ME5320 Advanced heat transfer

Lecturer : K. Badarinath

Lectures 4-5

17/8/2015 & 20/8/2015

2-D steady state conductionCartesian - examples

2

Exercise problemA rectangular plate defined by 𝑥 = 0; 𝑥 = 𝑊; 𝑦 = 0; 𝑦 = 𝐻is shown.

There are 4 boundaries for this 2-D geometry. The conditions are

𝑇 0, 𝑦 = 𝑇1𝑇 𝑊, 𝑦 = 𝑇1𝑇 𝑥, 0 = 𝑇1

𝑇 𝑥,𝐻 = 𝑓 𝑥

Solve for

1. 𝑇 𝑥,𝐻 = 𝑓 𝑥 =

𝑇𝑚 sin𝜋𝑥

𝑊+ 𝑇1

2. 𝑇 𝑥,𝐻 = 𝑓 𝑥 = 𝑇2. 3

------(BC)

Step-wise solution procedure (Exercise problems)

Three conditions are homogeneous if we replace 𝑇 with another variable 𝜃 = 𝑇 − 𝑇1.

Governing equation for this case is 𝜕2𝜃

𝜕𝑥2+𝜕2𝜃

𝜕𝑦2= 0

The variable with two homogeneous boundary conditions is 𝑥 as we have homogeneous BCs at 𝑥 = 0 ; 𝑥 = 𝐿. The BCs are

1. 𝜃 0, 𝑦 = 0; ℎ𝑜𝑚𝑜𝑔𝑒𝑛𝑒𝑜𝑢𝑠

2. 𝜃 𝑊, 𝑦 = 0; ℎ𝑜𝑚𝑜𝑔𝑒𝑛𝑒𝑜𝑢𝑠

3. 𝜃 𝑥, 0 = 0; ℎ𝑜𝑚𝑜𝑔𝑒𝑛𝑒𝑜𝑢𝑠

4. 𝜃 𝑥, 𝐻 = 𝑓 𝑥 − 𝑇1; 𝑛𝑜𝑛 − ℎ𝑜𝑚𝑜𝑔𝑒𝑛𝑒𝑜𝑢𝑠

4

Assume a solution of the form θ 𝑥, 𝑦 = 𝑋 𝑥 𝑌 𝑦 ≡ 𝑋𝑌

The solution must satisfy the governing equation. We get𝜕2(𝑋𝑌)

𝜕𝑥2+𝜕2(𝑋𝑌)

𝜕𝑦2= 0

𝑌𝑑2𝑋

𝑑𝑥2+ 𝑋

𝑑2𝑌

𝑑𝑦2= 0

1

𝑋

𝑑2𝑋

𝑑𝑥2= −

1

𝑌

𝑑2𝑌

𝑑𝑦2= ±𝜆𝑛

2

Since there are homogeneous BCs on 𝑥, we get the right sign on 𝜆𝑛2 as (also noting there are many possible solutions 𝜃𝑖 = 𝑋𝑖𝑌𝑖)

𝑑2𝑋𝑛𝑑𝑥2

+ 𝜆𝑛2𝑋𝑛 = 0;

𝑑2𝑌𝑛𝑑𝑦2

− 𝜆𝑛2𝑌𝑛 = 0

For the the case of 𝜆𝑛 = 0, we get𝑑2𝑋0𝑑𝑥2

= 0 𝑎𝑛𝑑𝑑2𝑌0𝑑𝑦2

= 0

5

The solution to the ODEs are given as𝑋𝑛 𝑥 = 𝐴𝑛 sin 𝜆𝑛𝑥 + 𝐵𝑛 cos 𝜆𝑛𝑥𝑌𝑛 𝑦 = 𝐶𝑛 sinh 𝜆𝑛𝑦 + 𝐷𝑛 cosh 𝜆𝑛𝑦

For 𝑋0 𝑎𝑛𝑑 𝑌0, we have𝑋0 𝑥 = 𝐴0𝑥 + 𝐵0𝑌0 𝑦 = 𝐶0𝑦 + 𝐷0

The complete solution is a linear summation of all solutions 𝜃𝑖= 𝑋𝑖𝑌𝑖

𝜃 𝑥, 𝑦 = 𝑋0 𝑥 𝑌0 𝑦 +

𝑛=1

∞

𝑋𝑛 𝑥 𝑌𝑛(𝑦)

We now apply the BCs and we get

BC 1 ∶ 𝜃 0, 𝑦 = 0 gives𝜃𝑛 0, 𝑦 = 𝑋𝑛 0 𝑌𝑛 𝑦 = 0𝜃0 0, 𝑦 = 𝑋0 0 𝑌0 𝑦 = 0

Since the variables 𝑌𝑛 𝑦 𝑎𝑛𝑑 𝑌0 𝑦 cannot vanish, we have 𝑋𝑛 0 = 0; 𝑋0 0 = 0→ 𝐵𝑛 = 𝐵0 = 0

6

(1)

(2)

(3)

(4)

(SOL)

Similarly BC 2: 𝜃 𝑊, 𝑦 = 0; gives𝜃𝑛 𝑊,𝑦 = 𝑋𝑛 𝑊 𝑌𝑛 𝑦 = 0𝜃0 𝑊,𝑦 = 𝑋0 𝑊 𝑌0 𝑦 = 0

Since the variables 𝑌𝑛 𝑦 𝑎𝑛𝑑 𝑌0 𝑦 cannot vanish, we have 𝑋𝑛 𝑊 = 0𝑋0 𝑊 = 0

Substituting from equations 1 and 3 we get𝑋𝑛 𝑊 = 𝐴𝑛 sin 𝜆𝑛𝑊 = 0

Since 𝐴𝑛 cannot be zero, we havesin 𝜆𝑛𝑊 = 0

𝜆𝑛 =𝑛𝜋

𝑊, 𝑛 = 1, 2, 3,…

Also for 𝑋0 𝑊 ,𝑤𝑒 ℎ𝑎𝑣𝑒𝑋0 𝑊 = 𝐴0𝑊 = 0 → 𝐴0 = 0

With 𝐴0 = 𝐵0 = 0, the solution corresponding to 𝜆𝑛 = 0vanishes

7

------(5)

Applying BC 3: θ 𝑥, 0 = 0 gives𝜃𝑛 𝑥, 0 = 𝑋𝑛 𝑥 𝑌𝑛 0 = 0

Since the variable 𝑋𝑛 𝑥 cannot vanish, we get𝑌𝑛 0 = 0

Substituting in Equation 2, we get 𝐷𝑛 = 0

Looking at the result so far, we have

𝐴0 = 𝐵0 = 𝐵𝑛 = 𝐷𝑛 = 0, and the solution 𝑇0 vanishes.

So we have𝑋𝑛 𝑥 = 𝐴𝑛 sin 𝜆𝑛𝑥𝑌𝑛 𝑦 = 𝐶𝑛 sinh 𝜆𝑛𝑦

The solution then becomes (from Equation (SOL))

𝜃 𝑥, 𝑦 =

𝑛=1

∞

𝐴𝑛𝐶𝑛 sin 𝜆𝑛𝑥 sinh 𝜆𝑛𝑦

8

------(SOL-b)

• The product 𝐴𝑛𝐶𝑛 can be replaced by a single constant 𝑎𝑛We now apply the BC 4 to determine 𝑎𝑛We get θ 𝑥,𝐻 = 𝑓 𝑥 − 𝑇1 which gives

𝜃 𝑥,𝐻 = 𝑓 𝑥 − 𝑇1 =

𝑛=1

∞

(𝑎𝑛 sinh 𝜆𝑛𝐻) sin 𝜆𝑛𝑥

We get then (for case 1)

𝑇𝑚 sin𝜋𝑥

𝑊=

𝑛=1

∞

𝑎𝑛 sinh𝑛𝜋𝐻

𝑊sin

𝑛𝜋𝑥

𝑊

This requires that 𝑎𝑛 = 0 for 𝑛 > 1 and

𝑎1 =𝑇𝑚

sinh𝜋𝐻𝑊

So , we have

𝑇 𝑥, 𝑦 = 𝜃 𝑥, 𝑦 + 𝑇1 =𝑇𝑚 sinh

𝜋𝑦𝑊

sinh𝜋𝐻𝑊

sin𝜋𝑥

𝑊+ T1

9

• Similarly for case 2, we have by applying BC 4

We get θ 𝑥,𝐻 = 𝑓 𝑥 − 𝑇1 which gives

𝜃 𝑥,𝐻 = 𝑓 𝑥 − 𝑇1 =

𝑛=1

∞

(𝑎𝑛 sinh 𝜆𝑛𝐻) sin 𝜆𝑛𝑥

We get then

𝑇2 − 𝑇1 =

𝑛=1

∞

𝑎𝑛 sinh𝑛𝜋𝐻

𝑊sin

𝑛𝜋𝑥

𝑊

This is a Fourier sine series and the values of the coefficients 𝑎𝑛 may be determined by expanding the constant temperature difference 𝑇2 − 𝑇1in a Fourier series over the interval 0 < 𝑥 < 𝑊. This series is

𝑇2 − 𝑇1 = 𝑇2 − 𝑇12

𝜋

𝑛=1

∞−1 𝑛+1 + 1

𝑛sin

𝑛𝜋𝑥

𝑊

→ 𝑎𝑛 = 𝑇2 − 𝑇12

𝜋

−1 𝑛+1 + 1

𝑛

1

sinh𝑛𝜋𝐻𝑊

10

• The final solution can then be expressed as

𝜃 𝑥, 𝑦 =

𝑛=1

∞

𝑎𝑛 sin 𝜆𝑛𝑥 sinh 𝜆𝑛𝑦

Which leads to

𝑇 − 𝑇1𝑇2 − 𝑇1

=2

𝜋

𝑛=1

∞−1 𝑛+1 + 1

𝑛sin

𝑛𝜋𝑥

𝑊

sinh𝑛𝜋𝑦𝑊

sinh𝑛𝜋𝐻𝑊

Note that the temperature is expressed in terms of a difference and is in non-dimensional form.

In the subsequent slides, we list 3 more examples and will solve one of them in class.

The other 2 are assignment problems, different for students with odd and even roll numbers. Due date : Monday, 24th August 2015, 5:00 PM.

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Example 1 (Assignment 2 – odd)

• Semi-infinite plate with surface convection

Consider two-dimensional conduction in the semi-infinite plate shown in Figure below. The plate exchanges heat by convection with an ambient fluid at 𝑇∞. The heat transfer coefficient is ℎand the plate width is 𝐿. Determine the temperature distribution and the total heat transfer rate from the upper and lower surfaces.

12

Example 2 (Assignment 2 – even)

• Plate with two insulated surfaces

Consider two-dimensional conduction in the rectangular plate shown in Figure below. Two opposite sides are insulated and the remaining two sides are at specified temperatures. Determine the temperature distribution in the plate.

13

Example 3 (In-class discussion)

• Moving plate with surface convection

A semi-infinite plate with thickness 2L moves through a furnace with velocity U and leaves at temperature 𝑇0. The plate is cooled outside the furnace by convection. The heat transfer coefficient is ℎ and the ambient fluid temperature is 𝑇∞. Determine the two-dimensional temperature distribution in the plate.

14

Generalized diffusion with moving volume• In Lecture 2, we had discussed the generalized heat diffusion

equation. An implicit assumption in deriving that equation was that the control volume is stationary.

• However, before we can discuss Example 3, we need to develop amodified form of the diffusion equation to include moving surfaces.

• Recall that the general form (with constant k) we got was,𝜕2𝑇

𝜕𝑥2+𝜕2𝑇

𝜕𝑦2+𝜕2𝑇

𝜕𝑧2+

𝑞

𝑘=1

𝛼

𝜕𝑇

𝜕𝑡

• The most general form is an energy balance equation for a boundary moving with a generalized velocity vector (including acceleration). This is the energy equation (used in CFD). For this example, however, we derive a simpler form of heat diffusion with moving boundaries but assuming a uniform velocity field

𝑉 = 𝑈 𝑖 + 𝑉 𝑗 +𝑊 𝑘 15

Consider an infinitesimal control volume 𝑑𝑥 𝑑𝑦 𝑑𝑧 in a region moving

with velocity 𝑉

Apply the principle of conversation of energy in this infinitesimal control volume 𝑑𝑥 𝑑𝑦 𝑑𝑧. We then have,

𝐸𝑖𝑛 + 𝐸𝑔 − 𝐸𝑜𝑢𝑡 = 𝐸𝑠𝑡

We make the following assumptions : uniform velocity, constant pressure, constant density, and negligible changes in potential energy

Energy is exchanged with the element by two modes - conduction and mass motion. Both the modes are shown for a 2-D differential volume 𝑑𝑥 𝑑𝑦 in the next slide.

17

Energy transfer by conduction mode

Energy transfer by

mass motion mode

Mode 1 – Heat conduction

This is governed by the Fourier’s law and the heat flux in and out ofthe control volume are as shown.

Mode 2- Energy carried by the mass motion

The energy carried by the mass motion is obtained by multiplying the

specific enthalpy ℎ (enthalpy per unit mass) with the correspondingmass. The mass flow entering the element in 𝑥 direction is given as𝜌𝑈𝑑𝑦𝑑𝑧 and similarly in the 𝑦 direction is given as 𝜌𝑉𝑑𝑥𝑑𝑧 . The rateof energy carried is obtained by multiplying the mass flow rate withthe specific enthalpy.

So considering all 3 inlet faces (in 3-D), we have

𝐸𝑖𝑛 = 𝑞𝑥′′𝑑𝑦𝑑𝑧 + 𝑞𝑦

′′𝑑𝑥𝑑𝑧 + 𝑞𝑧′′𝑑𝑦dx + 𝜌𝑈 ℎ𝑑𝑦𝑑𝑧 + 𝜌𝑉 ℎ𝑑𝑥𝑑𝑧 +

𝜌𝑊 ℎ𝑑𝑦𝑑𝑥

The energy stored by the differential volume is given as 𝐸𝑠𝑡 = 𝑞 𝑑𝑥 𝑑𝑦 𝑑𝑧

Where 𝑞 is the rate of energy generated per unit volume.

The energy going out of the volume (in 3-D) can be written as a sum of the energy from conduction and mass motion. It is given as (ref. slide 15) 𝐸𝑜𝑢𝑡

= (𝑞𝑥′′+

𝜕𝑞𝑥′′

𝜕𝑥𝑑𝑥) 𝑑𝑦𝑑𝑧 + (𝑞𝑦

′′+𝜕𝑞𝑦

′′

𝜕𝑦𝑑𝑦) 𝑑𝑥𝑑𝑧 + (𝑞𝑧

′′+𝜕𝑞𝑧

′′

𝜕𝑧𝑑𝑧) 𝑑𝑦𝑑𝑥

+ 𝜌𝑈 ℎ +𝜕 ℎ

𝜕𝑥𝑑𝑥 𝑑𝑦𝑑𝑧 + 𝜌𝑉 ℎ +

𝜕 ℎ

𝜕𝑦𝑑𝑦 𝑑𝑥𝑑𝑧

+ 𝜌𝑊 ℎ +𝜕 ℎ

𝜕𝑧𝑑𝑧 𝑑𝑦𝑑𝑥

Note here that 𝑈, 𝑉,𝑊 are constant since material motion is assumed uniform.

The energy stored in the differential volume is given as

𝐸𝑠𝑡 = 𝜌𝜕 ℎ

𝜕𝑡𝑑𝑥 𝑑𝑦 𝑑𝑧

where 𝑢 is the integral energy per unit mass and 𝑡 is time.

Now applying the energy balance and dividing all the terms by 𝑑𝑥 𝑑𝑦 𝑑𝑧 we get,

𝐸𝑖𝑛 + 𝐸𝑔 − 𝐸𝑜𝑢𝑡 = 𝐸𝑠𝑡

i.e.

−𝜕𝑞𝑥

′′

𝜕𝑥−𝜕𝑞𝑦

′′

𝜕𝑦−𝜕𝑞𝑧

′′

𝜕𝑧− 𝜌𝑈

𝜕 ℎ

𝜕𝑥− 𝜌𝑉

𝜕 ℎ

𝜕𝑦− 𝜌𝑊

𝜕 ℎ

𝜕𝑧+ 𝑞 = 𝜌

𝜕 𝑢

𝜕𝑡

Now ℎ is defined as

ℎ = 𝑢 +𝑃

𝜌

Where 𝑃 – constant pressure. Thus we can write

𝜕 𝑢

𝜕𝑡=𝜕 ℎ

𝜕𝑡

Substituting in the energy balance equation to express

everything in terms of ℎ, and also applying Fourier’s law for each

of the terms 𝑞𝑥′′ = −k

𝜕𝑇

𝜕𝑥and so forth, we get

𝜕

𝜕𝑥𝑘𝜕𝑇

𝜕𝑥+

𝜕

𝜕𝑥𝑘𝜕𝑇

𝜕𝑥+

𝜕

𝜕𝑥𝑘𝜕𝑇

𝜕𝑥+ 𝑞

= 𝜌𝜕 ℎ

𝜕𝑡+ 𝑈

𝜕 ℎ

𝜕𝑥+ 𝑉

𝜕 ℎ

𝜕𝑦+𝑊

𝜕 ℎ

𝜕𝑧

Now enthalpy change at constant pressure is given as

𝑑 ℎ = 𝑐𝑝𝑑𝑇

Substituting this and assuming constant 𝑘 we get𝜕2𝑇

𝜕𝑥2+𝜕2𝑇

𝜕𝑦2+𝜕2𝑇

𝜕𝑧2+

𝑞

𝑘=1

𝛼

𝜕𝑇

𝜕𝑡+ 𝑈

𝜕𝑇

𝜕𝑥+ 𝑉

𝜕𝑇

𝜕𝑦+𝑊

𝜕𝑇

𝜕𝑧

where

𝛼 =𝑘

𝜌𝑐𝑝

The above equation (slide 21) is also applicable for incompressible flow with variable velocity as long as dissipation (work done by viscous forces) is negligible .

We now go back to our example 3 (Refer slide 14).

Observations

1. Temperature distribution is symmetrical about the 𝑥 axis.

2. At 𝑥 = ∞ ,the plate reaches ambient temperature.

3. All four BC’s are non-homogeneous but by defining 𝜃 = 𝑇 −𝑇∞, 3 of the conditions can become homogeneous in 𝜃.

Formulation

i. Assumptions : 2-D,steady, constant properties, uniform velocity only in ‘x’, 𝑞 = 0 and uniform ℎ and 𝑇∞, no heat generation

22

ii. Governing equation. The general equation is

𝜕2𝑇

𝜕𝑥2+𝜕2𝑇

𝜕𝑦2+𝜕2𝑇

𝜕𝑧2+

𝑞

𝑘=1

𝛼

𝜕𝑇

𝜕𝑡+ 𝑈

𝜕𝑇

𝜕𝑥+ 𝑉

𝜕𝑇

𝜕𝑦+𝑊

𝜕𝑇

𝜕𝑧

Using the assumptions, we get

𝜕2𝑇

𝜕𝑥2+𝜕2𝑇

𝜕𝑦2−𝜌𝑐𝑝𝑈

𝑘

𝜕𝑇

𝜕𝑥= 0

We substitute 𝜃 = 𝑇 − 𝑇∞ and replace 𝛽 =𝜌𝑐𝑝𝑈

2𝑘

which gives 𝜕2𝜃

𝜕𝑥2+𝜕2𝜃

𝜕𝑦2− 2𝛽

𝜕𝜃

𝜕𝑥= 0

iii. Boundary conditions : Here the independent variable with 2 homogenous BC’s is ‘y’ . Noting symmetry the four BC’s are (PTO)

-------(A)

1.𝜕𝜃 𝑥,0

𝜕𝑦= 0 (Symmetry conditions on ‘x’ axis)

2. −𝑘𝜕𝜃 𝑥,𝐿

𝜕𝑦= hθ(𝑥, 𝐿)

3. 𝜃 ∞, 𝑦 = 0

4. 𝜃 0, 𝑦 = 𝑇0 − 𝑇∞Solution : We assume a product solution of the form 𝜃 𝑥, 𝑦 =𝑋 𝑥 𝑌(𝑦). Substituting this in the Equation (A), we have

𝜕2(𝑋𝑌)

𝜕𝑥2+𝜕2(𝑋𝑌)

𝜕𝑦2− 2𝛽

𝜕 𝑋𝑌

𝜕𝑥= 0

We then get

𝑌𝑑2𝑋

𝑑𝑥2+ 𝑋

𝑑2𝑌

𝑑𝑦2− 2𝛽𝑌

𝑑𝑋

𝑑𝑥= 0

Dividing through out by XY and rearranging we get,1

𝑋

𝑑2𝑋

𝑑𝑥2− 2𝛽

𝑑𝑋

𝑑𝑥= −

1

𝑌

𝑑2𝑌

𝑑𝑦2= ±𝜆2

We get then,𝑑2𝑋𝑛𝑑𝑥2

− 2𝛽𝑑𝑋𝑛𝑑𝑥

± 𝜆𝑛2𝑋𝑛 = 0

and𝑑2𝑌

𝑑𝑦2∓ 𝜆𝑛

2𝑌𝑛 = 0

Since 2 HBCs are on variable ‘y’, we select the sign as positive for ‘y’ 𝑑2𝑋𝑛𝑑𝑥2

− 2𝛽𝑑𝑋𝑛𝑑𝑥

− 𝜆𝑛2𝑋𝑛 = 0

And𝑑2𝑌

𝑑𝑦2+ 𝜆𝑛

2𝑌𝑛 = 0

For the case 𝜆 = 0 , the equation are 𝑑2𝑋0𝑑𝑥2

− 2𝛽𝑑𝑋0𝑑𝑥

= 0

𝑑2𝑌0𝑑𝑦2

= 0

----(1)

----(2)

----(3)

----(4)

The solution to the Equations (1) to (4) are 𝑋𝑛 𝑥

= 𝐴𝑛 exp 𝛽𝑥 + 𝛽2 + 𝜆𝑛2 𝑥 + 𝐵𝑛 𝑒𝑥𝑝 𝛽𝑥 − 𝛽2 + 𝜆𝑛

2 𝑥

𝑌𝑛 𝑦 = 𝐶𝑛𝑠𝑖𝑛𝜆𝑛𝑦 + 𝐷𝑛 𝑐𝑜𝑠𝜆𝑛𝑦

𝑋0 𝑥 = 𝐴0𝑒𝑥𝑝 2𝛽𝑥 + 𝐵0

𝑌0 𝑦 = 𝐶0𝑦 + 𝐷0

The complete solution becomes

𝜃 𝑥, 𝑦 = 𝑋0 𝑥 𝑌0 𝑦 +

𝑛=1

∞

𝑋𝑛 𝑥 𝑌𝑛 𝑦

Now we apply the four boundary condition to determine the constants 𝐴0, 𝐵0, 𝐶0,𝐷0 𝑎𝑛𝑑 𝐴𝑛, 𝐵𝑛, 𝐶𝑛,𝐷𝑛 and also the characteristics values 𝜆𝑛.

BC 1 : 𝜕𝜃

𝜕𝑦𝑥, 0 = 0 from equation 2 we get,

𝐶𝑛𝜆𝑛 cos(𝜆𝑛𝑦) − 𝐷𝑛𝜆𝑛𝑠𝑖𝑛 (𝜆𝑛𝑦) 𝑋𝑛 𝑥 = 0

→ 𝐶𝑛𝜆𝑛𝑋𝑛 𝑥 = 0

Since 𝑋𝑛 𝑥 cannot vanish , we get𝐶𝑛 = 0

Similarly from Equation 4, we get𝐶0 = 0

BC2 : −𝑘𝜕𝜃

𝜕𝑦𝑥, 𝐿 = ℎ𝜃(𝑥, 𝐿) gives (from Equations 2 and 4)

𝑘𝐷𝑛𝜆𝑛𝑠𝑖𝑛 (𝜆𝑛𝐿)𝑋𝑛 𝑥 = ℎ𝑋𝑛(𝑥)𝐷𝑛 cos (𝜆𝑛𝐿)

Which gives 𝐷𝑛𝑋𝑛 𝑥 k𝜆𝑛𝑠𝑖𝑛 (𝜆𝑛𝐿 − ℎ cos (𝜆𝑛𝐿)) = 0

tan 𝜆𝑛𝐿 =ℎ

𝑘𝜆𝑛→ 𝜆𝑛𝐿 tan 𝜆𝑛𝐿 = 𝐵𝑖

Where 𝐵𝑖 is the Biot number defined as

𝐵𝑖 =ℎ𝐿

𝑘The above equation is the characteristic equation whose roots give the value of 𝜆𝑛.

Applying BC2 to Equation 4 gives𝑘𝐶0𝑋0 𝑥 = ℎ𝐶0𝐿𝑋0(𝑥)

→ 𝐷0 = 0

BC3 : 𝜃 ∞, 𝑦 = 0 on equations 1 and 3 leads to 𝐴𝑛 exp ∞ + 𝐵𝑛 exp(−∞) = 0

→ 𝐴𝑛 exp∞ = 0 → 𝐴𝑛 = 0

The result till now: We have 𝐶0 = 𝐷0 = 𝐴𝑛 = 𝐶𝑛 = 0

Since 𝐶0 = 𝐷0 = 0, 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑋0𝑌0 = 0. So the solution is

𝜃 𝑥, 𝑦 =

𝑛=1

∞

𝐵𝑛 exp 𝛽 − 𝛽2 + 𝜆𝑛2𝑥 𝐷𝑛 cos 𝜆𝑛𝑦

We rewrite the solution as

𝜃 𝑥, 𝑦 =

𝑛=1

∞

𝑎𝑛 exp 𝛽 − 𝛽2 + 𝜆𝑛2 𝑥 cos 𝜆𝑛𝑦

We now apply BC 4 : 𝜃 0, 𝑦 = 𝑇0 − 𝑇∞

𝜃 0, 𝑦 = 𝑇0 − 𝑇∞ =

𝑛=1

∞

𝑎𝑛𝑐𝑜𝑠 𝜆𝑛𝑦

We now invoke the Sturm-Lioville orthogonality condition.

The governing equation for 𝑌𝑛 is𝑑2𝑌𝑛𝑑𝑦2

+ 𝜆𝑛2𝑦𝑛 = 0

Comparing with SL problem (Lec 3) we have 𝑎1 = 𝑎2 = 0 and 𝑎3 = 1

This gives𝑝 = 𝑤 = 1 and 𝑞 = 0

Since the boundary conditions on y at 𝑦 = 0 and 𝑦 = 𝐿 are homogeneous, the solution 𝑌𝑛 = cos 𝜆𝑛𝑦 , 𝑛 = 1,2,3… are orthogonal with respect to the weighting function 𝑤 = 1.

We also have

𝑇0 − 𝑇∞ =

𝑛=1

∞

𝑎𝑛 cos 𝜆𝑛𝑦

Multiplying both sides by cos 𝜆𝑚𝑦 𝑑𝑦 and taking the integral from 𝑦 =0 to 𝑦 = 𝐿, we get

0

𝐿

(𝑇0−𝑇∞) cos 𝜆𝑚𝑦 𝑑𝑦 = 0

𝐿

𝑛=1

∞

𝑎𝑛 cos 𝜆𝑛𝑦 cos 𝜆𝑚𝑦 𝑑𝑦

Invoking orthogonality, implies that for 𝑚 ≠ 𝑛, the integral on RHS goes to zero. We get

(𝑇0−𝑇∞) 0

𝐿

cos 𝜆𝑛𝑦 𝑑𝑦 = 𝑎𝑛 0

𝐿

cos2(𝜆𝑛𝑦) 𝑑𝑦

→ 𝑎𝑛 =2 𝑇0 − 𝑇∞ sin(𝜆𝑛𝐿)

(𝜆𝑛𝐿) + sin(𝜆𝑛𝐿) cos(𝜆𝑛𝐿)

• The final solution in terms of T is 𝑇 𝑥, 𝑦 = 𝜃 𝑥, 𝑦 + 𝑇∞

= 𝑇∞ +

𝑛=1

∞

𝑎𝑛 exp 𝛽 − 𝛽2 + 𝜆𝑛2 𝑥 cos 𝜆𝑛𝑦

Where

𝑎𝑛 =2 𝑇0 − 𝑇∞ sin(𝜆𝑛𝐿)

(𝜆𝑛𝐿) + sin(𝜆𝑛𝐿) cos(𝜆𝑛𝐿)

31

2-D steady state conductionCylindrical co-ordinates

32

Cylindrical 2-D conduction

• Steady state conduction in cylindrical co-ordinates depends on the variables 𝑟, 𝜙, 𝑧 . In 2-D conduction, the temperature is a function of any two of the three variables i.e.

𝑇 = 𝑇 𝑟, 𝜙 ; 𝑇 = 𝑇 𝑟, 𝑧 ; 𝑇 = 𝑇(𝜙, 𝑧)

General form of the heat diffusion equation is

1

𝑟

𝜕

𝜕𝑟𝑘𝑟

𝜕𝑇

𝜕𝑟+

1

𝑟2𝜕

𝜕𝜙𝑘𝜕𝑇

𝜕𝜙+

𝜕

𝜕𝑧𝑘𝜕𝑇

𝜕𝑧+ 𝑞 = 𝜌𝑐𝑝

𝜕𝑇

𝜕𝑡

33

Bessel differential equations and Bessel functions

• A class of linear ordinary differential equations with variable coefficients is known as Bessel differential equations. A general form of Bessel differential equations is given by

𝑥2𝑑2𝑦

𝑑𝑥2+ 1 − 2𝐴 𝑥 − 2𝐵𝑥2

𝑑𝑦

𝑑𝑥+ 𝐶2𝐷2𝑥2𝐶 + 𝐵2𝑥2 − 𝐵 1 − 2𝐴 𝑥 + 𝐴2 − 𝐶2𝑛2 𝑦 = 0

Note the following

1. It is a linear second order differential equation with variable coefficients. That is, the coefficients of the dependent variable and its first and second derivatives are functions of the independent variable 𝑥.

2. 𝐴, 𝐵, 𝐶, 𝐷 𝑎𝑛𝑑 𝑛 are constants. Their values vary depending on the equation under consideration.

3. 𝑛 is called the order of the differential equation.

4. 𝐷 can be real or imaginary. 34

Solutions : Bessel functions • The general solution to a Bessel DE can be constructed in the

form of infinite power series. Two linearly independent solutions are needed since the equation is a second order one. The form of the solution depends on the constants 𝑛 𝑎𝑛𝑑 𝐷. There are four possible combinations

1. 𝑛 𝑖𝑠 𝑧𝑒𝑟𝑜 𝑜𝑟 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟, 𝐷 𝑖𝑠 𝑟𝑒𝑎𝑙

The solution is 𝑦 𝑥 = 𝑥𝐴 exp 𝐵𝑥 𝐶1𝐽𝑛 𝐷𝑥𝐶 + 𝐶2𝑌𝑛 𝐷𝑥𝐶

Where

𝐶1, 𝐶2 are constants of integration

𝐽𝑛 𝐷𝑥𝐶 = Bessel function of order n of the first kind

𝑌𝑛 𝐷𝑥𝐶 = Bessel function of order n of the second kind

Note

i. The term 𝐷𝑥𝐶 is the argument of the Bessel function.

ii. Values of the Bessel functions are tabulated.35

2. 𝑛 𝑖𝑠 𝑛𝑒𝑖𝑡ℎ𝑒𝑟 𝑧𝑒𝑟𝑜 𝑛𝑜𝑟 𝑎 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟, 𝐷 𝑖𝑠 𝑟𝑒𝑎𝑙

The solution is 𝑦 𝑥 = 𝑥𝐴 exp 𝐵𝑥 𝐶1𝐽𝑛 𝐷𝑥𝐶 + 𝐶2𝐽−𝑛 𝐷𝑥𝐶

3. 𝑛 𝑖𝑠 𝑧𝑒𝑟𝑜 𝑜𝑟 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟, 𝐷 𝑖𝑠 𝑖𝑚𝑎𝑔𝑖𝑛𝑎𝑟𝑦

The solution is 𝑦 𝑥 = 𝑥𝐴 exp 𝐵𝑥 𝐶1𝐼𝑛 𝑝𝑥𝐶 + 𝐶2𝐾𝑛 𝑝𝑥𝐶

Where

𝑝 =𝐷

𝑖, where 𝑖 is imaginary, 𝑖 = −1

𝐼𝑛 𝑝𝑥𝐶 =modified Bessel function of order n of the first kind

𝐾𝑛 𝑝𝑥𝐶 =modified Bessel function of order n of the second kind

4. 𝑛 𝑖𝑠 𝑛𝑒𝑖𝑡ℎ𝑒𝑟 𝑧𝑒𝑟𝑜 𝑛𝑜𝑟 𝑎 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑖𝑛𝑡𝑒𝑔𝑒𝑟, 𝐷 𝑖𝑠 𝑟𝑒𝑎𝑙

The solution is 𝑦 𝑥 = 𝑥𝐴 exp 𝐵𝑥 𝐶1𝐼𝑛 𝑝𝑥𝐶 + 𝐶2𝐼−𝑛 𝑝𝑥𝐶

36

Forms of Bessel functions• The Bessel functions 𝐽𝑛, 𝑌𝑛, 𝐽−𝑛, 𝐼𝑛, 𝐼−𝑛, 𝐾𝑛 𝑎𝑛𝑑 𝐾−𝑛 are symbols

representing infinite power series. The series form depends on 𝑛. For example 𝐽𝑛 𝑚𝑥 represents the following infinite series

𝐽𝑛 𝑚𝑥 =

𝑘=0

∞−1 𝑘 𝑚𝑥

2

2𝑘+𝑛

𝑘! Γ(𝑘 + 𝑛 + 1)

Where Γ is the gamma function.

• Special forms : When 𝑛 =𝑜𝑑𝑑 𝑖𝑛𝑡𝑒𝑔𝑒𝑟

2

Bessel functions for 𝑛 =1

2take the following form

𝐽 1 2 𝑥 =2

𝜋𝑥sin 𝑥 ; 𝐽− 1 2 𝑥 =

2

𝜋𝑥𝑐𝑜𝑠 𝑥

Bessel functions of order 3 2 , 5 2 , 7 2,… are determined using

𝐽𝑘+ 1 2 𝑥 =2𝑘 − 1

𝑥𝐽𝑘− 1 2 𝑥 − 𝐽𝑘− 3 2 𝑥 ; 𝑘 = 1, 2, 3,… 37

Similarly modified Bessel functions for 𝑛 =1

2take the following form

𝐼 1 2 𝑥 =2

𝜋𝑥sinh 𝑥 ; 𝐼− 1 2 𝑥 =

2

𝜋𝑥cosh 𝑥

Modified Bessel functions of order 3 2 , 5 2 , 7 2,… are determined using

𝐼𝑘+ 1 2 𝑥 =2𝑘 − 1

𝑥𝐼𝑘− 1 2 𝑥 + 𝐼𝑘− 3 2 𝑥 ; 𝑘 = 1, 2, 3,…

• Special relations for 𝑛 = 1,2,3,…𝐽−𝑛 𝑥 = −1 𝑛𝐽𝑛 𝑥𝑌−𝑛 𝑥 = −1 𝑛𝑌𝑛 𝑥

𝐼−𝑛 𝑥 = 𝐼𝑛 𝑥𝐾−𝑛 𝑥 = 𝐾𝑛 𝑥

• A table of selected values of Bessel functions is given below

38

𝒙 𝑱𝟎(𝒙) 𝑱𝒏(𝒙) 𝑰𝟎(𝒙) 𝑰𝒏(𝒙) 𝒀𝒏(𝒙) 𝑲𝒏(𝒙)

0 1 0 1 0 −∞ ∞

∞ 0 0 ∞ ∞ 0 0

The formulae for derivatives of the Bessel functions are shown here. The symbol 𝑍𝑛represents certain Bessel functions of order 𝑛.

39

40

The integrals of Bessel functions where 𝑍𝑛(𝑥)represents 𝐽𝑛 𝑥 𝑜𝑟 𝑌𝑛(𝑥)are shown here. Note that the integral

𝑍0 𝑥 𝑑𝑥

cannot be evaluated in closed form.

Equidimensional (Euler) equation• Consider the following second order differential equation with

variable coefficients

𝑥2𝑑2𝑦

𝑑𝑥2+ 𝑎1𝑥

𝑑𝑦

𝑑𝑥+ 𝑎0𝑦 = 0

Where 𝑎0 𝑎𝑛𝑑 𝑎1 are constant. The coefficients have a pattern 𝑥2

multiplies the second derivative, 𝑥 multiplies the first derivative and 𝑥0

multiples the function 𝑦. This equation is a special case of a class of equations known as equidimensional or Euler equation.

The solution to the above equation depends on the roots of the following equation

𝑟1,2 =− 𝑎1 − 1 ± 𝑎1 − 1 2 − 4𝑎0

2

41

• There are three possibilities

1. If the roots are distinct, the solution takes the form 𝑦 𝑥 = 𝐶1𝑥

𝑟1 + 𝐶2𝑥𝑟2 .

2. If the roots are imaginary as 𝑟1,2 = 𝑎 ± 𝑏𝑖, where

𝑖 = −1, the solution is 𝑦 𝑥 = 𝑥𝑎[𝐶1 cos 𝑏 l𝑛 𝑥 + 𝐶2 sin(𝑏 l𝑛 𝑥)]

3. If there is only one root, the solution is𝑦 𝑥 = 𝑥𝑟(𝐶1 + 𝐶2 l𝑛 𝑥)

42