Page 1
Resistance Concept in Heat Transfer
Thermal Resistance Concept
The Fourier equation, for steady conduction through a constant area plane wall, can
be written:
n ' i a i j — T -,
Q c o ifd = — kA — = k.4---------
dx L
This can be re-arranged as:
O'cm = 7- ~ T ] (W)
R is o H
r, , = ± r c m
• Rwan is the thermal resistance of the wall against heat conduction or simply
the conduction resistance of the wall.
• The heat transfer across the fluid/solid interface is based on Newton’s law of
cooling:
Q'=hA(Ts - T j [W)
- ^ 7 CC/W)
hA •
• Rconv is the thermal resistance of the surface against heat convection or
simply the convection resistance of the surface. Thermal radiation between
a surface of area A at Ts and the surroundings at T°° can be expressed as:
Page 2
Resistance Concept in Heat Transfer
Thermal Resistance Concept
The Fourier equation, for steady conduction through a constant area plane wall, can
be written:
n ' i a i j — T -,
Q c o ifd = — kA — = k.4---------
dx L
This can be re-arranged as:
O'cm = 7- ~ T ] (W)
R is o H
r, , = ± r c m
• Rwan is the thermal resistance of the wall against heat conduction or simply
the conduction resistance of the wall.
• The heat transfer across the fluid/solid interface is based on Newton’s law of
cooling:
Q'=hA(Ts - T j [W)
- ^ 7 CC/W)
hA •
• Rconv is the thermal resistance of the surface against heat convection or
simply the convection resistance of the surface. Thermal radiation between
a surface of area A at Ts and the surroundings at T°° can be expressed as:
Q 'r a d = e a i{T ;-i:)= hra d A(Ts -T J =
T,~TX
m
1
ft™ -. - 4
J L _
where a = 5.67x10-8 [W/m2K4] is the Stefan-Boltzman constant.
Also 0 < e <1 is the emissivity of the surface. Note that both the temperatures must
be in Kelvin
Thermal Resistance Network
- \ V — # — # —
Consider steady, one-dimensional heat flow through two plane walls in series which
are exposed to convection on both sides, Under steady state condition
rate of heat = rate of heat = rate of heat = rate of heat
convection conduction conduction through convection from the
into the wall through wall 1 wall 2 wall
Q ? = M fc u - Ti)=M = M = M fa - Tmi)
L\ L2
o- = - t - r . _ j; - r, _ r a- r ^
\Ih\A L!kx A L/k2A 1 !h2A
C. _ T ^ ~ T ^ t-.-t, t, - t; r . - r , .
O' =
T -T
Jn.,l 1a s ,2
• Note that A is constant area for a plane wall. Also note that the thermal
resistances are in series and equivalent resistance is determined by simply
adding thermal resistances.
• The rate of heat transfer between two surfaces is equal to the temperature
difference divided by the total thermal resistance between two surfaces. It
can be written: AT = Q' R
Thermal Resistances in Parallel
Page 3
Resistance Concept in Heat Transfer
Thermal Resistance Concept
The Fourier equation, for steady conduction through a constant area plane wall, can
be written:
n ' i a i j — T -,
Q c o ifd = — kA — = k.4---------
dx L
This can be re-arranged as:
O'cm = 7- ~ T ] (W)
R is o H
r, , = ± r c m
• Rwan is the thermal resistance of the wall against heat conduction or simply
the conduction resistance of the wall.
• The heat transfer across the fluid/solid interface is based on Newton’s law of
cooling:
Q'=hA(Ts - T j [W)
- ^ 7 CC/W)
hA •
• Rconv is the thermal resistance of the surface against heat convection or
simply the convection resistance of the surface. Thermal radiation between
a surface of area A at Ts and the surroundings at T°° can be expressed as:
Q 'r a d = e a i{T ;-i:)= hra d A(Ts -T J =
T,~TX
m
1
ft™ -. - 4
J L _
where a = 5.67x10-8 [W/m2K4] is the Stefan-Boltzman constant.
Also 0 < e <1 is the emissivity of the surface. Note that both the temperatures must
be in Kelvin
Thermal Resistance Network
- \ V — # — # —
Consider steady, one-dimensional heat flow through two plane walls in series which
are exposed to convection on both sides, Under steady state condition
rate of heat = rate of heat = rate of heat = rate of heat
convection conduction conduction through convection from the
into the wall through wall 1 wall 2 wall
Q ? = M fc u - Ti)=M = M = M fa - Tmi)
L\ L2
o- = - t - r . _ j; - r, _ r a- r ^
\Ih\A L!kx A L/k2A 1 !h2A
C. _ T ^ ~ T ^ t-.-t, t, - t; r . - r , .
O' =
T -T
Jn.,l 1a s ,2
• Note that A is constant area for a plane wall. Also note that the thermal
resistances are in series and equivalent resistance is determined by simply
adding thermal resistances.
• The rate of heat transfer between two surfaces is equal to the temperature
difference divided by the total thermal resistance between two surfaces. It
can be written: AT = Q' R
Thermal Resistances in Parallel
• The thermal resistance concept can be used to solve steady state heat
transfer problem In parallel layers or combined series-parallel arrangements.
Ri Qi
O' = Qi + Qi
o 'J - ^
Ti-T, [ T - T :
i?i R -.
1 '1 i 1
__
I T - A
r A
\ ^
Thermal Resistance of Hollow Cylinders
Hollow cylinder
• ri = Internal radius of cylinder
• r2 = Outer radius of cylinder
• L = Length of cylinder
Thermal resistance of hollow cylinders is given as
R =
___ X
2-k.L
Heat transfer
_ r
~ R
Thermal Resistance of a Hollow Sphere
Page 4
Resistance Concept in Heat Transfer
Thermal Resistance Concept
The Fourier equation, for steady conduction through a constant area plane wall, can
be written:
n ' i a i j — T -,
Q c o ifd = — kA — = k.4---------
dx L
This can be re-arranged as:
O'cm = 7- ~ T ] (W)
R is o H
r, , = ± r c m
• Rwan is the thermal resistance of the wall against heat conduction or simply
the conduction resistance of the wall.
• The heat transfer across the fluid/solid interface is based on Newton’s law of
cooling:
Q'=hA(Ts - T j [W)
- ^ 7 CC/W)
hA •
• Rconv is the thermal resistance of the surface against heat convection or
simply the convection resistance of the surface. Thermal radiation between
a surface of area A at Ts and the surroundings at T°° can be expressed as:
Q 'r a d = e a i{T ;-i:)= hra d A(Ts -T J =
T,~TX
m
1
ft™ -. - 4
J L _
where a = 5.67x10-8 [W/m2K4] is the Stefan-Boltzman constant.
Also 0 < e <1 is the emissivity of the surface. Note that both the temperatures must
be in Kelvin
Thermal Resistance Network
- \ V — # — # —
Consider steady, one-dimensional heat flow through two plane walls in series which
are exposed to convection on both sides, Under steady state condition
rate of heat = rate of heat = rate of heat = rate of heat
convection conduction conduction through convection from the
into the wall through wall 1 wall 2 wall
Q ? = M fc u - Ti)=M = M = M fa - Tmi)
L\ L2
o- = - t - r . _ j; - r, _ r a- r ^
\Ih\A L!kx A L/k2A 1 !h2A
C. _ T ^ ~ T ^ t-.-t, t, - t; r . - r , .
O' =
T -T
Jn.,l 1a s ,2
• Note that A is constant area for a plane wall. Also note that the thermal
resistances are in series and equivalent resistance is determined by simply
adding thermal resistances.
• The rate of heat transfer between two surfaces is equal to the temperature
difference divided by the total thermal resistance between two surfaces. It
can be written: AT = Q' R
Thermal Resistances in Parallel
• The thermal resistance concept can be used to solve steady state heat
transfer problem In parallel layers or combined series-parallel arrangements.
Ri Qi
O' = Qi + Qi
o 'J - ^
Ti-T, [ T - T :
i?i R -.
1 '1 i 1
__
I T - A
r A
\ ^
Thermal Resistance of Hollow Cylinders
Hollow cylinder
• ri = Internal radius of cylinder
• r2 = Outer radius of cylinder
• L = Length of cylinder
Thermal resistance of hollow cylinders is given as
R =
___ X
2-k.L
Heat transfer
_ r
~ R
Thermal Resistance of a Hollow Sphere
A hollow sphere
• Consider a hollow cylinder of internal radius r-\ and external radius r2 with
respective internal and external temperatures of Ti and TO as shown in figure
R = r'- ~ ' :
4 -k ^ r,
where, k = Thermal conductivity
R
Heat Transfer through a Composite Cylinder
Assume heat transfer coefficient for liquid and inner surface of cylinder is hi and
for gas and outer surface of cylinder is hO then heat transfer will be as: Convection
-> conduction — ? conduction — ? convection
Heat transfer through a composite cylinder
O =
r-r.
Riq = R, + R: + R. + Ra
Rl = - ^ . R,= ! ! ^
1 h.2^1 * 2 -1 0 ,1
R - .=
In r, / r,
. * * =
hs2Trr:i
Heat Transfer through a Composite Sphere:
All the arrangements are same as given in composite cylinder only.
Page 5
Resistance Concept in Heat Transfer
Thermal Resistance Concept
The Fourier equation, for steady conduction through a constant area plane wall, can
be written:
n ' i a i j — T -,
Q c o ifd = — kA — = k.4---------
dx L
This can be re-arranged as:
O'cm = 7- ~ T ] (W)
R is o H
r, , = ± r c m
• Rwan is the thermal resistance of the wall against heat conduction or simply
the conduction resistance of the wall.
• The heat transfer across the fluid/solid interface is based on Newton’s law of
cooling:
Q'=hA(Ts - T j [W)
- ^ 7 CC/W)
hA •
• Rconv is the thermal resistance of the surface against heat convection or
simply the convection resistance of the surface. Thermal radiation between
a surface of area A at Ts and the surroundings at T°° can be expressed as:
Q 'r a d = e a i{T ;-i:)= hra d A(Ts -T J =
T,~TX
m
1
ft™ -. - 4
J L _
where a = 5.67x10-8 [W/m2K4] is the Stefan-Boltzman constant.
Also 0 < e <1 is the emissivity of the surface. Note that both the temperatures must
be in Kelvin
Thermal Resistance Network
- \ V — # — # —
Consider steady, one-dimensional heat flow through two plane walls in series which
are exposed to convection on both sides, Under steady state condition
rate of heat = rate of heat = rate of heat = rate of heat
convection conduction conduction through convection from the
into the wall through wall 1 wall 2 wall
Q ? = M fc u - Ti)=M = M = M fa - Tmi)
L\ L2
o- = - t - r . _ j; - r, _ r a- r ^
\Ih\A L!kx A L/k2A 1 !h2A
C. _ T ^ ~ T ^ t-.-t, t, - t; r . - r , .
O' =
T -T
Jn.,l 1a s ,2
• Note that A is constant area for a plane wall. Also note that the thermal
resistances are in series and equivalent resistance is determined by simply
adding thermal resistances.
• The rate of heat transfer between two surfaces is equal to the temperature
difference divided by the total thermal resistance between two surfaces. It
can be written: AT = Q' R
Thermal Resistances in Parallel
• The thermal resistance concept can be used to solve steady state heat
transfer problem In parallel layers or combined series-parallel arrangements.
Ri Qi
O' = Qi + Qi
o 'J - ^
Ti-T, [ T - T :
i?i R -.
1 '1 i 1
__
I T - A
r A
\ ^
Thermal Resistance of Hollow Cylinders
Hollow cylinder
• ri = Internal radius of cylinder
• r2 = Outer radius of cylinder
• L = Length of cylinder
Thermal resistance of hollow cylinders is given as
R =
___ X
2-k.L
Heat transfer
_ r
~ R
Thermal Resistance of a Hollow Sphere
A hollow sphere
• Consider a hollow cylinder of internal radius r-\ and external radius r2 with
respective internal and external temperatures of Ti and TO as shown in figure
R = r'- ~ ' :
4 -k ^ r,
where, k = Thermal conductivity
R
Heat Transfer through a Composite Cylinder
Assume heat transfer coefficient for liquid and inner surface of cylinder is hi and
for gas and outer surface of cylinder is hO then heat transfer will be as: Convection
-> conduction — ? conduction — ? convection
Heat transfer through a composite cylinder
O =
r-r.
Riq = R, + R: + R. + Ra
Rl = - ^ . R,= ! ! ^
1 h.2^1 * 2 -1 0 ,1
R - .=
In r, / r,
. * * =
hs2Trr:i
Heat Transfer through a Composite Sphere:
All the arrangements are same as given in composite cylinder only.
* 1 =
h f A , h A n r 2
J * J *
£ — r : ~ r i £ —
A - k r ^ r , ’ 3
R . =
M - r ;
r,~ r.
r,-rc
A - / ? , + R . + R ,
Critical Radius of Insulation
• To insulate a plane wall, the thicker the insulator, the lower the heat transfer
rate (since the area is constant).
• However, for cylindrical pipes or spherical shells, adding insulation results in
increasing the surface area which in turns results in increasing the
convection heat transfer.
• As a result of these two competing trends, the heat transfer may increase or
decrease
r . - r , _ J j-J ;
K r- 'r l . _ 1 _
liik L [2wjL)h
• The variation of Q‘ with the outer radius of the insulation reaches a maximum
that can be determined from dQ‘ / dr2 = 0. The value of the critical radius for
the cylindrical pipes and spherical shells are:
£
c y lin d e r ~ ^
Note
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