Question

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?

Short answer

Net work done $W_{\text{net}}$ on $0.400{\mathrm{m}}^3$ of the water as it moves from the left section to the right section is $-2.50\mathrm{J}$.

Step-by-step solution

Given information

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)

Understanding the concept of Bernoulli’s equation and equation of continuity

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.

Determining the net work done Wnor on 0.400 m3 of the water as it moves from the left section to the right section

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,

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

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

$=-2.492\mathrm{J}$$=-2.5\mathrm{J}$

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 $-2.5\mathrm{J}$.