# Bernoulli’s principle and equation

What is Bernoulli’s theorem ?

Asked on 8th December 2020 in

Bernoulli’s equation tells us about the relationship between pressure and velocity for an inviscid and incompressible flow. The equation is $p + \frac{1}{2}\rho {V^2} = {\text{constant}}$ The equation states that static pressure + dynamic pressure = total pressure. Here, the constant represents the total pressure of the flow.

To derive Bernoulli’s equation consider $$‘x ‘$$  component of momentum equation of an inviscid flow with no body forces acting on the flow, the equation is $\rho \frac{{Du}}{{Dt}} = – \frac{{\partial p}}{{\partial x}}$

This is $\rho \frac{{\partial u}}{{\partial t}} + \rho u\frac{{\partial u}}{{\partial x}} + \rho v\frac{{\partial u}}{{\partial y}} + \rho w\frac{{\partial u}}{{\partial z}} = – \frac{{\partial p}}{{\partial x}}$

The flow is steady, therefore $$\frac{{\partial u}}{{\partial t}}=0$$

and the equation now becomes $u\frac{{\partial u}}{{\partial x}} + v\frac{{\partial u}}{{\partial y}} + w\frac{{\partial u}}{{\partial z}} = – \frac{1}{\rho }\frac{{\partial p}}{{\partial x}}$

On multiplying both side with $$‘ dx ‘$$ , $u\frac{{\partial u}}{{\partial x}}dx + v\frac{{\partial u}}{{\partial y}}dx + w\frac{{\partial u}}{{\partial z}}dx = – \frac{1}{\rho }\frac{{\partial p}}{{\partial x}}dx$

For a streamline 3D flow $udz – wdx = 0$

$vdx – udy = 0$

On substituting it in above equation , it will become $u\left( {\frac{{\partial u}}{{\partial x}}dx + \frac{{\partial u}}{{\partial y}}dy + \frac{{\partial u}}{{\partial z}}dz} \right) = – \frac{1}{\rho }\frac{{\partial p}}{{\partial x}}dx$

Since, $du = \left( {\frac{{\partial u}}{{\partial x}}dx + \frac{{\partial u}}{{\partial y}}dy + \frac{{\partial u}}{{\partial z}}dz} \right)$

Therefore, the above equation will become, $udu = – \frac{1}{\rho }\frac{{\partial p}}{{\partial x}}dx$

$\Rightarrow \frac{1}{2}d\left( {{u^2}} \right) = – \frac{1}{\rho }\frac{{\partial p}}{{\partial x}}dx$

Similarly for $$y$$ and $$z$$ component of momentum equation for the inviscid flow with no body forces,

$\frac{1}{2}d\left( {{v^2}} \right) = – \frac{1}{\rho }\frac{{\partial p}}{{\partial y}}dy$

and

$\frac{1}{2}d\left( {{w^2}} \right) = – \frac{1}{\rho }\frac{{\partial p}}{{\partial z}}dz$

On adding these $$x,y,z$$ components of equations

$\frac{1}{2}d\left( {{u^2} + {v^2} + {w^2}} \right) = – \frac{1}{\rho }\left( {\frac{{\partial p}}{{\partial x}}dx + \frac{{\partial p}}{{\partial y}}dy + \frac{{\partial p}}{{\partial z}}dz} \right)$

${u^2} + {v^2} + {w^2} = {V^2}$

and also

$\frac{{\partial p}}{{\partial x}}dx + \frac{{\partial p}}{{\partial y}}dy + \frac{{\partial p}}{{\partial z}}dz = dp$

Therefore, the equation will become,

$\frac{1}{2}d\left( {{V^2}} \right) = – \frac{{dp}}{\rho }$

$\Rightarrow dp = – \rho VdV$

For a streamline flow, with density being constant on integrating the above equation between two points of the flow,

$\int\limits_{{p_1}}^{{p_2}} {dp = – \rho \int\limits_{{V_1}}^{{V_2}} {VdV} }$

The equation becomes ${p_2} – {p_1} = – \rho \left( {\frac{{V_2^2}}{2} – \frac{{V_1^2}}{2}} \right)$

$\Rightarrow {p_1} + \frac{1}{2}\rho V_1^2 = {p_2} + \frac{1}{2}\rho V_2^2$

This is the Bernoulli’s equation. If a flow is irrotational, then

$p + \frac{1}{2}\rho {V^2} = {\text{constant}}$

From, the above equation it can be deduced that if density is constant, then if velocity increases then the pressure decreases and if velocity decreases then pressure increases in the flow. This is also called Bernoulli’s theorem. Bernoulli’s equation can be used to derive the lift and drag force for a low speed airfoil where the flow is incompressible and density is constant. Surface of the airfoil, has a streamline flow, and velocity varies along this airfoil, therefore Bernoulli’s equation can be used to find the pressure by integrating over the entire surface of the airfoil.

Bernoulli’s equation is also used in pitot-static tube which is used in aircraft’s instrument to provide information about aircraft’s speed.