CHE332:Heat Transfer Operations

Lecture 2: Conduction

Dr. Abdul Razzaq

Assistant professor, Chemical Engineering,
CIIT Lahore.

1
 Contents

 Fourier’s Law of Heat Conduction

 Heat Diffusion Equation
 Thermal conductivity
 Thermal diffusivity
 Thermal contact resistance
 Heat conduction : Plane wall, Composite wall,
Pipe/Cylinder/Sphere
 Transient Heat conduction

2
Conduction

3
Conduction

4
 Conduction

Rate of heat conduction 

 Area  Temperature difference 
Thickness
Q/t α (A)(T1-T2) / ∆x
Q/t = k (A)(T1-T2) / ∆x
Q/t= -k A ∆T/∆x
In differential form,
qx = -kA (dT / dx)
[Fourier's Law]
Where,
Q = Rate of heat transfer
A = Heat flow area
-dT / dx = Temperature gradient
k = Thermal conductivity

5
 Thermal conductivity
 The thermal conductivity of a material is a measure of the ability
of the material to conduct heat.

 Thermal conductivity is defined as the rate of heat transfer (Q)

transmitted through a unit thickness (L) in a direction normal to a
surface of unit area (A) due to a unit temperature gradient (ΔT)
under steady state conditions

 Thermal Conductivity = (heat rate × distance) / (area ×

temperature gradient)
k = Q × L / (A × ΔT)
 Thermal conductivity: Solids

• Accordingly, transport of thermal energy may be due to two

effects: the migration of
– Free electrons and
– Lattice vibrational waves.

• When viewed as a particle-like phenomenon, the lattice vibration

quanta are termed phonons.

• In pure metals, the electron contribution to conduction heat

transfer dominates, while in nonconductors and semiconductors,
the phonon contribution is
Thermal conductivity: Solids
Thermal conductivity: Solids
Thermal conductivity: Fluids
Thermal conductivity: Fluids
 Thermal diffusivity

 It measures the ability of a material to conduct thermal energy

relative to its ability to store thermal energy. Materials of large
value will respond quickly to changes in their thermal
environment, while materials of small value will respond more
sluggishly, taking longer to reach a new equilibrium condition.

12
 Conduction: Plane wall

Rate of heat conduction 

 Area  Temperature difference 
Thickness

Q α (A)(T1-T2) / ∆x
Q = k (A)(T1-T2) / ∆x
Q = -k A ∆T/∆x
Qx = kA (-dT / dx)

Q = ∆T /R

13
 Plane wall

Q = -(kA/∆x) ∆T or
Qx = k A (-dT / dx)

kA/L = Conductance

Thus,

1/Conductance = Resistance

Q = ∆T /R
Conduction: Composite wall

Q = ∆T /RT = To-T3/ RT

Q = To-T1/ Ra + T1-T2/ Rb +
T2-T3/ Rc
Ra, Rb, Rc
 Thermal contact resistance

 An error caused in the measurement of thermal conductivity by

nature of bonds between hot and cold entities

 By presence of any fluid / air

 Almost impossible to exclude this error

 Examples:

1. Solid receives heat by contacting another solid, it is

impossible to exclude air between those solids

2. When a liquid contacts metal, the presence of minute pits or

surface roughness may permanently trap small bubbles of air.
Thermal Contact Resistance
 Pipe wall/Cylindrical system

Qx = k A (-dT / dx)

Q = k A (-dT / dr)

Q = -k 2πrL (dT / dr)

Q (1 / r) dr = -k 2πL dT

Integration r1 & r2, T1 & T2

Q ln (r2/r1) = 2πkL (T1-T2)

What about log?
Q = 2πkL (T1-T2) / ln (r2/r1)
And Diameters D1, D2?
T1 = T2 + Q ln r2/r1 / 2πk
Pipe wall: Composite
Pipe wall/Cylindrical system

20
Pipe wall/Cylindrical system

21
 Maximum heat loss: Critical Radius

Rb = 1/2πkb ln r/r1

Ra = 1/ha2πr

dR/dr = 0 = 1/2πkb d ln r/r1 +

1/ha2π d1/r

1/2πkbr - 1/ha2πr2 = 0

at maximum heat loss r = rc

rc = Kb/ha
Maximum heat loss: Critical Radius
General Conduction Equation
 General Conduction Equation
 Let, an elemental cube of volume, dV = dx.dy.dz receives a
differential quantity of heat dQ’1 (Btu) through its left yz plane. In
the same interval, differential quantity of heat dQ’2 (Btu) leaves
through its right yz plane.
dQ’ = dQ’1 – dQ2' -------------------------(1)

 From Fourier's law of heat conduction, the heat entering the left yz
plane is given by,
dQ'1/dθ = k dy dz (- ∂T/∂x) ------------(2)

 The temperature gradient (- ∂T/∂x) may vary with both time and
position in the cube.
Variation of (- ∂T/∂x) as a f(x) is ∂(- ∂T/∂x)/(∂x) = (-∂2T/∂x2)
 General Conduction Equation
 Over the distance dx from x to x+dx, the total change in
Temperature gradient is (- ∂2T/∂x2).dx

 Thus at x+dx, the Temperature gradient is (- ∂T/∂x - ∂2T/∂x2.dx)

Heat leaving can be written as

dQ’2/dθ = k dy dz (- ∂T/∂x - ∂2T/∂x2.dx)----------(3)

 When the flow of heat in and out of the cube is constant as in the
steady state, (∂T/∂x) is constant then (∂2T/∂x2) = 0
Equation (3) can be written as,
dQ’/dθ = k dy dz (- ∂T/∂x)
Qx = - kA dT / dx
 General Conduction Equation
 Putting values form equations 2 and 3 into equation 1 we have
dQ’/dθ = k dy dz (∂2T/∂x2).dx------(4)

Analysis based on elemental volume so the change in temperature

per unit time will be (dT/dθ) and over the time interval dθ, it will be
(dT/dθ).dθ

The volumetric specific heat (Cv) (Btu/ft3.oF), which is obtained

by multiplying weight specific heat (cp) by the density (ℓ).

The rate of heat transfer to raise the volume (dx.dy.dz) by

(dT/dθ).dθ in the cube is given by,
dQ’/dθ = cp.ℓ.dx.dy.dz. (∂T/∂θ) -------- (5)

By combining eq. (4) & (5)

cp.ℓ.dx.dy.dz. (∂T/∂θ) = k dy dz (∂2T/∂x2).dx
(∂T/∂θ) = (k/ℓ. cp).(∂2T/∂x2) ----------(6)
 General Conduction Equation
 The eq. (6) is Fourier’s General Conduction Equation in one
dimension (x-axis), which may be written as,
(∂T/∂θ) = (α).(∂2T/∂x2) ------------(7)

 α is known as thermal diffusivity of the medium.

 In the special case of heat propagation in an isotropic and
homogeneous medium in a 3-dimensional space (x,y,z), the
Fourier’s General Conduction Equation in

T k T 2
T T 
2 2
  2 2  2 2  2 2 
 .c p  x  y  z 
29
30
A furnace is installed in the synthesis laboratory of chemical engineering department for the
purpose of high temperature oxidation. It is required to keep the inner temperature of the
furnace at 300 ºC, with the air inside having heat transfer coefficient of 5 W/m2·ºC. The
material of the furnace wall have thermal conductivity of 0.8 W/m·ºC. The temperature of the
outer air is 25 ºC with heat transfer coefficient of 7 W/m2·ºC. Consider a steady state heat
transfer in one dimension (x-axis direction) through one of furnace’s wall having thickness of
0.3 m, height of 0.8 m, and width of 0.5m then
(a) Sketch the problem with clear depiction of heat transfer modes involved in it
(b) Calculate the steady rate of heat transfer from the inner of furnace to outside
(c) If another wall of insulating material is placed over the furnace’s considered wall with
similar dimensions (height, width and thickness) and k = 0.1 W/m·ºC and thermal contact
resistance at interface to be 0.05 ºC/W, then what will be the rate of steady state heat
transfer? Will it be beneficial or not? Justify your logic.
31
32
Transient Heat Transfer

Lump system analysis

33
 Transient Heat Transfer: Lump system
• Consider a body of arbitrary shape initially at a uniform temperature Ti
• At time t = 0, the body is placed into a medium at temperature T∞, and
heat transfer takes place between the body and its environment, with a
heat transfer coefficient h

34
 Transient Heat Transfer: Lump system
• This heat transfer is due to convection heat transfer at the solid–liquid
interface.

• The essence of the lumped capacitance method is the assumption that the
temperature of the solid is spatially uniform at any instant during the
transient process. This assumption implies that temperature gradients
within the solid are negligible.

• This approximation implies that resistance to conduction is very small, a

compared to the resistance to heat transfer between the solid and its
surrounding.

• Applying energy balance:

35
Transient Heat Transfer: Lump system

36
 Transient Heat Transfer: Lump system

Where,

37
 Transient Heat Transfer: Lump system

Energy balance
• h As (T∞-Ti) dt = m Cp dT
• d (Ti- T∞) / (Ti- T∞) = -(hAs/ρVCp ) dt
• Integrating from t =0, where T=Ti and t=t where T=T
• ln[(T(t)- T∞) / (Ti- T∞)] = -(hAs/ρVCp ) dt
• [T(t)- T∞) / (Ti- T∞)] = e-bt
• b = hAs/ρVCp = 1/b = τ =Time constant

38
Transient Heat Transfer: Lump system

39
Transient Heat Transfer: Lump system

40
 Transient Heat Transfer

• One surface is maintained at a temperature Ts,1 and the other surface is

exposed to a fluid of temperature T∞ < Ts,1. The temperature of this surface
will be some intermediate value, Ts,2, for which T∞ < Ts,2 < Ts,1 , hence
under steady-state conditions the surface energy balance

41
Transient Heat Transfer

42
Transient Heat Transfer

43
Unsteady state heat transfer

44
 Unsteady state heat transfer
• The formulation of the problems for the determination of the one
dimensional transient temperature distribution T(x, t) in a wall
results in a partial differential equation, which can be solved using
advanced mathematical techniques.

• The solution, however, normally involves infinite series, which are

inconvenient and time-consuming to evaluate. Therefore, there is
clear motivation to present the solution in tabular or graphical
form.

• However, the solution involves the parameters x, L, t, k, ɑ, h, and

T, which are too many to make any graphical presentation of the
results practical. In order to reduce the number of parameters, we
nondimensionalize the problem by defining the dimensionless
quantities.
45
Unsteady state heat transfer

46
 Thank You
for
Your Attention

email: abdulrazzaq@ciitlahore.edu.pk

47