infinite solid.

Consider an infinite flat plate of uniform thickness 2L as shown in fig. 1.15, which is initially at

a uniform temperature of Ti. It is suddenly exposed to a large mass of fluid having a temperature

T . This temperature is assumed to be constant throughout the process of cooling or heating.

The plate is extended to infinity in the y and z directions.

The heat transfer co-efficient between the surface of the plate and the fluid on both sides

is assumed to be constant. The center of the plate is selected as the orgin.

The governing differential equation is

dx

dT

dx

Td

12

2

The boundary conditions are

Tt t = 0, T0 = Ti

Tt x = 0, 0dx

dT

Tt x = L, kA TThAdx

dT0

The solution of the above differential equation with this boundary condition is given by

2

0 ,,L

t

L

hL

L

xf

TT

TT

i

From this equation, we know that conduction resistance is not negligible. The

temperature history becomes a function of biot number ,

k

hLcFourier number

2L

tand the

dimension less parameter

L

xwhich indicates the location of point within the plate where

temperature is to be obtained. The dimensionless parameter

L

x is replaced by

R

rin case of

cylinders and spheres.

Heisler has prepared charts for graphical solutions of the unsteady state conduction

problems. These charts have been constructed in non-dimensional parameters. The charts are

suitable for problems with a finite surface and internal resistance. For such case the biot number

lies between 0 and 100.

These heiler charts were further extended and improved by grobe.

The heiler and grober charts are used to solve the problems of sudden immersion of plate,

cylinder or sphere into a fluid.

Note: For infinite solids,

Take Ti – Initial temperature – K

T - Final temperature – K

T0 - Center line temperature – K

T x - Intermediate temperature – K

The infinite solids, biot number value is in between 0.1 and 100 i.e., 0.1 < Bi < 100.

1. The wall of a cold room is composed of three layer. The outer layer is brick 30 cm thick.

The middle layer is cork 20 cm thick, the inside layer is cement 15 cm thick. Temperatures of

the outside air is 25˚C and on the inside air is - 20˚ C. The film co-efficient for outside air and

brick is 55.4 W / m2 K. Film co-efficient for inside air and cement is 17 W/m

2 K. Find heat

flow rate.

Take

k for brick = 2.5 W/mK

k for cork = 0.05 W/mK

k for cement = 0.28 W/mK

Given:

Thickness of brick, L3 = 30 cm = 0.3 m

Thickness of cork, L2 = 20 cm = 0.2 m

Thickness of cement, L1 = 15 cm = 0.15 m

Inside air temperature, Ta = -20˚ C + 273 = 253 K

Outside air temperature, Tb = 25˚ C + 273 = 293 K

Film co-efficient of r inner side, ha = 17 W/m2 K

kbrick = k3 = 2.5 W/mK

kcork = k2 = 0.05 W/mK

kcement = k1 = 0.28 W /mK

Inside Outside

Ta Tb

ha hb

L1 L2 L3

Cement

k1

Cork

k2

Brick

k3

To find:

Heat flow rate (Q/A)

Solution:

Heat flow through composite wall is given by

R

Q Toverall [From Equn no. 1.42 or HMT Date book page No. 43 and 44]

Where

4.55

1

5.2

3.0

05.0

2.0

28.0

15.0

17

1

298253/

11/

11

11

3

3

2

2

1

1

3

3

2

2

1

1

3

3

2

2

1

1

AQ

hk

L

k

L

k

L

h

TTAQ

AhAk

L

Ak

L

Ak

L

Ah

TTQ

AhAk

L

Ak

L

Ak

L

AhR

TTT

ba

ba

ba

ba

ba

ba

The negative sign indicates that the heat flows from the outside into the cold room.

Result:

Heat flow rate, Q/A = -9.5 W/m2

2. A wall of a cold room is composed of three layer. The outer layer is brick 20 cm thick, the

middle layer is cork 10 cm thick, the inside layer is cement 5 cm thick. The temperature of the

outside air is 25˚ C and that on the inside air is -20˚ C. The film co-efficient for outside air and

brick is 45.4 W/m2K and for inside air and cement is 17 W/m

2K.

Q/A = -9.5 W/m2

Take

k for brick = 3.45 W/mK

k for cork = 0.043 W/mK

k for cement = 0.294 W/mK

Given:

Inside Outside

Ta Tb

ha hb

L1 L2 L3

Thickness of brick, L3 = 20 cm = 0.2 m

Thickness of cork, L2 = 10 cm = 0.1 m

Thickness of cement, L1 = 5 cm = 0.05 m

Outside air temperature, Tb = 25˚ C + 273 = 293 K

Inside air temperature, Ta = -20˚ C + 273 = 253 K

Film co-efficient for outside air and brick, hb = 45.4 W/m2K

Film co-efficient for inside air and cement, ha = 17 W/m2K

k3 = 3.45 W/mK

k2 = 0.043 W/mK

k1 = 0.294 W /mK

Cement

k1

Cork

k2

Brick

k3

To find:

1. Heat flow rate

2. Thermal resistance of the wall

Solution:


RQ Toverall [From Equn no. (1.42) (or) HMT Date book page No. 43 and 44]

Where

4.45

1

45.3

2.0

043.0

1.0

294.0

05.0

17

1

298253/

11/

11

11

3

3

2

2

1

1

3

3

2

2

1

1

3

3

2

2

1

1

AQ

hk

L

k

L

k

L

h

TTAQ

AhAk

L

Ak

L

Ak

L

Ah

TTQ

AhAk

L

Ak

L

Ak

L

AhR

TTT

ba

ba

ba

ba

ba

ba

The negative sign indicates that the heat flows from the outside into the cold room.

Thermal Resistance

AhAk

L

Ak

L

Ak

L

AhR

ba

11

3

3

2

2

1

1

For Unit Area

ba hk

L

k

L

k

L

hR

11

3

3

2

2

1

1

Q/A = -17.081W/m2

W/m2

4.45

1

45.3

2.0

043.0

1.0

294.0

05.0

17

1 R

Result:

1. Heat flow rate, Q/A = -17.081 W/m2

2. Thermal resistance, R = 2.634 K/W

3. A wall is constructed of several layers. The first layer consists of masonary brick 20 cm

thick of thermal conductivity 0.66 W/mK, the second layer consists of 3 cm thick mortar of

thermal conductivity 0.6 W/mK, the third layer consists of 8 cm thick lime stone of thermal

conductivity 0.58 W/mK and the outer layer consists of 1.2 cm thick plaster of thermal

conductivity 0.6 W/mK. The heat transfer co-efficient on the interior and exterior of the wall

are 5.6 W/m2K and 11 W/m

2 K respectively. Interior room temperature is 22˚ C and outside air

temperature is -5˚C.

Calculate

Overall heat transfer co-efficient

Overall thermal resistance

The rate of heat transfer

The temperature at the junction between the mortar and the limestone

Given:

WKR /634.2

Thickness of masonary, L1 = 20 cm = 0.20 m

Thermal conductivity, k1 = 0.66 W/mK

Thickness of mortar, L2 = 3 cm = 0.03 m

Thermal conductivity of mortrar, k2 = 0.6 W/mK

Thickness of limestone, L3 = 8 cm = 0.08m


Thickness or Plaster, L4 = 1.2 cm = 0.012 m


Interior heat transfer, co-efficient ha = 5.6 W/m2K

Exterior heat transfer co-efficient hb = 11 W/m2K

Inside air temperature, Ta = 22˚C + 273 = 295 K

Outside air temperature, Tb = -5˚C + 273 = 268 K.

To fine:

a) Overall heat transfer co-efficient, U

b) Overall thermal resistance, (R)

c) Heat transfer/m2, (Q/A)

d) The temperature at the junction between the Mortar and the limestone, (T3)

Solution:


RQ Toverall [From Equn no. (1.42) or HMT Date book page No. 43 & 44]

Where

11

1

6.0

012.0

58.0

08.0

6.0

03.0

66.0

20.0

6.5

1

268295/

11

11

3

3

2

2

1

1

3

3

2

2

1

1

AQ

AhAk

L

Ak

L

Ak

L

Ah

TTQ

AhAk

L

Ak

L

Ak

L

AhR

TTT

ba

ba

ba

ba

We know that,

Heat transfer, ba TTUAQ [From equation no. 1.43]

Where, U – overall heat transfer co-efficient

268295

56.34

U

TTA

QU

ba

We know that,

Overall thermal resistance (R)

AhAk

L

Ak

L

Ak

L

Ak

L

AhR

ba

11

4

4

3

3

2

2

1

1

For unit Area

ba hk

L

k

L

k

L

k

L

hR

11

4

4

3

3

2

2

1

1

11

1

6.0

012.0

58.0

08.0

6.0

03.0

66.0

20.0

6.5

1

Heat transfer per unit area, Q/A = 34.56W/m2

Overall heat transfer co-efficient, U = 1.28 W/m2K

Interface temperature between mortasr and the limestorn, T3

Interface temperatures relation

b

b

a

a

R

TT

R

TT

R

TT

R

TT

R

TT

R

TTQ

5

4

54

3

43

2

32

1

211

a

a

R

TTQ 1

Ah

TQ

a/1

295 1

haARa

1

6.5/1

29556.34

/1

295/

1

1

T

h

TAQ

a

1

21

R

TTQ

Ak

L

TQ

1

1

28.288

Ak

LR

1

11

66.0

20.0

8.28856.34

8.288/

2

1

1

2

T

k

L

TAQ

R= 0.78 K/W

T1 = 288.8 K

KT 3.2782

2

32

R

TTQ

Ak

L

TQ

2

2

33.278

Ak

LR

2

2

2

6.0

03.0

3.27856.34

3.278/

3

2

2

3

T

k

L

TAQ

Temperature between Mortar and limestone (T3) is 276.5 K

Result:

Overall heat transfer co-efficient, U = 1.28 W/m2K

Overall thermal resistance, R = 0.78 K/W

Heat transfer, Q/A = 34.56 W/m2

Temperature between mortar and limestorne, (T3) = 276.5 K

4. An insulated steel pipe carrying a hot liquid. Inner diameter of the pipe is 25 cm, wall

thickness is 2 cm, thickness of insulation is 5 cm, temperature of hot liquid is 100˚C,

temperature of surrounding is 20˚C, inside heat transfer co-efficient is 730 W/m2K and outside

heat transfer co-efficient is 12 W/m2K. Calculate the heat loss per meter length of the pipe.

Take ksteel = 55W/mK, kinsulating material = 0.22 W/mK

Given:

T3=276.5 K

Inner diameter, d1 = 25 cm

Inner radius, r1 = 12.5 cm

Radius, r2 = r1 + thickness of wall

= 0.125 + 0.02

Temperature of hot liquid, Ta = 100˚C + 273

Ta = 373 K

Temperature of surrounding, Tb = 20˚C + 273

Tb = 293 K

Inside heat transfer co-efficient, ha = 730 W/m2K

Outside heat transfer co-efficient, hb = 12 W/m2K

ksteel = 55 W/mK

kinsulation = 0.22 W/mK

To find:

Heat loss per metre length

Solution:

Heat flow through composite cylinder is given by

RQ Toverall [From Equn no. 1.48 or HMT Date book page No. 43 & 44]

Where

r1 = 0.125 m

r3 = 0.195

m

195.12

1

22.0

145.

195.

55

125.

145.

125.730

1

2

1

293373

11

2

1

11

2

1

32

2

3

1

1

2

1

32

2

3

1

1

2

1

InInL

Q

rhk

r

rIn

k

r

rIn

rhL

TTQ

rhk

r

rIn

k

r

rIn

rhLR

TTT

ba

ba

ba

ba

Result:

Heat transfer per metre length, Q/L = 281. 178 W/m.

5. Air at 90˚C flows in a copper tube of 5 cm inner diameter with thermal conductivity 380

W/mK and with 0.7 cm thick wall which is heated from the outside by water at 120˚C. a scale

of 0.4 cm thick is deposited on the outer surface of the tube whose thermal conductivity is 1.82

W/mK. The air and water side unit surface conductance are 220 W/m2 K and 3650 W/m

2 K

respectively. Calculate

Overall water to air transmittance

Water to air heat exchange

Temperature drop across the scale deposit.

Given:

Q/L = 281. 178W/m

Inner air temperature, Ta = 90˚ + 273

Inner diameter of the copper, d1 = 5 cm

Radius, r1 = 2.5 cm

Thermal conductivity, k1 = 380 W/mK

Outer radius of the copper, r2 = inner radius + thickness of wall

r2 = 0.025 + 0.007 m

r3 = r2 + thickness of scale

= 0.032 + 0.004

Outside temperature of water, Tb = 120˚C + 273

= 393 K


Surface conductance of air, ha = 220 W/m2K

Surface conductance of water, hb = 3650 W/m2K

To find:

1. Overall heat transfer co-efficient, U

2. Water to air heat transfer, Q

3. Temperature drop across the scale deposit, (T3 - T2)

Solution:

Heat flow through composite cylinder is given by

Ta = 363 K

r1 = 0.025 m

r2 = 0.032 m

r3 = 0.036 m

R

Q Toverall [From Equn no. (1.48) or HMT Date book page No. 43 & 45]

Where

036.3650

1

82.1

032.

036.

380

025.

032.

025.220

1

2

1

393363

11

2

1

11

2

1

32

2

3

1

1

2

1

32

2

3

1

1

2

1

InInL

Q

rhk

r

rIn

k

r

rIn

rhL

TTQ

rhk

r

rIn

k

r

rIn

rhLR

TTT

ba

ba

ba

ba

[Negative sign indicates that heat flows from outside to inner side]

We know that,

Heat transfer, Q = UA T

Where

U – Overall heat transfer co-efficient

A – Area = 2 Lr3

T = Ta - Tb

Q/L = -739.79 W/m

393363036.0279.739

2

2

3

3

U

TTrUL

Q

TTLrUQ

ba

ba

==>

Interface temperatures

b

b

a

aba

R

TT

R

TT

R

TT

R

TT

R

TT

R

TQ

3

2

32

1

211 …….. (1)

(1) 2

32

R

TTQ

Where

82.1

032.0

036.0

2

1

3279.739

2

1

2

1

2

12

2

2

3

32

2

2

3

32

2

2

3

In

TT

k

r

rIn

TT

L

Q

k

r

rIn

L

TTQ

k

r

rIn

LR

U = 109.01 W/m2K

Overall heat transfer co-efficient, U = 109.01 W/m2K.

==> T2 – T3 = -7.6 K

==>

Result:

1. Overall heat transfer co-efficient, U = 109.01 W/m2K

2. Heat exchange, Q/L = -739.79 W/m

[Negative sign indicates that heat flows from outside to inner side]

3. Temperature drop across the scale deposit,

T3 – T2 = 7.6 K

6. An electrical wire of 10 m length and 1 mm diameter dissipates 200W in air at 25˚C. The

convection heat transfer co-efficient between the wire surface and air is 15W/m2K. The

thermal conductivity of wire is 0.582 W/mK. Calculate the critical radius of insulation and

also determine the temperature of the wire if it is insulated to the critical thickness of

insulation.

Given:

Length of the wire, L = 10 mm

Diameter of the wire, d = 1 mm

Radius of the wire, r = 0.5 mm = 0.510-3

m

Heat transfer, Q = 200 W

Surrounding temperature, Tb = 25˚C + 273 = 298 K

Convection heat transfer co-efficient between the wire surface and air, hb = 15 Wm2K.

Thermal conductivity of wire, k = 0.582 W/mk

T3 – T2 = 7.6 K

Temperature across the scale deposit, T3 – T2 = 7.6 K

To find:

1. Critical radius of insulation, rc

2. Temperature of wire, Ta

Solution:

We know that,

Critical radius of insulation, n

krc

15

582.0

Heat transfer through an insulated wire when critical radius is used is given by

146.0

298200

0388.015

1

582.0

0005.0

0388.0

102

1

298200

1

2

1

1

1

Ta

In

T

rhk

r

rIn

L

TTQ

a

cb

c

ba

==>

Result:

1. Critical radius of insulation, rc = 0.0388mm

2. Temperature of the wire, Ta = 327.28 K (or) 54.2˚C.

rc = 0.0388 m

Ta = 327.28 K

7. A wire of 6 mm diameter with 2 mm thick insulation (K = 0.11 W/mK). If the convective

heat transfer co-efficient between the insulating surface heat transfer co-efficient between the

insulating surface and air is 25 W/m2K, find the critical thickness of insulation and also find

the percentage of change in the heat transfer rate if the critical radius is used.

Given:

d1 = 6 mm

r1 = 3 mm

r2 = r1+2 = 3+2 = 5 mm = 0.005 m

k = 0.11W / mK

hb = 25 W/m2K

Solution:

1. Critical radius, h

krc [From. No. (1.50)]

mrc

3104.4

25

11.0

Critical thickness, tc = rc – r1

= 4.4 10-3

m

2. Heat transfer through an insulated wire is given by

mrc

3104.4

Critical thickness, tc = 1.4 10-3

m (or) 1.4 mm

64.12

2

005.025

1

11.0

003.0

005.0

2

1

2

1

1

21

1

2

1

ba

ba

b

ba

TTLQ

In

TT

rhk

r

rIn

L

TTQ

[From HMT data book page no. 43 & 45]

Heat flow through an insulated wire when critical radius is used is given by

572.12

2

104.425

1

11.0

003.0

104.4

2

1

2

1

2

3

3

1

1

2

ba

ba

cb

c

ba

TTLQ

In

TTL

rhk

r

rIn

L

TTQ

Percentage of increase in heat flow by using

Critical radius 1001

12

Q

QQ

%55.0

64.12

1

10064.12

1

57.12

1

Result:

1. Critical thickness, tc = 1.410-3

2. Percentage of increase in heat transfer by using critical radius = 0.55%

8. An electric current is passed through a composite wall made up of two layers. First layer is

steel of 10 cm thickness and second layer is brass of 8 cm thickness. The outer surface

temperature of steel and brass are maintained at 120˚C and 65˚C respectively. Assuming that

the contact between two slab is perfect and the heat generation is 1, 65,000 W/m3.

Determine:

1. Heat flux through the outer surface of brass slab

2. Interface temperature.

Take k for steel is 45 W/mK. K for brass is 80 W/mK.

Given:

Thickness of steel, L1 = 10 cm = 0.10 m

Thickness of brass, L2 = 8 cm = 0.08 m

Surface temperature of steel, T1 = 120˚C + 273 = 393 K

Outside surface temperature of brass, T3 = 25˚ C + 273 = 293 K

Heat generation, qg = 1, 65,000 W/m3

k1 = 45 W/mK

k2 = 80 W/mK

To find:

1. Heat flux through the surface of the brass slab, q2

2. Interface temperature, T2.

Solution:

Let

q1 – Heat flux through the surface of the steel slab

q2 – Heat flux through the surface of the brass slab

Heat generation qg = q1 + q2 ……. (1)

Heat transfer through steel,

Ak

L

TTQ

R

TQ

1

1

21

1

1

kA

LR

Let interface temperature T2 is greater than T1. So,

Ak

L

TTQ

1

1

12

1

…… (2)

Heat transfer through brass is given by

Ak

L

TTQ

R

TQ

2

2

32

2

2

……. (3)

Total heat transfer [Adding (2) + (3)]

2

2

32

1

1

12

2

2

32

1

1

12

211

/

1

k

L

TT

k

L

TTAQ

Ak

L

TT

Ak

L

TTQ

QQQ

Heat flux (or) Heat generation

KT

T

T

T

TT

TT

TTq

k

L

TT

k

L

TTAQqg

g

6.468

5.1454636,16,5000,65,1

636,16,554.1454

3380003.636,78,1100054.454

101

338

101102.2

393

102.2

101

338

102.2

393000,65,1

80

08.0

338

45

10.0

393

/

2

2

2

2

33

2

33

2

3

2

3

2

22

2

2

32

1

1

12

Heat transfer through steel, is given by

Interface temperature, T2 = 468.6K

45

10.0

3936.468

/

1

1

1

12

1

1

1

12

1

q

k

L

TTAQ

Ak

L

TTQ

From equation (1)

Heat generation, qg = q1+ q2

==> 1, 65,000 =34.020 +q2

==> q2 = 1, 65,000 – 34.020

q2 = 1, 30,980 W/m2

Heat flux through the

Surface of the brass slab, q2 = 1, 30,980 W/m2

Result:

(i) q2 = 1, 30,980 W/m2

(ii) T2 = 468.6 K.

9. A copper wire of 40 mm diameter carries 250 A and has a resistance of 0.25 10-4

cm/length surface temperature of copper wire is 250˚C and the ambient air temperature is

10˚C. If the thermal conductivity of the copper wire is 175 W/mK, calculate

1. Heat transfer co-efficient between wire surface and ambient air.

2. Maximum temperature in the wire.

Given:

Diameter, d = 40 mm = 0.040 m

Radius, r = 20 mm = 0.020 m

Current, I = 250A.

q1=34020 W/m2

Resistance, R = 0.2510-4 cm/length

Surface temperature, Tw = 205˚C + 273 = 523 K

Ambient air temperature, T

= 10˚C + 273 = 283 K

Thermal conductivity, k = 175 W/mK

To find:

1. Heat transfer co-efficient, h

2. Maximum temperature, Tmax.

Solution:

Heat transfer, Q = I2R

= (250)2 (0.2510-4

)

= 1.562 W/cm

= 1.56102 W/m

= 156W/m

We know that,

Heat generated, LrV

Qq

2

156

1020.0

1562

q

We know that,

Maximum temperature

k

qtTT w

4

2

max [From Equn no. (1.62)]

.07.523

1754

020.0124140523

2

K

q = 124140W/m3

Tmax = 523.07 K.

We know that,

Surface temperature, h

rqTTw

2 [From Equn no.(1.65)]

h

2

124140020.0283523

==>

Result:

1. Heat transfer co-efficient, h = 5.17W/m2K

2. Maximum temperature, Tmax = 523.07 K.

10. An aluminum alloy fin of 7 mm thick and 50 mm long protrudes from a wall, which is

maintained at 120˚C. The heat transfer coefficient and conductivity of the fin material are

140W/m2K and 55 W/mK respectively. Determine

1. Temperature at the end of the fin.

2. Temperature at the middle of the fin.

3. Total heat dissipated by the fin.

Given:

Thickness, t = 7 mm = 0.007 m

Length, L = 50 mm = 0.050 m

Base temperature, Tb = 120˚C + 273 = 393 K

Ambient temperature, T

= 22˚ + 273 = 295 K

Heat transfer co-efficient, h = 140 W/m2K

Thermal conductivity, k = 55 W/mK.

To find:

1. Temperature at the end of the fin.

2. Temperature at the middle of the fin

3. Total heat dissipated by the fin.

h = 5017 W/m2K.

Solution:

Since the length of the fin is 50 mm, it is treated as short fin. Assume end is insulated.

We known that,

Temperature distribution [Short fin, end insulated]

mL

xLm

TT

TT

b cosh

cosh

……. (1)

[From HMT data book page no. 49]

i) Temperature at the end of the fin, Put Lx

mLTT

TT

mL

LLm

TT

TT

b

b

cosh

1

cosh

cosh)1(

…… (2)

Where

kA

hPm

P = Perimeter = 2L (Approx)

= 2 0.050

A – Area = Length thickness = 0.0500.007

4105.355

1.0140

kA

hPm

P = 0.1 m

A = 3.5 10-4

m2

m = 26.96 m-1

8.47295

05.2

1

295393

295

05.2

1

050.09.26cosh

1)2(

T

T

TT

TT

TT

TT

b

b

==>

ii) Temperature at the middle of the fin.

Put Lx /2 in Equation (1)

6025.0295393

295

049.2

234.1

295393

295

050.09.26cosh

2

050.0050.09.26cosh

cosh

2/cosh)1(

T

T

TT

TT

mL

LLm

TT

TT

b

b

Temperature at the middle of the fin

iii) Total heat dissipated


050.09.26tanh295393105.3551.0140

tanh

21

4

21

mLTThPkAQ b

T = 342.8 K

Temperature at the end of the fin, T Lx = 342.8 K

T = 354.04 K

KTLx 04.354

2/

Q = 44.4 W

Result:

1. Temperature at the end of the fin, KT Lx 8.342

2. Temperature at the middle of the fin, KT Lx 04.3542/

3. Total heat dissipated, Q = 44.4 W

11. Ten thin brass fins (k=100 W/mK), 0.75 mm thick are placed axially on a 1 m long and 60

mm diameter engine cylinder which is surrounded by 27˚C. The fins are extended 1.5 m from

the cylinder surface and the heat transfer co-efficient between cylinder and atmospheric air is

15 W/m2K. Calculate the rate of heat transfer and the temperature at the end of fins when the

cylinder surface is at 160˚C.

Given:

Number of fins = 10


Thickness of the fin, t = 0.75 10-3

m

Length of engine cylinder, Lcy = 1 m

Diameter of the cylinder, d = 60 mm = 0.060 m

Atmosphere temperature, T

= 27˚C + 273 = 300 K

Length of the fin, Lf = 1.5 cm = 1.510-2

m


Cylinder surface temperature

Or

base temperature, Tb = 160˚C + 273 = 433 K

To find:

1. Rate of heat transfer Q

2. Temperature at the end of the fin

Solution:

Length of the fin is 1.5 cm. So, this is short fin. Assuming that the fin end is insulated

We know that,

Heat transferred, fb mLTThPkAQ tanh2

1

……… (1)

[From HMT data book page no.49]

Where

P – Perimeter = 2Length of the cylinder

= 21

A – Area = length of the cylinder thickness

= 10.7510-3

m

31075.0100

215

kA

hPm

P = 2 m

A = 10.7510-3

m

20 = m-1

WQ

Q

mLTThPkAQ fb

1.58

29.01335.1

105.120tanh3004331075.0100215

tanh)1(

1

1

221

3

21

1

Heat transferred per fin = 58.1 W

Heat transferred for 10 fins = 58.1 10 = 581 W

…….. (2)

Heat transfer from un finned surface due to convection is

TTLftdLh

ThAQ

bcy 10

2

[Area of unfinned surface = Area of cylinder – Area of fin]

300433105.11075.0101060.01523

……. (3)

So, total heat transfer, Q = Q1 + Q2

Q = 581 + 375.8

We known that,

Temperature distribution [short fin, end insulated]

f

f

b mL

xLm

TT

TT

cosh

cosh


We need temperature at the end of fin, so, put Lx

Q1 = 581 W

Q2 = 375.8 W

Total heat transfer, Q = 956.8 W

95.0

300433300

95.0

95.0

95.0

1

105.120cosh

1

cosh

cosh

2

TTTT

TTTT

TT

TT

mL

LLm

TT

TT

b

b

b

fb

Result:

1. Heat transfer, Q = 956.8 W

2. Temperature at the end of the fin, T = 440K.

12. An aluminum plate (k = 160 W/m˚C, = 2790 kg/m3, Cp = 0.88 KJ/kg˚C) of thickness L

= 3 cm and at a uniform temperature of 225˚C is suddenly immersed at time t = 0 in a well

stirred fluid maintained at a constant temperature T

= 25˚C. Take h = 320 W/m2C.

Determine the time required for the centre of the plate to reach 50˚C.

Given:

Thermal conductivity of aluminum, k = 160 W/m˚C

Density, = 2790 kg/m3

Specific heat, Cp = 0.88 KJ/kg˚C = 0.88103 J/kg˚C

Thickness, L = 3 cm = 0.03 m

Initial temperature, T0 = 225˚C + 273 = 498 K

Final temperature, T

= 25˚C + 273 = 323 K

Heat transfer co-efficient, h = 320 W/m2˚C

To find:

Time (t) required to reach 50˚C.

T = 440 K

Solution:

We know that, For slab,

Characteristic length, cL

2

03.0

2

L

Biot number Bi k

hLe

160

015.0320

Biot number value is less than 0.1. So, this is lumped heat analysis type problem.

For lumper parameter system.

tVCp

hA

eTT

TT

0


We know that,

Characteristics length, A

VLc

t

tIn

e

eTT

TT

t

tVCp

hA

00868.0079.2

2790015.01088.0

320125.0

298498

298323

)1(

3

2790015.01088.0

320

0

3

==>

Lc = 0.015 m

Bi = 0.03 < 0.1

t = 239.26 s

Result:

Time required to reach 50˚C is 239.26 s.

13. A steel ball (specific heat = 0.46 kJ/kgK. and thermal conductivity = 35 W/mK) having 5

cm diameter and initially at a uniform temperature of 450˚C is suddenly placed in a control

environment in which the temperature is maintained at 100˚C. Calculate the time required of

the ball to attained a temperature of 150˚C.

Take h = 10 W/m2K

Given:

Specific heat, Cp = 0.46 kJ/kg K = 460 J/kg K


Diameter of the sphere, D = 5 cm = 0.05 m

Radius of the sphere, R = 0.025 m

Initial temperature, T0 = 450˚C + 273 = 723 K

Final temperature, T

= 100˚C + 273 = 373 K

Intermediate temperature, T = 150˚C + 273 = 423 K


To fine:

Time required for the ball to reach 150˚C

Solution:

Density of steel is 7833 kg/m3

For sphere,

Characteristic length, Lc = 3

R

3

025.0

3/7833 mkg

We know that,

Biot number, Bi = k

hLe

1.01038.2

35

103.810

3

3

iB

Biot number value is less than 0.1. So, this is lumped heat analysis type problem.

For lumped parameter system,

tVCp

hA

eTT

TT

0 …… (1)


We know that,

Characteristics length, A

VLc

tIn

e

eTT

TT

t

tVCp

h

78331033.8460

10

373723

373423

373723

373423

)1(

3

78331033.8460

10

0

3

==>

Result:

Time required for the ball to reach 150˚C is 5840.54s.

14. A large wall 2 cm thick has uniform temperature 30˚C initially and the wall temperature is

suddenly raised and maintained at 400˚C. Find

1. The temperature at a depth of 0.8 cm from the surface of the wall after 10 s.

Lc = 8.33 10-3

m

t = 5840.54 s

2. Instantaneous heat flow rate through that surface per m2 per hour.

Take = 0.008 m2/hr, k = W/m˚C.

Given:

Thickness, L = 2 cm = 0.02 m

Initial temperature, Ti = 30˚C + 273 = 303 K

Surface temperature, T0 = 400˚C + 273 = 673 K

Thermal diffusivity, = 0.008 m2/h

= 2.2210-6

m2/s

Thermal conductivity, k = 6 W/m˚C.

Case (i)

Depth, x = 0.8 cm = 0.8 10-2 m

= 0.008 m

Time, t = 10 s

Case (ii)

Time, t = 1 h = 3600 s

To find:

1. Temperature (Tx) at a depth of 0.8 cm from the surface of the wall after 10s.

2. Instantaneous heat flow rate (q x ) through that surface per hour.

Solution:

In this problem heat transfer co-efficient h is not given. So take it as . i.e., h .

We know that,

Biot number, Bi = k

hLe

h

h

==>

Bi value is . So, this is semi infinite solid type problem.

Case (i)

For semi infinite solid,

t

x

i

x erfTT

TT 2

0

0


ZerfTT

TT

i

x

0

0

……. (1)

Where,

t

xZ

2

Put x = 0.008 m, t = 10 s, = 2.22 10-6

m2/s.

101022.22

008.0

6

Z

Z = 0.848, corresponding erf (Z) is 0.7706

==>

[Refer HMT data book page no. 59]

Z = 0.848

erf (Z) = 0.7706

7706.0370

673

7706.0673303

673

7706.0)1(0

0

x

x

i

x

T

T

TT

TT

Case (ii)

Instantaneous heat flow

t

x

i

x et

TTkq

40

2


t = 36000 s (Given)

36001022.24

008.0

36001022.2

30367366

2

6eq x

Result:

Intermediate temperature, T x = 387.85 K

Heat flux, q x = 13982.37 W/m2

15. A semi infinite slab of aluminum is exposed to a constant heat flux at the surface of 0.25

MW/m2. Initial temperature of the slab is 25˚C. Calculate the surface temperature after 10

minutes and also find the temperature at a distance of 30 cm from the surface after 10

minutes.

Given:

Heat flux, q0 = 0.25 MW/m2

q0 = 0.25106 W/m

2

KTx 85.387

2/37.13982 mWq x

Initial temperature, Ti = 25˚C + 273 = 298 K

Distance, x = 30 cm = 0.30 m

Time, t = 10 minutes = 600 s

To find:

1. Surface temperature (T0) after 10 minutes.

2. Temperature (T x ) at a distance of 30 cm from the surface

Solution:

Heat flux,

t

TTkq i

0

0 ……. (1)


[HMT data book page no. 1]

Properties of aluminum

Thermal diffusivity, = 84.1810-6

m2/s

Thermal conductivity, k = 204.2 W/mK.

6001018.84

2982.2041025.0)1(

6

06

T

(ii) For semi infinite solid,

t

xerf

TT

TT

i

x

20

0


ZerfTT

TT

i

x

0

0

……. (2)

Where

T0 = 785.68 K

6001018.842

30.0

2

6

0

Z

t

xZ

Z = 0.667, corresponding erf (Z) is 0.65663


65663.068.785298

68.785

65663.0)2(0

0

x

i

x

T

TT

TT

==>

Temperature at a distance at 30 cm is 465.45 K

Result:

1. Surface temperature, T0 = 785.65 K

2. Temperature at a distance of 30 cm, T x = 465.45 K

Z = 0.667

erf (Z) = 0.65663

T x = 465. 45 K

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