Introduction: -A uniform homogenous rod is made of infinite number of molecules. They are

interconnected by cohesive force. When heat is applied to a rod through a one side of it, they tends to

get expand, because of that they gets even more interconnected and hence heat flows molecule to

molecule more rapidly. The flow of heat in this way in a uniform of rod is known as Heat Conduction.

One-Dimensional Heat Flow: -

α R1 R2

0 x x X

Consider a homogeneous bar of uniform cross section α (cm2.) Suppose that the sides are covered with a

material impervious to heat so that streamlines of heat-flow are all parallel and perpendicular to area.

Take one end of the bar as the origin and the direction of flow as the positive x-axis. Let be the density

(gr/cm3), s the specific heat (cal/gr. deg.) and k the thermal conductivity (cal/gr. deg. sec).

Let u(x, t)be the temperature at a distance x from O. If u be the temperature change in a slab

of thickness of x then the quantity of heat in this slab=Sαxu. Hence the rate of increase of heat in

this slab i.e. s ραδ x

∂ u∂ t =R1-R2, where R1 and R2 are respectively the rate (cal/sec.) of inflow and outflow

of heat.

Now R1=−kα (∂ u

∂ x )x and R2=

−kα (∂ u∂ x )

x+Δx

The negative sign appearing as result of decrease of temperature as t increase

Hence s ραδ x

∂ u∂ t =

−kα (∂ u∂ x )

x +kα (∂u

∂ x )x+Δx

ONE DIMENSIONAL HEAT EQUATION

i.e.

∂u∂ t =

ksρ [ (∂u

∂ x )x+Δx

−(∂u∂ x )

x

δx ]Writing

ksρ

=c2

, called the diffusivity of the substance (cm2/sec.,) and taking limit x0, we get

is called the one-dimensional Heat-flow equation.

Solution of One-Dimensional Heat Equation by the method of separating

variables: (with insulated faces)

We first consider the temperature in a long thin bar or thin wire of constant cross section

and homogeneous material, which is oriented along x-axis and is perfectly insulated laterally, so

that heat flows in the x-direction only. Then u depends only on x and time t, and the heat

equation becomes the one-dimensional heat equation

-------------------------------------------------(1)

We take the case that both ends x=0 and x=L are kept at zero degree temperature, we have the

boundary conditions

-------------------------------(2)

and the temperature in the bar is f(x), so that we have the initial condition

-------------------------------(3)

Method of separating the variables:

In the method of separating variables or product method, we determine the solution of

Heat equation (1) of the form,

u(0,t)=0, u(L, t)=0 for all t

u(x, 0)=f(x)

-------------------------------(4)

Where F(x) is only a function of x and G(t) is only a function t only, on differentiating equation

(4), we obtain,

Where dot denotes the partial derivatives with respect to t and primes derivatives with respect to

x. By inserting this into differential equation (1), we get

On dividing by c2FG we obtain,

The left-hand side the right hand side of the above equation are function t and x respectively and

since both x & t are both independent variables, they must be equal to some constant k=-p2.

We assume k = - p2 because the solution for k =0 and k = p2 is not consistent with the physical

problem i.e. it will not give the transient solution where the temperature decreases as time

inceases.

This yields two differential equations, namely

----------------------------------------------------------(5)

And

---------------------------------------------------(6)

Satisfying the boundary conditions: -

Solving (5) the general solution is given by F(x)=Acospx+Bsinpx. ---------(i)

From the boundary (2) it follows that

u(0, t)=F(0)G(t)=0 and u(L, t)=F(L)G(t)=0.

Since G0 would gives u0. Which is of no interest. Hence G0. This gives F(0)=0 and

F(L)=0.

u(x, t)=F(x)G(t)

F”+ p2F=0.

.

From (i) F(0)=0 0 = A F(x) = B sinpx

F(L)=0 0 = B sin(pL) sin(pL) = 0 ( B 0)

pL = nπ ( zeros of sine)

p = nπ/ L, n=0,1,2,-------

Applying above conditions, we obtain the infinitely many solutions of F(x) are given by

-----------------------------------------------------(7)

Solving (6), The constant k is now restricted to the values k=-p2=-( nπ

L )2

. Inserting it into the

equation (6) we get

Its general solution is given by Gn ( t )=Cne− λn

2 t n=1,2,3----- where Cn is constant. Hence the

functions

-------(8)

are solutions of the heat equation (1) satisfying (2). These are the Eigenfunctions of the problem,

corresponding to the Eigen values λn=

nπcL

.

Since the equation (1) is linear and homogeneous applying principle of superposition we

obtain the general solution of (1) is given by

----(9)

Applying (3) on the above equation (9), we obtain

u ( x , 0 )=∑n=1

∞an sin( nπx

L )=f ( x ) .

Fn( x )=Bnsin( nπx

L )

⇒un ( x ,t )=Fn( x )Gn (t )=ansin( nπxL )e

− λn2 t(n=1,2,3-----) where an =BnCn

⇒u ( x , t )=∑n=1

∞un ( x ,t )=∑

n=1

∞

ansin ( nπxL )e

− λn2 t

( n=1,2,3-----) where λn =nπcL

So, f(x) is Half-range Fourier series with an’s as Fourier coefficients and is given by

------------------------------------------------------------(10)

Examples

1. A rod of length l with insulated sides is initially at a uniform temperature u0. Its ends are

suddenly cooled at 00 C and are kept at that temperature. Find the temperature function

u(x,t).

Solution:

The temperature function u(x,t) satisfies the differential equation

∂u∂ t

=c2 ∂u

∂ x2

The boundary conditions associated with the problem are

u(0,t)=0, u(l,t)=0

The initial condition is u(x,0)=u0

Its solution is

Since u(x,0)=u0 we have

which is the half-range sine series for u0 where

Hence the temperature function is

2. (a) An insulated rod of length l has its ends A and B maintained at 00C and 1000C

respectively until steady state conditions prevail. If B is suddenly reduced to 0 0C and

maintained at 00C, find the temperature at a distance x from A at time t.

an=2L∫0

L

f (x ) sin( nπxL )dx

(b) Find also the temperature if the change consists of raising the temperature of A to

200C and that of 800C.

Solution:

(a) The temperature function u(x,t) satisfies the differential equation

.......(1)

The boundary conditions associated with the problem are

u(0,t)=0, u(l,t)=0

When t=0, the heat flow is independent of time( steady state condition) and so eqn.(1)

becomes

Its general solution is given by u= ax + b …….(2)

where a and b are arbitrary.

Since u= 0 for x=0 and u=100 for x=l we get from eqn.(2), b=0 and .

Thus the initial condition is expressed by u(x,0)= .

The solution of (1) with the respective boundary conditions is given by

Since u(x,0)= ,

Where =

= =

Hence

(b) The boundary conditions associated in this case are

u(0,t)=20, u(l,t)=80 for all values of t.

The initial condition is same as in part(a) i.e. u(x,0)= .

Since the boundary values are non-zero we modify the procedure.

Let u(x,t) = us(x)+ut(x,t) ……(1)

where us(x) is the steady state solution and ut(x,t) may be regarded as the transient

part of the solution which decreases with increase of time.

We shall obtain the solution for ut(x,t).

So we need the boundary conditions and initial condition associated with ut(x,t).

Boundary conditions for ut(x,t)

From (1) u(0,t)=us(0)+ut(0,t) ut(0,t) = u(0,t) – us(0) = 20 – us(0). ….(2)

ut(l,t) = u(l,t) – us(l) = 80 – us(0) ….(3)

Now, us(x) = ax + b ( steady state condition)

u(0,t)=20, u(l,t)=80 a= , b= 20

us(x) = x + 20 us(0) = 20 and us(l) = 80.

Thus, (1) gives ut(0,t) = 20 – 20 = 0 …..(3)

ut(l,t) = 80 – 80 = 0 ….(4)

Initial condition for ut(x,t)

From (1) u(x,0) = us(x) + ut(x,0)

ut(x,0) = u(x,0) – us(x)

= - x + 20

ut(x,0) = - 20 …..(5)

Since the boundary values for ut are zero at both the ends, we can write the solution

for ut(x,t) as

wgw

ww

Hence,

Thus the required solution is

u(x,t) = us(x) + ut(x,t)

u(x,t) = x + 20

3. Determine the solution of one-dimensional heat equation

Where u(0,t)=u(1,t)=0(t > 0) and the initial condition u(x,0) = x, l being the length of bar.

(Ans: u(x,t) = x + 20 )

4. A homogeneous rod of conducting material of length 100cm has its ends kept at zero

temperature and the temperature initially is u ( x , 0 )=¿ { x 0 ≤x≤50 ¿ ¿¿¿

Find the temperature at any time.

(Ans. )

5. A bar 10 cm long with insulated sides has its ends A & B maintained at the temperature

500 C & 1000 C until steady-state conditions prevails. Then both ends are suddenly

reduced to 00 C and maintained at the same temperature. Find the temperature distribution

in the bar at any time.

(Ans.