A Siphon at the Bar: Jane's DIY Drink Mixing Trick

[sreya]

Homework Statement

Jane goes to a juice bar with her friend Neil. She is thinking of ordering her favorite drink, 7/8 orange juice and 1/8 cranberry juice, but the drink is not on the menu, so she decides to order a glass of orange juice and a glass of cranberry juice and do the mixing herself. The drinks come in two identical tall glasses; to avoid spilling while mixing the two juices, Jane shows Neil something she learned that day in class. She drinks about 1/8 of the orange juice, then takes the straw from the glass containing cranberry juice, sucks up just enough cranberry juice to fill the straw, and while covering the top of the straw with her thumb, carefully bends the straw and places the end over the orange juice glass. After she releases her thumb, the cranberry juice flows through the straw into the orange juice glass. Jane has successfully designed a siphon.
Assume that the glass containing cranberry juice has a very large diameter with respect to the diameter of the straw and that the cross-sectional area of the straw is the same at all points. Let the atmospheric pressure be pa and assume that the cranberry juice has negligible viscosity.

Consider the end of the straw from which the cranberry juice is flowing into the glass containing orange juice, and let h0 be the distance below the surface of cranberry juice at which that end of the straw is located: (Figure 1) . What is the initial velocity v of the cranberry juice as it flows out of the straw? Let g denote the magnitude of the acceleration due to gravity.

Homework Equations

Bernoulli's: $p_1 + \frac{1}{2}\rho v_1^2+\rho gy_1 = p_2 + \frac{1}{2}\rho v_2^2 + \rho gy_2$

The Attempt at a Solution

I tried using Bernoulli's but I really don't understand it at all. I'm bad with fluids, can anyone walk me through it?

 

[paisiello2]

You can make some assumptions and greatly simplify Bernoulli's equation.

 

[sreya]

What would those be? I know that you could make y1=0 ,but I'm not sure what else you could do.

 

[paisiello2]

Ok, where is that point located on the diagram?

 

[sreya]

The top of the cranberry juice would make it easier I think.

 

[paisiello2]

I agree. So if that is your datum where y1=0 then what and where is y2?

 

[sreya]

h_0, where the juice is falling out

 

[paisiello2]

Ok, so what else can you set to 0 or cancel out?

 

[sreya]

Okay I think I screwed up a little. The placement is till going to be at the top of the juice but this is the formula

The formula for Bernoulli's when applied to this problem becomes this

$p1 + \frac{1}{2}\rho v_1^2 - \rho g d = p2 + \frac{1}{2}\rho v_2^2 - \rho g h_0$
$p_a +\rho gd - \rho gd = p_a + \frac{1}{2}\rho v_2^2 - \rho g h_0$

p_a's cancel and the rho*gd's cancel, move rho*gh over and solve for v.

Pressure can be described as atmospheric plus rho*gd and the p2 is just atmospheric because it's outside the juice.

Questions: V1 goes to zero because the diameter of the straw is small relative to the glass thus it allows the velocity to go to zero. I'm not sure why, I've just seen this trend. Perhaps you could explain why?

 

[paisiello2]

You got it.

As for your first question, you actually gave the answer yourself. Maybe assume some typical value for diameters of the straw and glass, then use conservation of mass equation to estimate v1. Then realize that taking the square of a small number is a really, really small number.

For your 2nd question, what is the formula for pressure in a fluid consisting of multiple layers?

 

[sreya]

I'm not actually sure. I just know that in a situation like this you can describe the pressure of a place in water as atmospheric pressure + the density*gravity*deepness of the point. I'm not sure what it would be for multiple layers

 

[paisiello2]

Well, again, you gave the answer. The pressure for multiple layers is just summation of each layer.

Since the top of the liquid only has atmospheric pressure on it then the summation is only the atmosphere. And since the point at which the juice exits only has atmospheric pressure on top of it then again the summation is only the atmosphere.

The difference in elevation h0 is negligible compared to the height of the atmosphere. So therefore we can say p1=p2.