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.