Fluids-blow across top of straw to draw up olive oil

[ryann_117]

Homework Statement

You try to remove olive oil from a glass by blowing across the top of a vertical straw immersed in the olive oil. What is the minimum speed you must give the air at the top of the straw to draw olive oil upward through a height of 1.5 cm?
(Note: The density of olive oil, if you need it is 920 kg per cubic meter.)

Homework Equations

P1 + .5*density*v1^2 + density*g*y1 = P2 + .5*density*v2^2 + density*g*y2

The Attempt at a Solution

P1 + .5*density*v1^2 + density*g*y1 = P2 + .5*density*v2^2 + density*g*y2
P1 = .5*density*v2^2 + density*g*y2
V2 = 14.83 m/s
V2 is close, but not right; I'm not sure how to find P2. What do I need to do different?

 

[ideasrule]

I don't understand your calculations. How did you get from P1 = .5*density*v2^2 + density*g*y2 to v2=14.83 m/s?

You can calculate the pressure difference between the top and bottom of the straw that's needed to support the olive oil column; it's just rho*g*h. Then, you can apply the equation P1 + .5*density*v1^2 + density*g*y1 = P2 + .5*density*v2^2 + density*g*y2 with the left side representing the bottom of the straw and the right side representing the top. v1 would be 0, since the air at the bottom isn't moving, and you've just calculated P1-P2.

 

[kerimek]

To determine the minimum speed required for the air at the top of the straw to draw up olive oil, we can use the Bernoulli's equation. This equation relates the pressure, density, and velocity of a fluid at two different points in a fluid flow. In this case, the two points are the top and bottom of the straw, where the fluid (air) is in motion.

Using the given information, we can set up the Bernoulli's equation as follows:

P1 + 0.5*density*v1^2 + density*g*y1 = P2 + 0.5*density*v2^2 + density*g*y2

Where P1 is the pressure at the top of the straw, P2 is the pressure at the bottom of the straw, v1 and v2 are the velocities at the top and bottom of the straw respectively, y1 and y2 are the heights at the top and bottom of the straw respectively, and g is the acceleration due to gravity.

Since we are trying to find the minimum speed, we can assume that the pressure at the top and bottom of the straw are equal (P1 = P2). This means that the pressure terms cancel out from the equation.

0.5*density*v1^2 + density*g*y1 = 0.5*density*v2^2 + density*g*y2

We are given the density of olive oil (920 kg/m^3) and the height through which the oil needs to be drawn (1.5 cm = 0.015 m). We can also assume that the velocity at the bottom of the straw is zero (v2 = 0) since the olive oil is at rest at the bottom.

Substituting these values into the equation, we get:

0.5*density*v1^2 + density*g*y1 = 0.5*density*0^2 + density*g*0.015

Solving for v1, we get:

v1 = √(2*g*y1)

Plugging in the values, we get:

v1 = √(2*9.8 m/s^2 * 0.015 m) = 0.39 m/s

Therefore, the minimum speed required for the air at the top of the straw to draw up olive oil through a height of 1.5 cm is 0.39 m/s.