Force of reaction due to ejection of liquid – Derivation, Equation and Solved examples

Consider an example shown in figure 23.1 As the water comes out from the vessel with some speed ( generally ( 2gh )1/2 ) , it has some momentum, which was initially almost zero , when it is in the container. This change in momentum is due to a force on water ejecting in forward direction and the reaction of this force must be experienced by the container and the liquid inside .

Fig. 23.1

If the hole has a cross section ‘ a ‘ then according to Torricelli’s theorem and liquid is coming with a speed ( 2gh )1/2 .

Velocity of efflux at that instant
v = ( 2gh )1/2

Change is mass = vaρ
(Change in mass per unit time)

dm/dt = vaρ
we know that ,
F = dp/dt
F = mdv/dt + vdm/dt

Velocity for that particular small instant is constant ,
mdv/dt = 0
So , F = vdm/dt
F = v( vaρ )
F = ρav²

F = ρav² = 2ρagh

Note : If the velocity of a liquid (or gas) of density p coming out through an opening of area of cross section a is v, then there will be a thrust force F = ρav² due to liquid coming out the opening. The direction of the thrust force will be just opposite to the velocity direction .

Solved Examples

  • Problem 1 . In a wide vertical vessel filled with water, two identical holes are opened on the opposite sides .  Area of cross – section of each hole is ‘ a ‘ . Find the resultant force of reaction of water flowing out of vessel if the height difference between them is equal to h .

Solution. Let A and B be two openings, each of area ‘a’ the velocity of water coming out of opening are VA and VB , respectively .

Volume of water emerging per second through A is aVA

Volume of water emerging per second through B is aVB

Force of reaction due to water coming out at A is given by

FA = Rate of change of momentum per second at A
= (aVAρ)VA = aρVA²

Similarly, force of reaction at B is given by FB = aρVB²

Fig. 23.2

Therefore , net force on vessel is

F = FB – FA = aρ( VB² – VA² ) ….( i )

We can find value of ( VB² – VA² ) by direct writing the value of VB and VA with help of velocity of efflux ,
or we can solve by Bernoulli’s equation.

Now , Applying Bernoulli’s theorems at A ans B , we get

Pa + ρVA²/2 + ( h2 + h )ρg = Pa + ρVB²/2+ ρgh2 ….( ii )

Pressure at both the efflux will be equal to atmospheric pressure because both efflux is open in atmosphere .
Here Pa is atmospheric pressure in equation ( ii )

VB² – VA² = 2gh

Substituting this value of ( VB² – VA² ) in equation ( i ) , we get

F = aρ2gh

or F = 2ρgah

  • Problem 2 .The side wall of a wide vertical cylindrical vessel of height h = 75 cm has a narrow vertical slit running all the way down to the bottom of the vessel. The length of the slit is = 50 cm and the width b = 1.0 mm. With the slit closed, the vessel is filled with water. Find the resultant force of reaction of water flowing out the vessel immediately after the slit is opened.

Solution – Consider an element of length dx of the slit as shown in Figure 23.3

Fig. 23.3

Area of the slit= bdx

Discharge per sec through this area =ρ(bdx)v

Force of reaction due to element dx
dF= – ρ(bdx)v² ….. ( i )

Negative sign is used because this is opposite to direction of v

Applying Bernoulli’s theorem at point A , we have

Pa = Pa + ρg(h – x) + ρv²/2

v² = – 2ρg(h – x) …. ( ii )

Substituting the value of v² from equation-( ii ) in equation-( i ) we get

dF =ρ(bdx) 2ρg (h – x)

F =2ρgb ∫(h – x) dx
[ limit of integration is from x = 0 to x = l ]

F=2ρgb [ hl – ( l²/2 )]

F = ρgbl ( 2h – l )

Substituting the given values, we get

F = (1000)(9.8)(1×10³)(0.5)[2×0.75 -0.5]

F =5N

  • Problem 3 – A large open top container of negligible mass and uniform cross-sectional area A has a small hole of cross-sectional area A/100 in its side wall near the bottom . The container contains a liquid of density ρ and mass Mo and kept on a smooth horizontal floor. Assuming that the liquid starts flowing
    out horizontally through the hole at t=0 , calculate
    (a) the acceleration of the container and
    (b) its velocity when 75% of the liquid has drained out .

Solution- Liquid is escaping from the container. Let y be the height of the liquid at any instant t after start .

Fig. 23.4

(a) Then velocity of escape is given by v= (2gy)1/2

Mass of the liquid flowing in time dt= avρdt , where a = area of the hole = A/100.

Force is the rate of change of momentum. Hence ,

F= dp/dt = (avρdt/dt)v = av²ρ
F = 2agyρ

As F = mass of container × acceleration = Ma’

a’ = F/M

But M=mass of container = Ayρ

a’ = 2agyρ/M = [(A/100)2gyρ]/Ayρ = g/50

(b) When 75% of the liquid has drained out, the height of the liquid y is given by

Ayp =25Mo/50

y = Mo/4Aρ

But v = ( 2gy )1/2

v = ( 2gMo/4Aρ)1/2

v = ( Mog/2Ay )1/2

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