This problem is a combinations of a manometer
problem as well as a Bernoullis problem.
If you not it says state all assumptions.
Since we are going to use Bernoullis equation
we have to assume that the system is at steady
state.
We also assume the fluid is incompressible
as well as inviscid and it is along a stream
line.
Let's say that we have some velocity at the
top.
We know that that velocity at the top is going
to be very much less than the velocity in
the pipe because D1 is very much greater than
D2.
We will assume that our V1 is about 0.
We are going to make z2 equal to 0 therefore
z1 equals 12 feet.
The first thing we are going to need to do
after making all of those assumptions is to
find P2.
We are looking right here which is 2 and the
best way to do it is to start at that point
and basically do a manometer approach.
We start with P2 and then we are going to
go down 2 feet to here in the water.
you know as we go down pressure increases.
It increases by the specific weight of water
times the height which is 2 feet.
We are then going to go back to right here.
As you know static fluid at the same height
and same fluid have the same pressure.
We go across to here and then we go up which
means we are subtracting the specific weight
of water times the specific gravity of the
mercury times the height which is 0.5 feet.
That is going to equal our P1.
This goes into the atmosphere and we are going
to work with gauge pressure so our P1 is actually
going to be equal to 0.
We are looking for P2.
That is the gamma of mercury times 0.5 feet
minus the gamma of water times 2 feet.
You might look at this and say that this could
be negative which means that it is a vacuum
which it isn't because the difference in height,
the two feet in water as opposed to the half
foot in the mercury.
What you should know is that the specific
weight of mercury is considerably greater
than the specific weight of water.
When we put these numbers in to our equation
we are able to solve for P2 which is 298.7
lb/ft^2.
Why do we need P2?
What we are looking at is what is the velocity
at this point 2.
Now we are going to use up here and we will
call this point 3.
We are going to do a Bernoullis between point
2 and point 3.
Our Bernoullis says gamma of water times z1
equals P2 plus 1/2 times rho times V2^2.
How did I simplify the Bernoullis equation?
If we go back up here again.
If you notice V1 is very much less than V2
so our V1 was 0 on the left hand side.
Our z2 was zero and our z1 is 12 feet.
Now we can rewrite this such that V2 is the
square root of 2 times gamma times z1 minus
P2 all over rho.
When we solve for V2 we find that this is
21.5 ft/s.
That is the first thing that we are asked
to find.
The next thing is the flow rate or the volumetric
flow rate as exits from the tank.
So Q is just V2 times A2.
We solve using our 2" diameter to find the
area.
Our A2 is just pi times r^2.
In feet that becomes 0.022 ft^2.
When we multiply 21.5 ft/s times 0.022 ft^2
we come up with a Q of 0.469 ft^3/s.
