water is flowing through a pipe with a velocity of 0.212m/s and 0.477m/s at section 1-1 (bottom end) and section 2-2(upper end) respectively as shown in figure-03. The pressure at section 1-1 is 350KN/m². Determine pressure at section 2-2, if the difference in elevation between section 1-1 and section 2-2 is 83.5ft.
Answers
Step-by-step explanation:
Velocity of water at lower end, \sf{v_1=0.212\ m\,s^{-1}.}v
1
=0.212 ms
−1
.
Velocity of water at upper end, \sf{v_2=0.477\ m\,s^{-1}.}v
2
=0.477 ms
−1
.
Pressure at lower end, \sf{P_1=350\ kN\,m^{-2}=3.5\times10^5\ N\,m^{-2}.}P
1
=350 kNm
−2
=3.5×10
5
Nm
−2
.
Height difference between the ends, \sf{\Delta h=83.5\ ft=25.45\ m.}Δh=83.5 ft=25.45 m.
By Bernoulli's Theorem,
\sf{\longrightarrow P_1+\dfrac{1}{2}\,\rho(v_1)^2=P_2+\dfrac{1}{2}\,\rho(v_2)^2+\rho g\Delta h}⟶P
1
+
2
1
ρ(v
1
)
2
=P
2
+
2
1
ρ(v
2
)
2
+ρgΔh
where \sf{\rho=10^3\ kg\,m^{-3}}ρ=10
3
kgm
−3
is density of water and \sf{g=10\ m\,s^{-2}}g=10 ms
−2
is acceleration due to gravity.
Then, pressure at upper end,
\sf{\longrightarrow P_2=P_1+\dfrac{1}{2}\,\rho[(v_1)^2-(v_2)^2]-\rho g\Delta h}⟶P
2
=P
1
+
2
1
ρ[(v
1
)
2
−(v
2
)
2
]−ρgΔh
\sf{\longrightarrow P_2=3.5\times10^5+\dfrac{1}{2}\times10^3[0.212^2-0.477^2]-10^3\times10\times25.45}⟶P
2
=3.5×10
5
+
2
1
×10
3
[0.212
2
−0.477
2
]−10
3
×10×25.45
\sf{\longrightarrow\underline{\underline{P_2=95.4\ kN\,m^{-2}}}}⟶
P
2
=95.4 kNm
−2
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