Physics · Ch 2 — Mechanical Properties of Fluids

Bernoulli's Equation

2.9

Bernoulli's Equation

Observing a river, one notices that the speed of the water decreases in a wider stretch of the river and increases in a narrower stretch. From this observation, one might guess that the pressure within the river's water is greater in the narrower stretches — but this guess is exactly backwards: the pressure within the fluid is actually lower in the narrower part of the river and higher in the wider part. This was discovered experimentally by the Swiss scientist Daniel Bernoulli (1700-1782), working with fluid flowing inside pipes: he observed that the speed of a fluid increases in a narrow region of a pipe, while the internal pressure of the fluid in that same narrow region simultaneously decreases. This phenomenon is called Bernoulli's principle.

Bernoulli's equation relates the speed of a fluid at a point, the pressure at that point, and the height of that point above some chosen reference level; it is, at its core, an application of the work-energy theorem to a fluid in flow. Because Bernoulli's principle is fully consistent with the general principle of conservation of energy, it can be derived directly from energy conservation.

Consider the flow of an ideal fluid through a tube of varying cross-section and height, and focus on an element of fluid lying between an entry cross-section P and an exit cross-section R. Let $v_1$ and $v_2$ be the fluid's speed at the lower end P and the upper end R respectively; $A_1$ and $A_2$ the cross-sectional area at P and R; $P_1$ and $P_2$ the pressure of the fluid at P and R; and $d_1$ , $d_2$ the distances travelled at P and R respectively, over a small time interval dt, by fluid moving at $v_1$ and $v_2$ . The forces $P_1A_1$ and $P_2A_2$ act on the fluid at areas $A_1$ (at P) and $A_2$ (at R) respectively. Because the fluid is incompressible, the volume dV of fluid passing through any cross-section during time dt is the same everywhere along the tube:

dV = A_1 d_1 = A_2 d_2 \qquad \text{--- (2.41)}

Since the fluid is ideal, there is no internal friction at all within it (and even for a real fluid like water, the energy lost to viscous friction is negligible in practice) — so the only non-gravitational force doing work on this fluid element as it moves from P to R is the pressure exerted by the surrounding fluid. The net work W done on the element by this surrounding pressure, over the move from P to R, is:

W = P_1A_1d_1 - P_2A_2d_2

(the second term carries a negative sign because the pressure force at R opposes the displacement of the fluid). Using Eq. (2.41), this becomes:

W = P_1\,dV - P_2\,dV = (P_1 - P_2)\,dV \qquad \text{--- (2.42)}

Because this work W arises purely from forces other than the conservative force of gravity, it must equal the resulting change in the fluid element's total mechanical energy — its kinetic energy plus its gravitational potential energy:

W = \Delta K.E. + \Delta P.E. \qquad \text{--- (2.43)}

At the start of the time interval dt, the mass and kinetic energy of the fluid element between P and the neighbouring section Q are $\rho A_1 d_1$ and $\tfrac{1}{2}\rho A_1 d_1 v_1^2$ respectively; at the end of dt, the kinetic energy of the fluid now between the corresponding sections R and S is $\tfrac{1}{2}\rho A_2 d_2 v_2^2$ . So the net change in kinetic energy during dt is:

\Delta K.E. = \tfrac{1}{2}\rho A_2 d_2 v_2^2 - \tfrac{1}{2}\rho A_1 d_1 v_1^2 = \tfrac{1}{2}\rho\,dV(v_2^2 - v_1^2) \qquad \text{--- (2.44)}

(using $A_1d_1 = A_2d_2 = dV$ from Eq. 2.41). Similarly, at the start of dt, the gravitational potential energy of the mass m of fluid between P and Q is $mgh_1 = \rho\,dV\,g h_1$ , and at the end of dt, the potential energy of the same mass, now between R and S, is $mgh_2 = \rho\,dV\,g h_2$ ; so the net change in potential energy during dt is:

\Delta P.E. = \rho\,dV\,g(h_2 - h_1) \qquad \text{--- (2.45)}

Substituting Eqs. (2.42), (2.44) and (2.45) into Eq. (2.43):

(P_1 - P_2)\,dV = \tfrac{1}{2}\rho\,dV(v_2^2 - v_1^2) + \rho\,dV\,g(h_2 - h_1)

Dividing throughout by dV:

P_1 - P_2 = \tfrac{1}{2}\rho(v_2^2 - v_1^2) + \rho g(h_2 - h_1) \qquad \text{--- (2.46)}

This is Bernoulli's equation: it states that the work done per unit volume of a fluid, by the surrounding fluid, equals the sum of the changes in kinetic and potential energy per unit volume that occur during the flow. Equation (2.46) can equally be rewritten, collecting P, Q terms on each side, as:

P_1 + \tfrac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \tfrac{1}{2}\rho v_2^2 + \rho g h_2 \qquad \text{--- (2.47)}

\boxed{P + \tfrac{1}{2}\rho v^2 + \rho g h = \text{constant}} \qquad \text{--- (2.48)}

Dimensionally, pressure is an energy per unit volume, and both terms on the right-hand side of Eq. (2.46) also have the dimensions of energy per unit volume — so the left-hand side, P, is often referred to as the pressure energy per unit volume, or pressure head; the term $\tfrac{1}{2}\rho v^2$ is called the velocity head, and $\rho g h$ the potential head. In other words, Bernoulli's principle is simply the principle of conservation of energy, applied directly to a flowing fluid.

Applications of Bernoulli's equation:

a) Speed of efflux (Torricelli's theorem). The word 'efflux' means fluid outflow. Torricelli discovered that the speed of efflux from an open tank is given by a formula identical to that of a freely falling body. Consider a liquid of density ρ filled in a tank of large cross-sectional area $A_1$ , with an orifice of cross-sectional area $A_2$ at the bottom, where $A_2 \ll A_1$ . Let $v_1$ and $v_2$ be the speeds of the liquid at $A_1$ and $A_2$ respectively. Since both the inlet (free surface) and the outlet (orifice) are exposed to the atmosphere, the pressure at both equals atmospheric pressure $p_0$ . If h is the height of the free surface above the orifice, Bernoulli's equation gives:

p_0 + \tfrac{1}{2}\rho v_1^2 + \rho g h = p_0 + \tfrac{1}{2}\rho v_2^2 \qquad \text{--- (2.49)}

Using the continuity equation, $v_1 = v_2(A_2/A_1)$ , and substituting into Eq. (2.49):

v_2^2\left(1 - \dfrac{A_2^2}{A_1^2}\right) = 2gh

v_2 = \sqrt{\dfrac{2gh}{1 - A_2^2/A_1^2}}

and if $A_2 \ll A_1$ , this reduces to:

\boxed{v_2 = \sqrt{2gh}} \qquad \text{--- (2.50)}

This is the speed of a liquid flowing out through an orifice at depth h below its free surface — identical to the speed a particle would reach falling freely under gravity through the same height h.

b) Venturi tube. A Venturi tube is used to measure the speed of flow of a fluid inside a pipe; it has a constriction built into the tube, and as fluid passes through this constriction its speed increases (per the equation of continuity), so its pressure correspondingly decreases (per Bernoulli's equation). If a fluid of density ρ flows through a Venturi tube with cross-sectional area $A_1$ (speed $v_1$ , pressure $p_1$ ) at the wide part and $A_2$ (speed $v_2$ , pressure $p_2$ ) at the constriction, Bernoulli's equation (with the tube horizontal, so height terms drop out) gives:

p_1 + \tfrac{1}{2}\rho v_1^2 = p_2 + \tfrac{1}{2}\rho v_2^2

p_1 - p_2 = \tfrac{1}{2}\rho(v_2^2 - v_1^2) \qquad \text{--- (2.51)}

Two vertical tubes connected to the Venturi tube at $A_1$ and $A_2$ show a visible difference h in liquid-column height, related to the pressure difference by $p_1 - p_2 = \rho g h$ ; substituting into Eq. (2.51):

\rho g h = \tfrac{1}{2}\rho(v_2^2 - v_1^2) \qquad \text{--- (2.52)}

Using the continuity equation, $A_1v_1 = A_2v_2$ , to substitute $v_1$ in terms of $v_2$ (or vice versa) in Eq. (2.52), the fluid's actual rate of flow can be calculated purely from the two known areas $A_1$ and $A_2$ and the measured level difference h. …

Figure 2.34Fig. 2.34: Flow of fluid through a tube of varying cross section and height — element between P (speed v₁, area A₁, height h₁, force P₁A₁) and R (speed v₂, area A₂, height h₂, force P₂A₂)