Conduction &
Convection
Quiz 9 – 2014.01.27
A flat furnace wall is constructed with a 4.5-inch layer of refractory
brick (k = 0.080 Btu/ft·h·F) backed by a 9-inch layer of common
brick (k = 0.800 Btu/ft·h·F) and a 2-inch layer of silica foam (k =
0.032 Btu/ft·h·F). The temperature of the inner face of the wall is
1200F, and that of the outer face is 170F.
a. What is the temperature of the interface between the
refractory brick and the common brick?
b. What would be the temperature of the outer face if the silica
foam is placed between the two brick layers?
TIME IS UP!!!
Outline
2. Conduction Heat Transfer
2.1. Series/Parallel Resistances
2.2. Geometric Considerations
3. Convection Heat Transfer
3.1. Heat Transfer Coefficient
3.2. Dimensionless Groups for HTC
Estimation
Geometric Considerations
Heat Conduction Through Concentric Cylinders
Geometric Considerations
Heat Conduction Through Concentric Cylinders
Geometric Considerations
Heat Conduction Through Concentric Cylinders
Geometric Considerations
Heat Conduction Through Hollow Spheres
𝑄
𝑑𝑇
= −𝑘
𝐴
𝑑𝑟
𝐴 = 4𝜋𝑟 2
𝑄
𝑑𝑟 = −𝑘
2
4𝜋𝑟
Integrating
both sides:
1 1
4𝜋
−
𝑄=−
(𝑇2 − 𝑇1 )
𝑟1 𝑟2
𝑘
𝑑𝑇
Geometric Considerations
Heat Conduction Through Hollow Spheres
𝑄
𝑑𝑇
= −𝑘
𝐴
𝑑𝑟
𝐴 = 4𝜋𝑟 2
𝑄
𝑑𝑟 = −𝑘
2
4𝜋𝑟
Rearranging:
4𝜋𝑟1 𝑟2 𝑇2 − 𝑇1
𝑄=−
𝑘
𝑟2 − 𝑟1
𝑑𝑇
Geometric Considerations
Heat Conduction Through Hollow Spheres
Define a geometric mean area:
𝐴𝐺𝑀 = 4𝜋𝑟1 𝑟2
4𝜋𝑟1 𝑟2 𝑇2 − 𝑇1
𝑄=−
𝑘
𝑟2 − 𝑟1
…and a geometric mean radius:
𝑟𝐺𝑀 = 𝑟1 𝑟2
2
𝐴𝐺𝑀 = 4𝜋𝑟𝐺𝑀
𝑄 = −𝑘𝐴𝐺𝑀
*Final form
𝑇2 − 𝑇1
𝑟2 − 𝑟1
Shell Balance
Plane Wall/Slab
Shell Balance
Plane Wall/Slab
Shell Balance
Plane Wall/Slab
Shell Balance
Plane Wall/Slab
Shell Balance
Cylinder
Shell Balance
Sphere
Heat Transfer Coefficient
Convection Heat Transfer
Where:
Q = heat flow rate
A = heat transfer area
h = heat transfer coefficient
Tw = temperature at solid wall
Tf = temperature at bulk fluid
Useful
Conversion:
𝑄 = ℎ𝐴 𝑇𝑤 − 𝑇𝑓
𝐵𝑡𝑢
𝑊
1
= 5.6783 2
2
ℎ𝑟 𝑓𝑡 °𝐹
𝑚 𝐾
Heat Transfer Coefficient
Convection Heat Transfer
𝑄 = ℎ𝐴 𝑇𝑤 − 𝑇𝑓
Where:
Q = heat flow rate
Driving force
A = heat transfer area
h = heat transfer coefficient
𝑇𝑤 − 𝑇𝑓
𝑄=
Tw = temperature at solid wall
1
Tf = temperature at bulk fluid
ℎ𝐴
Thermal
Resistance
Heat Transfer Coefficient
Dimensionless Groups
Thermal conductivities are easy to determine by
calorimetric experiments, but heat transfer coefficients
require the analysis of transfer mechanisms.
Mechanism Ratio Analysis
1. Heat, mass, and momentum transport are
described by differential equations of change.
e.g.

Dv
Dt
  p   g    v
2
NavierStokes Eq’n
Dimensionless Groups
Thermal conductivities are easy to determine by
calorimetric experiments, but heat transfer coefficients
require the analysis of transfer mechanisms.
Mechanism Ratio Analysis
2. However, these equations are complex and most
of the time difficult to solve/integrate.

Dv
Dt
  p   g    v
2
𝒗 = 𝒗 𝑥, 𝑦, 𝑧, 𝑡 ? ?
Dimensionless Groups
Thermal conductivities are easy to determine by
calorimetric experiments, but heat transfer coefficients
require the analysis of transfer mechanisms.
Mechanism Ratio Analysis
3. But still, valuable information is described in
these equations, relating the different forces.
Pressure Forces
Inertial

Forces
Dv
Dt
  p   g    v
2
Viscous
Forces
Dimensionless Groups
Thermal conductivities are easy to determine by
calorimetric experiments, but heat transfer coefficients
require the analysis of transfer mechanisms.
Mechanism Ratio Analysis
4. In the Mechanism Ratio Analysis, solving the
equations of change is replaced by empiricism.
Inertial
Pressure
Viscous
=
+
Forces
Forces
Forces
Dimensionless Groups
Thermal conductivities are easy to determine by
calorimetric experiments, but heat transfer coefficients
require the analysis of transfer mechanisms.
Mechanism Ratio Analysis
5. This is done by just taking
the ratio of the mechanisms
and making them into
dimensionless groups.
Pressure
Forces
Inertial
Forces
+
Viscous
Forces
Inertial
Forces
Dimensionless Groups
Reynolds Number, Re – the ratio
of inertial to viscous forces.
𝐷𝑣𝜌 𝑖𝑛𝑒𝑟𝑡𝑖𝑎𝑙
𝑅𝑒 =
=
𝜇
𝑣𝑖𝑠𝑐𝑜𝑢𝑠
Euler Number, Eu – the ratio of
pressure to inertial forces.
𝑃
𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒
𝐸𝑢 = 2 =
𝜌𝑣
𝑖𝑛𝑒𝑟𝑡𝑖𝑎𝑙
Pressure
Forces
Inertial
Forces
+
Viscous
Forces
Inertial
Forces
Dimensionless Groups
Reynolds Number, Re – the ratio
of inertial to viscous forces.
𝐷𝑣𝜌
𝑅𝑒 =
𝜇
Euler Number, Eu – the ratio of
pressure to inertial forces.
𝑃
𝐸𝑢 = 2
𝜌𝑣
“If the phenomenon is so complex that negligible
knowledge can be gained from the investigation of
the differential equations, then empirical processes
are available for evolving dimensionless groupings
of the involved variables.” – Foust, 1980
Dimensionless Groups
Useful dimensionless groups for Heat Transfer:
Dim. Group
Prandtl, Pr
Nusselt, Nu
Ratio
Equation
molecular diffusivity of momentum /
molecular diffusivity of heat
𝑐𝑃 𝜇
𝑘
heat convection / heat conduction
cP = specific heat (J/kg-K)
μ = viscosity (Pa-s)
D = characteristic length (diameter) (m)
k = thermal conductivity (W/m-K)
h = heat transfer coefficient (W/m2-K)
ℎ𝐷
𝑘
Dimensionless Groups
Correlations for Heat Transfer Coefficients:
Dittus-Boelter Equation
(for forced convection/
turbulent, horizontal tubes)
Sieder-Tate Equation
𝑁𝑢 = 0.023𝑅𝑒 0.8 𝑃𝑟 𝑛
n = 0.4 when fluid is heated
n = 0.3 when fluid is cooled
𝑁𝑢 = 0.023𝑅𝑒 0.8 𝑃𝑟1/3 𝜙𝑣
(for forced convection/ turbulent,
Re > 10000 & 0.5 < Pr < 100)
𝜇
𝜙𝑣 =
𝜇𝑤
0.14
Dimensionless Groups
Exercise!
An organic liquid enters a 0.834-in. ID horizontal steel
tube, 3.5 ft long, at a rate of 5000 lb/hr. You are given
that the specific heat, thermal conductivity, and
viscosity of the liquid is 0.565 Btu/lb-°F, 0.0647 Btu/hrft-°F, and 0.59 lb/ft-hr, respectively. All these properties
are assumed constant. If the liquid is being cooled,
determine the inside-tube heat transfer coefficient
using the Dittus-Boelter Equation.