Ideal Internal Flow
One would think that what engineers really want is a REAL flow and not
an Ideal Flow. But, unexpectedly perhaps, that's NOT what they really
want. With the apparatus I've been working on, what the engineers want
is the flow field, calculated in such a way that temperature stresses
cannot be worse in reality than they are in the calculations. We call
such calculations conservative. An insightful moment of thinking has
revealed that not a realistic flow simulation but rather an Ideal Flow
simulation has the desired properties.
Consider Fuid Flow in three dimensions. The flow is assumed to be ideal,
which means that it is steady, incompressible and irrotational.
The latter means that Vorticity $\,\vec{\omega}\,$ is zero, everywhere in the fluid.
Let $(x,y,z)$ be Cartesian coordinates and $(u,v,w)$ be flow velocity components, then:
$$
\mbox{incompressible :} \quad \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z} = 0 \\
\mbox{irrotational :} \quad \begin{cases}
\large \omega_x = \frac{\partial w}{\partial y} - \frac{\partial v}{\partial z} = 0 \\
\large \omega_y = \frac{\partial u}{\partial z} - \frac{\partial w}{\partial x} = 0 \\
\large \omega_z = \frac{\partial v}{\partial x} - \frac{\partial u}{\partial y} = 0 \end{cases}
$$
It can be shown that an irrotational flow is consistent with the existence of a potential function $\phi$.
Define $\phi$ as follows:
$$
\frac{\partial \phi}{\partial x} = u \quad ; \quad
\frac{\partial \phi}{\partial y} = v \quad ; \quad
\frac{\partial \phi}{\partial z} = w
$$
A necessary condition for this to be possible is that the second order cross derivatives of $\phi$ are unique:
$$
\frac{\partial (\partial \phi / \partial x)}{\partial y} = \frac{\partial (\partial \phi / \partial y)}{\partial x}
\quad \Longleftrightarrow \quad \frac{\partial u}{\partial y} - \frac{\partial v}{\partial x} = 0 \\
\frac{\partial (\partial \phi / \partial y)}{\partial z} = \frac{\partial (\partial \phi / \partial z)}{\partial y}
\quad \Longleftrightarrow \quad \frac{\partial v}{\partial z} - \frac{\partial w}{\partial y} = 0 \\
\frac{\partial (\partial \phi / \partial z)}{\partial x} = \frac{\partial (\partial \phi / \partial x)}{\partial z}
\quad \Longleftrightarrow \quad \frac{\partial w}{\partial x} - \frac{\partial u}{\partial z} = 0
$$
Therefore a potential function $\phi$ exists if and only if the flow is irrotational.
If the flow is incompressible as well then we can derive another equation, which is the well known Laplace equation:
$$
\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z} = 0
\quad \Longrightarrow \quad \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} + \frac{\partial^2 \phi}{\partial z^2} = 0
$$
We conclude that ideal flow is exactly the same as Potential flow .
Now consider the following expression, for the kinetic energy of ideal flow in the 3-D domain $D$ of interest:
$$
E = \frac{1}{2} \rho \iiint_D \left[u^2+v^2+w^2\right] dV \quad \Longleftrightarrow \\
E \sim \iiint_D \left\{\left(\frac{\partial \phi}{\partial x}\right)^2+\left(\frac{\partial \phi}{\partial y}\right)^2
+\left(\frac{\partial \phi}{\partial z}\right)^2\right\}dV
$$
It is already known that the flow under consideration is irrotational, otherwise the function $\phi$ would not even exist.
One is asking for other conditions eventually, namely such that the kinetic energy of the fluid is minimal:
$$
\iiint_D \left\{\left(\frac{\partial \phi}{\partial x}\right)^2+\left(\frac{\partial \phi}{\partial y}\right)^2
+\left(\frac{\partial \phi}{\partial z}\right)^2\right\}dV = \mbox{minimum}(\phi)
$$
Let $\phi = \psi+\epsilon.f$ , with $\epsilon$ a "small" disturbance and $fx,y,z)$ a completely arbitrary function,
which is zero at the boundaries $\partial D$ of the domain of interest. In this way the three-dimensional integral
has become an ordinary one-dimensional function $E(\epsilon)$ , which can be simply differentiated to find the minimum,
especially at $\epsilon = 0$ , where $\psi = \phi$ :
$$
\left.\frac{d}{d\epsilon}\right|_{\Large \epsilon=0}\;\iiint_D \left\{\left[\frac{\partial (\psi+\epsilon.f)}{\partial x}\right]^2
+\left[\frac{\partial (\psi+\epsilon.f)}{\partial y}\right]^2+\left[\frac{\partial (\psi+\epsilon.f)}{\partial z}\right]^2\right\}dV = 0
\quad \Longleftrightarrow \\
2\iiint_D \left\{\frac{\partial \psi}{\partial x}\frac{\partial f}{\partial x}
+\frac{\partial\psi}{\partial y}\frac{\partial f}{\partial y}+\frac{\partial \psi}{\partial z}\frac{\partial f}{\partial z}\right\}dV = 0
$$
With the rues for differentiation of a product of functions:
$$
\frac{\partial}{\partial x}\left(f\frac{\partial \psi}{\partial x}\right) =
\frac{\partial f}{\partial x}\frac{\partial\psi}{\partial x} + f\,\frac{\partial^2\psi}{\partial x^2} \\
\frac{\partial}{\partial y}\left(f\frac{\partial \psi}{\partial y}\right) =
\frac{\partial f}{\partial y}\frac{\partial\psi}{\partial y} + f\,\frac{\partial^2\psi}{\partial y^2} \\
\frac{\partial}{\partial z}\left(f\frac{\partial \psi}{\partial z}\right) =
\frac{\partial f}{\partial z}\frac{\partial\psi}{\partial z} + f\,\frac{\partial^2\psi}{\partial x^2}
$$
Giving:
$$
\iiint_D \left\{\frac{\partial \psi}{\partial x}\frac{\partial f}{\partial x}
+\frac{\partial\psi}{\partial y}\frac{\partial f}{\partial y}+\frac{\partial \psi}{\partial z}\frac{\partial f}{\partial z}\right\}dV = \\
\iiint_D f\left\{\frac{\partial^2\psi}{\partial x^2}+\frac{\partial^2\psi}{\partial y^2}+\frac{\partial^2\psi}{\partial z^2}\right\}dV \\
- \iiint_D \left\{\frac{\partial}{\partial x}\left(f\frac{\partial \psi}{\partial x}\right)
+\frac{\partial}{\partial y}\left(f\frac{\partial \psi}{\partial y}\right)+\frac{\partial}{\partial z}\left(f\frac{\partial \psi}{\partial z}\right)\right\}dV
$$
The last integral can be simplified with help of the divergence theorem:
$$
\iiint_D \left\{\frac{\partial}{\partial x}\left(f\frac{\partial \psi}{\partial x}\right)
+\frac{\partial}{\partial y}\left(f\frac{\partial \psi}{\partial y}\right)
+\frac{\partial}{\partial z}\left(f\frac{\partial \psi}{\partial z}\right)\right\}dV = \\
\bigcirc\kern-1.4em\iint_{\partial D} f\left\{\left(\frac{\partial \psi}{\partial x}\right)dA_x
+\left(\frac{\partial \psi}{\partial y}\right)dA_y+\left(\frac{\partial \psi}{\partial z}\right)dA_z\right\} = 0
$$
This area integral is zero because the test function $f$ is zero a the boundaries. Furthermore $\psi = \phi$ , so we are left with:
$$
\iiint_D \left\{\left(\frac{\partial \phi}{\partial x}\right)^2+\left(\frac{\partial \phi}{\partial y}\right)^2
+\left(\frac{\partial \phi}{\partial z}\right)^2\right\}dV = \mbox{minimum}(\phi) \quad \Longleftrightarrow \\
\iiint_D f\left\{\frac{\partial^2\phi}{\partial x^2}+\frac{\partial^2\phi}{\partial y^2}+\frac{\partial^2\phi}{\partial z^2}\right\}dV = 0
$$
For an anbitrary function $f(x,y,z)$. Meaning that:
$$
\frac{\partial^2\phi}{\partial x^2}+\frac{\partial^2\phi}{\partial y^2}+\frac{\partial^2\phi}{\partial z^2} = 0
$$
This means that the Laplace equation is fulfilled, which is equivalent with the statement that the fluid is incompressible.
Conclusion:Ideal Internal Flow has the lowest global kinetic energy of all possible internal flows.
There is an argument why ideal internal flow is litteraly IDEAL for calculating temperature fields in heat exchangers.
Let's repeat now the equations for heat transfer in a Fluid Tube Continuum, three dimensional form:
$$
c.G_P\left(u\frac{\partial T_P}{\partial x}+v\frac{\partial T_P}{\partial y}+w\frac{\partial T_P}{\partial z}\right)
+a\left(T_P-T_S\right) = 0 \\
c.G_S\frac{\partial T_S}{\partial z}+a\left(T_S-T_P\right) = 0
$$
The total power transferred is expressed by the volume integral at the left hand side, in the following nice application
of Schwartz inequality. The worded result is apart from constants.
$$
(\vec{a}\cdot\vec{b})^2\le(\vec{a}\cdot\vec{a})\,(\vec{b}\cdot\vec{b}) \quad \Longrightarrow \\
\left[\iiint_D a\left(T_P-T_S\right)\,dV \right]^2 =
\left[\iiint_D c.G_P\left(u\frac{\partial T_P}{\partial x}+v\frac{\partial T_P}{\partial y}+w\frac{\partial T_P}{\partial z}\right) dV \right]^2 \\
\le \frac{1}{2} \rho \left[\iiint_D\left\{u^2+v^2+w^2\right\}dV\right]\frac{(c.G_P)^2}{\rho/2}
\left[\iiint_D\left\{\left(\frac{\partial T_P}{\partial x}\right)^2+\left(\frac{\partial T_P}{\partial y}\right)^2
+\left(\frac{\partial T_P}{\partial z}\right)^2\right\}dV\right] \quad \Longrightarrow \\
\frac{\left(c.G_P\right)^2}{\rho/2}\left[\frac{\iiint_D\left\{\left(\partial T_P/\partial x\right)^2+\left(\partial T_P/\partial y\right)^2
+\left(\partial T_P/\partial z\right)^2\right\}dV}{\iiint_D dV}\right] \le
\frac{\left[\iiint_D a\left(T_P-T_S\right)\,dV/\iiint_D dV\right]^2}{\rho/2 \left[\iiint_D\left\{u^2+v^2+w^2\right\}dV\right]/\iiint_D dV} \\
(\mbox{ mean squares of temperature gradients }) \le \frac{\left(\mbox{ mean power transferred }\right)^2}{(\mbox{ mean kinetic energy of flow })}
$$
Potential flow has the least kinetic energy of any flow having the same normal velocity on the boundary and satisfying
the condition of conservation of mass. As a consequence, the denominator at the right hand side is minimal for Ideal Flow.
If the power delivered by the apparatus is predescribed, then we may conclude: the mean temperature gradients are maximal
for Ideal Internal Flow. The calculations are thus safe, because they give rise to maximal temperature stresses with
given operating conditions. With other words: our calculations are conservative: reality cannot be worse than theory.
We conclude that, in this case, what's good and ideally in mathematics, is also good and ideally for engineers !