N2L Along Streamline

We use $s$-$n$ coordinates, where $s$ is the direction along the streamline and $n$ is the direction normal to the streamline. Let us consider a small particle of size $\delta s$ by $\delta n$.

Note that we assume:

• Steady flow
• Inviscid flows
• Incompressible flows
• Along a streamline

In the $s$-direction, applying $F=ma$ gets us:

$$[p_s-(p_s+\delta p_s)]\delta n\delta y-\delta W\sin \theta =\rho \delta \forall a_s$$

Writing $\sin \theta =\frac{\delta z}{\delta s}$ give us

$$-\delta p_s\delta n\delta y-\gamma \delta \forall \frac{\delta z}{\delta s}=\rho \delta \forall a_s$$

Since we are following a streamline, by definition, there is no dependence of $p_s$​ on $n$ because $p_s$​ only changes along $s$:

$$\begin{array}{rl}{p}_s& =p_s(s,n)\\ d{p}_s& =\frac{\partial p_s}{\partial s}ds+\text{cancelto}0\frac{\partial p_s}{\partial n}dn\end{array}$$

Then, we can substitute to get:

$$-\frac{\partial p_s}{\partial s}\underset{d\forall }{\underbrace{\delta s\delta n\delta y}}-\gamma \delta \forall \frac{\delta z}{\delta s}=\rho \delta \forall a_s$$

Cancelling out all the $\delta \forall$ terms gives us

$$-\frac{\partial p_s}{\partial s}-\gamma \frac{\delta z}{\delta s}=\rho a_s$$

We can write the acceleration as

$$a_s=\frac{dV}{dt}=\frac{\partial V}{\partial s}\frac{ds}{dt}=V\frac{\partial V}{\partial s}$$

Note that we can write the derivative of $V^2$:

$$\frac{d}{ds}(V^2)=2V\frac{\partial V}{\partial s}$$

So we can write

$$a_s=V\frac{\partial V}{\partial s}=\frac{1}{2}\frac{d{V}^2}{ds}$$

Substituting $a_s$ back into our main equation gives

$$\begin{array}{r}-\frac{dp}{ds}-\gamma \frac{dz}{ds}=\frac{\rho }{2}\frac{d{V}^2}{ds}\\ dp+\gamma dz+\frac{\rho }{2}d{V}^2=0\end{array}$$

Since we assumed incompressibility, $\rho$ is a constant, so we have:

$$\boxed{p+\frac{1}{2}\rho V^2+\gamma z=c}$$

which is the Bernoulli Equation.

Example

Let’s determine the pressure on a bicyclist due to airflow. We analyze two points:

• At point (1) the air velocity is $V_1=V_0$
• At point (2) the air velocity is 0 (air comes to rest on cyclist’s body)

Applying Bernoulli’s equation between the two points:

$$p_1+\frac{1}{2}\rho V_1^2+\gamma z_1=p_2+\frac{1}{2}\rho V_2^2+\gamma z_2$$

Using $z_1=z_2$, $V_1=V_0$ and $V_2=0$ lets us simplify to

$$\begin{array}{r}{p}_1+\frac{1}{2}\rho V_0^2=p_2+0\\ p_2=p_1+\frac{1}{2}\rho V_0^2\end{array}$$

At gage pressure, $p_1=0$, we have

$$p_2=\frac{1}{2}\rho V_0^2$$

We can multiply both sides of the equation by volume $\forall$ to get

$$p_2\forall =\frac{1}{2}\rho \forall V_0^2$$

which is nice since $p_2\forall$ tells us the total energy due to pressure over a given volume and the right side gives the total kinetic energy.