91Ó°ÊÓ

According to Bernoulli's principle, when a non-viscous incompressible fluid passing through non uniform cross-sectional area, the sum of pressure energy $p$, kinetic energy $\left(\frac{1}{2}\mathrm{ÒÏ}{v}^2\right)$and potential energy $\left(\mathrm{ÒÏ}gh\right)$per unit mass of fluid between two points remains constant. That is $P+\frac{1}{2}\mathrm{ÒÏ}{v}^2+\mathrm{ÒÏ}gv=\text{Constant}$is the density of the liquid, $\mathit{v}$is the speed of the fluid flow, $g$is the acceleration due to gravity $\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)$is the height of the fluid.

Unit conversion:

Convert the units of diameter of the pipe from centimeter to meter

$d_1=16.8\mathrm{cm}\left(\frac{1.0\mathrm{m}}{100\mathrm{cm}}\right)$

$=16.8×{10}^{−2}\mathrm{m}$

$d_2=20\mathrm{cm}\left(\frac{1.0\mathrm{m}}{100\mathrm{cm}}\right)$

$=20×{10}^{−2}\mathrm{m}$

Convert the unit of pressure from kilo-Pascal's to Pascal's

$P_0=50k\mathrm{ka}\left(\frac{{10}^3\mathrm{Pa}}{1.0k\mathrm{ka}}\right)$

$=5×{10}^4\mathbf{Pa}$

The following diagram shows the complicated system of liquid flow from point A to Point B. But the Bernoulli's Equation dictates the parameters between point A and point B.

a.) The Bernoulli's equation gives us:

$p_0+\frac{1}{2}\mathrm{ÒÏ}{v}_0^2=p_1+\frac{1}{2}\mathrm{ÒÏ}{v}_1^2$

Here $p_o$is the pressure narrower section; $v_o$is the liquid flow speed in the narrower section; $p_1$is the pressure wider section; $v_1$is the liquid flow speed in the wider section; $\mathrm{ÒÏ}$is the mass density of the liquid.

The idea of continuity has the following equation:

$v_0\mathrm{Ï€}{\left(q^d/2\right)}^2=v_1\mathrm{Ï€}{\left(\frac{d}{2}\right)}^2$

Here $d_0$is the diameter of pipe in the narrower section; $d_1$the diameter of pipe in the wider section. Divide the above equation by $\frac{\mathrm{Ï€}}{4}$, we get:

$v_0{d}_0^2=v{d}_1^2$

$u_1=\frac{v_0{d}_0^2}{v_1^2}$

Substitute $u_1=\frac{u_0{d}_0^2}{d_1^2}$in equation (1)

$p_0+\frac{1}{2}{\mathrm{ÒÏ}}_0^2=p_1+\frac{1}{2}\mathrm{ÒÏ}{\left(\frac{v_0{d}_0^2}{d_1^2}\right)}^2$

$p_1=p_0+\frac{1}{2}\mathrm{ÒÏ}{v}_0^2\left(1−\frac{d_0^4}{d_1^4}\right)$

Hence the pressure$p_0+\frac{1}{2}\mathrm{ÒÏ}{v_0}^2\left(1−\frac{d_0^4}{d_1^4}\right)$

Substitute $5×{10}^4{\mathrm{Pa}}^{\text{for}}{\mathit{p}}_0,16.8×{10}^{−2}\mathrm{m}\text{for}{\mathit{d}}_1,20×{10}^{−2}\mathrm{m}\text{for}{\mathit{d}}_2\mathrm{cm}\text{and}$${\text{1000kgfm}}^3\text{for}\mathit{ÒÏ}\text{and}10\mathrm{m}/\mathrm{s}\text{for}{\mathit{v}}_{\mathbf{0}}$in equation

$p_1=\left(50×{10}^3\mathrm{Pa}\right)+\frac{1}{2}\left({10}^3{\mathrm{kgim}}^3\right)(10\mathrm{m}/\mathrm{s})\left(1−\frac{{\left(16.8×{10}^{−2}\mathrm{m}\right)}^4}{{\left(20×{10}^{−2}\mathrm{m}\right)}^4}\right)$

$=75×{10}^3\mathrm{Pa}\left(\frac{1.0\mathbf{kPa}}{{10}^3\mathrm{Pa}}\right)$

$=75\mathbf{kPa}$