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An oil gusher shoots crude oil 25.0 m into the air through a pipe with a 0.100-m diameter. Neglecting air resistance but not the resistance of the pipe, and assuming laminar flow, calculate the gauge pressure at the entrance of the 50.0-m-long vertical pipe. Take the density of the oil to be 900 kg/m3 and its viscosity to be 1.00 (N/m2 )⋅s (or 1.00 Pa⋅s). Note that you must take into account the pressure due to the 50.0-m column of oil in the pipe

User Gratzy
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1 Answer

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Answer:3764.282 KPa

Step-by-step explanation:

Given gusher shoots oil at h=25 m

i.e. the velocity of jet is

v=\sqrt{2gh}[/tex]

v=22.147 m/s

Now the pressure loss in pipe is given by hagen poiseuille equation


\Delta P=(32L\mu v)/(D^2)


\Delta P=(32* 50* 22.147* 1)/(10^(-2))


\Delta P=3543.557 KPa

For 25 m head in terms of Pressure


\Delta P_2=\rho * g* h=220.725 KPa

Total Pressure=
\Delta P+\Delta P_2=3543.557+220.725=3764.282 KPa

User AxelWass
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