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The Bernoulli Equation for an

In 1738 Daniel Bernoulli (1700-1782) formulated the famous equation for fluid flow that bears his name. The Bernoulli Equation is a statement derived from conservation of energy and work-energy ideas that come from Newton's Laws of Motion.

An important and highly useful special case is where friction is ignored and the fluid is incompressible. This is not as unduly restrictive as it might first seem. The absence of friction means that the fluid flow is steady. That is, the fluid does not stick to the pipe sides and has no turbulence. Most common liquids such as water are nearly incompressible, which meets the second condition.

Consider the case of water flowing though a smooth pipe. Such a situation is depicted in the figure below. We will use this as our working model and obtain Bernoulli's equation employing the work-energy theorem and energy conservation.

We examine a fluid section of mass $m$ traveling to the right as shown in the schematic above. The net work done in moving the fluid is

 $\large{W = W_1 + W_2 = F_1 x_1 + F_2 x_2}$ (1)

where $F$ denotes a force and an $x$ a displacement. The second term picked up its negative sign because the force and displacement are in opposite directions.

Pressure is the force exerted over the cross-sectional area, or $P = F/A$. Rewriting this as $F = PA$ and substituting into Eq.(1) we find that

 $\large{\Delta W = P_1 A_1 x_1 - P_2 A_2 x_2.}$ (2)

The displaced fluid volume $V$ is the cross-sectional area $A$ times the thickness $x$. This volume remains constant for an incompressible fluid, so

 $\large{V = A_1 x_1 = A_2 x_2.}$ (3)

Using Eq.(3) in Eq.(2) we have

 $\large{\Delta W = \left( P_1 - P_2 \right) V.}$ (4)

Since work has been done, there has been a change in the mechanical energy of the fluid segment. This energy change is found with the help of the next diagram.

The energy change between the initial and final positions is given by

 $\large{\Delta E = E_2 - E_1 = \left( U_2 + K_2 \right) - \left( U_1 + K_1 \right)}$ (5)

where $U$ and $K$ are the potential and kinetic energy, respectively.

The potential energy of each fluid segment $U = mgh$ where $g$ is the acceleration of gravity, and $h$ is average fluid height. The kinetic energy is $K = \frac{1}{2}mv^2$ where $m$ is the fluid mass and $v$ is the speed of the fluid. Substituting into Eq. (5), we write

 $\large{\Delta E = \left(m_2gh_2 + \frac{1}{2}m_2 v_2^2 \right) - \left(m_1gh_1 + \frac{1}{2}m_1 v_1^2 \right).}$ (6)

The work-energy theorem says that the net work done is equal to the change in the system energy. This can be written as

 $\large{\Delta W = \Delta E.}$ (7)

Substitution of Eq.(4) and Eq.(6) into Eq.(7) yields

 $\large{\left( P_1 - P_2 \right) V = \left( m_2gh_2 + \frac{1}{2} m_2 v_2^2 \right) - \left( m_1gh_1 + \frac{1}{2}m_1 v_1^2 \right).}$ (8)

Dividing Eq.(8) by the fluid volume, $V$ gives us

 $\large{ P_1 - P_2 = \left( \rho g h_2 + \frac{1}{2} \rho v_2^2 \right) - \left( \rho g h_1 + \frac{1}{2} \rho v_1^2 \right)}$ (9)
where
 $\large{ \rho = \frac{m}{V}}$ (10)

is the constant fluid mass density. To complete our derivation, we reorganize Eq.(9).

 $\large{ P_1 + \rho g h_1 + \frac{1}{2} \rho v_1^2 = P_2 + \rho g h_2 + \frac{1}{2} \rho v_2^2. }$ (11)

Finally, note that Eq.(11) is true for any two positions. Therefore,

 $\large{ P + \rho g h + \frac{1}{2} \rho v^2 = \text{constant.} }$ (11)

Equation (11) is commonly referred to as Bernoulli's equation. Keep in mind that this expression was restricted to incompressible fluids and smooth fluid flows.