Impact of Jets MCQ Quiz in తెలుగు - Objective Question with Answer for Impact of Jets - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

Concept: 

Force exerted by a jet of water on a flat plate:

V = velocity of jet, d = diameter of the jet, a = area of x – section of the jet

\(a=\frac{\pi d^2} {4}\)

ρ = Density of water
The force exerted by the jet on the plate in the direction of jet.

F_X = ρaV²

Calculation:

Given Data:

V= 20 m/s

d= 100mm

ρ = 1000 Kg/m³

\(a=\frac {\pi\times0.1^2}{4}\)

 \(a= \frac{\pi}{400}\)

now, F_X =  ρaV2

\(F_X =\frac{1000\times \pi \times20\times 20}{400} \)

F_X = 3.14kN

Concept:

Force exerted by a jet on the vertical moving plate in direction of the jet with velocity u is given by:

\(F = \rho a(v-u)^2\)

where ρ = Density of Liquid, v = velocity of Jet, u = velocity of plate, a = area of jet

Calculation:

Given:

v = 25 m/s, u = -15 m/s, a = 0.03 m², Relative Density of Jet Liquid = 0.8

Density of Jet Liquid, ρ = 0.8 × 1000 = 800 Kg/m3

F = 800 × 0.03 (25 + 15)²

F = 38400 N

F = 38.4 KN

Concept:

Force exerted by the jet in the direction of jet,

F_x = Mass per sec × [V_1x - V_2x]

F_x = ρaV[V - (-V cos θ)] = ρaV[V + V cos θ]

= ρaV²[1 + cos θ]

F_y = Mass per sec × [V_1y - V_2y]

F_y = ρaV[0 - V sin θ] = -ρaV² sin θ

Calculation:

∴ Area, \(a = \frac{\pi }{4}{\left( {0.5} \right)^2} = 0.001963\;{m^2}\) 

Velocity of jet, V = 40 m/s

Angle of deflection = 120° = 180° - θ or θ = 60°

Force exerted by the jet on the curved plate in the direction of the jet:

F_x = ρaV² [1 + cos θ]

Explanation:

Let a jet of water strikes an inclined plate, which is moving with a uniform velocity in the direction of the jet

Let
V = absolute velocity of the jet of water

u = velocity of the plate in the direction of the jet

A = cross-sectional area of jet

θ = angle between jet and plate

The relative velocity of the jet of water = (V - u)

The velocity with which jet strikes = (V - u)

Mass of water striking per second

= ρ x A x (V - u)

If the plate is smooth and the loss of energy due to the impact of the jet is assumed zero, the jet of water will leave the inclined plate with a velocity equal to (V - u).
The force exerted by the jet of water on the plate in the direction normal to the plate is given as

F_n = Mass striking per second × [ Initial velocity in the normal direction with which jet strikes - Final velocity]

= ρ A (V - A) [(V - u) sin θ - 0 ] = ρ A (V - u)² sin θ 

At circular opening the jet velocity will be given by:

\(v = \sqrt {2gh}\) 

h: height from opening centre to free surface

\(v = \sqrt {2 \times 10 \times 6.2} = \sqrt {124} \;m/s\) 

Force exerted by jet on the block will be:

F = ρav²

\(= {10^3} \times \frac{\pi }{4}{\left( {0.3} \right)^2} \times {\left( {\sqrt {124} } \right)^2}\) 

= 10³ × 8.7650

F = 8.7650 × 10³ N

Now make free body diagram of block & wall:

Under stationary condition:

∑F_H = 0

F = N

N = 8.7650 × 10³ N

∑F_V = 0

μN = mg

m × 10 = 0.228 × 8.7650 × 10³

Concept:

• There is no energy interaction between an external source and a flow or any dissipation of mechanical energy in the flow.
• Fluid mass rotates due to the conservation of angular momentum.
• Velocity is inversely proportional to the radius.
• For the case of constant angular momentum in fluid flow, the flow will be free vortex flow.

v ∝ 1/r

⇒ v1r1 = v2r2

Calculation:

Given:

v₁ = 1.2 m/sec, r₁ = 3 m, r₂ = 1.5 m, v₂ = ?

⇒ v₁r₁ = v₂r₂

⇒  1.2  × 3 = v2  × 1.5

⇒  v2 =  2.4 m/s

Additional Information 

Vortex flow:

The motion of a fluid in a curved path is known as vortex flow.

When a cylindrical vessel containing some liquid is rotated about its vertical axis, the vortex flow will be followed by liquid.

Vortex motion is of two types:

1. Forced vortex:

(i) In the forced vortex, fluid moves on the curve under the influence of external torque.

(ii) Due to the external torque, a forced vortex is a rotational flow.

(iii) As there is a continuous expenditure of energy, Bernoulli's equation is not valid for forced vortex.

(iv) For forced vortex, v = rω is applicable.

(v) Examples: 

• The flow of water through a runner of the turbine.
• Rotation of water in the washing machine.
• Flow of liquid inside the impeller of a centrifugal pump
• A vertical cylinder containing liquid which is rotated about its central axis with a constant angular velocity

2. Free vortex:

(i) When no external torque is required to rotate the fluid mass, that type of flow is called a free vortex.

(ii) As there is no torque in the free vortex, so free vortex is an irrotational flow.

(iii) For free vortex, a moment of momentum is constant i.e. 

vr = constant      ....(1)

(iv) Examples:

• The flow of liquid through a hole provided at the bottom of a container.
• Draining the bathtub.

Concept:

V_r = relative velocity of jet with respect to the plate

• When plate is moving (u) in the direction of jet (V) i.e. away from the jet: V_r = V - U
• When plate is moving (U) in the opposite direction of jet (V) i.e. towards the jet: V_r = V + U

Force exerted by jet on the plate in the direction normal to the plate:

\(\begin{array}{l} {F_n} = \dot m\left[ {{V_{1n}} - {V_{2n}}} \right] = \rho A{V_r}\left[ {{V_r}\sin \theta - 0} \right]\\ {F_n} = \rho AV_r^2\sin \theta \end{array}\)

Force exerted by jet on plate in the direction of jet (F_x) and in the perpendicular to the direction of flow (F_y)

\(\begin{array}{l} {F_x} = {F_n}\sin \theta = \rho AV_r^2{\sin ^2}\theta \\ {F_y} = {F_n}\cos \theta = \rho AV_r^2\sin \theta \cos \theta \end{array}\)

Application:

So, force exerted by jet in the direction of jet (F_x) on flat plate moving with velocity u, towards the jet of velocity V is

F_x = ρA (v + u)² sin²θ

Concept:

Force exerted on the blades by a jet is given by

F = ρAv(v - u)

u is the velocity of the plate. If the plate is stationary then u = 0

For stationary plate:

F = ρAv²

The power obtained, P = F × v

Calculation:

Given:

v = 8 m/s, F = 120 N

P = F × v = 120 × 8 = 960 N = 0.96 kN

Concept:

Force on the plate due to impact, F = ρav2

where, ρ = density, a = Area of jet, v = velocity of jet

Calculation:

Velocity at the inlet of jet, \({V_1} = \frac{Q}{{{A_1}}} = \frac{{0.25}}{{\frac{\pi }{4} \times {{0.3}^2}}}\)

V₁ = 3.536 m/s

from continuity equation, A₁V₁ = A₂V₂

\(\frac{\pi }{4} \times {0.3^2} \times 3.536 = \frac{\pi }{4} \times {0.15^2} \times {V_2}\)

V₂ = 14.14 m/s

Force required to hold the plate normally

\(F = \rho AV_2^2\)

\(F = {10^3} \times \frac{\pi }{4} \times {0.15^2} \times {(14.14)^2}\)

F = 3530 N

Explanation:

If the plate is smooth and if it is assumed that there is no loss of energy due to impact of the jet, than jet will move over the plate after striking with a velocity equal to initial velocity i.e., with a velocity V. Let us find the force exerted by the jet on the plate in the direction normal to the plate. Let this force is represented by F_n

Then

F_n = mass of jet striking per second × {Initial velocity of jet before striking in the direction of n - Final velocity of jet after striking in the direction of n)

= ρaV [V sin θ - 0] = ρaV² sin θ

This force can be resolved into two components, one in the direction of the jet and other perpendicular to the direction of flow. Then we have,

F_x = component of F_n in the direction of flow

= F_n cos (90° - θ) = F_n sin θ = ρA V² sin θ × sin θ (∴ F_n = ρav sin θ)

= ρAV² sin² θ

F_y = component of Fr, perpendicular to flow

= F_n sin (90° - θ) = F_n cos θ = ρAV² sin θ cos θ.

NOTE:

Let us apply the impulse-momentum equation in the direction normal to the plate