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HEAT TRANSFER

Heat transfer is a discipline of thermal engineering that concerns the

generation, use, conversion, and exchange of thermal energy or heat between

physical systems. Heat transfer is classified into various mechanisms, such as

thermal conduction, thermal convection, thermal radiation, and transfer of

energy by phase changes. Engineers also consider the transfer of mass ofdiffering chemical species to achieve heat transfer.

Mechanisms

The fundamental modes of heat transfer are:

Conduction or diffusion

The transfer of energy between objects that are in physical contact.

Convection

The transfer of energy between an object and its environment, due tofluid motion.

Radiation

The transfer of energy to or from a body by means of the emission or

absorption of electromagnetic radiation.

Advection

The transfer of energy from one location to another as a side effect ofphysically moving an object containing that energy.

Advection

By transferring matter, heat or energy is moved by the physical transfer of a hot

or cold object from one place to another. This can be as simple as placing hot

water in a bottle and heating a bed, or the movement of an iceberg in changingocean currents. A practical example is thermal hydraulics. This can be described

by the formula

where Q is heat flux (W/m), is density (kg/m), cp is heat capacity at constantpressure (J/(kg*K)), T is the change in temperature (K), v is velocity (m/s).
http://en.wikipedia.org/wiki/Thermal_engineeringhttp://en.wikipedia.org/wiki/Thermal_energyhttp://en.wikipedia.org/wiki/Heathttp://en.wikipedia.org/wiki/Thermal_conductionhttp://en.wikipedia.org/wiki/Convective_heat_transferhttp://en.wikipedia.org/wiki/Thermal_radiationhttp://en.wikipedia.org/wiki/Phase_changeshttp://en.wikipedia.org/wiki/Thermal_hydraulicshttp://en.wikipedia.org/wiki/Thermal_hydraulicshttp://en.wikipedia.org/wiki/Phase_changeshttp://en.wikipedia.org/wiki/Thermal_radiationhttp://en.wikipedia.org/wiki/Convective_heat_transferhttp://en.wikipedia.org/wiki/Thermal_conductionhttp://en.wikipedia.org/wiki/Heathttp://en.wikipedia.org/wiki/Thermal_energyhttp://en.wikipedia.org/wiki/Thermal_engineering


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1. ConductionHeat conduction is occurred by two mechanisms are

i) By molecular interaction whereby the energy exchanges takesplace by the kinetic motion or direct impact of molecules.

ii) By the draft of free electrons as in the case of metallic solids. Themetallic alloys have a different concentration of free electrons, and

their ability to conduct heat is directly proportional to theconcentration of free electrons in them. The free electronconcentration of non-metals is very low. Hence materials that are

good conductors (pour meal, viz., copper, silver etc.) are good

conductors of heat. Pure conduction is found only in solids.

Fourier law of heat conduction: It states that the rate of heat conduction is

proportional to the area measured normal to the direction of heat flow, and tothe temperature gradient in that direction.

Q= or q = (were, A is Area, is change of temperature and is difference of length)

K is the constant of proportionality also called coefficient of thermalconductivity

General differential equation of heat conduction


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The energy balanced is obtained from the first law of thermodynamics as-

Net heat conducted into element dxdydz per unit time (Hc) + Internal heat

generated per unit time(Hg) =Increase in internal energy per unit time.(I)

Hc=[

(Kx+ (Ky+ (Kz]dxdydz

This is obtained by using Fouriers law of heat conduction.

Now, let q be the internal heat generation per unit time and per unit volume,

the rate of energy generation in theer element then

Hg=qdxdydz

The change in internal energy for the element over a period of time dt is given

as

I=CpdxdydzPutting the values of Hc, Hg and I in the balance equation we get-

[ (Kx+ (Ky+ (Kz]+q= Cp Now, assuming Kx=Ky=Kz=K (if constant)

Therefore, general three-dimensional heat conduction equation becomes

+

+

+

= Cp = ............................(B)

The quantity

is known as Thermal Diffusivity, and is denoted by .Thermal diffusivity gives the information that how fast heat will be propagatedor it will be diffused through the material. It has got unit m

2/s


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DIFFERENT FORMS OF HEAT CONDUCTION EQUATION

Case I: Uniform thermal conductivityWhen the thermal conductivity of the material is independent of position and

temperature, equation (B) takes the form

+

+

+

= Cp =

or,2T + = where

2T = ++ is called the laplacian (operator)

of T

Case-II: Steady state conditions

If the temperature at any point in the material does not change with time, i.e.

Then the equation (B) reduce to+

+

+

=0

or,2T +

=0

The above equation is also known as Poisson Equation

Case-III: No heat sourcesIn the absence of any heat generation or release of energy within the body,

equation (B) become

++= ...........(i)The equation (i) is also known as Diffusion equation.

[Where is called Thermal Diffusivity of material (m2

/sec): It is the ratio

between the thermal conductivity and the heat capacity per unit volume of an

object.

Mathematically,


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, where k is thermal conductivity, is density and cp is specific heatNote that the thermal conductivity (k) represented how well a material conducts

heat, and the heat capacity (

represents how much energy a material stores

per unit volume. The larger the diffusivity means faster the propagation of heatinto the medium. A small value of thermal diffusivity means that heat is mostly

absorbed by the material and a small amount of heat will be conducted further.]

If the process is in steady state, then the heat conduction equation becomes

+

+

=0

or,2

T =0 ..........(i)

Equation (ii) is also known as Laplace equation.

Case IV: One dimensional heat conductionIf the temperature varies only in the x-direction, then the term 2 T reduces to

=

in all above case.

For a steady state case with no heat generation, heat conduction equation is Heat conduction equation in cylindrical co-ordinates


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Consider a small cylindrical element having sides dr, dz and r dAs shown in fig. The rate of heat flow in r direction is:

Qr=-krddzThe rate of heat flow out of the element in r direction at r+dr is:Qr+dr=Qr+

dr

Then, the net heat entering the element in r direction is:

Qr-Qr+dr= dr=

dr=k ( )

= k

= k

Similarly, = [ rd]

= rd= rd= d

And, = [ d]= d=

dd=

d

Therefore, balancing the energy as equation-

k* + + (

) d+

d=q d+ d

Or, k

*

++

(

)+

=q

+


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Or, * ++(

)+

= +

Or, + +

=+

Heat conduction equation takes the form +

) +

+

+

=

For one dimension case (considering heat conduction in r direction only): + =

+

Or,

( ) =

+

Heat conduction equation in spherical coordinate in spherical co-ordinate

system

Consider a small spherical element having sides dr, rd and rsindAs shown in fig

The rate of heat flow in r direction is:

Qr=-k

rddThe rate of heat flow out of the element in r direction at r+dr is:


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Qr+dr=Qr+ dr

Then, the net heat entering the element in r direction is:

Qr-Qr+dr=

dr=k

r

2

ddd

=k(r2+2r

dd

=k(+

dd

Similarly,

Q- Q+d= =

*

+ And,

= [ rsind]rd==

rsind= ( )

(

) dTherefore, balancing the energy as equation-

k* + + *+ + (

) d

=q +

Or, k* ++ *++ (

)=q+ Or, * ++ *

++(

)=

+

For one dimension case (considering heat conduction in r direction only):

+

=

+


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ONE DIMENTION STEADY STATE HEAT CONDUCTIONTemperature field in the system is described in terms of only one space co-

ordinate.

1. Heat conduction through plane walls

Considering a plane wall of a material of uniform thermal conductivity

K as shown above. The wall is extended to infinite in y and z directions.

The general heat conduction equation is

+

+

+

=

For above case =0( for steady state)=

=0 (for one dimensional)

And also =0 (since no heat generation)

The conduction equation transform to =0 or

=0 or,

c1.........(1)

or, T=xc1+c2....................(2)

Solving above equation and taking boundary condition as


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T=T1 at x=0...............(a)

And T=T2 at x=L......................(b)

Putting the boundary condition of (a) into the equation (2) we get c2=T1 and also

from equation (1) we get c1= =

So the equation (2) become, T= (

)x +T1

From Fouriers law,

Q=-KA

Therefore, Q=-KA(

)= KA(

)=

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

Where Q= quantity of heat, which must be supplied to the left face of the wall

to maintain a temperature difference (T1-T2) across it.Thermal resistance for plane wall is given as

Rpln,Th =

2.Redial heat conduction through cylindrical system

Consider a long cylinder of inside radius r1, outside r0 and length L as shownbelow

The inside and outside surfaces are kept at constant temperature Ti and Torespectively.

The general heat conduction equation in cylindrical co-ordinates is


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+

) +

+

+

=

Assuming that heat flows only in a radial direction, the above equation under

steady state (without heat generation) takes the form

+ )=0 [as =0

Or,

=0 or, r

=0 or

=0 ]

Or, =c1

Or,

=c1

or,T=C1lnr+C2.............................................(1)

Boundary conditions are

T=Ti at r=ri......................................................(a)

T=To at r=ro....................................................(b)

Using condition (a) into the equation (1) we get, Ti=C1lnri+C2....................................(2)Also using condition (b) into the equation (1) we get, To=C1lnro+C2......................(3)

Subtracting the equation (2) and (3) we get, Ti -To =C1lnri+ C1lnro= C1(lnri+lnro)

Or, C1=

Putting the value of c1 in the equation (2) we get Ti=

lnri+C2

Or, C2=Ti lnri=

Substituting the value of C in the equation (1) , we get

T=

lnr+

Heat flow, Q=-k =

Therefore, the thermal resistance for the hollow cylinder is

Rcyl,th=


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Log mean area:

It can be used to transfer a cylinder into an equivalent slab.

It is assumed that the heat transfer through the cylinder and the slab are equal.

Am = log meanarea of cylinder= Where Ai and Ao are the inside and outside surface area of the cylinder.

If


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Boundary conditions are

T=Ti at r=ri......................................................(a)

T=To at r=ro....................................................(b)

Using condition (a) into the equation (1) we get, Ti= - C1+C2....................................(2)

Also using condition (b) into the equation (1) we get, To= - C1+C2......................(3)

Subtracting the equation (2) and (3) we get, Ti -To = - C1 ( -

)

Or, C1=

Putting the value of C1 in the equation (2) we get C2=Ti

Substituting the value of C in the equation (1), we get

T= +Ti

= Ti

)=

= =

( )

Therefore, Heat flow, Q = - k = =

Therefore, the thermal resistance for the hollow sphere is

Rsph,th=

Geometric mean area:

It can be used to transfer a sphere into an equivalent slab.

It is assumed that the heat transfer through the sphere and the slab are equal.

Ag = Geometric meanarea of cylinder==Where Ai and Ao are the inside and outside surface area of the cylinder.

Thermal resistance is given as

Rsph,th=

=

=


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The thermal resistance of a hollow sphere is of exactly the same as that for a

slab except that the logarithmic mean area is used for the sphere.

Example-1

Determine the heat transfer through the composite is shown in fig. Take

the conductivities of A, B, C, D and E as 50, 10, 6.67, 20 and 30W/mk

respectively and assume one dimensional heat transfer.

Solution: The various wall resistances, as shown in figure are

Ra=

=

=110-3 K/W

Rb=

=2 -2 K/WRc=

=3 -2 K/W

Rd=

=2 -3 K/WRe=

=1.67

-3

K/W


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If the equivalent resistance for Rb and Rc isRbc

Then

=

or,

=

12

K/W

Therefore, Ra + Rbc + Rd + Re = (1+12+2.5+1.67)10-3=17.17-3So, Q =

=

= 4.07104 W= 40.7 KW

Example-2

A 10cm O.D. steam at 1373 KPa(1.373MN/m2) is lagged to 20 cm diameter

with magnesia (k=0.07 W/mK) and further lagged with laminated asbestosof thermal conductivity 0.08 W/mK to 25 cm diameter. The whole pipe is

further protected by layer of canvas. If the temperature under the canvas is

20c, find the mass of steam condensed in 12 hours on 150 meter length of

pipe. Neglect thermal conductivity effect of the pipe material.

Solution: At 137 kpa, T, the saturation temperature of stem=467.14K=194.14Chfg= Latent heat of steam =1963.15 kJ/Kg

Q=

=

= 12916.5W

=12.916 KJ/s = 12.916 46497.6 kJ/hSo, mass of steam condensed per hour =

= 23.685 Kg

Mass of steam condensed in 12hours = 284.22 Kg

Example-5

The temperature distribution across a large concrete slab 50 cm thick

heated from one side as measured by thermocouples approximates to the

following relation

T=60-50x+12x2+20x3-15x4

Where T is in C and x is in meters. Considering an area of 5 m2, compute

(i) the heat entering and leaving the slain unit time.(ii) the heat energy stored in unit time(iii) the rate of temperature change at both side of the slab(iv) the point where the rate of heating or cooling is maximum

Take the following data for concrete: K=1.2 w/mK, 1.7710-3 m2/hSllotion:

Here T=60-50x+12x2+20x

3-15x

4


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= - 50+24x+60x

2+60x

3

and= - 24+120x-180x

2

(i) Q= - kA() = - KA(- 50+24x+60x2+60x3)Heat entering the slab = Qo=-Ka()x=0

=-(1.2 (-50)=300WHeat leaving the slab = QL=-KA()x=0.5

=(- 1.2 (-50+12+15-7.5)=183 WExample-6

A wall of a furnace is made up of inside layer of silica brick 120 mm thick

converted with layer of magnesite brick 240 mm thick. The temperature at

inside surface silica brick wall and outside surface of magnesite brick wall

are 725C and 100C respectively. The contact thermal resistance between

the two walls at the interface is 0.0035C /W per unit wall area. If thermal

conductivity of silica and magmesite brick are 1.7W/mC and 5.8 W/mC.

Calculate

(a)The rate of heat loss per unit of area of walls.(b)Temperature drop at the interface.

Solution: Given data

Lo=0.12 m

Lb=0.24 m

Kd= 1.7 W/mCKb= 5.8 W/mC

The thermal resistance between silica bricks and magnesite brick=0.0035C/W

Temperature of the inside surface of silica brick wall t1=725C

Temperature at the outside surface of the magnesite brick wall t4=110C(a)The rate of heat loss per unit area of wall.

q=

=

=

=

=5324.67 W/


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(b)The temperature drop at the interface ( )As the same heat flows through each layer of composite wall-

q=

= or, 5324.67= or, =349.15C

Similarly, T3=330.33 C

The temperature drop at the interface = 349.15 - 330.33=18.82C

Example-8

A plane wall is made by fire clay brick. The wall thickness L=300mm. The

temperature of wall surface T1=1550C and T2=50C. Thermal

conductivity of brick K=0.96(1+0.0008T) W/mC. Calculate and represent

graphically the temperature distribution through wall.

Solution: Given data:

T1=115CT2=50C


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K=0.96(1+0.0008T) W/mC

The Fouriers equation-

Q=-KA

Or, Q=-K0(1+ Or,

-k0

Or,L =-[( )+()].........................(1)

Or,L =-[1+()] )

Or,L =-[1+] )

Or, Q=-[1+] Further if t is the temperature of the surface at a distance x from the left surfacethen- = -ko[(t - )+(

[ =-ko[(t - )+( ...............................(2)

Putting the value of from (1) in to (2) we get,

-[( )+()]=-ko[(t - )+( Or, [( )+()]=[(t - )+( Or, t=

[(1+ )2{(1+)2(1+)2}] -

Or, t=

[(1+)2{(1+)2(1+)2}] - = 2562CExample: A wall is made of plastered material of 25 cm thickness and

followed by concrete of 5 cm thickness. Thermal conductivity of plastered

material is 0.69 W/m and that of concrete is 0.93 W/mK the temperature ofexposed plastered surface is 30C and that of concrete is 5C. Find out the

heat loss through the wall of area 50m2.

Solution: Given data: A=50m2, tp=30C, Lp=25cm, Lc=5cm,Kp=0.69W/mk

kc=0.93W/mk.

The rate of heatloss of wall:

Q=

=

=3 kW


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Example: For a hot solid cylinder of radius ro with uniform rate of heat

generation q per unit volume conducting heat radially and losing heat from

its surface to the ambient (at the temperature T) by convection with heat

transfer co-efficient h then prove that

= Where T=temperature of the cylinder at a distance at a distance r from as

axis. and =axis temperatureSolution: Qr(heat conduction in at radius r)=-k2r Qg(heat generation in the element)=(2)q(heat generated out at radius )=+ ()drFor steady state conduction of heat flow-

= + ()drOr, ()drOr, (2)q = (-k2r

)dr

Or, (2r)q =

(-k2r)

Or, = (r)

Or,

(r

)=

Or, r = +C1Or,

= .............................(1)

Or, t= lnr+....................(2)Boundary conditions are:

When r =0, =0...................................(a)

When r=, t=..............................(b)Using the boundary condition (a) in to the equation (1) we get, C1=0

Using the boundary condition (b) and C1=0 in to the equation (1) we get,

C2=tw+

Thus the equation (2) become,

t=tw+

......................(i)As t=tmax when r=0, so


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tmax=tw +

...........................(ii)From (i) and (ii) we get,

=

=1-

= = =

Or,

Putting the value of, we get =

ht donot give now 8.8.13

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