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11TH•Physics•Solids And Fluid Dynamics•Bernoulli's Equation

5.11 Bernoulli's Equation

P + ½ρv² + ρgh is constant for an ideal fluid in steady flow.

5.11 Bernoulli's Equation

The sum of pressure, K.E. per unit volume and P.E. per unit volume of an ideal fluid throughout its steady flow remains constant.

As the fluid moves through a pipe of varying cross-section and height, the pressure will change along the pipe. Bernoulli's equation is the fundamental equation in fluid dynamics that relates pressure to fluid speed and height.

In deriving Bernoulli's equation, we assume that the fluid is incompressible, non-viscous and flows in a steady state manner. Let us consider the flow of the fluid through the pipe in time tt, as shown in Fig. 5.17.

The force on the upper end of the fluid is P1A1P_{1}A_{1}, where P1P_{1} is the pressure and A1A_{1} is the area of cross-section at the upper end. The work done on the fluid, by the fluid behind it, in moving it through a distance Δx1\Delta x_{1}, will be:

W1=F1Δx1=P1A1Δx1W_{1} = F_{1}\Delta x_{1} = P_{1}A_{1}\Delta x_{1}

Similarly, the work done on the fluid at the lower end is:

W2=−F2Δx2=−P2A2Δx2W_{2} = -F_{2}\Delta x_{2} = -P_{2}A_{2}\Delta x_{2}

where P2P_{2} is the pressure, A2A_{2} is the area of cross-section of lower end and Δx2\Delta x_{2} is the distance moved by the fluid in same time interval tt. The work W2W_{2} is taken to be −-ve as this work is done against the fluid force. The net work done will be:

W=W1+W2W = W_{1} + W_{2}

or

W=P1A1Δx1−P2A2Δx2…(5.17)W = P_{1}A_{1}\Delta x_{1} - P_{2}A_{2}\Delta x_{2} \quad \ldots (5.17)

If v1v_{1} and v2v_{2} are the velocities at the upper and lower ends respectively, then:

W=P1A1v1t−P2A2v2tW = P_{1}A_{1}v_{1}t - P_{2}A_{2}v_{2}t

From equation of continuity, A1v1=A2v2A_{1}v_{1} = A_{2}v_{2}

Hence A1v1t=A2v2t=VA_{1}v_{1}t = A_{2}v_{2}t = V (volume)

So

W=(P1−P2)V…(5.18)W = (P_{1} - P_{2})V \quad \ldots (5.18)

If mm is the mass and ρ\rho is the density, then V=m/ρV = m/\rho. So, Eq. (5.18) becomes:

W=(P1−P2)m/ρ…(5.19)W = (P_{1} - P_{2})m/\rho \quad \ldots (5.19)

A part of this work is utilized by the fluid in changing its K.E. and a part is used in changing its gravitational P.E.

Change in K.E.:

ΔK.E.=12mv22−12mv12…(5.20)\Delta\text{K.E.} = \tfrac{1}{2}mv_{2}^{2} - \tfrac{1}{2}mv_{1}^{2} \quad \ldots (5.20)

Change in P.E.:

ΔP.E.=mgh2−mgh1…(5.21)\Delta\text{P.E.} = mgh_{2} - mgh_{1} \quad \ldots (5.21)

where h1h_{1} and h2h_{2} are the heights of the upper and lower ends respectively.

Applying the law of conservation of energy to this volume of the fluid, we have:

(P1−P2)m/ρ=12mv12−12mv22+mgh1−mgh2…(5.22)(P_{1} - P_{2})m/\rho = \tfrac{1}{2}mv_{1}^{2} - \tfrac{1}{2}mv_{2}^{2} + mgh_{1} - mgh_{2} \quad \ldots (5.22)

Rearranging Eq. (5.22), we have:

P1+12ρv12+ρgh1=P2+12ρv22+ρgh2P_{1} + \tfrac{1}{2}\rho v_{1}^{2} + \rho gh_{1} = P_{2} + \tfrac{1}{2}\rho v_{2}^{2} + \rho gh_{2}

This is Bernoulli's equation and is often expressed as:

P+12ρv2+ρgh=constantP + \tfrac{1}{2}\rho v^{2} + \rho gh = \text{constant}

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