Chapter 10, Problem 27P

To determine

(a)

The characteristic scale for v, y-component of velocity.

Answer to Problem 27P

The characteristic scale for v, y-component of velocity is $\text{v}\sim \frac{{\text{Vh}}_{\text{0}}}{\text{L}}$.

Explanation of Solution

First, we need to use continuity equation for velocity component.

From continuity equation,

We have,

  $\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0$

Now, to obtain order of magnitude. We need to use characteristic scale separately.

We have,

  $\begin{array}{l}\frac{\partial \text{u}}{\partial \text{x}}=\frac{\text{V}}{\text{L}}\\ \frac{\partial \text{v}}{\partial \text{y}}=\frac{\text{v}}{{\text{h}}_{\text{0}}}\end{array}$

Now, characteristic scale v is,

  $\begin{array}{l}\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}=0\\ \Rightarrow \text{v}\sim \frac{{\text{Vh}}_{\text{0}}}{\text{L}}\end{array}$

To determine

(b)

The order of magnitude of inertial term to that of viscous and pressure term.

Answer to Problem 27P

The characteristic scale for v, y-component of velocity is $\text{v}\sim \frac{{\text{Vh}}_{\text{0}}}{\text{L}}$.

Explanation of Solution

First, we need to use momentum of x-component.

  $\rho u\frac{\partial u}{\partial x}+\rho v\frac{\partial v}{\partial y}=-\frac{\partial P}{\partial x}+\mu \frac{{\partial }^{2}u}{\partial x^2}+\mu \frac{{\partial }^{2}u}{\partial y^2}$

As we know that the flow is carried out in two-dimensional, so the z component will be zero.

Now, to obtain order of magnitude. We need to use characteristic scale separately.

We have,

  $\begin{array}{l}\rho u\frac{\partial u}{\partial x}+\rho \text{v}\frac{\partial v}{\partial y}=-\frac{\partial P}{\partial x}+\mu \frac{{\partial }^{2}u}{\partial x^2}+\mu \frac{{\partial }^{2}u}{\partial y^2}\\ \text{ρu}\frac{\partial u}{\partial x}=\frac{\rho {\text{V}}^2}{\text{L}}\\ \text{ρv}\frac{\partial \text{u}}{\partial \text{y}}=\text{ρ}\frac{{\text{Vh}}_{\text{0}}}{\text{L}}\frac{\text{V}}{{\text{h}}_{\text{0}}}-\frac{\rho {\text{V}}^2}{\text{L}}\\ \frac{\partial P}{\partial x}=\frac{\mu V}{{\text{h}}_{\text{0}}{}^2}\\ \mu \frac{{\partial }^{2}u}{\partial x^2}=\frac{\mu \text{V}}{{\text{L}}^2}\\ \mu \frac{{\partial }^{2}u}{\partial y^2}=\frac{\mu \text{V}}{{\text{h}}_{\text{0}}{}^2}\end{array}$

Now,

The second term of viscosity is very smaller as compared to the second viscous term. So, we neglect this term (h_o<

Now, we need to multiply all the order of magnitude with a factor of $\frac{\text{L}}{\left(\rho {\text{V}}^2\right)}$. The magnitude of order becomes,

  $\begin{array}{l}\text{ρu}\frac{\partial u}{\partial x}=\frac{\rho {\text{V}}^2}{\text{L}}\times \frac{\text{L}}{\left(\rho {\text{V}}^2\right)}=1\\ \text{ρv}\frac{\partial \text{u}}{\partial \text{y}}=\left(\text{ρ}\frac{{\text{Vh}}_{\text{0}}}{\text{L}}\frac{\text{V}}{{\text{h}}_{\text{0}}}-\frac{\rho {\text{V}}^2}{\text{L}}\right)\times \frac{\text{L}}{\left(\rho {\text{V}}^2\right)}=1\\ \frac{\partial P}{\partial x}=\frac{\mu V}{{\text{h}}_{\text{0}}{}^2}\times \frac{\text{L}}{\left(\rho {\text{V}}^2\right)}=\frac{\mu }{\rho {\text{Vh}}_{\text{0}}}\frac{\text{L}}{{\text{h}}_{\text{0}}}\\ \mu \frac{{\partial }^{2}u}{\partial y^2}=\frac{\mu \text{V}}{{\text{h}}_{\text{0}}{}^2}\times \frac{\text{L}}{\left(\rho {\text{V}}^2\right)}=\frac{\mu }{\rho {\text{Vh}}_{\text{0}}}\frac{\text{L}}{{\text{h}}_{\text{0}}}\end{array}$

Now,

  $\rho u\frac{\partial u}{\partial x}+\rho \text{v}\frac{\partial v}{\partial y}=-\frac{\partial P}{\partial x}+\mu \frac{{\partial }^{2}u}{\partial y^2}$

We can see that the order of magnitude consists a factor of $\left(\frac{1}{\mathrm{Re}}\right)$. These pressure and viscous term contain a length scale of $\frac{\text{L}}{{\text{h}}_{\text{0}}}$. Both are very large for the flow to be creeping. Thus, we can say that the inertial terms are comparatively very smaller than the viscosity and pressure term. So, they can be neglected.

To determine

(c)

If the value of Reynolds number is less than 1 and (h_o<

Explanation of Solution

First, we need to use momentum of x-component.

  $\rho u\frac{\partial u}{\partial x}+\rho v\frac{\partial v}{\partial y}=-\frac{\partial P}{\partial x}+\mu \frac{{\partial }^{2}u}{\partial x^2}+\mu \frac{{\partial }^{2}u}{\partial y^2}$

As we know that the flow is carried out in two-dimensional, so the z component will be zero.

Now, to obtain order of magnitude. We need to use characteristic scale separately.

We have,

  $\begin{array}{l}\rho u\frac{\partial u}{\partial x}+\rho \text{v}\frac{\partial v}{\partial y}=-\frac{\partial P}{\partial x}+\mu \frac{{\partial }^{2}u}{\partial x^2}+\mu \frac{{\partial }^{2}u}{\partial y^2}\\ \text{ρu}\frac{\partial u}{\partial x}=\frac{\rho {\text{V}}^2}{\text{L}}\\ \text{ρv}\frac{\partial \text{u}}{\partial \text{y}}=\text{ρ}\frac{{\text{Vh}}_{\text{0}}}{\text{L}}\frac{\text{V}}{{\text{h}}_{\text{0}}}-\frac{\rho {\text{V}}^2}{\text{L}}\\ \frac{\partial P}{\partial x}=\frac{\mu V}{{\text{h}}_{\text{0}}{}^2}\\ \mu \frac{{\partial }^{2}u}{\partial x^2}=\frac{\mu \text{V}}{{\text{L}}^2}\\ \mu \frac{{\partial }^{2}u}{\partial y^2}=\frac{\mu \text{V}}{{\text{h}}_{\text{0}}{}^2}\end{array}$

Now,

The second term of viscosity is very smaller as compared to the second viscous term. So, we neglect this term. (h_o<

Now,

  $\rho u\frac{\partial u}{\partial x}+\rho \text{v}\frac{\partial v}{\partial y}=-\frac{\partial P}{\partial x}+\mu \frac{{\partial }^{2}u}{\partial y^2}$

We can see that the order of magnitude consists a factor of $\left(\frac{1}{\mathrm{Re}}\right)$. These pressure and viscous term contain a length scale of $\frac{\text{L}}{{\text{h}}_{\text{0}}}$. Both are very large for the flow to be creeping. If the value of Reynolds number is 10 and (h_o<