Bernoulli’s equation describes how pressure energy, kinetic energy, and gravitational potential energy trade places in an ideal fluid moving along a streamline.
It is one of the most useful equations in fluid mechanics.
1. Bernoulli’s equation
For steady, incompressible, non-viscous flow along a streamline,
\[
\begin{aligned}
P+\frac{1}{2}\rho v^2+\rho gz &= \text{constant}.
\end{aligned}
\]
Between two points this becomes
\[
\begin{aligned}
P_1+\frac{1}{2}\rho v_1^2+\rho gz_1
&=
P_2+\frac{1}{2}\rho v_2^2+\rho gz_2.
\end{aligned}
\]
2. Meaning of each term
| Term |
Name |
Meaning |
| \(P\) |
Static pressure |
Pressure energy per unit volume |
| \(\frac{1}{2}\rho v^2\) |
Dynamic pressure |
Kinetic energy per unit volume |
| \(\rho gz\) |
Hydrostatic elevation term |
Gravitational potential energy per unit volume |
3. Total head form
Dividing the Bernoulli equation by \(\rho g\) gives the head form:
\[
\begin{aligned}
\frac{P}{\rho g}+\frac{v^2}{2g}+z &= H.
\end{aligned}
\]
Here \(H\) is the total head, measured in meters.
The three parts are:
\[
\begin{aligned}
\text{pressure head} &= \frac{P}{\rho g},\\
\text{velocity head} &= \frac{v^2}{2g},\\
\text{elevation head} &= z.
\end{aligned}
\]
4. Bernoulli equation with head loss
Real pipes lose energy because of viscosity, friction, fittings, bends, and turbulence.
A simple correction is to include a head loss term \(h_L\):
\[
\begin{aligned}
P_1+\frac{1}{2}\rho v_1^2+\rho gz_1
&=
P_2+\frac{1}{2}\rho v_2^2+\rho gz_2+\rho gh_L.
\end{aligned}
\]
In head form:
\[
\begin{aligned}
H_1 &= H_2+h_L.
\end{aligned}
\]
5. Horizontal pipe
If the pipe is horizontal, then \(z_1=z_2\). Bernoulli’s equation becomes
\[
\begin{aligned}
P_1+\frac{1}{2}\rho v_1^2
&=
P_2+\frac{1}{2}\rho v_2^2.
\end{aligned}
\]
Solving for the pressure change:
\[
\begin{aligned}
P_1-P_2
&= \frac{1}{2}\rho(v_2^2-v_1^2).
\end{aligned}
\]
If the speed increases, the pressure decreases.
6. Vertical pipe
If the downstream point is higher, then \(z_2>z_1\). Some energy must go into elevation head.
If speed stays the same, the pressure drop is
\[
\begin{aligned}
P_1-P_2 &= \rho g(z_2-z_1).
\end{aligned}
\]
If both height and speed change, both effects must be included:
\[
\begin{aligned}
P_1-P_2
&= \frac{1}{2}\rho(v_2^2-v_1^2)+\rho g(z_2-z_1)+\rho gh_L.
\end{aligned}
\]
7. Combining Bernoulli with continuity
In incompressible pipe flow, the continuity equation is
\[
\begin{aligned}
A_1v_1 &= A_2v_2.
\end{aligned}
\]
Therefore,
\[
\begin{aligned}
v_2 &= \frac{A_1}{A_2}v_1.
\end{aligned}
\]
This is useful in Venturi-tube problems, where a narrower section causes the fluid speed to increase and the pressure to decrease.
8. Circular pipe areas
For a circular pipe with diameter \(D\),
\[
\begin{aligned}
A &= \frac{\pi D^2}{4}.
\end{aligned}
\]
If radius \(R\) is used instead,
\[
\begin{aligned}
A &= \pi R^2.
\end{aligned}
\]
Since area depends on diameter squared, a pipe whose diameter is halved has one-fourth the area.
For incompressible flow, the speed then becomes four times larger.
9. Solving for downstream pressure
Starting with Bernoulli’s equation with optional head loss:
\[
\begin{aligned}
P_1+\frac{1}{2}\rho v_1^2+\rho gz_1
&=
P_2+\frac{1}{2}\rho v_2^2+\rho gz_2+\rho gh_L.
\end{aligned}
\]
Solve for \(P_2\):
\[
\begin{aligned}
P_2
&= P_1+\frac{1}{2}\rho(v_1^2-v_2^2)+\rho g(z_1-z_2-h_L).
\end{aligned}
\]
10. Solving for downstream speed
If \(P_2\) is known and \(v_2\) is unknown:
\[
\begin{aligned}
v_2
&= \sqrt{
v_1^2+\frac{2(P_1-P_2)}{\rho}+2g(z_1-z_2-h_L)
}.
\end{aligned}
\]
The expression under the square root must be nonnegative. If it is negative, the chosen pressures, heights, speeds, and losses are not physically compatible with the ideal model.
11. Worked example: horizontal speed increase
Suppose water flows in a horizontal pipe, so
\[
\begin{aligned}
z_1 &= z_2.
\end{aligned}
\]
Let
\[
\begin{aligned}
\rho &= 1000\ \mathrm{kg/m^3},\\
P_1 &= 200\ \mathrm{kPa},\\
v_1 &= 2\ \mathrm{m/s},\\
v_2 &= 8\ \mathrm{m/s}.
\end{aligned}
\]
For no head loss, downstream pressure is
\[
\begin{aligned}
P_2
&= P_1+\frac{1}{2}\rho(v_1^2-v_2^2)\\
&= 200000+\frac{1}{2}(1000)(2^2-8^2)\\
&= 200000+500(4-64)\\
&= 170000\ \mathrm{Pa}.
\end{aligned}
\]
Therefore,
\[
\begin{aligned}
P_2 &= 170\ \mathrm{kPa},\\
P_1-P_2 &= 30\ \mathrm{kPa}.
\end{aligned}
\]
The speed increased, so the pressure dropped.
12. Formula summary
| Goal |
Formula |
Use |
| Bernoulli equation |
\(P+\frac{1}{2}\rho v^2+\rho gz=\text{constant}\) |
Ideal energy balance along a streamline |
| Head form |
\(H=P/(\rho g)+v^2/(2g)+z\) |
Energy per unit weight |
| Downstream pressure |
\(P_2=P_1+\frac{1}{2}\rho(v_1^2-v_2^2)+\rho g(z_1-z_2-h_L)\) |
Find pressure after speed, height, and loss changes |
| Downstream speed |
\(v_2=\sqrt{v_1^2+2(P_1-P_2)/\rho+2g(z_1-z_2-h_L)}\) |
Find speed from pressure and height change |
| Continuity equation |
\(A_1v_1=A_2v_2\) |
Connect pipe area change to speed change |
| Circular pipe area |
\(A=\pi D^2/4\) |
Convert diameter to cross-sectional area |
13. Assumptions
- The flow is steady.
- The fluid density is constant.
- The flow follows a streamline.
- Viscous losses are ignored unless a head loss \(h_L\) is entered.
- The average speed is used at each pipe cross-section.
- The equation is not intended for highly compressible gas flow.
Key idea: in ideal flow, increasing speed usually reduces pressure, while increasing height also requires pressure or velocity energy.
