Chapter 13, Problem 63P

To determine

To Derive:The Bernoulli Equation in more general way

To Show: $P_1+\rho g{h}_1+\frac{1}{2}\rho v_1^2=P_2+\rho g{h}_2+\frac{1}{2}\rho v_2^2$ .

Explanation of Solution

Introduction:

Bernoulli’s equation describes the quantitative relationship between the pressure and velocity in fluids. This is derived by Daniel Bernoulli, a Swiss Physicist and Mathematician in the year 1738.

  

FIGURE: 1

Consider the fluid flowing through a tube which varies in elevation as well as in cross-sectional area, as shown in the figure 1.The work-energy theorem can be applied to a parcel of fluid between the points 1 and 2. Let $\Delta t$ be the time taken by this parcel of liquid to move along the tube to the region between the points $1\text{'}$ and $2\text{'}$ .Let $\Delta V$ be the volume of the fluid passing through the point $1\text{'}$ in time $\Delta t$ . The same volume of fluid passes through the point $2\text{'}$ in the same time. Let $\rho$ be the density of the fluid. Then,the mass of the fluid $\left(\Delta m\right)$ with volume $\Delta V$ can be written as $\Delta m=\rho \Delta V$ .

The expression for work-energy theorem can be written as:

  $W_{total}=\Delta U+\Delta K$  $\left(1\right)$

Where, $\Delta U$ is change in the potential energy of the parcel which is given by,

  $\Delta U=\left(\Delta m\right)g{h}_2-\left(\Delta m\right)g{h}_1$

  $=\left(\Delta m\right)g\left(h_2-h_1\right)$

Substituting for $\Delta m$ ,

  $\Delta U=\rho \Delta Vg\left(h_2-h_1\right)$  $\left(2\right)$

Where, $g$ is acceleration due to gravity, $g=9.81{\text{ m/s}}^2$

  $\Delta K$ is change in kinetic energy of the parcel which is given by,

  $\Delta K=\frac{1}{2}\left(\Delta m\right)v_2^2-\frac{1}{2}\left(\Delta m\right)v_1^2$

  $=\frac{1}{2}\left(\Delta m\right)\left(v_2^2-v_1^2\right)$

Substituting for $\Delta m$ ,

  $\Delta K=\frac{1}{2}\rho \Delta V\left(v_2^2-v_1^2\right)$  $\left(3\right)$

The fluid behind the parcel (that is fluid in parcel’s left in the diagram) pushes on the parcel with a force of magnitude $F_1=P_1{A}_1$ and hence work done by the fluid behind the parcel can be expressed as,

  $\begin{array}{l}{W}_1=F_1\Delta x_1\\ W_1=P_1{A}_1\Delta x_1\end{array}$

Since area $\times$ length = volume, above expression becomes,

  $W_1=P_1\Delta V$  $\left(4\right)$

Where, $P_1$ is the pressure at the point $1$ .

The fluid in the front of the parcel (that is fluid in parcel’s left in the diagram) pushes on the parcel with a force of magnitude $F_2=P_2{A}_2$ and hence work done by the fluid in the front of the parcel can be expressed as,

  $\begin{array}{l}{W}_2=-F_2\Delta x_2\\ W_2=-P_2{A}_2\Delta x_2\end{array}$

Since area $\times$ length = volume, above expression becomes,

  $W_2=-P_2\Delta V$  $\left(5\right)$

Where, $P_1$ is the pressure at the point $2$ .

This work done by the fluid in the front of the parcel is negative because the applied force and displacement are in the direction opposite to the flow of fluid.

Thus, the total work done is,

  $\begin{array}{l}{W}_{total}=W_1+W_2\\ W_{total}=P_1\Delta V-P_2\Delta V\\ W_{total}=\left(P_1-P_2\right)\Delta V\end{array}$  $\left(6\right)$

Substituting for $W_{total}$ , $\Delta U$ , and $\Delta K$ in equation $\left(1\right)$ ,

  $\left(P_1-P_2\right)\Delta V=\rho \Delta Vg\left(h_2-h_1\right)+\frac{1}{2}\rho \Delta V\left(v_2^2-v_1^2\right)$

  $\Rightarrow \left(P_1-P_2\right)=\rho g\left(h_2-h_1\right)+\frac{1}{2}\rho \left(v_2^2-v_1^2\right)$

On rearranging,

  $P_1+\rho g{h}_1+\frac{1}{2}\rho v_1^2=P_2+\rho g{h}_2+\frac{1}{2}\rho v_2^2$

This is the Bernoulli’s equation for steady, non-viscous flow of an incompressible fluid.