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Proof of Bernoulli's theorem Consider a fluid of negligible viscosity moving with laminar flow, as shown in Figure 1. Let the velocity, pressure and area of the fluid column be v1, P1 and A1 at Q and v2, P2 and A2 at R. Let the volume bounded by Q and R move to S and T where QS = L1, and RT = L2. If the fluid is incompressible: A1L1 = A2L2 The work done by the pressure difference per unit volume = gain in k.e. per unit volume + gain in p.e. per unit volume. Now: Work done = force x distance = p x volume Net work done per unit volume = P1 - P2 k.e. per unit volume = ½ mv2 = ½ Vρ v2 = ½ρv2 (V = 1 for unit volume) Therefore: k.e. gained per unit volume = ½ ρ(v22 - v12) p.e. gained per unit volume = ρg(h2 – h1) where h1 and h2 are the heights of Q and R above some reference level. Therefore: P1 - P2 = ½ ρ(v12 – v22) + ρg(h2 - h1) P1 + ½ ρv12 + ρgh1 = P2 + ½ ρv22 + rgh2 Therefore: P + ½ ρv2 + ρgh is a constant For a horizontal tube h1 = h2 and so we have: P + ½ ρv2 = a constant This is Bernoulli's theorem You can see that if there is a increase in velocity there must be a decrease of pressure and vice versa. No fluid is totally incompressible but in practice the general qualitative assumptions still hold for real fluids.
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