91Ó°ÊÓ

i) Fluid is fresh water, $\mathrm{ÒÏ}-1000\mathrm{kg}/{\mathrm{m}}^3$

ii) The scale on the vertical axis, $\mathrm{Δ}{P}_s=300\mathrm{kN}/{\mathrm{m}}^2=300×{10}^3\mathrm{N}/{\mathrm{m}}^2$

iii) The values of $\mathrm{Δ}P$from the graph,

$\mathrm{Δ}P=0$at $A_1^{-2}=16{\mathrm{m}}^{-4}$

$\mathrm{Δ}P=-300\mathrm{kN}/{\mathrm{m}}^2=-300×{10}^3\mathrm{N}/{\mathrm{m}}^2$at${A_1}^{-2}=0{\text{m}}^{-\text{4}}$

$\mathrm{Δ}P=300\mathrm{kN}/{\mathrm{m}}^2=300×{10}^3\mathrm{N}/{\mathrm{m}}^2{A}_1^{-2}=32{\mathrm{m}}^{-4}$

From the graph, note that the pressures are equal when the value of area square is inverse ${\mathit{A}}_{\mathbf{1}}^{\mathbf{-}\mathbf{2}}$is $\mathbf{16}{\mathbf{}\mathbf{m}}^{\mathbf{-}\mathbf{4}}$, and using this condition, find the value of ${\mathit{A}}_{\mathbf{2}}$. By using Bernoulli's equation, find the speed ${\mathit{V}}_{\mathbf{1}}$of the flow of fresh water through the pipe in section 1 . Finally, by using the value of ${\mathit{V}}_{\mathbf{1}}$, find the value of the volume rate flow $\mathit{Q}$. 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, $\mathrm{β}\mathrm{V}\frac{1}{2}\mathrm{ÒÏ}{g}^{2}y+$ constant

ii) Equation of continuity, $av=AV=$ constant

iii) The volume rate flow $Q,Q=VA$

Where, $P$is pressure, $V$and $V$are velocities, $y$is distance, $g$is an acceleration due to gravity, $h$is height, $A,a$are areas, $Q$is the rate of flow, and $\mathrm{ÒÏ}$is density.

From the graph, note that, $\mathrm{Δ}P=0$at $A_1^{-2}=16{\mathrm{m}}^{-4}$:

$A_1^2=\frac{1}{16}$

$A_1=\frac{1}{\sqrt{16}}$

$A_1=\frac{1}{4}$

$A_1-0.25{\mathrm{m}}^2$

By applying Bernoulli's equation,

${\mathrm{β}}_1\sqrt{}+\frac{1}{2}\mathrm{ÒÏ}{g}_y^2+P{}_1={\mathrm{ÒÏ}}_{2}V+\frac{1}{2}\mathrm{ÒÏ}{g}^2+$constant

role="math" localid="1657639855451" $\{P_1=P_2$and $\left.y_1=y_2\right\}$

$\frac{1}{2}\mathrm{ÒÏ}{V}_1^2=\frac{1}{2}\mathrm{ÒÏ}{V}_2^2$

$V_1=V_2$

By using the equation of continuity,

$V_1{A}_1=V_2{A}_2$

$A_1=A_2$

$A_2=A_1$

Hence, the value of $A_2$ is $0.25{\mathrm{m}}^2$

From the graph,

$\mathrm{Δ}P=-300\mathrm{kN}/{\mathrm{m}}^2=-300×{10}^3\mathrm{N}/{\mathrm{m}}^2$at $A_1^{-2}=0{\mathrm{m}}^{-4}$

By applying Bernoulli's equation,

$\mathrm{ÒÏ}∨4\frac{1}{2}\mathrm{ÒÏ}{g}_{\mathrm{ν}}^2+P_1={\mathrm{ÒÏ}}_2/4\frac{1}{2}\mathrm{ÒÏ}{g}_y^{\hat{y}}+$2 $=$constant

$P_1+\frac{1}{2}{P}_1^2={\mathrm{ÒÏ}}_2/4+\frac{1}{2}$ $2^2$

$P_{2}V-P_1={\mathrm{ÒÏ}}_2^{1}V{1}^2-\frac{1}{2}{2}^2$

$\mathrm{Δ}Pâˆ\P Ä\frac{1}{2}\left(i^2-2^2\right)$

By applying the equation of continuity,

$\left\{A_1^{-2}=0{\mathrm{m}}^{-4}\right\}$

$V_1{A}_1=V_1,A$

$V_1=0$

Putting this value in Bernoulli’s equation,

$\mathrm{Δ}\mathrm{β}V=\frac{1}{2}{2}^2$

$V_2^2=\frac{2\mathrm{Δ}P}{\mathrm{ÒÏ}}$

$V_2=\sqrt{\frac{2\mathrm{Δ}P}{\mathrm{ÒÏ}}}$

$=\sqrt{\frac{2×300×{10}^3\mathrm{Pa}}{1000\mathrm{kg}/{\mathrm{m}}^3}}$

$=24.4949\mathrm{m}/\mathrm{s}$

$\left\{\begin{array}{c}\text{B}{V}_2\text{is speed of water,}\\ ∴V_2\text{is positive}\end{array}\right\}$

The volume rate flow $Q$is,

$Q=V_2{A}_2$

$=24.4949\mathrm{m}/\mathrm{s}×0.25{\mathrm{m}}^2$

$=6.1237{\mathrm{m}}^3/\mathrm{s}$

Hence, the volume flow rate $Q$ is $6.12{\mathrm{m}}^3/\mathrm{s}$