Chapter 4, Problem 4.2CP

To determine

(a)

Value of nonzero velocity, $v\left(x\right)$.

Answer to Problem 4.2CP

The value of nonzero velocity is,

$v\left(x\right)=\frac{\rho g}{2\mu }{x}^2-\frac{\rho gh}{\mu }x+V$

Explanation of Solution

Assume the flow to be parallel,

$u=w=0$

Due to this, the momentum equation along x and z directions can be neglected.

Consider the y momentum equation,

$\rho \left(u\frac{dv}{dx}+v\frac{dv}{dy}+w\frac{dv}{dz}\right)=-\frac{dv}{dy}-\rho g+\mu \left(\frac{d^{2}v}{d{x}^2}+\frac{d^{2}v}{d{y}^2}+\frac{d^{2}v}{d{z}^2}\right)$

Substitute 0 for u, 0 for w, 0 for $\frac{\partial v}{\partial y}$, 0 for $\frac{{\partial }^{2}v}{\partial y^2}$ and 0 for $\frac{{\partial }^{2}v}{\partial z^2}$.

$\rho \left(0+0+0\right)=-0-\rho g+\mu \left(\frac{d^{2}v}{d{x}^2}+0+0\right)$

$\frac{d^{2}v}{d{x}^2}=\frac{\rho g}{\mu }$

Integrate the second-order linear differential equation with respect to y, it gives:

$\int \frac{d^{2}v}{d{x}^2}dx=\int \frac{\rho g}{\mu }dx$

Δ $\frac{du}{dx}=\frac{\rho g}{\mu }x+K_1$ ……Eq.(1)

Integrate again gives:

$\int \frac{du}{dx}dx=\int (\frac{\rho g}{\mu }x+C_1)dx$

Δ $v\left(x\right)=\frac{\rho g}{2\mu }{x}^2+K_{1}x+K_2$ ………Eq. (2)

By applying the 1^st boundary condition, ${\frac{du}{dx}|}_{x=h}=0$ to Eq. (1) yields:

$0=\frac{\rho g}{\mu }\left(h\right)+K_1$

Δ $K_1=-\frac{\rho gh}{\mu }$

By applying the2^nd boundary conditions, $u\left(0\right)=V$ to Eq. (2) gives:

$V=\frac{\rho g}{2\mu }{(0)}^2+K_1\left(0\right)+K_2$

Δ $K_2=V$

Substitute, $K_1=-\frac{\rho gh}{\mu }$ and $K_2=V$ into Eq. (2) gives the formula which describes the velocity distribution of the fluid flow:

$v\left(x\right)=\frac{\rho g}{2\mu }{x}^2-\frac{\rho gh}{\mu }x+V$.

To determine

(b)

The average velocity $,V_{avg.}$.

Answer to Problem 4.2CP

The average velocity is, $V_{avg.}=V-\frac{\rho g{h}^2}{3\mu }$

Explanation of Solution

Based on the knowledge of calculus, the average value on an interval [a,b] of an integrable function f is given by:

$avg.\left(f\right)=\frac{1}{b-a}\underset{a}{\overset{b}{{\displaystyle \int }}}f\left(x\right)dx$

Likewise, the average velocity, $V_{avg.}$ of the film can be obtained by integrating the function of the velocity distribution, u(y), which derived in part (a) across the film:

$V_{avg.}=\frac{1}{h-0}\underset{0}{\overset{h}{{\displaystyle \int }}}v\left(x\right)dx=\frac{1}{h}\underset{0}{\overset{h}{{\displaystyle \int }}}(\frac{\rho g}{2\mu }{x}^2-\frac{\rho gh}{\mu }x+V)dx=\frac{1}{h}\left(\frac{\rho g}{6\mu }{h}^3-\frac{\rho gh}{2\mu }{x}^2+Vx\right)|\begin{array}{c}h\\ 0\end{array}=\frac{1}{h}\left\{\left(\frac{\rho g}{6\mu }{h}^3-\frac{\rho gh}{2\mu }{h}^2+Vh\right)-0\right\}=V-\frac{\rho g{h}^2}{3\mu }$

Therefore, $V_{avg.}=V-\frac{\rho g{h}^2}{3\mu }$.

To determine

(c)

Velocity, $V_c$ for which there is no net flow either up or down.

Answer to Problem 4.2CP

Velocity, $V_c$ for which there is no net flow either up or down is, $V_C=\frac{\rho g{h}^3}{3\mu }$

Explanation of Solution

With the relationship of the velocity distribution obtained in part (a), the equation of flow-rate per unit width, Q, can be formulated from relation,

$Q=\underset{0}{\overset{h}{{\displaystyle \int }}}v\left(x\right)dx=\underset{0}{\overset{h}{{\displaystyle \int }}}\left(\frac{\rho g}{2\mu }{x}^2-\frac{\rho gh}{\mu }x+V\right)dy=\left(\frac{\rho g}{2\mu }{x}^2-\frac{\rho gh}{\mu }x+V\right)|\begin{array}{c}h\\ 0\end{array}=Vh-\frac{\rho g{h}^3}{3\mu }$

Since, there is no net flow either up or down, therefore, $V_C=\frac{\rho g{h}^3}{3\mu }$.

To determine

(d)

Sketch the $v\left(x\right)$ for case (c).

Explanation of Solution

The equation which describe the velocity distribution, u(y), which derived in part (a) can be expressed in dimensionless form by dividing both sides of the equation by the velocity, V.

$\frac{u}{V}=\frac{\rho g}{2\mu V}{x}^2-\frac{\rho gh}{\mu V}\frac{x}{h}+1$

$\frac{u}{V}=\frac{\rho g{h}^2}{2\mu V}.\frac{x^2}{h^2}-\frac{\rho g{h}^2}{\mu V}.\frac{x}{h}+1$

$\frac{u}{V}=\frac{\rho g{h}^2}{2\mu V}.{(\frac{x}{h})}^2-\frac{\rho g{h}^2}{2\mu V}\mathrm{.2.}\left(\frac{x}{h}\right)+1$

$\frac{u}{V}=A{(\frac{x}{h})}^2-2A\left(\frac{x}{h}\right)+1$

Where, $A=\frac{\rho g{h}^2}{2\mu V}$

The velocity distribution of the film at various y-location which has the domain range from $\frac{x}{h}=0$ to $\frac{x}{h}=1$ are computed by introducing the variable A from A=0 to A=2.

The results are tabulated below:

x/h	u/V
A=0	A=0.5	A=1	A=1.5	A=2
0	1	1	1	1	1
0.05	1	0.95125	0.9025	0.85375	0.805
0.1	1	0.905	0.81	0.715	0.62
0.15	1	0.86125	0.7225	0.58375	0.445
0.2	1	0.82	0.64	0.46	0.28
0.25	1	0.78125	0.5625	0.34375	0.125
0.3	1	0.745	0.49	0.235	-0.02
0.35	1	0.71125	0.4225	0.13375	-0.155
0.4	1	0.68	0.36	0.04	-0.28
0.45	1	0.65125	0.3025	-0.04625	-0.395
0.5	1	0.625	0.25	-0.125	-0.5
0.55	1	0.60125	0.2025	-0.19625	-0.595
0.6	1	0.58	0.16	-0.26	-0.68
0.65	1	0.56125	0.1225	-0.31625	-0.755
0.7	1	0.545	0.09	-0.365	-0.82
0.75	1	0.53125	0.0625	-0.40625	-0.875
0.8	1	0.52	0.04	-0.44	-0.92
0.85	1	0.51125	0.0225	-0.46625	-0.955
0.9	1	0.505	0.01	-0.485	-0.98
0.95	1	0.50125	0.0025	-0.49625	-0.995
1	1	0.5	0	-0.5	-1

Dimensionless Velocity Distribution (Velocity Profile):

As illustrated on the plot of velocity profile of fluid flow shown above, there are still some portions of the fluid flow downward ( $\raisebox{1ex}{$u$}\!\left/ \!\raisebox{-1ex}{$V$}\right.<0)$ although the belt is moving upward with a velocity V as a result of small viscosity or slow belt speed. Apart from that, there will be a net upward flow of fluid (positive V) if $V>\raisebox{1ex}{$\rho g{h}^2$}\!\left/ \!\raisebox{-1ex}{$3\mu $}\right..$

Conclusion:

It can be concluded that in order to lift a fluid with small viscosity, a relatively large belt speed is required.