Torricelli’s theorem ( Velocity of efflux ) : Principle , Derivation and sloved examples

Torricelli’s Theorem 

This concept is used to evaluate the velocity of liquid flowing out from a hole in a container .

According to Torricelli’s theorem – The velocity of efflux of a liquid issuing out of an orifice is the same as it would attain if allowed to fall freely through the vertical height between the liquid surface and orifice .

Derivation for velocity of efflux

The example can be taken as shown in figure 24.1 , If from the surface of the liquid at a depth h , a hole is made of small cross section. The liquid will come out from this hole with some speed, say v. To evaluate this speed, we apply Bernoulli’s theorem at two points A and B, just inside and outside the hole .

Fig. 24.1

At point A as the cross-section of the vessel is large, we can consider the velocity of liquid particles close to zero and the
pressure at A is given as

PA = Patm + ρgh

And at B the pressure is only atmospheric .

Applying Bernoulli’s Theorem at A and B

0 + 0 + ( Patm + ρhg ) = ρv2/2 + 0 + Patm

v = ( 2gh )1/2 …. ( i )

Equation ( i ) is known as Torecelli’s Theorem and the velocity with which liquid comes out , is called efflux velocity.

This equation is used in problems concerned with liquid draining out from a vessel and in cases of conservation of momentum (cases of variable mass).

Note – The velocity of efflux is the velocity of escaping liquid relative to the container (but not necessarily relative to ground when the container moves ) .

Horizontal Range  of  the escaping liquid 

Water stands at a depth H in a tank whose sides are vertical . A hole is made in one of the walls at a depth h below the water surface [ or ( H-h) above from bottom of container , here in fig. 24.2 we use h’ in place of ( H-h ) for simplicity ]. Let container remains at rest .

Fig. 24.2

When the liquid stream emerges out of the hole, it goes along a parabolic path ( because water have some initial velocity V acceleration due to gravity is constan ) .

So time taken by water to fall through a height of (H – h) is given as

( H-h ) = 0xt + gt²/2

t = [ 2(H-h)/g ]1/2

Horizontal range is given by

horizontal range = velocity in horizontal direction × time of flight

x = v x t = (2gh)1/2[ 2(H-h)/g ]1/2

x = [ 4h(H-h) ]1/2

x = 2( hh’ )1/2 …. ( i )
or
x = [ 4h(H-h) ]1/2

x = [ H²-( 2h – H )² ]1/2

For the range to be maximum,
(2h – H)²= 0

2h – H=0
h=H/2

Note – for maximum range of escaping liquid we can find condition by using
dx/dt = 0

So the maximum horizontal range x [ Put h’ = H/2 in equation ( i ) ] = H

The formula = 2( hh’ )1/2 tells us if we make two holes at equal vertical distances from top and bottom , both liquids jets will strike the same spot ( but not simultaneously )

Solved Examples

  • Problem1. There is a small hole at the bottom of tank filled with water. If total pressure at the bottom is 3 atm ( 1 atm = 10⁵ N/m² ), then find the velocity of water flowing from hole.

Solution –
Pressure at bottom
Po + ρgh = 3 atm
ρgh = 2 atm
ρgh = 2× 10⁵ N/m²
gh = 200 N/m²

Velocity of efflux = (2gh)1/2
= ( 400 )1/2
= 20 m/s

  • Problem 2 . In the figure shown 24.3 , Find the range R 
Fig. 24.3

Solution – Applying Bernoulli’s equation at point 1 and 2 .

P1 + ( 2ρ )v1² + ( 2ρ )gh1 = P2 + ( 2ρ )v2² + ( 2ρ )gh2

Here , v1 ≈ 0 , v2 = v
h1 = h2 , p2 = po

and p1 = Po + ρgH + 2ρhH
p1 = Po + 3ρgH

Substuting in the above equation we have ,

po + 3ρgH = po + ρv²

v = ( 3gH )1/2

Now , this velocity is horizontal . So , time taken by the liquid to fall to the ground is free fall time or

t = ( 2h/g )1/2
( where , h=H )

t = ( 2H/g )1/2

R =vt = ( 3gH )1/2( 2H/g )1/2

R = (6)1/2H

  • Problem 3. On a platform of height 5 m , a cylindrical tank of radius 1m is at rests , In starting the tank is filled with water up to a height of 5 m . A plug whose area is 10-4 m² is removed from an orifice on the side of the tank at the bottom .

    (a) Find out initial speed with which the s water flows from the orifice
    (b) Determine initial speed with which the water strikes the ground

Solution – As show in the figure 24.4

Fig. 24.4

( a ) speed of efflux is given by
v= ( 2gh )1/2
v= ( 2×10×5 )1/2
v = 10 m/s = vH
( This is horizontal speed of efflux)

( b ) As initial vertical speed of water is zero , so its vertical speed when it hits the ground

vv = ( 2gh )1/2
= ( 2×10×5 )1/2
= 10 m/s

So the initial speed with
which water strikes the ground .
v = ( vH² + vv² )1/2
= 10(2)1/2
= 14.1 m/s

  • Problem 4. In cylindrical container water is filled up to height 3 m. The ratio of the cross-sectional area of the orifice and the beaker is 0.1 . The square of the the liquid coming out from the orifice is (g = 10 m/s²)
Fig. 24.5

Solution-
Here we can’t direct apply the Torricelli’s theorem [ that the velocity of efflux will be ( 2gh )1/2 ] , Because area of efflux is not very less as compared to area of container . So we can’t consider velocity of decreasing of water level of container to zero .

So , we will solve this from basic equations.

Apply Bernoulli’s equation at uper surface of liquid and just out the efflux

po + ρv1²/2 + ρgh = po + ρv2²/2 + 0

ρv1²/2 + ρgh = ρv2²/2

v1²/2 + 10×2.475 = v2²/2

v2² – v1² = 49.5 …( i )

From equation of continuity ,
A1v1 = A2v2

v1/v2 = A2/A1 = 0.1 …( ii )

From equation ( i ) and ( ii )

v2² = 50 m²/s²

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