Bernoulli’s Equation

A pressure force F1 acts on the lower cylinder due to fluid to its left , and a pressure for F₁ acts on the upper cylinder in the opposite direction. The net work done on the system by F1 and F2 is

W = F₁Δl₁ – F₂Δl₂ = P₁A₁Δl₁ – P₂A₂Δl₂ = (P₁- P₂)ΔV

where we have used the relations F = PA and ΔV = A₁Δl₁ = A₂Δl₂. The net effect of the motion of the system is to raise the height of the lower cylinder of mass Δm and to change its speed. The changes in the potential and kinetic energies are

ΔU = Δm g(y₂ – y₁)

ΔK = 1/2 Δm (v₂₂ – v₁₂)

These changes are brought about by the net work done on the system,

W = ΔU + ΔK

(P₁ – P₂)ΔV = Δmg(y₂ – y₁) + 1/2 Δm (v₂₂ – v₁₂)

Since the density is ρ = Δm/ΔV, we have

p₁ + ρgy₁ + 1/2 ρv₁₂ = p₂ + ρgy₂ + 1/2 ρv₂₂

Since the points 1 and 2 can be chosen arbitrarily, we can express this result as Bernoulli’s Equation

p + ρgy + 1/2ρv₂ = constant (12.30)

Example: 12.20

Solution

Assume datum at the free surface of the liquid.

(a) Applying Bernoulli’s equation on point 1 and 2 , as shown in the figure.

Here p₁ = p₂ = p₀ = 105 N/m² ; y¹ = 0, y² = -5 m

Since area of the tube is very small as compared to that of the reservoir, therefore,

v1 << v₂ . thus

or p_B = 10⁵ - 1/2 (900)(10)² - (900)(10)(1.5) = 41.5 kN/m²

(c) Applying Bernoulli’s equation at 1 and A

pA = p₁ + ρg (y1 - yA)

orpA = 10⁵ + (90⁰)(10)(1) = 10⁹ kN/m².

Applying Bernoulli’s equation at 1 and C,

pC = p1 - 1/2

= 10⁵ - 1/2 (900)(10)² - (90⁰)(10)(-1) = 10⁵ - 45000 + 9000 = 64 kN/m².

(d) The velocity of flow is independent of the density of the liquid , therefore, the discharge would remain the same.

(e) Since the pressure at B is less than atmospheric, the liquid, therefore, has a tendency to get vapourised if the pressure becomes equal to the vapour pressure of it. Thus , pB > pvapour.

(f) The velocity of flow depends on the depth of the point D, below the free surface

For working of siphon , H ≠ 0, and H should not be high enough so that pB may not reduce to vapour pressure.

Example: 12.21

A garden hose has an inside cross-sectional area of 3.60 cm², and the opening in the nozzle is 0.250 cm2. If the water velocity is 50 cm/s in a segment of the hose that lies on the ground

(a) with what velocity does the water come from the nozzle when it is held 1.50 m above the ground and

(b) What is the water pressure in the hose on the ground?

Solution

(a) The problem is illustrated in figure. We first apply the Equation of Continuity. Equation (), to find the velocity of the fluid at the nozzle.

v2 =

(b) We next apply Bernoulli’s Equation to find the pressure p1. We know that

h₁ = 0 and h₁ = 1.5 m.

The pressure at the nozzle is atmospheric pressure

p² = 1.01 × 10⁵ Pa.

Solving for p1 and using the density of water

ρ = 1 × 10³ kg/m³, we have

p₁ = p₂ + 1/2 ρ

= (1.01 × 105) + 1/2(1 × 103)[(7.2)² – (0.50)2]

+ (1 × 10³)(9.8)(1.5 – 0) = 1.41 × 105 Pa

Example: 12.22

Let v be the instantaneous velocity of the tank, and c be the instantaneous velocity of efflux with respect to the tank.

Thrust exerted on the tank is

F = ρac²

where a is the cross-sectional area of the orifice.

c = √2gh

where h is the instantaneous height of water in the tank.

or    (1)

In a time dt if the water level falls by dh, then according to the conservation of mass.

Equation (1) can be written as

On integrating

v = 2√2gH

Since H = 4 m,

therefore v = 2√2(10)(4) = 17.9 m/s