Transcript Force exerted by the jet on the moving plate.

IMPACT OF JETS
PREPARED BY
KIRIT S DAYMA
PRUTHVI GADHVI
SHAIL GADHVI
BRIGESH GADIYA
JAY GAJJAR
JOYAL GAJJAR
GURBANI PAVAN
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(130260119023)
(130260119024)
(130260119025)
(130260119027)
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The liquid comes out in the form of a jet from the outlet of a nozzle,
which is fitted to a pipe through
which the liquid if flowing under pressure. If some plate, which may be
fixed or moving, is placed in
the path of the jet, a force is exerted by the jet on the plate. This force
is obtained from Newton’s
second law of motion or from impulse-moment equation. Thus impact
of jet means the force exerted
by the jet on a plate which may be stationary or moving. In this
chapter, the following cases of the
impact jet i.e., the force exerted by the jet on a plate, will be
considered:
(a) Force exerted by a jet on a stationary plate when
1. Plate is vertical to the jet,
2. Plate is inclined to the jet, and
3. Plate is curved.
(b) Force is exerted by the jet on the moving plate, when
1. Plate is vertical to the jet,
2. Plate is inclined to the jet, and
3. Plate is curved
1. Force exerted by a jet on a stationary vertical plate.
Consider a jet of water coming out of the nozzle, strikes a flat vertical
plate as shown in the figure.1.
(2) Force exerted by a jet on a stationary inclined flat plate.
Let a jet of water, coming out from the nozzle; strike an inclined flat plate
as shown in the figure.2.2
Then
Fn= Mass of the jet striking per second × [initial velocity of the jet
before striking in the direction of
n - final velocity of the jet after striking in the direction of n]
= pav[v sinθ – 0] =pav² sinθ
If the force can be resolved into two components, one in the direction of the jet and the other
perpendicular to the direction of the flow. Then we have,
Fx =pav² sin²θ
(along the direction of the flow) and Fy = pav² sinθ cos θ (perpendicular to flow)
(3) Force exerted by a jet on a stationary curved plate.
1.Jet strikes the curved plate at the centre. Let a jet of water strike a
fixed curved plate at the centre as shown in figure.3.
2.The jet after striking the plate comes out with the same velocity if
the plate is smooth and there is no loss of energy due to impact of
the jet, in the tangential direction of the curved plate.
The velocity at the outlet of the plate can be resolved in to two
components, one in the direction of the jet and the other
perpendicular to the direction of the jet.
Component of velocity In the direction of the jet = - v cos Ɵ
(-ve sign is taken as the velocity at the outlet is in the opposite
direction of the jet of water coming out from nozzle).
Component of the velocity perpendicular to the jet = V sin Ɵ
Force exerted by the jet In the direction of the jet,
Fx = Mass per sec × [V1x – V2x]
Where V1x = final velocity in the direction of the jet = V
V2x = final velocity in the direction of the jet = - V cos Ɵ
∴ Fx = ρaV [V- (-V cos Ɵ)] = ρaV[V + V cos Ɵ]
= pav²[1+cosθ]
similarly Fy = Mass per sec × [V1y – V2y]
Where V1y = initial velocity in the direction of y = 0
V2y = final velocity in the direction of y = V sin Ɵ
Fy = ρaV [0- V sin Ɵ] = - ρaV² sin Ɵ
-ve sign means the force is acting in the downward direction. In this case the angle of deflection of the
jet = (180° –Ɵ)
(4) Jet strikes the curved plate at on e end tangentially when
the plate is symmetrical.
Let the jet strike the curved plate at one end tangentially as shown in
the figure.4. Let the curved is symmetrical about x-axis.
Then the angle made by the tangents at the two ends of the plate will
b same.
Let V = Velocity of the jet of water
θ = angle made by jet with x-axis at inlet tip of the curved plate.
If the plate is smooth and loss of energy due to impact is zero, then
the velocity of the water at the outlet tip of the curved plate will be
equal to V. The force exerted by the jet of water in the direction of
x and y are
Fx = Mass per second × [V1x – V2x]
= ρaV[V cos θ - (-V cos θ)]
= ρaV[V cos θ + V cos θ]
Fy = ρaV [V1y – V2y]
= ρaV [V sin θ – Vsin θ] = 0
FIGURE 4.
(5) Jet strikes the curved plate at one end tangentially when
the plate is unsymmetrical.
When the curve plate is unsymmetrical about x axis, then the angles
made by the tangents drawn at the inlet and outlet tips of the plate
with x-axis will be different.
Let Ɵ = angle made by the tangent at the tip with the x – axis,
Φ = angle made by the tangent at the outlet tip with x-axis.
The two component of the velocity at inlet are
V1x = V cos Ɵ and v1y = V sin Ɵ
The two component of the velocity at outlet are
V2x = - V cos Φ and V2y = V sin Φ
∴ The force exerted by the jet of water in the direction o x and y are
Fx = mass rate of flow ( v1 – v2 )
= ρaV [ V cosθ –( - V cosφ)]
= ρaV ²( cosθ + cosφ)
Fy = mass rate of flow ( V1 – V2 )
= ρaV ( V sinθ - V sinφ)
= ρaV²(sinθ - sinφ)
(6) Force exerted by the jet on the moving plate.
1st Case:- Force on flat moving plate in the direction of jet
Consider, a jet of water strikes the flat moving plate moving with
a uniform
velocity away from the jet.
V = Velocity of jet
a = area of x-section of jet
U = velocity of flat plate
Relative velocity of jet w.r.t plate = V – u
Mass of water striking/ sec on the plate = ρa(V - u)
Force exerted by jet on the moving plate in the direction of jet
Fx = Mass of water striking/ sec x [Initial velocity – Final velocity]
= ρa(V - u) [(V - u) – 0]
= ρa(V - u) ²
In this case, work is done by the jet on the plate as the plate is moving,
for stationary plate the w.d is zero.
Work done by the jet on the flat moving plate
W.d.= Force x Distance in the direction of force/ Time
= ρa(V - u) ² * u
2nd Case: Force on inclined plate moving in the direction of jet.
Consider, a jet of water strikes the inclined plate moving in the
direction of jet
with a relative velocity.
V = Velocity of jet
a = area of x-section of jet
U = velocity of flat plate
Relative velocity of jet w.r.t plate = V – u
If the plate is smooth, it is assumed that the loss of energy due to
impact of jet is
zero, then the jet of water leaves the inclined plate with a velocity (V
– u ).
Force exerted by jet on the inclined plate in the direction normal to the jet
Fn = Mass of water striking/ sec x [Initial velocity – Final velocity]
= ρa(V- u) [(V- u) sinθ – 0]
= ρa(V - u) ² sinθ
This normal force can be resolved into two components one in the direction of jet and other perpendicular to the direction of jet.
Component of Fn in the direction of jet.
Fx= ρa(V - u) ² sin²Ɵ
Component of Fn in the direction perpendicular to the direction of jet
Fy= ρa(V - u) ² sinƟcosƟ
Work done by the jet on the flat moving plate
= Force x Distance in the direction of force/ Time
W.d.= ρa(V - u) ² sin²Ɵ * u
(7) Force exerted by the jet of water on series of vanes.
Force exerted by jet of water on single moving plate (Flat or
curved) is not feasible
one, it is only theoretical one.
Let,
V = Velocity of jet
a = area of x-section of jet.
u = velocity of vane
In this, mass of water coming out from the nozzle/s is always in
constant with
plate. When all plates are considered.
Mass of water striking/s w.r.t plate = ρaV
Jet strikes the plate with a velocity = V – u
Force exerted by the jet on the plate in the direction of motion of plate
= Mass/sec x (Initial velocity – Final velocity)
Fx = ρaV(V - u)
Work done by the jet on the series of plate/sec,
Work done = Fx * u
Work done = ρaV(V - u) * u
K.E. of jet/second = 1/2 mv²
= 1/2(ρaV) v²
K.E. of jet/second = 1/2 ρaV³
Efficiency ƞ= Work done by jet/sec/K.E. of jet/sec.
= ρaV(V - u) * u/1/2(ρaV) v²
ƞ=2u(V-u) /V²
Condition for Max Efficiency,
Efficiency is maximum when, dƞ / du=0
u=V/2
Put the value of u in ƞ=2u(V-u)/ V²
Ƞmax=1/2=50%