In Figure, water flows steadily from the left pipe section (radius ${\mathit{r}}_{\mathbf{1}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathit{R}$ ), through the middle section (radius $\mathit{R}$), and into the right section (radius localid="1657690173419" ${\mathit{r}}_{\mathbf{3}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\mathit{R}$). The speed of the water in the middle section is localid="1657690185115" $\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$. What is the net work done on localid="1657690178609" $\mathbf{0}\mathbf{.}\mathbf{400}{\mathbf{}\mathbf{m}}^{\mathbf{3}}$of the water as it moves from the left section to the right section?

i) Radius $r_1=2.00R$ (at left pipe section).

ii) Radius $r_3=3.00R$ (at right pipe section).

iii) Radius $R-R$ (at the middle section).

iv) Speed of water at the middle section, $V_m=0.500\mathrm{m}/\mathrm{s}$

v) The volume of the water, $V=0.400{\mathrm{m}}^3$ (it is constant for a steady flow of fluid)

By applying the equation of continuity, find the speed of the water flow at the left pipe section and right pipe section (i.e. ${}^{{\mathit{v}}_{\mathbf{t}}}$and ${\mathit{v}}_{\mathbf{3}}$). By putting the calculated values of ${\mathit{v}}_{\mathbf{7}}$and localid="1657686257320" ${\mathit{v}}_{\mathbf{3}}$in Bernoulli's equation, find the change in the pressure$\mathit{\Delta }\mathit{P}$. Finally, using the value of $\mathit{\Delta }\mathit{P}$in the formula for work done at the steady flow of the fluid, find the net work done $\left(W_{\text{net}}\right)$on $\mathbf{0}\mathbf{.}\mathbf{400}{\mathbf{}\mathbf{m}}^{\mathbf{3}}$of water as it moves from the left section to the right section. Also, by putting the values of ${\mathit{v}}_{\mathbf{1}}$and ${\mathit{v}}_{\mathbf{3}}$in the formula for network done, i.e.${\mathit{w}}_{\mathbf{\text{not}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{M}\left(v_3^2-v_1^2\right)$,find the net work done on the water. According to Bernoulli's equation, as the speed of a moving fluid increases, the pressure within the fluid decreases.

Formulae are as follows:

i) Bernoulli's equation,$P\left(4\frac{1}{2}\rho gy+\right.$constant

ii) Equation of continuity, $av=A{V}_s=$constant

iii) Work done at the steady flow of fluid, $W=\Delta P\times V$.

iv) Net work done on the water is, $W_{\text{net}}=W_1+W_2$.

$W_{\text{net}}=\frac{1}{2}M\left(v_3^2-v_1^2\right)$

Where, $P$is pressure, $V,V$are velocities, $y$is distance, $g$is an acceleration due to gravity, h is height, A, a are areas, W is work done, M is mass and $\rho$ is density.

According to the equation of continuity,

$a_1{V}_1=A_m{V}_m=a_3{V}_3=\text{constant}$

For $a_1,A_m$and $a_3$

$a_1=\pi r_1^2$

$=\pi (2R{)}^2$

$=4\pi R^2$

Here $A_m=\pi R^2$

Then,

$a_3=\pi r_3^2$

$=\pi (3R{)}^2$

$=9\pi R^2$

Putting the values,

$a_1{V}_1=A_m{V}_m$

$V_1=\frac{A_m{V}_m}{a_1}$

$-\frac{\pi R^2\times 0.500}{4\pi R^2}$

$=0.125\mathrm{m}/\mathrm{s}$

Similarly,

$a_m{V}_m=A_3{V}_3$

$V_3=\frac{A_m{V}_m}{a_3}$

$-\frac{\pi R^2\times 0.500}{49\pi R^2}$

$=0.056\mathrm{m}/\mathrm{s}$

According to Bernoulli's equation,

$\text{As}\left\{\begin{array}{lll}{y}_1& Y_{im}& y_3\end{array}\right\}$

$p_{\psi }/+\frac{1}{2}{P}_1^2=\rho V+\frac{1}{2}{p}_m^2=\rho v+\frac{1}{2}{3}^2=$constant

Simplifying the above expression,

$P\forall p_1=V\frac{1}{2}\left({}_1{}^2-{{}^2}^2\right)$

$\Delta \theta \forall \frac{1}{2}\left(1^2-m^2\right)$

$=\frac{1}{2}\times 1000\mathrm{kg}/{\mathrm{m}}^3\times \left((0.125\mathrm{m}/\mathrm{s}{)}^2-(0.500\mathrm{m}/\mathrm{s}{)}^2\right)$

$\Delta P=\frac{1}{2}\times 1000\mathrm{kg}/{\mathrm{m}}^3\times \left(0.015625(\mathrm{m}/\mathrm{s}{)}^2-0.25(\mathrm{m}/\mathrm{s}{)}^2\right)$

$=\frac{1}{2}\times 1000\times (-0.2344)\mathrm{Pa}$

$=-117.1875\mathrm{Pa}$

Work done can be calculated as:

$W_1=\Lambda P\times V$

$=(117.1875)\times 0.400$

$=46.875\mathrm{J}$

Here, the water flows steadily and $\Delta P$is negative, therefore, $W$is positive, and work is done by the water,

$W_1-46.875\mathrm{J}$

Similarly, according to Bernoulli's equation,

$PV\frac{1}{2}{p}_m^2={\rho }_3\vee +\frac{1}{2}{3}^2$

$p_3\vee P=v^1\left(m^2-3^2\right)$

$\Delta \beta \forall \frac{1}{2}\left(v_m{}^2-3^2\right)$

$=\frac{1}{2}\times 1000\times \left((0.500{)}^2-(0.056{)}^2\right)$

$\Delta P=\frac{1}{2}\times 1000\times (0.246864)$

$=+123.432\mathrm{Pa}$

Work done can be calculated as:

$W_2=\Delta P\times V$

$=(+123.432\mathrm{Pa})\times 0.400{\mathrm{m}}^3$

$=+49.373\mathrm{J}$

Here, water flows steadily and $\Delta P$is positive, therefore, $W$is negative, and work is done on the water,

$W_2=-49.373\mathrm{J}$

Net work done on the water is,

$W_{\text{net}}=W_1+W_2$

$=46.875\mathrm{J}-49.373\mathrm{J}$

$=-2.498\mathrm{J}$

$=-2.5\mathrm{J}$

There is another way to find the net work done on the water by using the following relation,

$W_{\text{not}}=\frac{1}{2}M\left(v_3^2-v_1^2\right)$

It is known that,$0.400{\mathrm{m}}^3$of water has a mass of $M=399\mathrm{kg}$and putting the values, $v_1=0.125\mathrm{m}/\mathrm{s}$and $v_3=0.056\mathrm{m}/\mathrm{s}$in above relation,

$W_{\text{net}}=\frac{1}{2}\times 399\mathrm{kg}\times \left((0.056\mathrm{m}/\mathrm{s}{)}^2-(0.125\mathrm{m}/\mathrm{s}{)}^2\right)$

$=\frac{1}{2}\times 399\times (-0.012489)$

Hence, net work done $\left(W_{n6t}\right)$ on $0.400{\mathrm{m}}^3$ of the water as it moves from the left section to the right section is $-2.5\mathrm{J}$.