Figure 14-25 shows four arrangements of pipes through which water flows smoothly toward the right. The radii of the pipe sections are indicated. In which arrangements is the net work done on a unit volume of water moving from the leftmost section to the rightmost section (a) zero, (b) positive, and (c) negative?

Radii for input and output sections are shown in the figure.

This problem is based on the work-energy theorem. The input & output section of the pipe is placed at the same level, so work done is due to a change in velocity only.

Formula:

The net work done by the body due to change in energy, $W=∆PE+∆KE$ (i)

From the equation of continuity through a pipe, $A_i{v}_i=A_0{v}_0$ (ii)

According to work-energy theorem that is from equation (i), we can get the net work as follows:

$$\begin{gathered} W=∆K.E. \\ =\left(\frac{1}{2}m{v}_0^2\right)-\left(\frac{1}{2}m{v}_i^2\right).................\left(a\right) \end{gathered}$$

Now, using the velocity equation from equation (ii), we get that

$v_0=A_i{v}_i/A_0$

Now, substituting the above value in equation (a), we get the net work equation as follows:

role="math" localid="1657250271274" $$\begin{gathered} W=\left(\frac{A_i}{A_0}\right)\left(\frac{1}{2}m{v}_i^2\right)-\left(\frac{1}{2}m{v}_i^2\right) \\ =K.E{.}_i\left(\frac{r_i^4}{r_0^4}\right).........................\left(b\right)(\because A={\mathrm{\pi r}}^2) \end{gathered}$$

For cases1and4, input area is the same as the output area.

Thus, velocities would be the same considering equation (b), that is$W_1=W_4=0J.$

Therefore, the net work done for cases 1 AND 4 is zero.

For case 2:$A_i>A_0$

Thus, from equation (b), we can see that the value of work done is positive.

Therefore, work done would be positive for situation 2.

For case 3: $A_I<A_0$

Thus, from equation (b), we can see that the value of work done is negative.

Therefore, work done would be negative for situation 3.