Bernoulli’s Equation
Solids And Fluids of Class 11
Bernoulli’s Equation
Bernoulli’s EquationLet us for our attention on the motion of the shaded region. This is our “system”. the lower cylindrical element of fluid of length Δl_{1} and area A_{1} is at height y_{1}, and moves at cylindrical element of fluid of length Δl1 and area A1 is at height y_{1}, and moves speed v1. After some time, the leading section of our system fills the upper cylinder of fluid of length Δl_{2} and area A_{2} at height y_{2}, and is then moving speed v_{2}. 

A pressure force F1 acts on the lower cylinder due to fluid to its left , and a pressure for F_{1} acts on the upper cylinder in the opposite direction. The net work done on the system by F1 and F2 is
W = F_{1}Δl_{1} – F_{2}Δl_{2 }= P_{1}A_{1}Δl_{1} – P_{2}A_{2}Δl_{2} = (P_{1} P_{2})ΔV
where we have used the relations F = PA and ΔV = A_{1}Δl_{1} = A_{2}Δl_{2}. 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(y_{2} – y_{1})
ΔK = 1/2 Δm (v_{22} – v_{12})
These changes are brought about by the net work done on the system,
W = ΔU + ΔK
(P_{1} – P_{2})ΔV = Δmg(y_{2} – y_{1}) + 1/2 Δm (v_{22} – v_{12})
Since the density is ρ = Δm/ΔV, we have
p_{1 }+ ρgy_{1 }+ 1/2 ρv_{12} = p_{2} + ρgy_{2} + 1/2 ρv_{22}
Since the points 1 and 2 can be chosen arbitrarily, we can express this result as Bernoulli’s Equation
p + ρgy + 1/2ρv_{2} = constant (12.30)
Daniel Bernoulli derived this equation in 1738. It is applied to all points along a streamline in a nonviscous, incompressible fluid. 

Example: 12.20
A siphon tube is discharging a liquid of specific gravity 0.9 from a reservoir as shown in the figure(12.32). (a) Find the velocity of the liquid through the siphon (b) Find the pressure at the highest point B. (c) Find the pressure at the points A(out side the tube) and C. State and explain the following : (d) Would the rate of flow be more, less or the same if the liquid were water ? (e) Is there a limit on the maximum height of B above the liquid level in the reservoir. (f) Is there a limit on the vertical depth of the right limb of the siphon. 

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 p_{1 }= p_{2 }= p_{0} = 105 N/m^{2} ; y^{1} = 0, y^{2} = 5 m
Since area of the tube is very small as compared to that of the reservoir, therefore,
v1 << v_{2} . thus
Here, p1 = 105 N/m2 ; ; y_{1 }= 0, v_{B} = v_{2} = 10 m/s, y_{B} = 1.5 m ∴p_{B} = p_{1}  

or p_{B} = 10^{5}  1/2 (900)(10)^{2}  (900)(10)(1.5) = 41.5 kN/m^{2}
(c) Applying Bernoulli’s equation at 1 and A
pA = p_{1} + ρg (y1  yA)
orpA = 10^{5 }+ (90^{0})(10)(1) = 10^{9} kN/m^{2}.
Applying Bernoulli’s equation at 1 and C,
pC = p1  1/2
= 10^{5}  1/2 (900)(10)^{2}  (90^{0})(10)(1) = 10^{5}  45000 + 9000 = 64 kN/m^{2}.
(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 crosssectional area of 3.60 cm^{2}, 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
h_{1 }= 0 and h_{1} = 1.5 m.
The pressure at the nozzle is atmospheric pressure
p^{2 }= 1.01 × 10^{5} Pa.
Solving for p1 and using the density of water
ρ = 1 × 10^{3} kg/m^{3}, we have
p_{1 }= p_{2} + 1/2 ρ
= (1.01 × 105) + 1/2(1 × 103)[(7.2)^{2} – (0.50)2]
+ (1 × 10^{3})(9.8)(1.5 – 0) = 1.41 × 105 Pa
Example: 12.22
A tank, intially at rest, is filled with water to a height H = 4 m. A small orifice is made at the bottom of the wall. Find the velocity attained by the tank when it becomes completely empty. Assume mass of the tank to be negligible. Friction is negligible. Solution 

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 = ρac^{2}
where a is the crosssectional area of the orifice.
c = √2gh
where h is the instantaneous height of water in the tank.
Mass of the tank at any time t is m = ρAh A = crosssectional area of the tank. Using Newton’s second law 2ρgah 

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