Chapter two

2.1 Study state conduction in one dimension:

2.1.1 The Plane Wall

a. temperature destribution

FIGURE 2.1 Heat transfer through a plane wall. (a) Temperature distribution. (b) Equivalent thermal circuit.

Assumption:

Temperature varies with x-direction only no heat generation. So, the general equation is reduces to:

𝜕2𝑇𝜕𝑥2

= 0

First integration:

𝜕𝑇𝜕𝑥

= 𝐶1

Second integration:

𝑇(𝑥) = 𝐶1𝑥 + 𝐶2 2.1

this is the general solution

𝐶1 and 𝐶2 are constant of integration.

Baundary condition:

Substitute in general solution:

2.2

the heat transfer rate:

2.3

b. Thermal resistance

The thermal resistance for conduction is:

2.4

The thermal resistance for convection is:

2.5

Circuit representations provide a useful tool for both conceptualizing and quantifying heat transfer problems. The equivalent thermal circuit for the plane wall with convection surface conditions is shown in Figure 2.1b. The heat transfer rate may be

determined from separate consideration of each element in the network. Since qx is constant throughout the network, it follows that

2.1.2 The Composite Wall:

a. Material in series:

Equivalent thermal circuits may also be used for more complex systems, such as composite walls. Such walls may involve any number of series and parallel thermalresistances due to layers of different materials. Consider the series composite wall of Figure 2.2. The one-dimensional heat transfer rate for this system may be expressed as

FIGURE 2.2 Equivalent thermal circuit for a series composite wall.

*Overall heat transfer coefficient

With composite systems it is often convenient to work with an overall heat transfer coefficient, U, which is defined by an expression analogous to Newton’s law of cooling. Accordingly,

Or

𝒒𝒙 = 𝑼𝑨(𝑻∞𝟏 − 𝑻∞𝟒)

That means:

b. Material in parallel:

𝑞𝑥 = 𝑞1 + 𝑞2

𝑞1 = 𝑇1−𝑇2∆𝑥

𝑘1𝐴1

∆𝑥

𝑇2 𝑇1

𝑞

1

2

𝑞2 = 𝑇1−𝑇2∆𝑥

𝑘2𝐴2

𝑞𝑥 = (𝑇1 − 𝑇2) 1∆𝑥

𝑘1𝐴1

+ + 1∆𝑥

𝑘2𝐴2

= ∆𝑇𝑅𝑒

1𝑅𝑒

= 1𝑅1

+ 1𝑅2

Where:

𝑅𝑒 : Equavelant resistance.

Example 2.1: Find the heat transfer per unit area through the composite wall in Figure below. Assume one-dimensional heat flow.

Solusion:

𝑞 = ∆𝑇∑𝑅

𝑅 = ∆𝑥𝑘𝐴

𝑅𝐴 = 0.025(150)(0.1)

= 1.667 ∗ 10−3

𝑅𝐵 = 0.075(30)(0.05)

= 0.05

𝑅𝐶 = 0.05(50)(0.1)

= 0.01

𝑅𝐷 = 0.075(70)(0.05)

= 0.02143

𝑅 = 𝑅𝐴 + 𝑅𝐶 + 11𝑅𝐵

+ 1𝑅𝐷

= 2.667 ∗ 10−2

𝑞 = ∆𝑇𝑅

= 370−662.667∗10−2

= 11400 W

H.W CH2:1,2,3,4

1.1.2 Radial system a. Cylindrical

Consider a long cylinder of inside radius ri, outside radius ro, and length L, such as the one shown in Figure 2-3. We expose this cylinder to a temperature differential Ti − To and ask what the heat flow will be. For a cylinder with length very large compared to diameter, it may be assumed that the heat flows only in a radial direction, so that the only space coordinate needed to specify the system is r.

The general conduction equation in cylindrical coordinate:

2.6

Assumptions:

steady state. One dimension with radias only. No heat generation.

Equation 2.6 will reduced to:

1𝑟𝜕𝜕𝑟𝑘𝑟 𝜕𝑇

𝜕𝑟 = 0 2.7

We may determine the temperature distribution in the cylinder by solving Equation 2.7 and applying appropriate boundary conditions. Assuming the value of k to be constant, Equation 2.7 may be integrated twice to obtain the general solution

𝑟 𝜕𝑇𝜕𝑟

= 𝐶1

FIGURE 2-3: Hollow cylinder with convective surface conditions.

To obtain the constants of integration C1 and C2, we introduce the following boundary conditions:

Applying these conditions to the general solution, we then obtain

Solving for C1 and C2

𝐶1 = 𝑇𝑠,1−𝑇𝑠,2

ln (𝑟1 𝑟2 )

𝐶2 = 𝑇𝑠,2 −𝑇𝑠,1−𝑇𝑠,2

𝑙𝑛𝑟1 𝑟2 ∗ 𝑙𝑛(𝑟2)

and substituting into the general solution, we then obtain

2.8 The rate at which energy is conducted across any cylindrical surface in the solid may be expressed as

𝑞𝑟 = −𝑘𝐴 𝜕𝑇𝜕𝑟

= −𝑘(2𝜋𝑟𝐿) 𝜕𝑇𝜕𝑟

2.9

2.10

From this result it is evident that, for radial conduction in a cylindrical wall, the thermal resistance is of the form

*Composite Cylinder Wall

FIGURE 2-4: Temperature distribution for a composite cylindrical wall.

Consider now the composite system of Figure 2.4. Recalling how we treated the composite plane wall and neglecting the interfacial contact resistances, the heat transfer rate may be expressed as

2.12

2.11 K/W, ˚C/W

*Overall Heat Transfer Coefficient

The foregoing result may also be expressed in terms of an overall heat transfer coefficient. That is,

If U is defined in terms of the inside area, 𝐴1 = 𝟐𝝅𝒓𝟏𝑳, Equations 2.12 and 2.13 may be equated to yield

This definition is arbitrary, and the overall coefficient may also be defined in terms of A4 or any of the intermediate areas. Note that

2.15

Example 2.2:

Water flows at 50C inside a 2.5-cm-inside-diameter tube such that h1 = 3500 W/m2 · ˚C. The tube has a wall thickness of 0.8 mm with a thermal conductivity of 16 W/m · ˚C. The outside of the tube loses heat by free convection with h2 = 7.6 W/m2 · ˚C. Calculate the overall heat-transfer coefficient and heat loss per unit length to surrounding air at 20C.

Solution

There are three resistances in series for this problem, with L = 1.0 m, d1= 0.025 m, and d2 = 0.025 + (2)(0.0008) = 0.0266 m, the resistances may be calculated as:

𝑅1 = 12𝜋𝑟1𝐿ℎ1

= 1(3500)𝜋(0.025)(1)

= 0.00364 ˚C/W.

𝑅𝑡,𝑐𝑜𝑛𝑑 =𝑙𝑛 (𝑟2𝑟1

)

2𝜋𝑘𝐿=

𝑙𝑛 (0.02660.025 )

2𝜋(16)(1)= 0.00062 ˚C/W.

𝑅2 = 12𝜋𝑟2𝐿ℎ2

= 1(7.6)𝜋(0.0266)(1)

= 1.575 ˚C/W.

𝑞𝑟 = 𝑈1𝐴1𝑇∞,1 − 𝑇∞,2 = 𝑈2𝐴2𝑇∞,1 − 𝑇∞,2

𝑈1 = 1𝐴1𝑅𝑡𝑜𝑡𝑎𝑙

= 1(𝜋𝑑1𝐿)𝑅𝑡𝑜𝑡𝑎𝑙

𝑈1 = 1𝜋(0.025)(1)(0.00364+0.00062+1.575)

= 8.064

2.13

2.14

𝑈2 = 1𝐴2𝑅𝑡𝑜𝑡𝑎𝑙

= 1(𝜋𝑑2𝐿)𝑅𝑡𝑜𝑡𝑎𝑙

𝑈2 = 1𝜋(0.0266)(1)(0.00364+0.00062+1.575)

= 7.577

𝑞𝑟 = 8.064(𝜋)(0.025)(1)(50− 20) = 19 W

𝑞𝑟 = 7.577(𝜋)(0.0266)(1)(50 − 20) = 19 W

Example 2.3:

The heat flow is given by

*Critical Thickness of Insulation

Let us consider a layer of insulation which might be installed around a circular pipe, as shown in Figure 2-5. The inner temperature of the insulation is fixed at Ti, and the outer surface is exposed to a convection environment at T∞. From the thermal network the heat transfer is

Figure 2-5 Critical insulation thickness.

Now let us manipulate this expression to determine the outer radius of insulation ro, which will maximize the heat transfer. The maximization condition is

which gives the result

Equation (2.17) expresses the critical-radius-of-insulation concept. If the outer radius is less than the value given by this equation, then the heat transfer will be increased by adding more insulation. For outer radii greater than the critical value an increase in insulation thickness will cause a decrease in heat transfer.

Example 2.4:

Calculate the critical radius of insulation for asbestos [k = 0.17 W/m · C] surrounding a pipe and exposed to room air at 20C with h = 3.0 W/m2 · C. Calculate the heat loss from a 200C, 5.0-cm-diameter pipe when covered with the critical radius of insulation and without insulation.

2.16

2.17

Solution:

From Equation (2-17) we calculate ro as

The inside radius of the insulation is 5.0/2 = 2.5 cm, so the heat transfer is calculated from Equation (2.16) as

Without insulation the convection from the outer surface of the pipe is

So, the addition of 3.17 cm (5.67 − 2.5) of insulation actually increases the heat transfer by 25 percent. As an alternative, fiberglass having a thermal conductivity of 0.04 W/m · C might be employed as the insulation material. Then, the critical radius would be

Now, the value of the critical radius is less than the outside radius of the pipe (2.5 cm), so addition of any fiberglass insulation would cause a decrease in the heat transfer.

b. Spherial

Figure 2.6 show a hollow sphere of radius R1 at Ts,1 and R2 at Ts,2. For sphere the heat equation is gives by equation 1.9c

Assumption:

One dimension heat flow r-direction. Steady state. No heat generation.

For the control volume in the figure energy conservation required that

The approperiate form of Fouries low is

The general heat equation reduced to

1𝑟2

𝜕𝜕𝑟𝑘𝑟2 𝜕𝑇

𝜕𝑟 = 0

𝜕𝜕𝑟𝑘𝑟2 𝜕𝑇

𝜕𝑟 = 0

First integration:

𝑟2 𝜕𝑇𝜕𝑟

= 𝐶1

𝜕𝑇𝜕𝑟

= 𝐶1𝑟2

Second integration:

𝑻(𝒓) = −𝐶1𝒓

+ 𝐶2 ..........a

B.C: at 𝑟 = 𝑟1 , 𝑇 = 𝑇𝑠,1

at 𝑟 = 𝑟2 , 𝑇 = 𝑇𝑠,2

apply B.C in equation a

𝑇𝑠,1 = −𝐶1𝒓𝟏

+ 𝐶2 ..........b

𝑇𝑠,2 = −𝐶1𝒓𝟐

+ 𝐶2 .............c

Solve equation b and c we get:

𝐶1 = (𝑇𝑠,1−𝑇𝑠,2)1𝑟2− 1𝑟1

𝐶2 = 𝑇𝑠,1 −1𝑟1

(𝑇𝑠,1−𝑇𝑠,2)1𝑟1− 1𝑟2

Sub. 𝐶1& 𝐶2 in equation a, for temperature distribution:

Figure 2.6: conduction in spherical shell

𝑇(𝑟)−𝑇𝑠,1

𝑇𝑠,2−𝑇𝑠,1= 𝑟2

𝑟 𝑟−𝑟1𝑟2−𝑟1

𝜕𝑇𝜕𝑟

= 𝐶1𝑟2

, 𝐴 = 4𝜋𝑟2

𝑞𝑟 = −4𝜋𝑘𝑟2 (𝑻𝒔,𝟏−𝑻𝒔,𝟐)

𝒓𝟐( 𝟏𝒓𝟐− 𝟏𝒓𝟏

)

𝑞𝑟 = 4𝜋𝑘 (𝑻𝒔,𝟏−𝑻𝒔,𝟐)

( 𝟏𝒓𝟏− 𝟏𝒓𝟐

)

𝑅𝑡,𝑐𝑜𝑛𝑑 =1𝑟1− 1𝑟2

4𝜋𝑘

2.2 Conduction with thermal energy generation:

2.2.1 Plane wall with heat generation:

To investigate the effect of uniform internal heat generation in one dimension temperature distribution within the plane wall. Figure 2.7 shows the plane wall with internal uniform heat generation, .

Figure 2.7 Conduction in a plane wall with uniform heat generation with Symmetrical boundary conditions.

The general conduction equation is:

For study, one dimension:

𝜕2𝑇𝜕𝑥2

+ 𝑘

= 0

First integration gives:

𝜕𝑇𝜕𝑥

= −𝑞𝑘

+ 𝐶1

Second integration gives:

𝑇(𝑥) = − 2𝑘𝑥2 + 𝐶1𝑥 + 𝐶2 ................a

B.C: at 𝑥 = 𝐿,𝑇 = 𝑇𝑠 : at 𝑥 = −𝐿,𝑇 = 𝑇𝑠

Apply B.C to equation a:

𝑇𝑠 = − 2𝑘𝐿2 + 𝐶1𝐿 + 𝐶2 .........b

𝑇𝑠 = − 2𝑘𝐿2 − 𝐶1𝐿 + 𝐶2 ...........c

So:

𝐶2 = 𝑇𝑠 + 2𝑘𝐿2 and 𝐶1 = 0

Sub. 𝐶1&𝐶2 in equation a gives:

𝑇(𝑥)− 𝑇𝑠 = 2𝑘𝐿2 1 − 𝑥2

𝐿2 ..........d

This is a parabolic temperature distribution that is symmetry about x=0

Thus at x=0, 𝜕𝑇𝜕𝑥

= 0

And the maximum temperature axist at x=0

𝑇𝑚𝑎𝑥 − 𝑇𝑠 = 2𝑘𝐿2 .........e

From equation d

2𝑘𝐿2 = 𝑇(𝑥)−𝑇𝑠

1−𝑥2

𝐿2

Sub. In equation e we get:

𝑇(𝑥)−𝑇𝑠𝑇𝑚𝑎𝑥−𝑇𝑠

= 1 − 𝑥2

𝐿2

in case the temperature distribution is asymmetrical as shown in Figure 2.8 the temperature distribution is as below:

Figure 2.8: Conduction in a plane wall with uniform heat generation with Asymmetrical boundary conditions.

2.2.2 Cylinder with heat generation:

Heat generation may occur in a variety of radial geometries. Consider the long, solid cylinder of Figure 2.9, which could represent a current-carrying wire or a fuel element in a nuclear reactor. For steady-state conditions the rate at which heat is generated within the cylinder must equal the rate at which heat is convected from the surface of the cylinder to a moving fluid. This condition allows the surface temperature to be maintained at a fixed value of Ts.

FIGURE 2.9 Conduction in a solid cylinder with uniform heat generation.

To determine the temperature distribution in the cylinder, we begin with the appropriate form of the heat equation.

For constant thermal conductivity k, the heat conduction Equation reduces to:

1𝑟𝑑𝑑𝑟𝑟 𝑑𝑇

𝑑𝑟 +

𝑘= 0

Separating variables and assuming uniform generation, this expression may be integrated to obtain

𝑟 𝑑𝑇𝑑𝑟

= − 2𝑘𝑟2 + 𝐶1

Repeating the procedure, the general solution for the temperature distribution becomes

𝑇(𝑟) = − 4𝑘𝑟2 + 𝐶1𝑙𝑛𝑟 + 𝐶2 .........a

To obtain the constants of integration C1 and C2, we apply the boundary conditions

𝑑𝑇𝑑𝑟

= 0 𝑎𝑡 𝑟 = 0 , 𝑇 = 𝑇𝑠 𝑎𝑡 𝑟 = 𝑟𝑜

Therefore:

𝐶1 = 0

𝐶2 = 𝑇𝑠 + 4𝑘𝑟𝑜2

The temperature distribution is therefore

𝑇(𝑟) − 𝑇𝑠 = 𝑟𝑜2

4𝑘1 − 𝑟2

𝑟𝑜2 ...........b

𝑇 = 𝑇𝑚𝑎𝑥 𝑎𝑡 𝑟 = 0

Sub in equation

𝑇𝑚𝑎𝑥 − 𝑇𝑠 = 𝑟𝑜2

4𝑘 ..............c

Devided equation (b) to equation (c) we obtained

𝑇(𝑟)−𝑇𝑠𝑇𝑚𝑎𝑥−𝑇𝑠

= 1 − 𝑟𝑟𝑜2

From equation (b)

𝑑𝑇𝑑𝑟

= − 2𝑘𝑟

𝑞(𝑟) = −𝑘2𝜋𝑟𝑙 𝑑𝑇𝑑𝑟

𝑞(𝑟) = −𝑘2𝜋𝑟𝐿(− 2𝑘 𝑟) (heat conduction in the radius direction)

𝑞(𝑟) = 𝜋𝐿𝑟2 .........d

𝑞(𝑟𝑜) = 𝜋𝐿𝑟𝑜2

To relate the surface temperature, 𝑇𝑠, to the temperature of the cold fluid, 𝑇 , either a surface energy balance or an overall energy balance may be used.

𝑞𝑐𝑜𝑛𝑑 = 𝑞𝑐𝑜𝑛𝑣

𝜋𝐿𝑟𝑜2 = ℎ(2𝜋𝑟𝑜𝐿)(𝑇𝑠 − 𝑇∞)

𝑇𝑠 = 𝑇∞ + 𝑟𝑜2ℎ

H.W a similar analysis applies to a hollow cylinder as shown in the figure below:

Baundary condition: at 𝑟 = 𝑟2 𝑇(𝑟) = 𝑇𝑠,2 ∶ 𝑟 = 𝑟2 𝑑𝑇𝑑𝑟

= 0

Example:

A current of 200 A is passed through a stainless-steel wire [k = 19 W/m · C] 3 mm in diameter. The resistivity of the steel may be taken as 70 μΩ · cm, and the length of the wire is 1 m. The wire is submerged in a liquid at 110 C and experiences a convection heat-transfer coefficient of 4kW/m2 · C. Calculate the center temperature of the wire.

Solution:

All the power generated in the wire must be dissipated by convection to the liquid:

𝑃 = 𝐼2𝑅 = 𝑞 = ℎ𝐴(𝑇𝑠 − 𝑇∞) ..........a

The resistance of the wire is calculated from

𝑅 = 𝜌 𝐿𝐴𝑐

= (70𝑥10−6)(100)𝜋(0.15)2

= 0.099Ω

where ρ is the resistivity of the wire. The surface area of the wire is πdL, so from Equation (a)

(200)2(0.099) = 4000𝜋(3𝑥10−3)(1)(𝑇𝑠 − 110)

𝑇𝑠=215˚C.

The heat generated per unit volume is calculated from

𝑃 = 𝑉 = 𝜋𝑟𝑜2𝐿

So that,

= 3960𝜋(1.5𝑥10−3)2(1)

= 560.2 𝑀𝑊/𝑚3

Finally, the center temperature of the wire is calculated from Equation below:

𝑇𝑚𝑎𝑥 − 𝑇𝑠 = 𝑟𝑜2

4𝑘

𝑇𝑚𝑎𝑥 = (5.605𝑥108)(1.5𝑥10−3)2

(4)(19)+ 215 = 231.6

2.2.3 sphere with heat generation

The general heat conduction equation

For one dimension (r direction only) study state the general equation has been reduced to:

1𝑟2

𝑑𝑑𝑟𝑟2

𝑑𝑇𝑑𝑟 +

𝑘

= 0

Baundary condition: at 𝑟 = 0 𝑑𝑇𝑑𝑟

= 0 & 𝑎𝑡 𝑟 = 𝑟𝑜 𝑇 = 𝑇𝑠

𝑑𝑑𝑟𝑟2 𝑑𝑇

𝑑𝑟 = −

𝑘𝑟2

First integration gives:

𝑟2 𝑑𝑇𝑑𝑟

= − 3𝑘𝑟3 + 𝐶1

𝑑𝑇𝑑𝑟

= − 3𝑘𝑟 + 𝐶1

𝑟2

Second integration gives:

𝑇(𝑟) = − 𝑟2

6𝑘− 𝐶1

𝑟+ 𝐶2

H.W find the temperature distribution and heat transfer.

3.3 Contact Resistance

Although neglected until now, it is important to recognize that, in composite systems, the temperature drop across the interface between materials may be appreciable. This temperature change is attributed to what is known as the thermal contact resistance, Rt,c. The effect is shown in Figure 2.10, and for a unit area of the interface, the resistance is defined as

FIGURE 2.10 Temperature drop due to thermal contact resistance.