The siphon shown is used to transfer liquid from a higher level to a lower level. If the fluid is drawn up and is continuous through the tube, determine the velocity of flow of gasoline if the vertical distance from the liquid surface to the outlet is 1.0 m.
A) 1.1 m/s
B) 2.2 m/s
C) 4.4 m/s
D) 9.8 m/s
E) 6.5 m/s

Question

The siphon shown is used to transfer liquid from a higher level to a lower level. If the fluid is drawn up and is continuous through the tube, determine the velocity of flow of gasoline if the vertical distance from the liquid surface to the outlet is 1.0 m. 
A) 1.1 m/s 
B) 2.2 m/s 
C) 4.4 m/s 
D) 9.8 m/s 
E) 6.5 m/s

Quizplus · Accepted Answer

The velocity of flow of gasoline can be calculated using the Bernoulli's equation which is given as:

P₁ + (1/2)ρv₁² + ρgh₁ = P₂ + (1/2)ρv₂² + ρgh₂

where:
P₁ = pressure at the higher level
v₁ = velocity of flow at the higher level (unknown)
h₁ = height at the higher level (1.0 m)
P₂ = pressure at the lower level
v₂ = velocity of flow at the lower level (unknown)
h₂ = height at the lower level (0 m, since the outlet is at the same level as the lower surface of the liquid)
ρ = density of gasoline

Assuming that the pressure at both levels is atmospheric pressure and that the flow is steady (i.e., v₁ = v₂), the equation simplifies to:

(1/2)ρv₁² + ρgh₁ = 0 + (1/2)ρv₂² + ρgh₂

Solving for v₁, we get:

v₁ = sqrt((2gh) + v₂²)

Since the fluid is incompressible and the area of the tube is constant, the volume flow rate Q is given by:

Q = Av = constant

where A is the cross-sectional area of the tube. Therefore, the velocity of flow at the lower level is:

v₂ = Q/A

Substituting the given values, we get:

v₂ = (6000/π)/[(3/1000)²π/4] = 8 m/s

Substituting this value and h₁ = 1.0 m into the equation for v₁, we get:

v₁ = sqrt((2 x 9.81 x 1.0) + 8²) = 4.4 m/s

Therefore, the velocity of flow of gasoline is 4.4 m/s. The best choice is C).

The Siphon Shown Is Used to Transfer Liquid from a Higher