Conduction equation cartesian, Cylindrical, spherical (7).pptx

Heat transfer - Conduction
Dr. Raja Kathiravan

Heat transfer- review
• Heat transfer (or heat) is thermal energy in transit due to a spatial
temperature difference.

Conduction: Diffusion of heat due to random molecular motion.

Conduction:
The heat flux (W/m2), q’’ is the heat transfer rate in the x direction per unit area
perpendicular to the direction of transfer, and it is proportional to the temperature gradient, dT/dx, in
this direction. The parameter k is a transport property known as the thermal conductivity (W/m K) and is
a characteristic of the wall material. The minus sign is a consequence of the fact that heat is transferred
in the direction of decreasing temperature.
Under the steady-state conditions
and the heat flux
or

Problem.1 The wall of an industrial furnace is constructed from 0.15-m-thick fireclay brick having a
thermal conductivity of 1.7 W/m K. Measurements made during steadystate operation reveal
temperatures of 1400 and 1150 K at the inner and outer surfaces, respectively. What is the rate of heat
loss through a wall that is 0.5 m by 1.2 m on a side? (Ans: 1700W).
Heat Diffusion Equation:
The conduction heat rates
perpendicular to each of
the control surfaces at the
x, y, and z coordinate
locations are indicated by
the terms qx, qy, and qz,
respectively. The conduction
heat rates at the opposite
surfaces can then be expressed
as a Taylor series expansion
where, neglecting higher order terms,

Within the medium there may also be an energy source term associated with the rate of thermal energy
generation. This term is represented as
and the energy storage term may be expressed as
the general form of the conservation of energy requirement is

substitute all inthe conservation of energy equation
and therefore
The conduction heat rates may be evaluated from Fourier’s law
is the general form, in Cartesian coordinates, of the heat diffusion equation. This equation, often referred to as
the heat equation

if the thermal conductivity is constant, the heat equation is
there can be no change in the amount of energy storage
Cylidrical Co ordinates

dt
dr
r
T
dz
rd
k
qr .
)
(



 
 
r
r q
r
q



)
(
)
( r
r
r
r q
r
q
r
q
q














heat is transfered in the cylindrical coordinates are radial, angular and axial directions
apply the fourier law of conduction for this coordinates
heat transfer radial (dr) direction:
heat inflow for time rate ‘dt’
heat out flow is =
net heat accumulate or efflux in radial direction is = heat in - heat out =

 
dt
r
T
r
r
T
dz
rd
dr
k
or
r
T
r
T
r
dt
dz
kdrd
dt
r
T
r
r
dzdr
kd
dt
dr
r
T
dz
rd
k
r


















































1
)
.
.
(
)
1
(
.
.
.
(
2
2
2
2





dt
r
T
dz
rd
dr
k
dt
rd
r
T
drdz
k
q .
).
.
.
.(
.
)
(













Heat transfer in angular (rdθ) direction:
heat inflow , qθ,
heat out flow =
Net heat efflux in angular direction is =
heat transfer in axial direction:
heat inflow , qz,
heat out flow is =
 



q
r
q



)
(
)
( 





q
r
q
r
q
q














 
  dt
r
T
dz
rd
dr
k
dt
rd
r
T
drdz
k
r
.
.
.
.
.
.
.
2
2
2





























dt
dz
z
T
rd
dr
k
qz .
)
.
.(



 
 
z
z q
z
q




net heat efflux along axial (z) direction is =
Heat generated within the volume = qv =
net heat rate stored within the volume =
)
(
)
( z
z
z
z q
z
q
z
q
q














 
dt
z
T
dz
rd
dr
k
dt
z
T
z
dz
rd
dr
k
dt
dz
z
T
rd
dr
k
z
.
)
.
.
.(
)
.
.
.(
.
.
(
2
2




































dt
dz
rd
dr
q ).
.
.
.( 

dt
t
T
C
dz
rd
dr
dt
t
T
C
V
dt
t
T
C
m p
p
p








.
).
.
.
.(
.
.
.
.
. 



therefore the energy balance of the control volume
net Heat efflux within the volume + = rate of heat stored in the volume
Cancel the volume (dr.rdθ.dz) , on both sides and divide the thermal conductivity ‘k’ , then we get
and finally,
Where α is the thermal diffusivity =
  dt
t
T
C
dz
rd
dr
dt
dz
rd
dr
q
dt
z
T
dz
rd
dr
k
dt
r
T
dz
rd
dr
k
dt
r
T
r
r
T
dz
rd
dr
k p





































.
).
.
.
.(
).
.
.
.(
.
)
.
.
.(
.
.
.
.
.
1
)
.
.
( 2
2
2
2
2
2
2






 
t
T
k
C
k
q
z
T
r
T
r
T
r
r
T p




































 


2
2
2
2
2
2
2
1
t
T
k
q
z
T
r
T
r
T
r
r
T

















1
1
2
2
2
2
2
2
2

p
C
k


Heat transfer in spherical coordinates

net heat efflux in radial, angular and peripharal directions are
apply the energy balance
the above equation is divided by volume
take
now
     





 q
q
q
q
q
q d
d
r
dr
r 





 


         
t
T
C
d
r
rd
dr
d
r
rd
dr
q
q
q
q
q
q
q p
d
d
r
dr
r










 

 .
.
sin
.
.
.
sin
.
. 











 
     
t
T
C
q
d
r
rd
dr
q
q
d
r
rd
dr
q
q
d
r
rd
dr
q
q
p
d
d
r
dr
r














.
sin
.
.
.
sin
.
.
.
sin
.
.

















     












d
dq
d
q
q
d
dq
d
q
q
dr
dq
dr
q
q d
d
r
r
dr
r











 


;
;
t
T
C
q
d
dq
r
rd
dr
d
dq
d
r
r
dr
dr
dq
d
r
rd
p
r







 .
.
sin
.
.
1
.
.
sin
.
.
1
.
.
.
sin
.
1









































r
T
r
r
r
k
r
T
d
r
rd
k
r
d
r
rd
r
q
d
r
rd
r 2
2
.
sin
.
.
.
.
sin
.
1
.
.
sin
.
1









Similarly




































 T
r
k
r
T
d
r
dr
k
d
d
d
r
r
dr
d
dq
d
r
r
dr
sin
.
sin
.
sin
.
.
.
sin
.
.
1
.
sin
.
.
1
2
and
2
2
2
2
.
sin
sin
.
.
.
.
sin
.
.
1
.
sin
.
.
1












d
T
d
r
k
r
T
dr
rd
k
r
rd
dr
d
dq
r
rd
dr

















Now the General heat ewquation for spherical coordinate is
t
T
k
q
d
T
d
r
T
r
r
T
r
r
r
t
T
k
C
k
q
d
T
d
r
T
r
r
T
r
r
r
p


































































1
.
sin
1
sin
.
sin
1
1
.
.
sin
1
sin
.
sin
1
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2



Boundary conditions
• To determine the
in a medium, it is necessary to solve the
appropriate form of the heat equation.
• Such a solution depends on the physical
conditions existing at the boundaries of the
medium and, if the situation is time
dependent, on conditions existing in the
medium at some initial time.
• The first condition corresponds to a
situation for which the surface is
maintained at a fixed temperature Ts. It is
commonly termed a , or
a boundary condition of the first kind.
• The second condition corresponds to the
existence of a fixed or constant heat flux
qs’’ at the surface.It is termed a
is the
existence of convection heating or cooling
at the surface

Problems
2. Assume steady-state, one-dimensional heat conduction through the symmetric shape shown.
Assuming that there is no internal heat generation, derive an expression for the
thermal conductivity k(x) for these conditions: A(x)= (1 - x), T(x) = 300 (1 - 2x - x3),
and q = 6000 W, where A is in square meters, T in kelvins, and x in meters. Ans: k=
3. A one-dimensional plane wall of thickness 2L=100 mm experiences uniform thermal energy generation
of q=1000 W/m3and is convectively cooled at x= 50 mm by an ambient fluid characterized by Tα= 20oC. If
the steady-state temperature distribution within the wall is T(x) = a(L2-x2)+b where a= 10 C/m2 and b = 30
C, what is the of the wall? What is the value of the convection
?
4. In the two-dimensional body illustrated, the gradient at surface A is found
to be δT/δy= 30 K/m. What are δT/δy and δT/δx at surface B? Ans:60 K/m.
b. Consider the above problem , for the case where the thermal conductivity
varies with temperature as k= (ko+aT), where ko= 10W/m.K,
a=-10-3 W/m.K2, and T is in kelvins. The gradient at surface B is δT/δx=30 K/m.
What is δT/δy at surface A? Ans=14.85 K/m


5. At a given instant of time, the temperature distribution within an infinite homogeneous body is given
by the function
Assuming constant properties and no internal heat generation, determine the regions where the
temperature changes with time.[Ans: temp distribution is independant with time]
6.A cylindrical rod of stainless steel is insulated on its exterior surface except for the ends. The steady-
state temperature distribution is T(x)= a - bx/L, where a=305 K and b=10 K. The diameter and length of
the rod are D=20 mm and L=100 mm, respectively. Determine the heat flux along the rod, The mass of
the rod is M= 0.248 kg.[Ans=1490 W/m2]
7.Uniform internal heat generation at q=5x107W/m3 is occurring in a cylindrical nuclear reactor fuel rod
of 50 mm diameter, and under steady-state conditions the temperature distribution is of the form
T(r)=a+ br2, where T is in degrees Celsius and r is in meters, while a= 800oC and b= - 4.167 x105 oC/m2.
The fuel rod properties are k=30 W/m K, ρ=1100 kg/m3, and Cp= 800 J/kg K.
(a) What is the rate of heat transfer per unit length of the rod at r= 0 (the centerline) and at
r=25 mm (the surface)?
(b) If the reactor power level is suddenly increased to q2= 108 W/m3, what is the initial time
rate of temperature change at r= 0 and r=25 mm?

• For 1DConduction in a plane wall, temperature is a
function of the x-coordinate only and heat is transferred
exclusively in this direction.
• In Figure , a plane wall separates two fluids of different
temperatures. Heat transfer occurs by convection from
the hot fluid at Tα,1 to one surface of the wall at Ts,1, by
conduction through the wall, and by convection from the
other surface of the wall at Ts,2 to the cold fluid at Tα,2.
• We begin by considering conditions within the wall. We
first determine the temperature distribution, from which
we can then obtain the conduction heat transfer rate.
Temperature Distribution
• For steady-state conditions with no distributed source or
sink of energy within the wall, the appropriate form of the
heat equation is
Conduction- Plane wall

2
1
)
( C
x
C
x
T 

• equation may be integrated twice to obtain the general solution is
• To obtain the constants of integration, C1 and C2, boundary conditions must be introduced. We choose
to apply conditions of the first kind at x = 0 and x = L, in which case
• T(0) = Ts,1 and T(L) = Ts,2. therfore Ts,1 = C2 then the equation is re write T(x) = C1x +Ts,1, now apply
second boundary condition T(L) = Ts,2 ch gives Ts,2 = C1L+Ts,1; (Ts,2- Ts,1)/L = C2 substitute C2 value in
the comman equation we get
• From this result it is evident that, for one-dimensional, steady-state conduction in a plane wall with no
heat generation and constant thermal conductivity, the temperature varies linearly with x.
• this temperature distribution may use Fourier’s law
• then the heat flux is
  1
,
1
,
2
,
)
( Ts
x
L
Ts
Ts
x
T 



• Conduction and convection resistances are in series and may be summed, and
The Composite Wall
Alternatively
interms ovf overall heat transfer co efficient
in general

• Composite walls may also be characterized by
series–parallel configurations
Contact Resistance
• normally it is neglected, but, 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

Conduction equation cartesian, Cylindrical, spherical (7).pptx

Conduction : Porous Media
• In many applications, heat transfer occurs within porous media that are combinations of a
stationary solid and a fluid.
• When the either a , the resulting to be saturated.
in an unsaturated porous medium.
• A saturated porous medium that consists of a stationary solid phase through which a fluid flows is
referred to as a packed bed.
• heat rate may be expressed where keffis an effective thermal conductivity.
• effective thermal conductivity of a saturated porous medium consisting of an interconnected solid
phase within which a dilute distribution of spherical fluid regions exists, resulting in an expression
of the form
• Equation is valid for relatively small porosities (ɛ ≤ 0.25).

• Problem1. A thin silicon chip and an 8-mm-thick aluminum substrate are separated by a 0.02-mm-
thick epoxy joint. The chip and substrate are each 10 mm on a side, and their exposed surfaces are
cooled by air, which is at a temperature of 25oC and provides a convection coefficient of 100 W/m2 K.
If the chip dissipates 104 W/m2 under normal conditions, will it operate below a maximum allowable
temperature of 85oC? (take the epoxy conductivity resistance 0.9x10-4 m2K/W and the thermal
conducity of aluminium is 273W/mK).[Ans:Tg=75.3oC, it will work]
• Problem 2. A conical section fabricated from pyroceram ( K=3.46W/mK at 500K). It is of circular cross
section with the diameter D= ax, where a= 0.25. The small end is at x1 = 50 mm and the large end at
x2=250 mm. The end temperatures are T1=400 K and T2= 600 K, while the lateral surface is well
insulated.
1. Derive an expression for the temperature distribution T(x) in symbolic form, assuming one-
dimensional conditions. Sketch the temperature distribution.
2. Calculate the heat rate qx through the cone.

• Temperature distribution without and with heat geration: (derivation please refer class notes)
• without heat generation
• with heat gereation
and in dimensionless form
• Problem 3. 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 4
kW/m2 · ◦C. Calculate the center temperature of the wire
Conduction - Cylinder
 
2
2
2
1
2
1
ln
ln
)
( T
r
r
r
r
T
T
r
T 


















  w
T
r
R
k
q
r
T 

 2
2
4
)
(

2
0
1 









R
r
T
T
T
T
w
w

• Adding insulation will always increase the conduction resistance.
• But in the case of a cylinder and sphere ,when the total resistance is made up of both conduction
resistance and convection resistance, the addition of insulation in some cases may reduce the
convection resistance due to the increase in surface area and the total resistance may actually
decrease resulting in increased heat flow.
• It may be shown that the heat flow actually first increases and then decreases in certain cases.
• The thickness upto which heat flow increases and after which heat flow decreases is termed as
critical thickness
• In the case of cylinders and spheres it is called critical radius.
• Cylinder:
• Total resistance, R, for radius r =
Cancelling the common terms
Critical thickness of insulation
rL
h
kL
r
r

 2
1
2
1
ln







t
xcons
r
h
r
r
k
R tan
1
1
ln
1
1











2
1
1
1
1
r
h
r
k
dr
dR


h
k
rcr 
Equating to zero then we get
For sphere
Problem 3: Calculate the critical radius of insulation for asbestos [k =0.17 W/m◦C] surrounding a pipe
and exposed to room air at 20◦C with h=3.0 W/m2 ◦C. Calculate the heat loss from a 200◦C, 5.0-cm-
diameter pipe when covered with the critical radius of insulation and without insulation.

Conduction equation cartesian, Cylindrical, spherical (7).pptx

More Related Content

What's hot (20)

Similar to Conduction equation cartesian, Cylindrical, spherical (7).pptx (20)

Recently uploaded (20)

Conduction equation cartesian, Cylindrical, spherical (7).pptx