Question

plzzzzz help me out..... ​

Answer 1

GiVeN :-

• Water is flowing through a pipe with a velocity of 0.212m/s at bottom .

• Water is flowing through a pipe with a velocity of 0.477m/s at upper .

• The pressure at bottom portion is 350kN/m² .

• Height difference between the two ends is 83.5 ft. .

To FiNd :-

• The pressure at upper portion .

SoLuTiOn :-

We know that,

➤ By Bernoulli's Theorem,

$\\ \huge\star$ $\bf\green{\triangle{P}\:=\:\dfrac{\rho\:(v^2_l\:-\:v^2_u)}{2}\:-\:\rho\:g\:\triangle{h}\:}$

$\bf{\red{\longrightarrow}}\:P_u\:-\:P_l\:=\:\dfrac{\rho\:(v^2_l\:-\:v^2_u)}{2}\:-\:\rho\:g\:\triangle{h}\:$

Where,

$\bf{P_l}$ = 350 kN/m² = 35 × 10⁴ N/m²

$\bf{v_l}$ = 0.212 m/s

$\bf{v_u}$ = 0.477 m/s

$\bf{\rho}$ = 10³ kg/m³ [density of water ]

$\bf{g}$ = 10 m/s²

$\bf{\triangle{h}}$ = 83.5 ft. = 25.45 m

$\bf{\red{\longrightarrow}}\:P_u\:=\:P_l\:+\:\dfrac{\rho\:(v^2_l\:-\:v^2_u)}{2}\:-\:\rho\:g\:\triangle{h}\:$

$\bf{\red{\longrightarrow}}\:P_u\:=\:35\times{10^4}\:+\:\dfrac{10^3\:\Big((0.212)^2\:-\:(0.477)^2\Big)}{2}\:-\:\Big(10^3\times{10}\times{25.45}\Big)\:$

$\bf{\red{\longrightarrow}}\:P_u\:=\:95.4\:kN.m^{-2}\:$

∴ The pressure at upper portion is '95.4 kN/m²' .

Answer 2

Answer :-

Given :-

• Water is flowing through a pipe with a velocity of 0.212m/s at bottom .

• Water is flowing through a pipe with a velocity of 0.477m/s at upper .

• The pressure at bottom portion is 350kN/m² .

• Height difference between the two ends is 83.5 ft. .

\begin{gathered}\\\end{gathered}

To Find :-

• The pressure at upper portion .

Solution :-

We know that,

✦ By Bernoulli's Theorem,

$\: </p><p>{ \bigstar} \: \: </p><p> </p><p> \begin{gathered}\bf{\triangle{P}\:=\:\dfrac{\rho\:(v^2_l\:-\:v^2_u)}{2}\:-\:\rho\:g\:\triangle{h}\:} \\\end{gathered} </p><p>$

$\: \begin{gathered}\bf{{\longrightarrow}}\:P_u\:-\:P_l\:=\:\dfrac{\rho\:(v^2_l\:-\:v^2_u)}{2}\:-\:\rho\:g\:\triangle{h}\: \\\end{gathered} </p><p></p><p>$

Where,

$\: \bf{P_l}</p><p> = 350 kN/m² = 35 × 10⁴ N/m²</p><p>$

$\: \bf{v_l}</p><p> </p><p> = 0.212 m/s</p><p>$

$\: </p><p>\bf{v_u}</p><p> </p><p> = 0.477 m/s$

$\: \bf{\rho} = 10³ kg/m³ [density of water ]$

$\: \bf{g} = 10 m/s²</p><p></p><p>$

$\: \bf{\triangle{h}} = 83.5 ft. = 25.45 m$

$\: \begin{gathered}\bf{{\longrightarrow}}\:P_u\:=\:P_l\:+\:\dfrac{\rho\:(v^2_l\:-\:v^2_u)}{2}\:-\:\rho\:g\:\triangle{h}\: \\\end{gathered}</p><p>$

$\:\begin{gathered}\bf{{\longrightarrow}}\:P_u\:=\:35\times{10^4}\:+\:\dfrac{10^3\:\Big((0.212)^2\:-\:(0.477)^2\Big)}{2}\:-\:\Big(10^3\times{10}\times{25.45}\Big)\: \\\end{gathered} </p><p>$

$\: \begin{gathered}\bf{{\longrightarrow}}\:P_u\:=\:95.4\:kN.m^{-2}\: \\\end{gathered} </p><p>$

∴ The pressure at upper portion is '95.4 kN/m²' .