heat conduction differential equation of heat...

Heat Conduction

Differential Equation of Heat Conduction Temperature is not varied with one direction

Temperature is varying with time

Arbitrarily solid

Need differential equation

Differential Equation of Heat Conduction

Element inside the solid

1) Thermal conductivity is not directly dependent at any point (Isotropic material).

2) Assume that heat generated = (W/m3)

Fourier’s law of heat conduction - For first one dqx

dz.dy.x

Tkdq x

dz.dydx.x

Tk

xx

Tkdq dxx

q


- For first one dqy

....dqy

....dq dyy

- For first one dqz

....dqz

....dq dzz

Similarly

Now, what is the net rate at which heat is being conducted into the element?


1) Net amount of heated conducted into dxdydz per unit time

)dqdqdq()dqdqdq( dzzdyydxxzyx

dz.dy.dxz

Tk

zy

Tk

yx

Tk

x

2) Net amount of heated conducted into dxdydz per unit time

)dz.dy.dx(q

3) Rate of change of energy of the element

t

TC)dz.dy.dx.( p


1 + 2 = 3

t

TCq

z

Tk

zy

Tk

yx

Tk

xp

Therefore, the required differential equation is

Note: dx.dy.dz is cancelled

Apply the 1st law of Thermodynics of closed system

This is the most general form of differential equation for Cartesian coordinate system


t

TCq

z

T

y

T

x

Tk p2

2

2

2

2

2

If k is constant

In the absence of heat generation

t

T1

k

qT2

is called the thermal diffusivity (unit: m2/s)

t

T1T2

If there is a steady state

0T2 Depending of the nature of the problem

(Laplace’s equation)


Cylindrical Spherical

cosrx

sinry

zz

cossinrx

sinsinry

cosrz

Zenith angleAzimuth angle

Differential Equation of Heat ConductionDerivation of different equation of heat conduction in a 2D polar coordinate system ),r(

r

dr drrdq

d

rdq

dq

ddq

1.d.r.r

Tkdq r

ddr.

r

T.r.k

rr

T.r.kdq drr

1.dr.T

r

kdq

drd.r.T

kr

T

r

kdq d

Cylindrical

A

Adistance


1) Rate at which heat is conducted into the element

d.dr

Tk

r

1

r

Tr.k

r

2) Heat generated per unit time

1drdrq

3) Rate of change of energy of the element

t

TC)1.dr.d.r.( p

volume


1 + 2 = 3

t

TCq

Tk

r

1

r

Tkr

rr

1p2

Therefore, the required differential equation is


t

TCq

T

r

1

r

Tr

rr

1k p2

2

2

If k is constant

In the absence of heat generation

If there is a steady state

t

TC

T

r

1

r

Tr

rr

1k p2

2

2

0T

r

1

r

Tr

rr

1k

2

2

2

Differential Equation of Heat Conduction)z,,r( Derive the differential equation for the 3-D cylindrical system

For constant k

2

2

z

Tk

Additional term on left-hand side

t

TCq

z

TT

r

1

r

Tr

rr

1k p2

2

2

2

2


Spherical

t

TCq

T

sinr

1Tsin

sinr

1

r

)rT(

r

1k p2

2

2222

2

Now, what do we know?

Coordinate: Cartesian, cylindrical, spherical coordinatesState: steady state, unsteady state Direction: one, two, three directionsMaterial: isotropic material constant k, not constant k


Solving a problem needs initial and boundary conditions

1. Initial condition

At t = 0, temperature distribution in body is specified

2. Boundary condition1) Prescribed surface temperature2) Prescribed heat flux incident on the surface3) Prescribed heat transfer coefficient at the surface


1. Prescribed surface temperatureFor example, let the surface be a plane face (at x = L)

T0At x = L T = T0

2. Prescribed heat flux incident on the surface

0Lx A

q

x

Tk

3. Prescribed heat transfer coefficient at the surface

)TT(hx

Tk fLx

Lx

0A

q

hTf

x

x directionOpposite x direction

x = L


Heat generation

1. Electrical conductor2. Nuclear fuel element3. Setting of concrete4. Agricultural product


Heat generation (Infinite slab)

Heat generated at uniform rate

x2b

q

We would like to know the temperature distribution at steady state, or equation describing this temperature distribution.

Differential Equation of Heat ConductionLet put down differential equation (k is constant)

t

TCq

z

T

y

T

x

Tk p2

2

2

2

2

2

• One direction, infinite slab • Steady state

k

q

dx

Td2

2

Differential Equation of Heat ConductionBoundary conditions

x2b

q

Tf h

hTf

)TT(hdx

dTk fbx

bx

)TT(hdx

dTk bxf

bx

0dx

dT

0x

Differential Equation of Heat ConductionSolving

1Cxk

q

dx

dT

Integrating it twice

21

2

CxC2

x

k

qT

k

q

dx

Td2

2

Differential Equation of Heat ConductionTemperature distribution

h

bq)xb(

k2

qTT 22f

h

1

k2

bbqTT fmax

Maximum Temperature (x = 0)


Heat generation (Infinite solid cylinder)

t

TCq

Tk

r

1

r

Tkr

rr

1p2

rk

q

dr

dTr

dr

d

Reducing the form to

0dr

dT

0r

fRrRr

TThdr

dTk

Boundary condition


Integrating differential equation and using boundary conditions, we get

h2

Rq)rR(

k4

qTT 22f

h

1

k2

R

2

RqTT fmax

Differential Equation of Heat ConductionProblemA nuclear fuel element is in the form of a long solid rod (k = 0.85 W/mK) of diameter 14 mm. It generates heat at the uniform rate of 0.45 x 108 W/m3 because of nuclear fusion. The heat is transferred to pressurized cooling water at 300 ºC and the surface heat transfer coefficient is 4500 W/m2K. Calculate the maximum temperature in fuel rod in the steady state.

Differential Equation of Heat ConductionProblemHeat generated in a slab of 120 mm thickness with a conductivity of 200 W/mK at a rate of 106 W/m3. Determine the temperature at the mid and quarter planes if the surface of the solid on both sides are exposed to convection at 30ºC with a convective coefficient of 500W/m2K. Also find heat flow rate at these planes and the temperature gradients at these planes

heat conduction differential equation of heat...

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