91Ó°ÊÓ

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$\mathit{Δ}\mathit{P}$. Finally, using the value of $\mathit{Δ}\mathit{P}$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,$P\left(4\frac{1}{2}\mathrm{ÒÏ}gy+\right.$constant

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

iii) Work done at the steady flow of fluid, $W=\mathrm{Δ}P×V$.

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 $\mathrm{ÒÏ}$ 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 $a_3$

$a_1=\mathrm{Ï€}{r}_1^2$

$=\mathrm{Ï€}(2R{)}^2$

$=4\mathrm{Ï€}{R}^2$

Here $A_m=\mathrm{Ï€}{R}^2$

Then,

$a_3=\mathrm{Ï€}{r}_3^2$

$=\mathrm{Ï€}(3R{)}^2$

$=9\mathrm{Ï€}{R}^2$

Putting the values,

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

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

$-\frac{\mathrm{π}{R}^2×0.500}{4\mathrm{π}{R}^2}$

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

Similarly,

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

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

$-\frac{\mathrm{π}{R}^2×0.500}{49\mathrm{π}{R}^2}$

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

According to Bernoulli's equation,

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

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

Simplifying the above expression,

$Pâˆ\P Äp_1=V\frac{1}{2}\left({}_1{}^2-{{}^2}^2\right)$

$\mathrm{Δ}\mathrm{θ}âˆ\P Ä\frac{1}{2}\left(1^2-m^2\right)$

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

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

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

$=-117.1875\mathrm{Pa}$

Work done can be calculated as:

$W_1=\mathrm{Âì}P×V$

$=(117.1875)×0.400$

$=46.875\mathrm{J}$

Here, the water flows steadily and $\mathrm{Δ}P$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,

$PV\frac{1}{2}{p}_m^2={\mathrm{ÒÏ}}_3∨+\frac{1}{2}{3}^2$

$p_3∨P=v^1\left(m^2-3^2\right)$

$\mathrm{Δ}\mathrm{β}âˆ\P Ä\frac{1}{2}\left(v_m{}^2-3^2\right)$

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

$\mathrm{Δ}P=\frac{1}{2}×1000×(0.246864)$

$=+123.432\mathrm{Pa}$

Work done can be calculated as:

$W_2=\mathrm{Δ}P×V$

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

$=+49.373\mathrm{J}$

Here, water flows steadily and $\mathrm{Δ}P$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,

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

$=\frac{1}{2}×399×(-0.012489)$

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 $-2.5\mathrm{J}$.