Derivation of Bernoulli’s Equation
Variables used
|
A = cross-sectional area |
V = volumetric flow rate |
F = Force |
x = displacement |
|
P = pressure = F/A |
v = velocity |
W = work |
g = accel of gravity |
|
E = energy |
m = mass |
t = time |
h = height |
|
Assumptions
(1) the fluid is incompressible (2) viscosity is small enough to be ignored.
Relationships can be derived for inclusion of
these effects but they are more involved and are not covered in the AP
Physics B course.
My derivation utilizes the work-energy theorem, and the continuity equation.
The diagram on the right illustrates the variables, etc, used in the
derivation.
We look at a fluid section of mass m travelling to the right as shown.
The net work done in moving the fluid is
W = W1 + W2 = F1x1 –F2x2
The second term is negative because the force and displacement are
in opposite directions.
Substituting F = PA into the above equation results in
DW
= P1A1x1 - P2A2x2 |
 |
|
Using the continuity equation
V = A1x1 = A2x2
in the previous equation
DW
= P1A1x1 - P2A2x2
we arrive at the following
DW
= P1A1x1 - P2 A1x1
or
DW
= (P1- P2)V
|
|
The change in energy between the initial and final positions is given by
DE
= E2 – E1 = (PE)2 + (KE)2 - (PE)1
- (KE)1
DE
= (mgh2 + mv22/2) – (mgh1 + mv12/2)
And, from the work-energy theorem, we know that the net work done is equal
to the change in energy of the system.
DW
=
DE
Substituting the previously derived expressions for the left and right sides
of the above, we arrive at
(P1- P2)V = (mgh2 + mv22/2)
– (mgh1 + mv12/2)
dividing by V gives (reason will soon become clear)
P1- P2 = (mgh2 + mv22/2)/V
– (mgh1 + mv12/2)/V
now, using the fact that
r
= m/V we arrive at
(the reason)
P1- P2 =
rgh2
+
rmv22/2
-
rgh1
-
rmv12/2
Rearranging the above, we arrive at Bernoulli’s equation
P1 + rgh1 + 1/2rv12= P2 + rgh2 + 1/2rv22
