Water is moving with a speed of $5.0\mathrm{m}/\mathrm{s}$through a pipe with a cross-sectional area of$4.0\backslash mathrm\{~cm\}^\left\{2\right\}.$The water gradually descends $10\mathrm{m}$as the pipe cross-sectional area increases to $8.0{\mathrm{cm}}^2$.

(a) What is the speed at the lower level?

(b) If the pressure at the upper level is $1.5\times {10}^5\mathrm{Pa}$, What is the pressure at the lower level?

i) The speed of the water at the upper level, $v_u=5.0\mathrm{m}/\mathrm{s}$.

ii) The area of the cross-section at the upper level, $A_u=4.0{\mathrm{m}}^2$.

iii) The depth of the lower level, $D=10\mathrm{m}$.

iv) The area of the cross-section at the lower level,role="math" localid="1657547338914" $A\_\left\{1\right\}=8.0\backslash mathrm\{~m\}^\left\{2\right\}.$

v) The pressure at the upper level,$p_u=1.5\times {10}^5\mathrm{Pa}$.

Determine the speed of the water at the lower level using the equation of continuity. Then, using Bernoulli's equation, find the pressure of water at the lower level. According to Bernoulli's equation, as the speed of a moving fluid increases, the pressure within the fluid decreases.

Formulae are as follows:

$p\forall \frac{1}{2}\rho g^{2}h+$constant

$Av=$constant

where, p is pressure, v is velocity , h is height, g is an acceleration due to gravity, h is height , A is area and P is density .

The water flow through the pipe obeys the equation of continuity,

$A_u{V}_u=A_1{V}_1$

Hence,

$\backslash begin\left\{aligned\right\}v\_\left\{i\right\}\&=\backslash frac\left\{A\_\right\{u\}v\_\{u\left\}\right\}\left\{A\_\right\{l\left\}\right\}\backslash \backslash \&=\backslash frac\{4.0\backslash mathrm\{~m\}^\{2\}\backslash times5.0\backslash mathrm\{~m\}/\backslash mathrm\{s\left\}\right\}\{8.0\backslash mathrm\{~m\}^\{2\left\}\right\}\backslash \backslash \&=2.5\backslash mathrm\{~m\}/\backslash mathrm\left\{s\right\}\backslash end\left\{aligned\right\}$

Hence, the speed of the water at the lower level is$2.5\backslash mathrm\{~m\}/\backslash mathrm\left\{s\right\}$

The water flow obeys Bernoulli's principle.

So,

$py+\frac{1}{2}\rho gh+p_u=\rho y+\frac{1}{2}\rho gh+$

Simplifying,

$p_l\forall P_u*\frac{1}{2}\left(o{q}^2{h}_1^2\right)\ne \left({}_u-1\right)$

Where, $h_u-h_t=D=10\mathrm{m}$and $\rho =$Density of water$=1000\mathrm{kg}/{\mathrm{m}}^3$.

Thus, putting the values,

$\backslash begin\left\{aligned\right\}p\_\left\{i\right\}\&=\backslash left(1.5\backslash times10^\{5\}\backslash mathrm\{~Pa\}\backslash right)+\backslash frac\left\{1\right\}\left\{2\right\}1000\backslash mathrm\{~kg\}/\backslash mathrm\left\{m\right\}^\left\{3\right\}\backslash times\backslash left\left(\right(5\backslash mathrm\{~m\}/\backslash mathrm\left\{s\right\})^\{2\}-(2.5\backslash mathrm\{~m\}/\backslash mathrm\left\{s\right\})^\{2\}\backslash right)+1000\backslash mathrm\{~kg\}/\backslash mathrm\left\{m\right\}^\left\{3\right\}\backslash times\backslash left(9.8\backslash mathrm\{~m\}/\backslash mathrm\{s\}^\{2\}\backslash right)\backslash times(10\backslash mathrm\{~m\left\}\right)\backslash \backslash \&=2.57\backslash times10^\left\{5\right\}\backslash mathrm\{~Pa\}\backslash \backslash \&\backslash approx2.6\backslash times10^\left\{5\right\}\backslash mathrm\{~Pa\}\backslash end\left\{aligned\right\}$