9.3: Nusselt's Technique

Last updated
Save as PDF

Page ID: 766

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$ \newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$

( \newcommand{\kernel}{\mathrm{null}\,}\) $ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\id}{\mathrm{id}}$

$ \newcommand{\Span}{\mathrm{span}}$

$ \newcommand{\kernel}{\mathrm{null}\,}$

$ \newcommand{\range}{\mathrm{range}\,}$

$ \newcommand{\RealPart}{\mathrm{Re}}$

$ \newcommand{\ImaginaryPart}{\mathrm{Im}}$

$ \newcommand{\Argument}{\mathrm{Arg}}$

$ \newcommand{\norm}[1]{\| #1 \|}$

$ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$

$ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\AA}{\unicode[.8,0]{x212B}}$

$ \newcommand{\vectorA}[1]{\vec{#1}} % arrow$

$ \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$

$ \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vectorC}[1]{\textbf{#1}} $

$ \newcommand{\vectorD}[1]{\overrightarrow{#1}} $

$ \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} $

$ \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} $

$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $

$ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $

$\newcommand{\avec}{\mathbf a}$ $\newcommand{\bvec}{\mathbf b}$ $\newcommand{\cvec}{\mathbf c}$ $\newcommand{\dvec}{\mathbf d}$ $\newcommand{\dtil}{\widetilde{\mathbf d}}$ $\newcommand{\evec}{\mathbf e}$ $\newcommand{\fvec}{\mathbf f}$ $\newcommand{\nvec}{\mathbf n}$ $\newcommand{\pvec}{\mathbf p}$ $\newcommand{\qvec}{\mathbf q}$ $\newcommand{\svec}{\mathbf s}$ $\newcommand{\tvec}{\mathbf t}$ $\newcommand{\uvec}{\mathbf u}$ $\newcommand{\vvec}{\mathbf v}$ $\newcommand{\wvec}{\mathbf w}$ $\newcommand{\xvec}{\mathbf x}$ $\newcommand{\yvec}{\mathbf y}$ $\newcommand{\zvec}{\mathbf z}$ $\newcommand{\rvec}{\mathbf r}$ $\newcommand{\mvec}{\mathbf m}$ $\newcommand{\zerovec}{\mathbf 0}$ $\newcommand{\onevec}{\mathbf 1}$ $\newcommand{\real}{\mathbb R}$ $\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$ $\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$ $\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$ $\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$ $\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$ $\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$ $\newcommand{\laspan}[1]{\text{Span}\{#1\}}$ $\newcommand{\bcal}{\cal B}$ $\newcommand{\ccal}{\cal C}$ $\newcommand{\scal}{\cal S}$ $\newcommand{\wcal}{\cal W}$ $\newcommand{\ecal}{\cal E}$ $\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$ $\newcommand{\gray}[1]{\color{gray}{#1}}$ $\newcommand{\lgray}[1]{\color{lightgray}{#1}}$ $\newcommand{\rank}{\operatorname{rank}}$ $\newcommand{\row}{\text{Row}}$ $\newcommand{\col}{\text{Col}}$ $\renewcommand{\row}{\text{Row}}$ $\newcommand{\nul}{\text{Nul}}$ $\newcommand{\var}{\text{Var}}$ $\newcommand{\corr}{\text{corr}}$ $\newcommand{\len}[1]{\left|#1\right|}$ $\newcommand{\bbar}{\overline{\bvec}}$ $\newcommand{\bhat}{\widehat{\bvec}}$ $\newcommand{\bperp}{\bvec^\perp}$ $\newcommand{\xhat}{\widehat{\xvec}}$ $\newcommand{\vhat}{\widehat{\vvec}}$ $\newcommand{\uhat}{\widehat{\uvec}}$ $\newcommand{\what}{\widehat{\wvec}}$ $\newcommand{\Sighat}{\widehat{\Sigma}}$ $\newcommand{\lt}{<}$ $\newcommand{\gt}{>}$ $\newcommand{\amp}{&}$ $\definecolor{fillinmathshade}{gray}{0.9}$

The Nusselt's method is a bit more labor intensive, in that the governing equations with the boundary and initial conditions are used to determine the dimensionless parameters. In this method, the boundary conditions together with the governing equations are taken into account as opposed to Buckingham's method. A common mistake is to ignore the boundary conditions or initial conditions. The parameters that results from this process are the dimensional parameters which control the problems. An example comparing the Buckingham's method with Nusselt's method is presented. In this method, the governing equations, initial condition and boundary conditions are normalized resulting in a creation of dimensionless parameters which govern the solution. It is recommended, when the reader is out in the real world to simply abandon Buckingham's method all together. This point can be illustrated by example of flow over inclined plane. For comparison reasons Buckingham's method presented and later the results are compared with the results from Nusselt's method.

Example 9.10

Utilize the Buckingham's method to analyze a two dimensional flow in incline plane. Assume that the flow infinitely long and thus flow can be analyzed per width which is a function of several parameters. The potential parameters are the angle of inclination, $\theta$, liquid viscosity, $\nu$, gravity, $g$, the height of the liquid, $h$, the density, $\rho$, and liquid velocity, $U$. Assume that the flow is not affected by the surface tension (liquid), $\sigma$. You furthermore are to assume that the flow is stable. Develop the relationship between the flow to the other parameters.

Solution 9.10

Under the assumptions in the example presentation leads to following

\[ \label{dim:eq:buckinghamNusseltCompar}
\dot{m} = f\left( \theta, \nu, g, \rho, U\right)
\]
The number of basic units is three while the number of the parameters is six thus the difference is $6-3=3$. Those groups (or the work on the groups creation) further can be reduced the because angle $\theta$ is dimensionless. The units of parameters can be obtained in Table 9.3 the following table.

Parameter	Units	Parameter	Units
$\nu$	$L^2t^{-1}$	$g$	$L^{1}t{-2}$
$\dot{m}$	$M\,t^{-1}L^{-1}$	$\theta$	none
$U$	$L^{1}t^{-1}$	$\rho$	$M\,L^{3}$

The basic units are chosen as for the time, $U$, for the mass, $\rho$, and for the length $g$. Utilizing the building blocks technique provides

\[ \label{inclidePlane:ini}
\overbrace{\dfrac{M}{t\,L} }^{\dot{m}}
= \left( \overbrace{\dfrac{M}{L^3}}^{\rho} \right)^a
\left( \overbrace{\dfrac{L}{t^2} }^{g} \right)^b
\left( \overbrace{\dfrac{L}{t} }^{U} \right)^c
\]

The equations obtained from equation (2) are

\[ \label{inclidePlane:gov0}
\left.
\begin{array}{rrl}
\text{Mass}, M & a =& 1 \\
\text{Length}, L & -3a + b +c =& -1 \\
\text{time}, t & -2b -c =& - 1
\end{array}
\right\} \Longrightarrow
\pi_1 = \dfrac{\dot{m} \,g}{\rho\,\,U^3}
\]

\[ \label{inclidePlane:ini1}
\overbrace{\dfrac{L^2}{t} }^{\nu}
= \left( \overbrace{\dfrac{M}{L^3}}^{\rho} \right)^a
\left( \overbrace{\dfrac{L}{t^2} }^{g} \right)^b
\left( \overbrace{\dfrac{L}{t} }^{U} \right)^c
\] The equations obtained from equation (2) are

\[ \label{inclidePlane:gov}
\left.
\begin{array}{rrl}
\text{Mass}, M & a =& 0 \\
\text{Length}, L & -3a + b +c =& 2 \\
\text{time}, t & -2b -c =& - 1
\end{array}
\right\} \Longrightarrow
\pi_2 = \dfrac{\nu \,g}{U^3}
\] Thus governing equation and adding the angle can be written as

\[ \label{inclidePlane:}
0 = f\left(\dfrac{\dot{m} \,g}{\rho\,\,U^3} , \dfrac{\nu \,g}{U^3} ,\theta\right)
\] The conclusion from this analysis are that the number of controlling parameters totaled in three and that the initial conditions and boundaries are irrelevant.

A small note, it is well established that the combination of angle gravity or effective body force is significant to the results. Hence, this analysis misses, at the very least, the issue of the combination of the angle gravity. Nusselt's analysis requires that the governing equations along with the boundary and initial conditions to be written. While the analytical solution for this situation exist, the parameters that effect the problem are the focus of this discussion. In Chapter 8, the Navier–Stokes equations were developed. These equations along with the energy, mass or the chemical species of the system, and second laws governed almost all cases in thermo–fluid mechanics. This author is not aware of a compelling reason that this fact should be used in this chapter. The two dimensional NS equation can obtained from equation (??) as

\[ \label{dim:eq:twoDNSx}
\begin{array} {rll}
\rho \left(\dfrac{\partial U_x}{\partial t} +
\right. & U_x\dfrac{\partial U_x}{\partial x} +
U_y \dfrac{\partial U_x}{\partial y} + \left. U_z
\dfrac{\partial U_x}{\partial z}\right) = &\\
&-\dfrac{\partial P}{\partial x} + \mu \left(\dfrac{\partial^2 U_x}{\partial x^2} +
\dfrac{\partial^2 U_x}{\partial y^2} + \dfrac{\partial^2 U_x}{\partial z^2}\right) +
\rho g\,\sin\theta
\end{array}
\]

and

\[ \label{dim:eq:twoDNSy}
\begin{array} {rll}
\rho \left(\dfrac{\partial U_y}{\partial t} +
\right. & U_x\dfrac{\partial U_y}{\partial x} +
U_y \dfrac{\partial U_y}{\partial y} + \left. U_z
\dfrac{\partial U_y}{\partial z}\right) = &\\
&-\dfrac{\partial P}{\partial x} + \mu \left(\dfrac{\partial^2 U_y}{\partial x^2} +
\dfrac{\partial^2 U_y}{\partial y^2} + \dfrac{\partial^2 U_y}{\partial z^2}\right) +
\rho g\,\sin\theta
\end{array}
\] With boundary conditions

\[ \label{dim:eq:twoDNSbcBotton}
\begin{array}{l}
U_x (y=0) = U_{0x} f(x) \\
\dfrac{\partial U_x}{\partial x} (y=h) = \tau_0 f(x)
\end{array}
\] The value $U_0x$ and $\tau_0$ are the characteristic and maximum values of the velocity or the shear stress, respectively. and the initial condition of

\[ \label{dim:eq:twoDNSic}
U_x (x=0) = U_{0y}\, f(y)
\] where $ U_{0y}$ is characteristic initial velocity. These sets of equations (7-??) need to be converted to dimensionless equations. It can be noticed that the boundary and initial conditions are provided in a special form were the representative velocity multiply a function. Any function can be presented by this form.
In the process of transforming the equations into a dimensionless form associated with
some intelligent guess work. However, no assumption is made or required about whether or not the velocity, in the $y$ direction. The only exception is that the $y$ component of the velocity vanished on the boundary. No assumption is required about the acceleration or the pressure gradient etc.
The boundary conditions have typical velocities which can be used. The velocity is selected according to the situation or the needed velocity. For example, if the effect of the initial condition is under investigation than the characteristic of that velocity should be used. Otherwise the velocity at the bottom should be used. In that case, the boundary conditions are

\[ \label{dim:eq:twoDNScbles}
\begin{array}{l}
\dfrac{ U_x (y=0)}{ U_{0x}} = f(x) \\
\mu \dfrac{\partial U_x}{\partial x} (y=h) = \tau_0 \,g(x)
\end{array}
\] Now it is very convenient to define several new variables:

\[ \label{dim:eq:twoDNSdefVer1}
\begin{array}{rlcrl}
\overline{U} = & \dfrac{ U_x (\overline{x} ) }{ U_{0x}}\\
where:\
\overline{x} = & \dfrac{x}{h} &\qquad& \overline{y} &= \dfrac{y}{h} \\
\end{array}
\] The length $h$ is chosen as the characteristic length since no other length is provided. It can be noticed that because the units consistency, the characteristic length can be used for "normalization'' (see Example 9.11). Using these definitions the boundary and initial conditions becomes

\[ \label{dim:eq:twoDNScbles1}
\qquad
\begin{array}{l}
\dfrac{ \overline{U_x} (\overline{y}=0)}{ U_{0x}} = f^{'}(\overline{x}) \\
\dfrac{h \, \mu}{U_{0x}}\,
\dfrac{\partial \overline{U_x}}{\partial \overline{x}} (\overline{y}=1) = \tau_0\, g^{'}(\overline{x})
\end{array}
\] It commonly suggested to arrange the second part of equation (13) as

\[ \label{dim:eq:twoDNScbles2}
\dfrac{\partial \overline{U_x}}{\partial \overline{x}} (\overline{y}=1) =
\dfrac{\tau_0\,U_{0x}}{h \, \mu}\,
g^{'}(\overline{x})
\] Where new dimensionless parameter, the shear stress number is defined as

\[ \label{dim:eq:tauDef}
\overline{\tau_0} = \dfrac{\tau_0\,U_{0x}}{h \, \mu}
\] With the new definition equation (14) transformed into

\[ \label{dim:eq:twoDNScbles3}
\dfrac{\partial \overline{U_x}}{\partial \overline{x}} (\overline{y}=1) =
\overline{\tau_0} \, g^{'}(\overline{x})
\]

Example 9.11

Non–dimensionalize the following boundary condition. What are the units of the coefficient in front of the variables, $x$. What are relationship of the typical velocity, $U_0$ to $U_{max}$?

\[ \label{twoDNSbc:bc}
U_x (y = h) = U_0 \left( a\,x^2 + b\,xp(x) \right)
\]

Solution 9.11

The coefficients $a$ and $b$ multiply different terms and therefore must have different units. The results must be unitless thus $a$

\[ \label{twoDNSbc:aDef1}
L^0 = a \, \overbrace{ {L^2} }^{x^2}
\Longrightarrow a = \left[ \dfrac{1}{L^2} \right]
\]

From equation (18) it clear the conversion of the first term is $U_x = a \, h^2 \overline{x}$. The exponent appears a bit more complicated as

\[ \label{twoDNSbc:bDef1}
{L}^{0} = b \, xp\left( h\,\dfrac{x}{h}\right) =
b \, xp\left( h \right)
\, xp\left( \dfrac{x}{h}\right) =
b \, xp\left( h \right)
\, xp\left( \overline{x}\right)
\] Hence defining

\[ \label{twoDNSbc:bDef12}
\overline{b} = \dfrac{1}{xp{h}}
\] With the new coefficients for both terms and noticing that $y=h\longrightarrow \overline{y} =1$ now can be written as

\[ \label{twoDNSbc:}
\dfrac{ U_x (\overline{y} =1)}{U_{0}} = \overbrace{a\,h^2}^{\overline{a}}\,x^2 +
\overbrace{b \, xp\left( h \right) }^{\overline{b}}
\, xp\left( \overline{x}\right)
= \overline{a}\,\overline{x}^2 + \overline{b} xp{\overline{x}}
\] Where $\overline{a}$ and $\overline{b}$ are the transformed coefficients in the dimensionless presentation.

After the boundary conditions the initial condition can undergo the non–dimensional process. The initial condition (10) utilizing the previous definitions transformed into

\[ \label{dim:eq:twoDNSicDless}
\dfrac{U_x(\overline{x}=0)}{U_{0x}} = \dfrac{U_{0y}}{U_{0x}}
f(\overline{y})
\]

Notice the new dimensionless group of the velocity ratio as results of the boundary condition. This dimensionless number was and cannot be obtained using the Buckingham's technique. The physical significance of this number is an indication to the "penetration'' of the initial (condition) velocity. The main part of the analysis if conversion of the governing equation into a dimensionless form uses previous definition with additional definitions. The dimensionless time is defined as $\overline{t} = t\,U_{0x}/h$. This definition based on the characteristic time of $h/U_{0x}$. Thus, the derivative with respect to time is

\[ \label{dim:eq:twoDNSdefDevT}
\dfrac{\partial U_x}{\partial t} =
\dfrac{\partial \overbrace{\overline{U_x}}^{\dfrac{U_x}{U_{0x}} } U_{0x}}
{\partial \underbrace{\overline{t}}_{ \dfrac{t\,U_{0x}}{ h } }\dfrac{h}{U_{0x}} }
=
\dfrac