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.
