# 5.6: The Details Picture – Velocity Area Relationship

- Page ID
- 707

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*Fig. 5.8 Control volume usage to calculate local averaged velocity in three coordinates.*

The integral approach is intended to deal with the "big'' picture. Indeed the method is used in this part of the book for this purpose. However, there is very little written about the usability of this approach to provide way to calculate the average quantities in the control system. Sometimes it is desirable to find the averaged velocity or velocity distribution inside a control volume. There is no general way to provide these quantities. Therefore an example will be provided to demonstrate the use of this approach. Consider a container filled with liquid on which one exit opened and the liquid flows out as shown in Figure 5.8. The velocity has three components in each of the coordinates under the assumption that flow is uniform and the surface is straight . The integral approached is used to calculate the averaged velocity of each to the components. To relate the velocity in the \(z\) direction with the flow rate out or the exit the velocity mass balance is constructed. A similar control volume construction to find the velocity of the boundary velocity (height) can be carried out. The control volume is bounded by the container wall including the exit of the flow. The upper boundary is surface parallel to upper surface but at \(Z\) distance from the bottom. The mass balance reads

\[
\label{mass:eq:containerZs}

\int_V \dfrac{d\rho}{dt} \,dV + \int_A U_{bn}\,\rho\,dA + \int_A U_{rn}\,\rho\,dA = 0

\]

\[
\label{mass:eq:containerZrho}

\int_A U_{rn}\,\rho\,dA = 0

\]
In the container case for uniform velocity equation (??) becomes

\[
\label{mass:eq:Zu}

U_z\,A = U_{e}\,A_e \Longrightarrow U_z = - \dfrac{A_e}{A} U_e

\]
It can be noticed that the boundary is not moving and the mass inside does not change this control volume. The velocity \(U_z\) is the averaged velocity downward.

*Fig. 5.9 Control volume and system before and after the motion.*

The \(x\) component of velocity is obtained by using a different control volume. The control volume is shown in Figure 5.9. The boundary are the container far from the flow exit with blue line projection into page (area) shown in the Figure 5.9. The mass conservation for constant density of this control volume is

\[
\label{mass:eq:containerXs}

- \int_A U_{bn}\,\rho\,dA + \int_A U_{rn}\,\rho\,dA = 0

\]

\[
\label{mass:eq:containerXU}

\int_

```
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```

\] Where \({A_{x}}^{-}\) is the area shown the Figure under this label. The area \(A_{yz}\) referred to area into the page in Figure under the blow line. Because averaged velocities and constant density are used transformed equation (??) into

\[
\label{mass:eq:containerXUa}

\dfrac{A_e}{A} {A_{x}}^{-} U_e + U_{x}\,\overbrace{Y(x)\,h}^{A_{yz}} = 0

\]
Where \(Y(x)\) is the length of the (blue) line of the boundary. It can be notice that the velocity, Ux is generally increasing with \(x\) because \({A_{x}}^{-}\) increase with \(x\). The calculations for the \(y\) directions are similar to the one done for \(x\) direction. The only difference is that the velocity has two different directions. One zone is right to the exit with flow to the left and one zone to left with averaged velocity to right. If the volumes on the left and the right are symmetrical the averaged velocity will be zero.

*Fig. 5.10 Circular cross section for finding \(U_x\) and various cross sections.}*

Example 5.12

Calculate the velocity, \(U_x\) for a cross section of circular shape (cylinder).

Solution 5.12

The relationship for this geometry needed to be expressed. The length of the line \(Y(x)\) is

\[
\label{Ccontainer:yLength}

Y(x) = 2\,r\, \sqrt{ 1 - \left(1-\dfrac{x}{r}\right)^2 }

\]

\[
\label{Ccontainer:yLengthalpha}

Y(x) = 2 \, r\,\sin\alpha

\]
Since this expression is simpler it will be adapted. When the relationship between radius angle and \(x\) are

\[
\label{Ccontainer:rx}

x = r (1 -\sin\alpha)

\]
The area \({A_{x}}^{-}\) is expressed in term of \(\alpha\) as

\[
\label{Ccontainer:area}

{A_{x}}^{-} = \left( \alpha - \dfrac{1}{2\dfrac{}{}},\sin(2\alpha) \right) r^2

\]
Thus, the velocity, \(U_x\) is

\[
\label{Ccontainer:Uxs}

\dfrac{A_e}{A} \left( \alpha - \dfrac{1}{2\dfrac{}{}}\,\sin(2\alpha) \right) r^2\, U_e +

U_{x}\, 2 \, r\,\sin\alpha\,h = 0

\]

\[
\label{Ccontainer:Uxf}

U_x = \dfrac{A_e}{A} \dfrac{r}{h}

\dfrac{\left( \alpha - \dfrac{1}{2\dfrac{}{}}\,\sin(2\alpha) \right) }{\sin\alpha} \, U_e

\]
Averaged velocity is defined as

\[
\label{Ccontainer:UxaveDef}

\overline{U_x} = \dfrac{1}{S}\int_S U dS

\]
Where here \(S\) represent some length. The same way it can be represented for angle calculations. The value \(dS\) is \(r\cos\alpha\). Integrating the velocity for the entire container and dividing by the angle, \(\alpha\) provides the averaged velocity.

\[
\label{Ccontainer:Uxtotal}

\overline{U_x} = \dfrac{1}{2\,r} \int_0^{\pi}\dfrac{A_e}{A} \dfrac{r}{h}

\dfrac{\left( \alpha - \dfrac{1}{2\dfrac{}{}}\,\sin(2\alpha) \right) }{\tan\alpha} \, U_e\, r\, d\alpha

\]
which results in

\[
\label{Ccontainer:Uxtotala}

\overline{U_x} = \dfrac{\left( \pi -1\right)}{4} \dfrac{A_e}{A} \dfrac{r}{h} \, U_e

\]

Example 5.13

Fig. 5.11 \(y\) velocity for a circular shape

What is the averaged velocity if only half section is used. State your assumptions and how it similar to the previous example.

Solution 5.13

The flow out in the \(x\) direction is zero because symmetrical reasons. That is the flow field is a mirror images. Thus, every point has different velocity with the same value in the opposite direction. The flow in half of the cylinder either the right or the left has non zero averaged velocity. The calculations are similar to those in the previous to example 5.12. The main concept that must be recognized is the half of the flow must have come from one side and the other come from the other side. Thus, equation (??) modified to be

\[
\label{ene:eq:containeryU}

\dfrac{A_e}{A} {A_{x}}^{-} U_e + U_{x}\,\overbrace{Y(x)\,h}^{A_{yz}} = 0

\]

\[
\label{Ccontainer:Uytotal}

\overline{U_x} = \dfrac{1}{2\,r} \int_0^{\pi/2}\dfrac{A_e}{A} \dfrac{r}{h}

\dfrac{\left( \alpha - \dfrac{1}{2\dfrac{}{}}\,\sin(2\alpha) \right) }{\tan\alpha} \, U_e\, r\, d\alpha

\]
which results in

\[
\label{Ccontainer:Uytotala}

\overline{U_x} = \dfrac{\left( \pi -2\right)}{8} \dfrac{A_e}{A} \dfrac{r}{h} \, U_e

\]

## Contributors and Attributions

Dr. Genick Bar-Meir. Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or later or Potto license.