Radiation and convection cools the corners of the blocks
immediately, they have very low thermal mass, and there's such low
thermal conductivity the heat from the centre doesn't diffuse out
to burn you when you touch it.
We derived the energy balance equation. \begin{align*}
\rho C_p\frac{\partial T}{\partial t} =& -\rho C_p v_j
\nabla_j T - \nabla_i q_i - \tau_{ji} \nabla_j v_i -
p \nabla_i v_i +\sigma_{energy} \end{align*}
Remember!
This form assumes there is no pressure dependence of the
internal energy.
For solids, we have $\boldsymbol{v}=0$ and it reduces to the
heat equation. \begin{align*} \rho C_p\frac{\partial
T}{\partial t} =& - \nabla_i q_i +\sigma_{energy}
\end{align*}
We have solved the heat equation for
electrical heat generation in a wire
and found the solution was identical to the solution of the
momentum equation for flow in a pipe
!
We also solved the heat equation for
heat flow in a spherical nuclear fuel pellet.
But, the complexity of heat transfer arises from the boundary
conditions (think radiation and convection).
Constant temperature BC's
are not realistic for the majority of problems, so lets look at
an integral approach…
Lets consider an infinite flat plate which is hot on one side
and cool on the other.
We know from previous work that the total heat flux
through
the plate is given by \begin{align*} Q=U A \Delta T \end{align*}
where $U = k/X$ for conduction problems.
But
why?
How did we get this equation?
In the previous lectures, we have found that we could derive the
standard equations for fluid flow from the differential balance
equations.
For example,
Bernoulli's equation
and the
Hagen-Poiseuille equation.
So we should be able to get the heat flux expression for solids:
\begin{align*} Q=U A \Delta T \end{align*} from the differential
balance equation for heat: \begin{align*} \rho C_p\frac{\partial
T}{\partial t} =& - \nabla_i q_i +\sigma_{energy}
\end{align*}
Assuming we are at steady state, and no heat is generated in the
plate, we can cancel two terms. \begin{align*}
\cancelto{0}{\rho C_p\frac{\partial T}{\partial t}} =& -
\nabla_i q_i +\cancelto{0}{\sigma_{energy}} \end{align*}
We select a rectangular coordinate system and expand the
dot-product/sum. As the system is symmetric (infinite plate) we
know there are no fluxes in the $y$ or $z$ direction which
yields \begin{align*} \nabla_x q_x
+\nabla_y \cancelto{0}{q_y}+\nabla_z \cancelto{0}{q_z}&=0\\
\nabla_x q_x &=0 \end{align*}
\begin{align*} \nabla_x q_x &=0 \end{align*}
This is exactly like the continuity equation in all of the flow
problems we've studied, it's a statement that what comes in,
must go out.
We know that $\nabla_x=\frac{\partial }{\partial x}$, so we have
$\frac{\partial q_x}{\partial x}=0$ and therefore \begin{align*}
q_x &= \text{Constant} \end{align*}
Let's take Fourier's law $q_x = - k\,{\rm d} T/{\rm d} x$
We know that at $x=0$ the temperature is $T=T_{in}$.
at $x=X$ the temperature is $T=T_{out}$.
Remembering that $q_x$ is constant, we can perform a integral
using these limits. \begin{align*} \int_0^X \frac{q_x}{k} {\rm
d}x &= -\int_{T_{in}}^{T_{out}} 1 {\rm d}T\\ \frac{q_x}{k} X
&= T_{in}-T_{out} \end{align*}
So we have \begin{align*} q_x &= \frac{k}{X}
\left(T_{in}-T_{out}\right) \end{align*}
But we wanted to derive the expression \begin{align*}
Q=U A \Delta T \end{align*} where $U=k/X$.
There is a difference between $q_x$, which is the flux per unit
area, and $Q$ which is the total flux.
\begin{align*} q_x &= \frac{k}{X} \left(T_{in}-T_{out}\right)
& Q&=U A \Delta T \end{align*}
Just like the volumetric flow rate $\dot{V}$ is the integral of
the velocity $v$ over the flow area… \begin{align*}
\dot{V}_x=\int v_x {\rm d}y {\rm d}z \end{align*}
… So the total heat flux $Q$ is equal to the integral of the
heat flux $q$ over the area \begin{align*} Q_x=\iint q_x {\rm
d}y {\rm d}z \end{align*}
\begin{align*} q_x &= \frac{k}{X} \left(T_{in}-T_{out}\right)
& Q&=U A \Delta T \end{align*}
So we have \begin{align*} Q_x=\iint q_x {\rm d}y {\rm d}z
\end{align*}
As $q_x$ is constant, the integral just becomes equal to the
area \begin{align*} Q_x= q_x A &= \frac{k}{X} A
\left(T_{in}-T_{out}\right)\\ &= U A \Delta T \end{align*}
Again, these balance equations are shown to be the general
equations behind all of these transport equations.
The equations for heat flux in solids are usually quite simple,
so we will often use the simpler, integrated equations for the
heat flux. E.g. \begin{align*} Q=U A \Delta T \end{align*}
What about for a pipe?
Using the same arguments, the heat equation becomes
\begin{align*} \nabla_r q_r = \frac{1}{r} \frac{\partial
r q_r}{\partial r} = 0 \end{align*}
Integrating this equation, we have $q_r = C/r$.
Our heat flux $q_r$ is not constant any more, but we know from a
heat balance that $Q_{x,in}=Q_{x,out}$. What's different in this
case?
The difference is due to the changing cross sectional area of
the pipe (the diameter is a function of $r$ ).
We can solve for the constant using Fourier's law, we have
\begin{align*} q_r = \frac{C}{r} = -k \frac{\partial T}{\partial
r} \end{align*}
Integrating and using the limits that at $r=R_{inner}$, the
temperature is $T=T_{inner}$ and at $r=R_{outer}$ the
temperature is $T=T_{outer}$, we have \begin{align*} C
\int_{R_{inner}}^{R_{outer}}\frac{1}{r}{\rm d}r &=
-k\int_{T_{inner}}^{T_{outer}}1 {\rm d}T\\ C
\ln\left(\frac{R_{outer}}{R_{inner}}\right) &= - k
(T_{outer} - T_{inner}) \end{align*}
Our final equation for the heat flux becomes \begin{align*} q_r
= \frac{k}{r \ln\left(\frac{R_{outer}}{R_{inner}}\right)}
(T_{inner} - T_{outer}) \end{align*}
Again, the total heat flux $Q$ is given by the integral over the
cross sectional area ( $r {\rm d}\theta {\rm d}z$ ).
As the heat flux is independent of $\theta$ and $z$, the
integral becomes the surface area of a cylinder ( $2 \pi L r$ ),
and we get \begin{align*} Q_r = 2 \pi L r q_r =
\frac{2 \pi L k}{\ln\left(\frac{R_{outer}}{R_{inner}}\right)}
(T_{inner} - T_{outer}) \end{align*}
\begin{align*} Q&= U A \left(T_{in}-T_{out}\right)= \frac{k}{X}
A \left(T_{in}-T_{out}\right) =
\frac{2 \pi L k}{\ln\left(\frac{R_{outer}}{R_{inner}}\right)}
(T_{in} - T_{out}) \end{align*}
This is approach includes a “Log-Mean diameter” in the
calculation of the heat transfer resistance.
After these two examples, generating an expression for $Q$ for
spherical shells should be straightforward.
\begin{align*} Q &= U A \Delta T= \frac{k A}{X}\Delta T=
\frac{2 \pi L k}{\ln\left(\frac{R_o}{R_i}\right)}\Delta T
\end{align*}
All of these equations should be familiar to you, they are the
equations for the heat flux through a plate and a pipe.
Using these integrated forms of the energy balance, we can
rapidly solve heat transfer problems, such as the effect of
insulation on pipes.
In essence they link
heat transfer resistance, $\left(U A\right)^{-1}$, heat
flux, $Q$, and temperature differences,
$\Delta T$.
However, driving temperature differences are not known
individually over heat-transfer “layers”. For example, in a
study of heat transfer through the walls of a house, the
exterior and interior air temperature might be known, but the
temperature of each side of the wall is not known..
We need a method to combine multiple
layers
or
modes
of heat transport (e.g., convection with the wall, conduction
through the wall, etc.).
\begin{align*} Q &= U A \Delta T \\ Q_x &= \frac{k}{X}(T_i -
T_o)& Q_r &=
\frac{2 \pi L k}{\ln\left(\frac{R_o}{R_i}\right)} (T_i - T_o)
\end{align*}
It is common to talk about heat transfer in terms of resistance.
\begin{align*} R = 1 / U A \end{align*}
This is because we can combine thermal resistances like
electrical resistances in electronics. \begin{align*} R_{total}
&= R_1 + R_2 + …\\ &= 1/U_1 A_1 + 1/U_2 A_2 + …
\end{align*}
To demonstrate this, consdier heat transfer through a layered
wall.
If the above system is at steady state, the heat flux through
both layers must be equal, and we have \begin{align*} Q &=
\frac{k_1 A}{X_1}(T_2 - T_1) = \frac{k_2 A}{X_2}(T_3 - T_2)
\end{align*}
Rearranging to eliminate $T_2$, we obtain: \begin{align*} Q
&= \frac{1}{\frac{X_1}{A k_1} + \frac{X_2}{A k_2}}(T_3 -
T_1) \end{align*}
Here you can see the addition of resistances to calculate heat
flux. \begin{align*}
\left(U A\right)_{total}=1/R_{total}=1/(1/U_1 A_1 + 1/U_2 A_2) =
1 / \left(X_1/k_1 A + X_2/k_2 A\right) \end{align*}
Quick example, calculate the heat loss through an 8 cm brick
wall ($k=0.69$ W/ ${}^\circ$ C m) covered with a 2 cm plaster
board ( $k=0.48$ W/ ${}^\circ$ C m) and insulated with a 10 cm
layer of rock wool ( $k=0.04$ W/ ${}^\circ$ C m). \begin{align*}
\left(U A\right)_{total} &= 1 /
\left(1/(U A)_{brick}+1/(U A)_{insulation}
+1/(U A)_{plaster}\right)\\ &=A /
\left(X_{brick}/k_{brick}+X_{insulation}/k_{insulation}
+X_{plaster}/k_{plaster}\right)\\ &\approx 0.38\times A W /
{}^\circ C \end{align*}
This isn't a bad estimate for a filled cavity wall, which is
around $U=0.6 W /m^2 {}^\circ C$.
On a hot day in Aberdeen we have $T_o=10^\circ$ C and
$T_i=20^\circ$ C, so we have a heat loss of
$\frac{Q}{A}=0.38 (20-10)$ which is 3.8 W m ${}^2$.
But this analysis is missing an important form of heat transfer,
convective
heat transfer.
This is a difficult form of heat transfer to study. In theory,
we must solve both the continuity equation, the momentum balance
and the heat balance equation simultaneously: \begin{align*}
\frac{\partial \rho}{\partial t} &= -\nabla\cdot
\rho\boldsymbol{v}& \frac{\partial \boldsymbol{v}}{\partial
t} &= -\boldsymbol{v}\cdot\nabla \boldsymbol{v} -
\rho^{-1}\nabla\cdot\boldsymbol{\tau} - \rho^{-1}\nabla p +
\boldsymbol{g} \end{align*} \begin{align*}
\rho C_p\frac{\partial T}{\partial t} &= -\rho C_p v_j
\nabla_j T - \nabla_i q_i - \tau_{ji} \nabla_j v_i -
p \nabla_i v_i +\sigma_{energy} \end{align*}
This is a difficult task (not impossible).
Newton (a great engineer) was the first to make the definition
\begin{align*} Q_{convection}=
h A \left(T_{air}-T_{surface}\right) \end{align*}
This is
Newton's law of cooling; however, it is more of a
definition of the heat transfer coefficient, $h$, than
an actual law.
The heat transfer coefficient $h$ is in reality a complex
function of the temperature difference, the flow properties and
the heat transfer area, $A$.
In previous years, you may have found that you could derive the
relationship between the heat transfer coefficient $h$ and the
properties of the flow using
dimensional analysis
\begin{align*} h&=\frac{k \text{Nu}}{L}& \text{Nu}
&\approx C \text{Re}^m \text{Pr}^n \end{align*} where the
Prandtl number is given by $\text{Pr}=\mu C_p/k$, $m$ and $b$
are
theoretically
or
experimentally
derived coefficients.
This means that for
forced convection
(where you have a driven flow) you can find approximate
expressions for the heat transfer coefficient, $h$, in the
literature
without having to directly solve all of the balance
equations.
For example, for fully-developed turbulent flows in pipes with a
Prandtl number in the range of 0.6 to 100, a popular expression
by Dittus and Beolter is \begin{align*}
\text{Nu}=0.023 \text{Re}^{0.8} \text{Pr}^n \end{align*} where
$n=0.4$ for heating of the fluid and $n=0.3$ for cooling of the
fluid.
There are huge numbers of these relationships which you can look
up in standard texts (e.g. Perry's or C&R vol 6), depending
on the flow geometry, the viscosity behaviour etc.
We won't try to derive any of these relationships, but it is
important to note that the error in the calculated convective
heat transfer coefficient may be as much as $\pm 25\%$ !
There is still some difficulty with using Newton's law of
cooling.
As convective heat transfer may be the limiting heat transfer
resistance ($h_i\approx h_o \ll k_{insulation}\ll k_{plaster}\ll
k_{brick}$ ), the large error in $h$ may be bigger than all the
other terms combined!
This is true for forced convection
in
pipes, the situation is worse for more complex situations, such
as the
flow over a bank of pipes in a heat exchanger!
If we want to redo our insulated wall problem, we can add the
two typical heat transfer coefficients, obtained from the
literature, to the overall resistance.
\begin{multline*}
1/\left(U A\right)_{total} = 1 /(U A)_{o,Convection} +
1/(U A)_{brick}+1/(U A)_{insulation}\\ +1/(U A)_{plaster} + 1
/(U A)_{i,Convection} \end{multline*} \begin{multline*}
A/\left(U A\right)_{total} = 1/h_i +
X_{brick}/k_{brick}+X_{insulation}/k_{insulation}
+X_{plaster}/k_{plaster}+1/h_o \end{multline*}
Inserting the values, we obtain \begin{align*} U_{total} = 0.35
\text{W}/\text{m}^2 \end{align*}
Again, on a normal day in Aberdeen we have $T_o=10^\circ$ C and
$T_i=20^\circ$ C, so we have a heat loss of
$\frac{Q}{A}=U \Delta T=0.35 (20-10)$ which is 3.5 W m ${}^2$
(8% less than without convection).
Including convection on the wall in this case only resulted in a
8% change in the heat transfer rate.
This doesn't fit with the statement that
often the convective heat transfer coefficient $h$ is the
limiting resistance.
This is because the wall has insulation to limit the conduction
of heat.
If the insulation layer did not limit the heat transfer, it
would be ineffective.
In other systems, such as a heat exchanger, we will want to
minimise the thermal resistance as much as possible (by using
metal walls).
In systems where heat exchange is to be maximised between two
fluids, $h$ will indeed be the limiting factor.
There is one difficulty with the example above. On a still day,
there is no significant flow of air over the walls of a house.
In the case of no flow, where do we get the convective heat
transfer coefficients $h_i$ and $h_o$ from?
These coefficients cannot be determined from the current
definition of the Nusselt number for
forced convection
\begin{align*} \text{Nu}=0.023 \text{Re}^{0.8} \text{Pr}^n
\end{align*} as the flow velocity is $v=0$, therefore
$\text{Re}=0$.
However, the heat of the wall will cause
natural convection
to occur and the fluid will flow due to the density changes.
We must consider
natural convection
and radiative transfer in the following lectures.
Learning Objectives
We discovered that we can solve the heat balance equation to
give integrated expressions for the heat transfer:
\begin{align*} Q_x&= \frac{k}{X} A
\left(T_{in}-T_{out}\right) & Q_r &=
\frac{2 \pi L k}{\ln\left(R_{outer}/R_{inner}\right)} (T_{in} -
T_{out}) \end{align*}
We revised the general heat flux equation, and that resistances
may be summed. \begin{align*} Q &= U A \Delta T =
\frac{\Delta T}{R_{total}} & R_{total} &= R_1 + R_2 + …
= 1/U_1 A_1 + 1/U_2 A_2 + … \end{align*}
We covered
conduction
and derived the conductive heat transfer resistances in a plate
and pipe: \begin{align*} R_{plate} &= \frac{X}{k A}&
R_{pipe} &=
\frac{\ln\left(R_{outer}/R_{inner}\right)}{2 \pi L k}
\end{align*}
We covered
forced convection, and revised that we have relationships
for the heat transfer coefficient $h$ \begin{align*}
h&=\frac{k \text{Nu}}{L} & \text{Nu} &\approx
C \text{Re}^m \text{Pr}^n \end{align*}
But this expression is only valid for
forced
convection, and natural convection is more common for heat loss calculations.