Fourier Series: One Dimensional Equation of Heat Conduction - Steady state conditions and zero boundary conditions

ONE DIMENSIONAL EQUATION OF HEAT CONDUCTION: Steady state conditions and zero boundary conditions:

Problems based on Steady state conditions and zero boundary conditions

Example 1: What is meant by steady state condition in heat flow?

Solution: Steady state condition in heat flow means that the temperature at any point in the body does not vary with time.

i.e., it is independent of t, the time.

Example 2: In steady state conditions, derive the solution of one dimensional heat flow?

Solution: The p.d.e. of unsteady one dimensional heat flow is

   ................(1)

In steady state condition, the temperature u depends only on x and not time t

.. (1) reduces to ∂²u/∂x². Since u depends only on x, integrating w.r.to x twice, we get the general solution as u = ax + b where a, b are arbitrary.

Example 3: What is the basic difference between the solution of one dimensional wave equation and one dimensional heat equation?

Solution:

Example 4: Distinguish between steady and unsteady states in heat conduction problems.

Solution: In unsteady state, the temperature at any point of the body depends on the position of the point and also the time t. In steady state, the temperature at any point depends only on the position of the point and is independent of the time t.

Example 5: A rod 30 cm long has its ends A and B kept at 20° and 80° respectively until steady state conditions prevail. The temperature at each end is then suddenly reduced to 0° C and kept so. Find the resulting temperature function u(x, t) taking x = 0 at A.

Solution:

The temperature function u(x, t) is the solution of the one dimensional heat equation

......... (A)

when the steady state condition prevails ∂u/∂t = 0 and hence we get

 d²u/dx² = 0

 (A) reduces to d²u/dx² = 0, on integration.

 u(x) = ax + b            ... (B)

when x = 0, we get u(0) = b = 20

when x = 30, we get u(30) = 30a + b

80 = 30a + 20

60 = 30a

 a = 2

Thus u(x, 0) = f (x) = 2x + 20 by (B)

Hence, Boundary and initial conditions are

(i) u (0, t) = 0 for all t≥ 0

(ii) u (30, t) = 0 for all t≥0

(iii) u (x, 0) = f(x) = 2x + 20

Now, the suitable solution which satisfies our boundary conditions is given by

 u (x, t) = (A cos px + B sin px)  …….(1)

Applying condition (i) in (1), we get

u (0, t) = A  = 0

Here,  ≠ 0      [it is defined for all t]

  A = 0

Substitute, A = 0 in (1), we get

  u(x, t) = B sin px    ……..(2)

Now, applying condition (ii) in equation (2), we get

 u(30, t) = B sin 30p  = 0

Here,  ≠ 0

          ['. it is defined for all t]

B ≠ 0

             ['. If B = 0 already A = 0 then we get a trivial solution]

sin 30p = 0

sin 30p = sin nπ

         ['. sin nπ = 0]

 30p = nπ

    p =  nπ / 30

Substitute, p =  nπ / 30 in equation (2), we get

To find B_n expand 2x + 20 in a half range Fourier sine series in the interval (0,30)

Example 6: An insulated rod of length l has its ends A and B kept at a° celsius and b° celsius respectively until steady state conditions prevail. The temperature at each end is suddenly reduced to zero degree celsius and kept so. Find the resulting temperature at any point of the rod taking the end A as origin.

Solution: The temperature function u(x, t) is the solution of the one dimensional heat equation.

          ............(A)

when the steady state condition prevails ∂u/∂t = 0 and

hence, we get ∂²u/∂x² = 0

.. (A) reduces to d²u/dx² = 0

on integration, we get u(x) = ax + b

when x = 0, we get u(0)=b

(i.e.,)  a⁰ = b

when x = l we get u(l) = al + b

b° = al + a⁰

( b° ‒ a° ) / l = a

Hence, u (x, 0) = f(x) = (b-a / l) x + a

The boundary and initial conditions are

(i) u (0, t) = 0 for all t≥ 0

 (ii) u (l, t) = 0 for all t≥ 0

(iii) u (x, 0) = f(x) = (b-a / l) x + a

Now, the suitable solution which satisfies our boundary conditions is given by

 u(x, t) (A cos px + B sin px)  ….(1)

Applying condition (i) in equation (1), we get

u (0, t) = A sin pl  = 0

Here,  ≠ 0        [it is defined for all t]

 A = 0

Substitute, A = 0 in equation (1), we get

 u(x,t) = B sin px  …….(2)

Applying condition (ii) in equation (2), we get

 u(l, t) = B sin pl  = 0

Here,  ≠ 0        [it is defined for all t]

B ≠ 0  [ suppose B = 0 already A = 0 then we get a trival solution]

  sin pl = 0

sin pl = sinnπ

         [sinnπ = 0]

 pl = nπ

To find B_n expand f(x) in a half range Fourier sine series in the interval (0, l)