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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

Answer :

Answer:3764.282 KPa

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

[tex]\Delta P=\frac{32L\mu v}{D^2}[/tex]

[tex]\Delta P=\frac{32\times 50\times 22.147\times 1}{10^{-2}}[/tex]

[tex]\Delta P=3543.557 KPa[/tex]

For  25 m head in terms of Pressure

[tex]\Delta P_2=\rho \times g\times h=220.725 KPa[/tex]

Total Pressure=[tex]\Delta P+\Delta P_2[/tex]=3543.557+220.725=3764.282 KPa

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