Question

Darth Maul is getting rather advanced at Bernoulli's equations, so he decides to tackle a slightly more complicated situation. Suppose the gauge pressure on the top surface of the fluid in the vessel is given by P2. (Remember, gauge pressure is just how much higher the pressure is relative to the the pressure outside the vessel (i.e. 1 atm, in this case.) Darth Maul needs help determining a formula (in letters) for the speed, v1, at which liquid flows from the opening at the bottom into atmospheric pressure. He is willing to assume that the velocity of the liquid surface, v2, is approximately zero. Once you have this equation in letters, calculate the following: a) If P2 = 0.81 atm and y2 - y1 = 2.1 m, determine v1 for water on Earth. m/s b) If P2 = 1.19 atm and y2 - y1 = 4.2 m, determine v1 for dibromomethane (SG = 2.48) on the planet Tatooine (where g = 3.82 m/sec2). You can assume atmospheric pressure is the same on Tatooine as Earth. m/s c) If P2 = 0.58 atm and y2 - y1 = 4.9 m, determine v1 for gasoline (SG = 0.68) on the planet SpiceWorld (where g = 11.32 m/sec2). You can assume atmospheric pressure is the same on SpiceWorld as Earth. m/s

Answer 1

a )

given

P2 = 0.81 atm and y2 - y1 = 2.1 m

using Bernouli's principle

P₂ + $\rho$ g $\Delta$ y = 1/2 $\rho$ v₁²

0.81 X 10⁵ + ( 9.8 X 2.1 X 1000) = 0.5 X 1000 X v₁²  

v₁² = 203.16

v₁ = 14.25 m/sec

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b )

for dibromomethane

P2 = 1.19 atm

and y₂ - y₁ = 4.2 m = $\Delta$ y

SG = 2.48 on the planet Tatooine

so $\rho$ = 2.48 X 1000

g = 3.82 m/sec²  

( 1.19 X 10⁵ ) + ( 2.48 X 1000 X 3.82 X 4.2 ) = 1/2 X 2.48 X 1000 X v₁²

v₁² = 128.055

v₁ = 11.31 m/sec

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c)

for gasoline

P2 = 0.58 atm and y₂ - y₁ = 4.9 m = $\Delta$ y

SG = 0.68 on the planet SpiceWorld

g = 11.32 m/sec²   

0.58 X 10⁵ + ( 0.68 X 1000 X 11.32 X 4.9 ) = 1/2 X 0.68 X 1000 X v₁²

v₁² = 281.524

v₁ = 16.77 m/sec