Euler equation, Bernoulli and Pitot tube

(a) Euler equations

For an inviscid fluid ($\mu=0$) the Navier-Stokes equation loses the viscous term:

$$\rho\left(\frac{\partial\vec V}{\partial t}+(\vec V\cdot\nabla)\vec V\right)=-\nabla p+\rho\vec g$$

In components (x-direction): $\dfrac{\partial u}{\partial t}+u\dfrac{\partial u}{\partial x}+v\dfrac{\partial u}{\partial y}+w\dfrac{\partial u}{\partial z}=-\dfrac1\rho\dfrac{\partial p}{\partial x}+g_x$.

Bernoulli from Euler: take steady flow along a streamline of length $ds$. The momentum equation along $s$ becomes

$$V\frac{dV}{ds}=-\frac1\rho\frac{dp}{ds}-g\frac{dz}{ds}$$

Multiply by $ds$ and integrate with constant $\rho$:

$$\int\frac{dp}{\rho}+\int V\,dV+\int g\,dz=\text{const}\ \Rightarrow\ \frac p\rho+\frac{V^2}{2}+gz=\text{constant}$$

Equivalently $p+\tfrac12\rho V^2+\rho gz=$ const along a streamline. It is valid for steady, incompressible, inviscid flow with no heat or work addition. If the flow is also irrotational, the constant is the same for the whole field.

In CFD, Euler solvers are used for high-speed external flows where viscous effects outside the boundary layer are small, as they are cheaper than full N-S solvers.

(b) Pitot tube

At the stagnation port the velocity is zero, so

$$p_0-p=\tfrac12\rho V^2$$ $$V=\sqrt{\frac{2(p_0-p)}{\rho}}=\sqrt{\frac{2\times450}{1.2}}=27.4\ \text{m/s}$$

(The Mach number is about 0.08, so incompressible treatment is valid.)

Answer: $V=27.4$ m/s.