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Appendices 10.A & 10.B:An Educational Presentation

Presented By:Joseph Ash

Jordan BaldwinJustin Hirt

Andrea Lance

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History of Heat Conduction

Jean Baptiste Biot (1774-1862)

French Physicist Worked on analysis of

heat conduction Unsuccessful at dealing

with the problem of incorporating external convection effects in heat conduction analysis

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Jean Baptiste Joseph Fourier (1768 – 1830) Read Biot’s work 1807 determined how to solve the

problem Fourier’s Law

Time rate of heat flow (Q) through a slab is proportional to the gradient of

temperature difference

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Ernst Schmidt German scientist Pioneer in Engineering

Thermodynamics Published paper “Graphical Difference

Method for Unsteady Heat Conduction” First to measure velocity and

temperature field in free convection boundary layer and large heat transfer coefficients

Schmidt Number Analogy between heat and mass transfer that causes a dimensionlessquantity

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Derivation of the Heat Conduction Equation

A first approximation of the equations that govern the conduction of heat in a solid rod.

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Consider the following:

A uniform rod is insulated on both lateral ends. Heat can now only flow in the axial direction.

It is proven that heat per unit time will pass from the warmer section to the cooler one.

The amount of heat is proportional to the area, A, and to the temperature difference T2-T1, and is inversely proportional to the separation distance, d.

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The final consideration can be expressed as the following:

is a proportionality factor called the thermal conductivity and is determined by material properties

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Assumptions

The bar has a length L so x=0 and x=L Perfectly insulated Temperature, u, depends only on position, x,

and time, t Usually valid when the lateral dimensions are

small compared to the total length.

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The differential equation governing the temperature of the bar is a physical balance between two rates:

Flux/Flow term Absorption term

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Flux

The instantaneous rate of heat transfer from left to right across the cross sections x=x0 where x0 is arbitrary can be defined as:

The negative is needed in order to show a positive rate from left to right (hot to cold)

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Flux

Similarly, the instantaneous rate of heat transfer from right to left across the cross section x=x0+Δx where Δx is small can be defined as:

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Flux

The amount of heat entering the bar in a time span of Δt is found by subtracting the previous two equations and then multiplying the result by Δt:

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Heat Absorption

The average change in temperature, Δu, can be written in terms of the heat introduced, Q Δt and the mass Δm of the element as:

where s = specific heat of the material

ρ = density

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Heat Absorption

The actual temperature change of the bar is simply the actual change in temperature at some intermediate point, so the above equation can also be written as:

This is the heat absorption equation.

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Heat Equation

Equating the QΔt in the flux and absorption terms, we find the heat absorption equation to be:

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If we divide the above equation by ΔxΔt and allow both Δx and Δt to both go to 0, we will obtain the heat conduction or diffusion equation:

where

and has the dimensions of length^2/time and called the thermal diffusivity

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Boundary Conditions

Certain boundary conditions may apply to the specific heat conduction problem, for example: If one end is maintained at some constant

temperature value, then the boundary condition for that end is u = T.

If one end is perfectly insulated, then the boundary condition stipulates ux = 0.

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Generalized Boundary Conditions

Consider the end where x=0 and the rate of flow of heat is proportional to the temperature at the end of the bar. Recall that the rate of flow will be given, from left to right, as

With this said, the rate of heat flow out of the bar from right to left will be

Therefore, the boundary condition at x=0 iswhere h1 is a proportionality constant

if h1=0, then it corresponds to an insulated end

if h1 goes to infinity, then the end is held at 0 temp.

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Generalized Boundary Conditions

Similarly, if heat flow occurs at the end x = L, then the boundary condition is as follows:

where, again, h2 is a nonzero proportionality factor

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Initial Boundary Condition

Finally, the temperature distribution at one fixed instant – usually taken at t = 0, takes the form:

occurring throughout the bar

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Generalizations

Sometimes, the thermal conductivity, density, specific heat, or area may change as the axial position changes. The rate of heat transfer under such conditions at x=x0 is now:

The heat equation then becomes a partial differential equation in the form:

or

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Generalizations

Other ways for heat to enter or leave a bar must also be taken into consideration.

Assume G(x,t,u) is a rate per unit per time. Source

G(x,t,u) is added to the bar G(x,t,u) is positive, non-zero, linear, and u does not depend on t G(x,t,u) must be added to the left side of the heat equation

yielding the following differential equation

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Generalizations

Similarly, Sink

G(x,t,u) is subtracted from the bar G(x,t,u) is positive, non-zero, linear, and u does not

depend on t G(x,t,u) then under this sink condition takes the

form:

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Generalizations

Putting the source and sink equations together in the heat equation yields

which is commonly called the generalized heat conduction equation

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Multi-dimensional space

Now consider a bar in which the temperature is a function of more than just the axial x-direction. Then the heat conduction equation can then be written: 2-D:

3-D:

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Example 1: Section 10.6, Problem 9

Let an aluminum rod of length 20 cm be initially at the uniform temperature 25°C. Suppose that at time t=0, the end x=0 is cooled to 0°C while the end x=20 is heated to 60°C, and both are thereafter maintained at those temperatures.

Find the temperature distribution in the rod at any time t

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Find the temperature distribution, u(x,t)

α2uxx=ut, 0<x<20, t<0

u(0,t)=0 u(20,t)=60, t<0

u(x,0)=25, 0<x<20

From the initial equation we find that:

L=20, T1=0, T2=60, f(x)=25

We look up the Thermal Diffusivity of aluminum→α2=0.86

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Using Equations 16 and 17 found on page 614, we find that

where

( ) ( ) ∑∞

=

−

++−=

1112 sin,

2

222

n

L

tn

n L

xnecT

L

xTTtxu

παπ

( ) ( )∫

−−−=

L

n dxL

xnT

L

xTTxf

Lc

0 112 sin2 π

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Evaluating cn, we find that

( )

( ) ( )( )( )

( )( )ππ

πππππ

π

n

nc

n

nnnnc

dxxnx

c

n

n

L

n

50cos70

5sin12cos710

20sin0

2006025

20

2

2

0

+=

+−=

−−−= ∫

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Now we can solve for u(x,t)

( ) ( ) ( )( )

( ) ( )( ) ( )

∑

∑∞

=

−

∞

=

−

++=

+++−=

1

400

86.0

1

20

86.0

20sin

50cos703,

20sin

50cos700

20060,

2

2

222

n

tn

n

tn

xne

n

nxtxu

xne

n

nxtxu

πππ

πππ

π

π

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Derivation of the Wave Equation

Applicable for:

•One space dimension, transverse vibrations on elastic string

•Endpoints at x = 0 and x = L along the x-axis

•Set in motion at t = 0 and then left undisturbed

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Schematic of String in Tension

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Equation Derivation

Since there is no acceleration in the horizontal direction

However the vertical components must satisfy

where is the coordinate to the center of mass and the weight is neglected

Replacing T with V the and rearranging the equation becomes

0cos),()cos(),( =−∆+∆+ θθθ txTtxxT

),(sin),()sin(),( txxutxTtxxT tt∆=−∆+∆+ ρθθθ

x

),(),(),(

txux

txVtxxVttρ=

∆−∆+

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Derivation continued

Letting , the equation becomes

To express this in terms of only terms of u we note that

The resulting equation in terms of u is

and since H(t) is not dependant on x the resulting equation is

0→∆x

),(),( txutxV ttx ρ=

),()(tan)(),( txutHtHtxV x== θ

ttxx uHu ρ=)(

ttxx uHu ρ=

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Derivation Continued

For small motions of the string, it is approximated that

using the substitution that

the wave equation takes its customary form of

TTH ≈= θcos

ρ/2 Ta =

ttxx uua =2

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Wave Equation Generalizations

The telegraph equation

where c and k are nonnegative constants

cut arises from a viscous damping force

ku arises from an elastic restoring force

F(x,t) arises from an external force

The differences between this telegraph equation and the customary

wave equation are due to the consideration of internal elastic

forces. This equation also governs flow of voltage or current in a

transmission line, where the coefficients are related to the electrical

parameters in the line.

),(2 txFuakucuu xxttt +=++

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Wave Equations in Additional Dimensions

For a vibrating system with more than on significant space coordinate it may be necessary to consider the wave equation in more than one dimension.

For two dimensions the wave equation becomes

For three dimensions the wave equation becomes

ttyyxx uuua =+ )(2

ttzzyyxx uuuua =++ )(2

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Consider an elastic string of length L whose ends are held fixed. The string is set in motion from its equilibrium position with an initial velocity g(x). Let L=10 and a=1. Find the string displacement for any time t.

( )( )

−=

,4

,1

,4

L

xL

L

x

xg

LxL

Lx

L

Lx

≤≤

<<

≤≤

4

34

3

4

40

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From equations 35 and 36 on page 631, we find that

where

( ) ∑∞

=

=

1

sinsin,n

n L

atn

L

xnktxu

ππ

( )∫

=

L

n dxL

xnxg

Lk

L

an0

sin2 ππ

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Solving for kn, we find:

( )

( )( )

( )

+

=

−

+

=

−+

+

= ∫ ∫ ∫

4sin

4

3sin

8

sin4

sin4

3sin

42

sin4

sinsin42

3

2

4

0

4

3

4 4

3

πππ

πππππ

ππππ

nn

na

Lk

nnn

n

L

ank

dxL

xn

L

xLdx

L

xndx

L

xn

L

x

Lk

L

an

n

n

L L

L

L

Ln

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Now we can solve for u(x,t)

( )( )

( )

( ) ∑

∑

∑

∞

=

∞

=

∞

=

+

=

+

=

+

=

133

133

13

10sin

10sin

4sin

4

3sin

180,

sinsin4

sin4

3sin

18,

sinsin4

sin4

3sin

8,

n

n

n

tnxnnn

ntxu

L

atn

L

xnnn

n

Ltxu

L

atn

L

xnnn

na

Ltxu

πππππ

πππππ

πππππ

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THE END

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