Transcript File

Unit 42: Heat Transfer and
Combustion
Lesson 3: The Thermal Boundary
Layer
Aim
• LO2: Understanding Heat Transfer
Mechanisms and Coefficients
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The Thermal Boundary
Layer
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• In hydrodynamics, the boundary layer is
defined as that region of the flow where
viscous forces are felt.
• A thermal boundary layer may be defined as
that region where temperature gradients are
present in the flow.
• These thermal gradients would result from
heat-exchange process between the fluid and
the wall
Temperature Profile in the
Thermal Boundary Layer
y
x
T∞
δt
Tw
qw = -k ∂T
∂y
w
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The Thermal Boundary
Layer
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• The temperature of the wall is Tw, the
temperature of the fluid outside the thermal
boundary is T∞ and the thickness of the
thermal boundary layer is designated as δt.
• At the wall the velocity is zero and the heat
transfer takes place by conduction.
• Thus the heat flux per unit area q” is…
The Thermal Boundary
Layer
q = q” = - k ∂T
A
∂y
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w
From newton’s law of cooling,
q” = h(Tw – T∞)
Where h is the convection heat-transfer
coefficient. Thus combining the equations
gives…
The Thermal Boundary
Layer
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h = - k(∂T/∂y)wall
(Tw – T∞)
• So we need only find the temperature
gradient at the wall in order to evaluate the
heat-transfer coefficient.
• This means that we need to obtain an
expression for the temperature distribution
The Thermal Boundary
Layer
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• Thus consider the following 4-conditions…
T = Tw
at y = 0
∂T/∂y = 0
at y = δt
T = T∞
at y = δt
∂2T = 0
at y = 0, since the velocities at the wall is zero
∂y2
Temperature Profile in the
Thermal Boundary Layer
y
NDGTA
A
A
x
T∞
u∞
H
δ
u
δt
dx
Tw
1
2
dqw = -k ∂T
∂y
w
The Thermal Boundary
Layer
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• The four conditions may be fitted into a cubic
polynomial so that…
θ = T – Tw = 3 y – 1 y 3
θ∞ T∞ - Tw 2 δt 2 δt
• We now need to determine δt, the thermalboundary-layer-thickness.
• Consider the volume bounded by AA12
• Thus the heat given up to the fluid over the
length dx is dqw
The Thermal Boundary
Layer
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• Thus…
Energy convected in + viscous work within element + heat transfer at
wall = energy convected out.
Energy convected in through plane A1 is…
H
ρcp uTdy
0
Energy convected out of plane A2 is…
H
ρcp
uTdy
0
+ d ρcp uTdy dx
dx
The Thermal Boundary
Layer
The mass flow through the plane AA is…
H
d
dx
ρudy dx
0
And this carries with it an Energy equal to…
H
cpT∞ d
dx
ρudy dx
0
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The Thermal Boundary
Layer
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The net viscous work done within this element is…
μ
H
du 2dy dx
dy
0
And the heat transfer at the wall is…
dqw = -k dx ∂T
∂y
w
Combining these energy equations gives the integral energy equation
of the boundary layer for constant properties and constant free-stream
temperature T∞.
Handout 1 – p233 & 234 Heat Transfer 10th Edition Holtzman
The Prandtl Number
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• Note Pr = ν/α is called the Prandtl number after
Ludwig Prandtl – he introduced the concept of
boundary-layer theory
• Thus when the plate is heated over its entire length,
x0 = 0 and…
δt = ζ = 1 Pr-1/3
δ
1.026
The Prandtl Number
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• In the handout 1 given the analysis was based
fundamentally on ζ < 1. This assumption is satisfactory
for fluids having Prandtl numbers greater than 0.7
• Fortunately most gases and fluids fall within this
category.
• Liquid metals are a notable exception since they have
Prandtl numbers of the order 0.01
• The Prandtl number ν/α relates the relative thickness of
the hydrodynamic and thermal boundary layers.
The Prandtl Number
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• The kinematic viscosity of a fluid conveys information
about the rate at which momentum may diffuse
through the fluid because of molecular motion.
• The thermal diffusivity tells us the same thing in
regard to the diffusion of heat in the fluid
• Thus the ratio of these two quantities should express
the relative magnitudes of diffusion of momentum
and heat in a fluid.
The Prandtl Number
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• But these diffusion rates are precisely the quantities
that determine how thick the boundary layers will be
for a given external flow field; large diffusivities
mean that the viscous or temperature influence is
felt further out in the flow field.
• The Prandtl number is thus the connecting link
between the velocity field and the temperature field.
Pr = ν = μ/ρ = cpμ
α k/Ρcp
k
The Prandtl Number
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• In SI units typically μ is in kg/s/m; cp in kJ/kg/oC and k
in kW/m/oC.
h = -k(∂T/∂y)w = 3 k = 3 k
Tw – T∞
2 δt 2 ζδ
Substituting for the hydrodynamic-layer boundary
thickness gives…
The Nusselt Number
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hx = 0.332kPr1/3 (u∞/νx)1/2 (1 – (x0/x)3/4)-1/3
This equation may be non-dimensionalised by
multiplying both sides by x/k producing the
dimensionless group on the LHS…
Nux = hxx/k
Which is called the Nusselt number after William
Nusselt
The Nusselt Number
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Hence
Nux = 0.332Pr1/3 Rex1/2 (1 – (x0/x)3/4)-1/3
Or for a heated plate over its entire length, xo = 0
thus…
Nux = 0.332Pr1/3 Rex1/2
Constant Heat Flux
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• Most of the preceding analysis has considered the
laminar heat transfer from an isothermal surface.
• In many practical problems the surface heat is
essentially constant and the objective is to find the
distribution of the plate surface temperature for
given fluid-flow conditions.
• For constant heat flux it can be shown that the
Nusselt number is given by…
Nux = hx = 0.453Pr1/2 Rex1/3
Constant Heat Flux
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• This may be expressed in terms of the wall heat flux
and temperature difference…
Nux = qwx /[k(Tw – T∞)]
The average temperature difference along the plate for
constant heat flux condition may be obtained by
performing integration. The result being…
qw = (3/2)hx=L(Tw – T∞)