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Bernoulli

Bernoulli's principle

Bernoulli's principle is often used in fluid dynamics and it states that an increase in velocity of a fluid results in a decrease of the static pressure and or a decrease in the potential energy. The sum of the static pressure, kinetic energy and potential energy must be constant along a streamline:

p+12ρv2+ρgh=constant\begin{equation} p + \frac {1} {2}\rho v^{2} + \rho g h = constant \end{equation}

With pp [Nm2][\frac{N}{m^2}] equal to the pressure of the fluid, ρ\rho [kgm3][\frac{kg}{m^3}] to the density of the fluid, vv [ms][\frac{m}{s}] to the velocity of the fluid, gg [ms2][\frac{m}{s^2}] to the gravitation constant and hh [m][m] to the height of the point along the streamline.

Assumptions for the equation

  • The stream is steady, the velocity and density at any point can't change over time.
  • The flow is incompressible, the density must have the same value at any point of the stream.
  • Friction is negligible

Use of the equation

Equation (1) is mostly used to calculate the pressure or the velocity at which a fluid is flowing at a point along the streamline. In order to do this, there must be only 1 unknown in the equation which then can be solved for.

Example problem

In figure 1 a streamline with a constant diameter can be seen, where Bernoulli's principle can be applied. Let's say that the pressure of the fluid in point C needs to be calculated.

The following values are known:

  • pA=pD=patm=1105[Pa]p_{A} = p_{D} = p_{atm} = 1*10^5[Pa] (The pipe is open ended)
  • vA=vB=vC=vDv_{A} = v_{B} = v_{C} = v_{D} (More in mass flow rate)
  • hA=1[m]h_{A} = 1[m] and hC=0.5[m]h_{C} = 0.5[m]
  • ρ=1000[kgm3]\rho = 1000[\frac{kg}{m^{3}}]
  • g=9.81[ms2]g = 9.81[\frac{m}{s^{2}}]
Visualization of a Bernoulli streamline

Figure 1. Visualization of a Bernoulli streamline

The pressure in point C can be calculated by using equation (1) and filling it in for 2 points of which 1 point is point C and the other point is one where all values of the equation are known:

pA+12ρvA2+ρghA=pC+12ρvC2+ρghCp_{A} + \frac{1}{2}\rho v^2_A+\rho gh_A = p_{C} + \frac{1}{2}\rho v^2_C+\rho gh_C

The equation can be rewritten without 12ρv2\frac{1}{2}\rho v^2 since this is constant everywhere:

pA+ρghA=pC+ρghCp_{A} + \rho gh_A = p_C+\rho gh_C

pC=pAρg(hAhC)p_C=p_A - \rho g(h_A - h_C)

The known values can be filled in:

pC=110510009.81(10.5)p_C=1*10^5-1000*9.81(1-0.5)

Thus pC=95095[Pa]p_C = 95095[Pa]

Now there is shown how Bernoulli's principle is applied to simple streamlines, and that calculations can be done using this principle. See the next reading for more information about streamlines.

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