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2.7 Steady state conduction in a composite wall, cylinder and sphere	35

2.7.1.3 Thermal contact resistance

When two surfaces are kept together, because of the presence of a surface roughness at the contacting surfaces, the gaps are invariably filled with air, and this introduces convection resistance. This is called thermal contact resistance, which reduces the heat transfer through the composite wall.

Needless to say, for maximizing the heat transfer through the composite wall, the thermal contact resistance at the interface has to be reduced.

Methods of reducing thermal contact resistance:

1.By applying high thermal conductivity gel at the interface. A gel is typically a semisolid material. When the gel is exposed to a higher temperature, it gets converted into liquid and slowly drops out and can be used for low-temperature applications.

2.Placing thin soft and high thermal conductivity metallic sheets at the interface during the assembly.

3.By increasing the contact area, by applying more pressure during assembly.

Thermal contact resistance is quite a difficult quantity to measure and, in practice, can often be a “sore thumb” in thermal system design.

2.7.2 Composite cylinder

Consider one-dimensional steady-state conduction in a composite cylinder, schematic representation of which is in Fig. 2.17. The corresponding resistance network is presented in Fig. 2.18. Geometrical details of thermophysical properties of the materials making up composite cylinder are also shown in Fig. 2.17.

Invoking, the resistance analogy, presented earlier, the total heat transfer from the inside to the outside through the cylindrical shell is given by

Q =	T∞1 − T∞2	(2.92)
	Rtotal

FIGURE 2.17

Schematic representation of a composite cylinder undergoing one-dimensional steady-state conduction.

36 CHAPTER 2 One-dimensional, steady state heat conduction

FIGURE 2.18

Schematic representation of a resistance network one dimensional steady state conduction in a composite cylinder.

The total resistance Rtotal is given by

Rtotal = Rconv1 + Ra + Rb + Rc + Rconv 2

The individual resistances in Eq. (2.93) can be calculated as follows:

Rconv1 = 1

h1 A1

ln r2

Ra = 2π kr1L

ln r3

Rb = 2π kr2L

ln r4

Rc = 2π r3 kc L

Rconv2 = h 1A

2 1

(2.93)

(2.94)

(2.95)

(2.96)

(2.97)

(2.98)

Please note that A1= 2πr1L1 and A4= 2πr4L4. An overall heat transfer coefficient can now be defined for the cylindrical shell as follows.


Q =	T∞1 − T∞2	= U A (T		− T )		(2.99)
Q =		= U A (T		− T )		(2.99)
	1 1		∞	∞	2
	Rtotal		1		2
	Rtotal

Here U1 is the overall heat transfer coefficient, based on inner area A1, and is given by

(2.100)

A1Rtotal

= A

(2.101)

2π r1 Lh1

2π ka L 2π kb L 2π kc L h2 2π r4 L


2.7 Steady state conduction in a composite wall, cylinder and sphere	37

Please note, as opposed to a plane wall, the overall heat transfer coefficient varies across the cylinder, and its specification is incomplete without a declaration of the area on which it is based.

Example 2.3: Consider a very long† cylindrical rod of 100 mm radius that consists of a nuclear-reacting material (k = 0.05 W/mK), generating 24,000 W/m3 uniformly throughout the volume (qg). This rod is enclosed within a tube having an outer radius of 200 mm, whose thermal conductivity is 4 W/mK. The outer surface is surrounded by a fluid at 100 °C and h = 20 W/m2K. Find the temperature at the interface of the two cylinders and the outer surface. A schematic representation of the given problem is shown in Fig. 2.19.

Solution:

From energy balance we have

qg × volume = hA(T2 − T∞2 )

(2.102)

24000 ×

× 0.22 × L = 20 × 2π × 0.2L(T2 −100)

(2.103)

T2 = 130 °C

Under steady state

qg × volume =

T1 − T2

(2.104)

2π kL

24000 × π

× 0.22 × L =

T1 −130

(2.105)

200

ln 100

2π × 4 ×1

T1 = 150.8 °C

FIGURE 2.19

Schematic representation of the cylindrical rod under consideration in Example 2.3.

†In the direction perpendicular to the plane of the paper.

38CHAPTER 2 One-dimensional, steady state heat conduction

Example 2.4: Revisit Example 2.3 and calculate the center temperature.

Solution:

1 d				dT		+	qg		= 0
		r
				dr				k
r dr
d			dT		= −			qg	r
	r
								k
dr			dr

r dTdr = − qkg × r22 + c1

dTdr = − qkg × 2r + cr1

Boundary condition 1: At r = 0;	dT	= 0
	dr
Therefore

c1 = 0

Integrating Eq. (2.128) with respect r on both sides, we have

T = qkg × r42 + c2

(2.106)

(2.107)

(2.108)

(2.109)

(2.110)

(2.111)

Boundary condition 2: At r = 0.1m;T = 150.8 °C

Therefore, c2 can be determined as follows

150.8 = − 240000.5 × 0.412 + c2 c2 = 270.8 °C

At the center r = 0, the temperature is given by

T = Tcenter = c2 = 270.8 °C

This example shows how we can use a “divide-and-conquer approach” to solve problems involving the third kind of boundary condition and a volumetric heat generation. Such problems are called conjugate problems, particularly when “h” at the surface also needs to be determined from, say, first-principles modeling.

2.7.3 Composite sphere

The next logical extension, as done before, is to work out the heat transfer rate from a composite spherical shell.

The cross-sectional view of a composite spherical shell is shown in Fig. 2.20 along with all the necessary details for us to be able to compute the heat transfer rate from the composite spherical shell.


2.7 Steady state conduction in a composite wall, cylinder and sphere	39

FIGURE 2.20

Schematic representation of a cross-sectional view of a composite sphere.

Under steady state conditions, and by invoking an electrical analogy, we can write out an expression for the heat transfer rate, Q, across the shell as

Q =

T∞1

− T1

− T2

− T3

− r1

− r2

h1 4πr12

4π kar1r2

4π kbr2r3

− T4

T4 − T∞2

(2.112)

− r3

4π kcr3r4

h4 4πr42

Using the Dividendo componendo rule, Eq. (2.112) may be written as

Q =

T∞ − T∞

(2.113)

Rtotal

Where, R

r2 − r1

r3 − r2

(2.114)

h1 4πr12

4π kar1r2

4π kbr2r3

total

r4 − r3

4π kcr3r4

h4 4πr42

By using Eq. (2.112), one can calculate any interface temperature.

Example 2.5: Consider a composite spherical shell of inner diameter 50 cm, made of lead and stainless steel with outer diameters 60 and 62 cm respectively (Fig. 2.21). The cavity contains radioactive waste that generates heat volumetrically (qv) and uniformly at the rate of 4.8 × 105 W/m3. The melting point of lead is 600 K;

klead = 35.3W/mK and kss = 15.1W/mK . The shell is kept in ambient at a temperature of 30 °C, with a convection coefficient of h = 500 W/m2K. What is the maximum tem-

perature in the lead? Will the lead melt?

Solution:

Given r1 = 25 cm, r2 = 30 cm, r3 = 31 cm

40 CHAPTER 2 One-dimensional, steady state heat conduction

FIGURE 2.21

Schematic representation of a cross-sectional view of the composite sphere under consideration in Example 2.5.

Energy balance equation:

qv × volume = hA3 (T3 − T∞ )

4.8 × 105 × 43π × 0.253 = 500 × 4π × 0.312 (T3 − 30)

T3 = 82 °C

T3 is the outermost surface temperature of the composite shell Under steady state

− T3 =

T3 − T∞

− r2

4π kssr2r3

h4πr32

T2 − 82

82 −

0.31 − 0.3

4π × 15.1 × 0.31 × 0.3

500 × 4π × 0.312

T2 = 99.8

°C

Under steady state

− T3 =

− T2

− r2

− r1

4π kssr2r3

4π klead r1r2

99.8 − 82

T1 − 99.8

0.31 − 0.3

0.3 − 0.25

4π × 15.1 × 0.3 × 0.31

4π × 30 × 0.3 × 0.25

T1 = 155.3 °C

(2.115)

(2.116)

(2.117)

(2.118)

(2.119)

T1 is the temperature at r1 = 0.25.m.

The melting point of lead is given as 600 K. Since all the temperatures inside the composite sphere are less than 600 K, there is no chance of the lead melting.


2.8 One-dimensional, steady state heat conduction with heat generation	41

Example 2.6: Revisit Example 2.5 and take k for radioactive waste as 7 W/mK. Calculate the maximum temperature in the cavity.

= 0

= −

2 dT

= − qv

+ c1

= − qv

Boundary condition 1: At r = 0; dT /dr = 0

c1 = 0

T = − qv

+ c2

Boundary condition 2: At r = r1; T = T1

= − qv

+ c2

155.3 = − 4.8 × 105

0.252 + c2

c2 = 869.6 °C

At r = 0;

T = Tcenter = Tmax = c2 = 869.6 °C

(2.120)

(2.121)

(2.122)

(2.123)

(2.124)

(2.125)

(2.126)

(2.127)

Please note that we again used “divide and conquer” to solve the problem. We tried to use every bit of information that was presented in advance to simplify the problem as much as possible. As a general rule, we have to examine first if we can employ energy balance to simplify the boundary conditions. The revelation here is that without solving the original governing equation for the shell, we are not only able to obtain the overall heat transfer but are also able to get estimates of the intermediate temperatures. This is classic heat transfer engineering practice and is very quick!

2.8 One-dimensional, steady state heat conduction with heat generation

2.8.1 Plane wall with heat generation

Consider the plane wall, shown in Fig. 2.22 with a uniform volumetric heat generation, given by qv W/m3 with Dirichlet boundary condition at x = ±L. Such a situation arises in nuclear fuel rods or when chemical reactions (exothermic) take place in a medium.

42 CHAPTER 2 One-dimensional, steady state heat conduction

FIGURE 2.22

Schematic representation of a plane wall with heat generation.

We apply the following simplifying assumptions

1.One dimensional conduction

2.Steady state

3.Constant properties

The governing equation for the problem under consideration becomes

d		dT	+ qv = 0
	k

dx		dx
		d 2T	+	qv	= 0
		dx2		k

			d 2T		= − qv
			dx2		k

On integrating once, we get

dTdx = − qkv x + c1

Boundary condition 1: At x = 0;	dT	= 0
	dx
From Eq. (2.131),

c1 = 0

(2.128)

(2.129)

(2.130)

(2.131)

(2.132)


2.8 One-dimensional, steady state heat conduction with heat generation	43

Integrating Eq. (2.131), we have

T = − qv x2 + c2 k 2

Boundary condition 2: At x = ±L;T = T1

From Eq. (2.133), we solve for c2

c2 = T1 + qv L2 k 2

Substituting for c2 in Eq. (2.133), we have


T = − qv				x2	+ T1 + qv					L2

	k		2					k 2
T = T1 +			qv		2		− x	2
					(L				)
			2k
T − T1 =	qv		2		− x	2
		(L					)
	2k

(2.133)

(2.134)

(2.135)

(2.136)

(2.137)

Here (T – T1) is excess temperature due to heat generation inside the slab.

Eq. (2.137) tells us that the temperature distribution is quadratic. A slightly more involved version of the above problem is when we have a situation where there is a plane wall with heat generation and with convection boundary conditions, as shown in Fig. 2.23.

FIGURE 2.23

Schematic representation of a plane wall with heat generation with convection boundary conditions on both the sides.

44 CHAPTER 2 One-dimensional, steady state heat conduction

The governing equation remains the same as before. The boundary conditions, though, are different.

Boundary condition 1: At x = 0; dT /dx = 0 From Eq. (2.131)

c1 = 0

(2.138)

Integrating Eq. (2.131), we have

T = − qv

+ c2

(2.139)

Boundary condition 2: At x = L; qconduction = qconvection

−k dT = h(T − T )

(2.140)

∞

−q

−k

L = h(T − T )

(2.141)

∞

T1 = T∞ + qv L

(2.142)

we know

c2 = T1 +

(2.143)

Therefore,

T = − qv

+ T1 + qv

(2.144)

T = T +

qv L

qv L2

(2.145)

1−

∞

Similarly, the heat transfer in a cylinder and sphere with internal heat generation can be derived.

Example 2.7: Consider a one-dimensional wall that uniformly generates heat at the rate of qv W/m2 whose thickness is 2L = 50 mm. The wall is made of a material whose thermal conductivity is 50 W/mK.Temperature measurements show the following relation, T(°C) = 200-2000x2. In the above expression, x is in meters. (1) Determine the volumetric heat generation rate qv in the wall. (2) Determine the heat fluxes at the two walls. (3) How are these related to the qv?

Solution:

Fig. 2.22 can be used for this problem.

Temperature symmetry about the mid-plane is clear from the given temperature distribution

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