As a fluid moves, a shearing stress will develop. This is due to the viscosity of the fluid. However, for some common fluids, such as air or water, the viscosity of the fluid is small. Therefore, we can assume that the effects from viscosity is negligible. Hence, it becomes an inviscid flow. This means that the flow becomes non-viscous, or in other words frictionless. As a result, the only stress on the fluid will be a normal stress that is independent of direction. In addition, the pressure, $p$, will negative to the normal stress since the normal stress is compressive.

**(Eq 1) **$-p=σ_{xx}=σ_{yy}=σ_{zz}$

**Euler’s Equation**

As mentioned above for an inviscid flow the shearing stresses will be zero. You can replace the normal stresses with $-p$. In turn, this will result in the following general equations of motion.

**(Eq 2)** $ρg_x-\frac{∂p}{∂x}=ρ\left(\frac{∂u}{∂t}+u\frac{∂u}{∂x}+ν\frac{∂u}{∂y}+w\frac{∂u}{∂z}\right)$

**(Eq 3)** $ρg_y-\frac{∂p}{∂y}=ρ\left(\frac{∂ν}{∂t}+u\frac{∂ν}{∂x}+ν\frac{∂ν}{∂y}+w\frac{∂ν}{∂z}\right)$

**(Eq 4)** $ρg_z-\frac{∂p}{∂z}=ρ\left(\frac{∂w}{∂t}+u\frac{∂w}{∂x}+ν\frac{∂w}{∂y}+w\frac{∂w}{∂z}\right)$

These equations are the Euler’s equations. The Euler’s equations can be combined into vector notation.

**(Eq 5) **$ρg-∇p=ρ\left[\frac{∂v}{∂t}+(v·∇)v\right]$

$ρ$ = fluid densisty

$g$ = gravitational constant

$v$ = fluid velocity

Be aware that Euler’s equations are nonlinear partial differential equations that do not have general method to solve. However, for certain circumstances, you can use them to obtain information for inviscid flow.

**Bernoulli Equation**

In order to derive Bernoulli’s equation a direct application of Newton’s second law will need to occur. Unlike previous pages on this website that talk about Bernoulli equation, Euler’s equations will be used to derive it. The reason why Euler’s equations can be used is because they represent a general form on Newton’s second law. To do this we will analyze a steady flow. As a result, Euler’s vector equation will become the following.

**(Eq 6)** $ρg-∇p=ρ(v·∇)v$

Next, we will have to express a gravity vector. This vector will be located on the z-axis and will result in the following equation.

**(Eq 7) **$g=-g_z∇z$

In addition to the above, it is convenient to use the following vector identity.

**(Eq 8) **$(v·∇)v=\frac{1}{2}∇(v·v)-v×(∇×v)$

As a result, by using this vector identity, equation 6 can rewritten into the following form.

**(Eq 9) ** $-ρg∇z-∇p = \frac{ρ}{2}∇(v·v)-ρv×(∇×v)$

In turn, the above equation will be rearranged.

**(Eq10)** $\frac{∇p}{ρ}+\frac{1}{2}∇(v^2)+g∇z=v×(∇×v)$

This will now allow us to take the dot product using the differential length $ds$ which represents the differential length along a streamline.

**(Eq 11)** $\frac{∇p}{ρ}·ds+\frac{1}{2}∇(v^2)·ds+g∇z·ds=[v×(∇xv)]·ds$

$ds$ represents the direction along the streamline. As a result, $ds$ is parallel to $v$. However, when considering the vector $v×(∇×v)$ the vector is perpendicular to $v$.· Hence,

**(Eq 12) **$[v×(∇×v)]·ds=0$

When you perform a dot product of a gradient of a scalar and a differential length, you will determine the differential change of the scalar in the direction of the differential length. Hence, if $ds = dx\hat{i}+dy\hat{j}+dz\hat{k}$ than $∇p·ds =(∂p/∂x)dx+(∂p/∂y)dy+(∂p/∂z)dz=dp$. In turn, this will allow us to write the equation in the following form.

**(Eq 13)** $\frac{dp}{ρ}+\frac{1}{2}d(v^2)+gdz=0$

In the above equation, the change in $p$, $v$, and $z$ will be along a streamline. Next, we will integrate equation 13.

**(Eq 14)** $\int{\frac{dp}{ρ}}+\frac{v^2}{2}+gz = c$

Equation 14 is stating that the sum of the three terms on the left hand side of the equation must be equal to some constant. It can be used for both compressible and incompressible flows. However, the fluid has to be inviscid. In addition, for compressible flows the variation of $ρ$ and $p$ has be known to evaluate the first term of the equation. However, if the fluid is an ideal fluid, that is incompressible, than the integration sign for the first term can be removed.

**(Eq 15)** $\frac{p}{ρ}+\frac{v^2}{2}+gz = c$

As a result, equation 15 is the Bernoulli equation. The Bernoulli equation is normally used to analyze a fluid between two points. To do this will will first divide equation 15 by $g$. Than we will express it in the following form.

**(Eq 16)** $\frac{p_1}{γ}+\frac{v^2_1}{2g}+z_1 = \frac{p_2}{γ}+\frac{v^2_1}{2g}+z_2$

In order to use this form of Bernoulli equation you will be restricted to the following conditions. First, the flow must be inviscid. Second, the flow must be steady. Third, the fluid must be incompressible. Finally, fourth, the flow that is being analyzed must be along a streamline.