In a previous screencast we determined how

long it would take to fill a coffee cup given information about this coffee urn. In that

problem we assumed steady state. In this problem we are interested to determine how long it

would take to drain the coffee pot if we left that nozzle completely open. In this problem

we have a unsteady state Bernoulli equation. With any problem we should probably start

by labeling our diagram with information given in the problem. We have a 2 foot tall coffee

urn that dispenses coffee through a nozzle that is 5″ above the table surface. The diameter

of the urn and nozzle are 10″ and 0.5″ respectively. We want to know how long it is going to take

to drain. At this point in the problem we need to make our assumptions to help us solve.

To use the Bernoulli equation we are going to assume that the fluid is inviscid and that

there are no viscid effects. We are going to assume that the flow follows specific stream

lines and lastly we are going to assume that it is incompressible. Obviously since it is

unsteady state we are not going to assume steady state conditions. This allows us to

write the simplified form of the Bernoulli equation. specific weight. Point 1 will be

the nozzle and point 2 is the liquid level. The pressure at point 1 over the specific

weight plus the velocity at point 1 squared over 2 times g plus the height of point 1

is equal to some constant which we will therefore be equal to the same at point 2. At this point

we are going to assume that the coffee pot is open to atmosphere. We know that the nozzle

is going to be open to atmosphere so we will cross out our pressure terms since they will

just be equal to each other and therefore atmosphere. We are also going to say that

our height at point 1 is a value of 0 so we will cross that out. Rewriting this equation

we get the following. We know at point 2 this is just going to be the height of the fluid

as a function of time so we can plug that into our equation. There is one last assumption

we are going to make to solve this problem. That being that the velocity at point 2 that

is changing with time is going to be much less than the velocity at point 1. We make

this assumption because the diameter is much less than the diameter of the coffee urn or

that of point 2. We say that the velocity at point 2 is therefore negligibly here and

we can model the velocity at point 1 as a free jet. At this point I think I know what

you are thinking which is that we said this is an unsteady state problem but now we have

just taken the Bernoulli equation and made it steady state and gotten a velocity at point

1. All though we have said the velocity at point 2 is much smaller it is not equal to

0 and we can write the velocity at point 2 as a differential. It is going to be the change

in height with time. We call this Quasi steady state. It is not quite steady state but we

are going to assume that it is pretty close to it. When we do this we take into consideration

the continuity equation which basically says that the mass flow rate at point 2 must equal

the mass flow rate at point 1. If we assume an incompressible flow we could rewrite it

as the following. I have rewritten these as volumetric flow rates where it is the velocity

times the cross sectional area. We know that we can calculate the cross sectional area

at point 1 given the diameter of the nozzle and the cross sectional area at point 2 given

the diameter of the tank. We plug this into our areas and we take our differential that

we have rewritten for point 2 into our free jet analysis for point1. At this point I have

written little r and big R so that I do not start plugging in numbers quite yet. Little

r is the radius of the nozzle and big R is the radius of the coffee pot. If I separate

out the variables so that we get the differential on the left side. I have grouped the variables

on the left side here as one since these are all constant. We are going to go ahead and

call k which is going to be equal to negative square root of 2 times g times little r squared

over big R squared. This allows us to rewrite it and integrate as following. We need parameters

to integrate from so we will say some height initial and some height final and time is

0 to some time of t. Integrating this we are let with the following. Since we are interested

in the time it takes we are going to rewrite this so time is on the left side and i get

the following. At this point again I have not plugged in numbers because this gives

us an equation that depending on any cylindrical system that looks like this urn and follows

the same assumptions we can plug in the two radius, our gravitational constant and our

initial and final height to figure out how long it would take to get there. In this specific

case let’s go ahead and plug in some numbers. You can see that I have used a large radius

of 5″, an initial height of 2′ and then we see that the nozzle is at approximately 5″.

We did not write the height of that actual nozzle because we know that as the coffee

flows it will stop when it reaches this point. Let’s go ahead and assume 5 “. The 2′ minus

5″ gives us the 19″ and then the radius of the nozzle being 0.25”. This takes a time

of 15.5 seconds. Recall for this unsteady state problem that we assume that the fluid

is inviscid, incompressible, and we used a Quasi steady state approximation to get V1

and the continuity equation to get V2. This gave us an expression to calculate the time

to go from one height to another in a cylindrical system where we could make these assumptions.

Plugging in our values for the coffee urn we got 15.5 seconds. One of the problems obviously

that would come up with this is that this diameter of the coffee pot is not quite that

much larger in magnitude than the diameter of the nozzle and we assumed 5″ is where we

do know there is going to be a little bit of a difference between the bottom of the

nozzle and where the coffee is going to stop pouring. Some things to consider next time

you are doing this type of analysis.

What if you consider its viscosity? I can't find anything about this type of calculation for viscous fluids, and I need it really bad for a lab 🙁

How to solve if V2 is not neglected? ?

That is in the eqn sqrt (2gh) there will be another term. Please do share the solution for it.

@Nelson Klein Thank you for your comment. In this situation, k has been defined to include the negative sign. I believe it has been accounted for in all of the written equations.

why hf is 19 in shouldn't it be 5 in?

@Alaa Taha

Thank you for your comment! Yes, it should be 5 in. We will correct this very soon. Thank you for paying attention!

thank you soooo much (emphasis on the "so")

If the datum is at point 0, ho should be 2ft-(5/12ft) and hf should be 0

why do you use the bernoulli equation if flow is not steady? Why not use the unsteady bernoulli equation? Thank you.

thank u very very much sir. I was stuck with the integral but with ur videos help I did it. thank you

What if the nozzle has an indicated length? How to approach the problem?

How do you account for viscosity it for the same problem you had heavy oil instead of water? 800cp viscosity

Thank you finally I understand it!

the answer its 15,5 or 76,6 ????

How can you model same problem when the opening of nozzle is a function of time?

Hey, hate to throw another correction in there but I'm fairly sure neither answer is correct. In using the Bernoulli Equation we say that the height of point 1 is zero, and height is defined as distance above this point. Because we use this equation to define V2, and this V2 is used in the continuity equation, when we differentiate with respect to h, we must continue to use point 1 as zero. The final equation looks the same except where hf is zero and ho is 19/12 ft. This gives an answer if 125 s. If I'm missing something or just plain wrong, please let me know. But as it is now, it implies the water is starting two feet above and stopping 5 in above the nozzle.

If there are two nozzles at different heights in the urn, how can we find the amount of coffee drawn from each nozzle, or the ratio between the two amounts of coffee emptied by the two nozzles after the level of coffee reaches the level of the lower nozzle ?

would this also apply if the cylinder was horizontal or would we need to take into consideration the fact that the radius is changing with accord with time

shouldn’t ho be 19 and hf be 0 since your reference for point 1 is the nozzle?