Fluids and Bernoulli’s Principle

General information about the variables involved:

P =pressure = F Cosq/Area       Units:   N/m² = Pa

r = Density = mass/volume       Units:  Kg/m³

F_{at bottom of column of water} = height x area x density _{of water} x g

P _{at bottom of column of water} = density _{of water} x g x height      (notice, Area canceled)

Formulas for Hydraulic Lifts:

F_input/A_input = F_output/A_output         or         F_out = (F_i/A_i) A_o

Bottom line:  You want a small area of input and a large area of output.  That will give you a large force output for a relatively small force input.

However, this increase of force output comes at the expense of the distance over which you must push in the piston.  For example, you will have to push the piston into the hydraulic lift ten times further than it comes out in order for you to produce 10 times the force output:      F_out = (x_in/x_out) F_in     or F_oX_o = F_iX_i

Buoyant Force:

F_buoyant = r_fluid x g x volume           (volume = height x area)

F_buoyant = (r_fluid/r_object) x weight_object          where weight = mg

      Therefore, if the r_object > r_fluid/air then F_buoyant is small or insignificant.

The net force on any object must take into account the buoyant force.  Usually, however, we can ignore the buoyant force because objects are much denser than air.

F_net = F_g – F_b     where Fb tends toward zero usually.

Specific Gravity:

Specific gravity has no units.  It is the ratio between the density of the object in question and water.  Since water has a density of 1 g/mL, the specific gravity of any object is equal to its own density (just sans units).

Specific gravity = r_object/r_water

      NOTE:   If you find the specific gravity of an object NOT IN WATER, but in say, oil, then its specific gravity will obviously NOT be equal to its density.

Bernoulli Principle:

Q = flow rate = Area x Velocity = Volume/time

Work of a fluid = Dm/r_liquid ( Pressure₁ – Pressure ₂) = DE = KE + U = 1/2 mv² + mgh

Equation of continutity:     A₁Velocity₁ = A₂ Velocity₂

Therefore, as area of a pipe increases, the velocity will decrease.  Flow rate will remain constant for an incompressible fluid flowing through a pipe of any diameter.  Example:  Water flowing into the hose at some rate Q.  You put your finger on the other end of the hose to decrease the area.  Water will shoot out at a high rate of speed.  This is because the water rate, Q, must be maintained by the same mass of water entering the pipe as leaving it.  This is accomplished by decreasing its escape area (with your finger) and so the liquid speeds up to get more mass through in a shorter amount of time.

Bernoulli’s equation:  P₁ + 1/2rv₁² + rgh₁ = P₂ + 1/2rv₂² + rgh₂

      Notice how Bernoulli’s equation is born out of the work of a fluid equation above which involves KE and U and Pressure.  P means pressure, and v means velocity.  r is the density of the fluid in question and g is an old friend!  h is the drop in height

      Things to note:  If there is no drop in height (eg. The pipe is horizontal) then h₁ = h₂ and you can drop rgh₁ and rgh₂ from the equation.  Realize this is the potential energy portion of the equation.

     If there is no change in the area of the pipe, then v₁ = v₂ and you can drop the KE portion of the equation.  1/2rv₁²  and 1/2rv₂²

     Bottom line:  Bernoulli’s equation takes into account both the potential and kinetic energies involved in moving fluids and how that affects pressure.  If there is no height drop, then there is no potential.  If there is no area change, then there is no kinetic energy change. 

     If the just the area decreases, such that the velocity increases, the pressure will Decrease.  (height stays same)

     If the just the height drops, Pressure will increase (area must remain constant)

     Example problem #75 on p. 330 is an excellent example to look at before the exam:

1.  No two holes have the same pressure.  BUT…..Each hole maintains its own constant pressure because the water is being replaced in the top at the same rate it is escaping out of the holes.

2.  The bottom hole has the most pressure on it.  The top has the least.

3.  The distance from the top of the container to each hole will give you the pressure of that hole. 

4.  Since the pressure is constant out of a particular hole, you can use the following equation to determine the velocity of the water shooting out of a particular hole

V_water =     where h is the distance the hole is from the top.

5.  Once you know the velocity of the water out of a hole, you can use Newton’s laws to determine the downrange distance of that water

     remember y = 1/2gt² so you can solve for t by re-arranging:  t =

     You can find range in the horizontal direction by x = V_xt.  Simply substitute for t and you can determine range x by

X = V_x