Impact of jets

 Introduction about impact of jet

 Introduction about impact of jet


 Force exerted on stationary plate held
normal to jet.
Consider a jet of water strike normally on the fixed plate
held perpendicular to flow direction of jet as shown in
fig.the jet after striking the plate will deflected through
90°.so final velocity of fluid in the direction of the jet after
striking plate will be zero.

normal to jet.
Let, V = velocity of jet,
d = diameter of jet,
ρ = density of fluid,
A = cross section area of jet,
m = mass flow rate of fluid,

normal to jet.

 Force exerted on fixed inclined flat
plate to jet.
Consider a jet striking on an incline fixed plate as shown
in fig.
Let , V = velocity of jet.
A = cross section area of jet.
Ɵ = inclination of the plate with the jet.

Assuming no loss of energy due to impact of jet then jet
will move over the plate after striking with velocity
equal to initial velocity of jet.
Let Fn = force exerted by the jet on the plate in
direction normal to the plate.
Fx = force exerted by the jet on the plate in
direction to the jet.
Fy = force exerted by the jet on the plate in
direction perpendicular to the jet.
plate to jet.

plate to jet.

 Force exerted on single curved plate.
(1)When the jet strikes at the centre of the symmetrical
blade :
consider jet of the water striking on the curved
fixed blade at the centre of blade as shown in fig.
let V = velocity of liquid jet
A = area of cross section of jet

 The plate is smooth and there is no loss of energy due
to impact of jet. Hence liquid leaving the plate with
velocity V in he tangential direction of the curved plate.
Force exerted by jet in the direction of jet,
Fx = (mass of water/sec) x (V1x - V2x)
where V1x =initial velocity of water jet in direction of jet = V
V2x =final velocity water in direction of jet = -VcosƟ
Fx = (ρAV)x[V-(-VcosƟ]
Fx = ρAV²[1+cosƟ]

Force exerted by jet on curved fixed plate in vertical direction
Fy = (ρAV)x[V1y - V2y]
where V1y =initial velocity of water jet in vertical direction=0
V2y =final velocity water in vertical direction=VsinƟ
Fy = (ρAV)x[0 -VsinƟ ]
Fy = -ρAV²sinƟ
Angle of deflecton = 180°- Ɵ
(2) When jet strikes tangentially at one end of symmetrical
plate :
consider a water jet striking on symmetrical curved plate
tangentially at one end as shown in fig

let V = velocity of water jet
Ɵ = angle between jet and x-axis at the tip of plate at
inlet
Fx = (ρAV) x (V1x - V2x)
= (ρAV) x [VcosƟ -(-VcosƟ)]
= ρAV2VcosƟ = 2ρAV²cosƟ

Fy = (ρAV) x [V1y - V2y]
= (ρAV) x [VsinƟ -VsinƟ ] = 0
Angle of deflection = 180°- 2Ɵ
(3) When jet strikes tangentially at one end of
unsymmetrically plate:
consider a water jet striking on unsymmetrical curved plate
tangentially at one end as shown in fig.
let α = angle between water jet and x-axis at of inlet tip.
β = angle between water jet and x-axis at of outlet tip.

Fx = (ρAV)x (V1x - V2x)
= ρAV x [Vcosα -Vcosβ]
= ρAV²[cosα +cosβ]
Fy = (ρAV)x[V1y - V2y]
= (ρAV)x[Vsinα -Vsinβ]
= ρAV²[sin α-sinβ]
Resultant force F =√Fx²+Fy²
Resultant force inclination with Horizontal is ᴓ = tan^-1(Fy / Fx)
Angle of deflecton = 180°- (α +β)

 Force exerted on series of curved plate.
consider a series of curved vanes
mounted on wheel as shown in
fig. in a radial curved vane the
radius of vane at inlet and outlet
is not same.hence the tangential
velocities of the radial vanes is
not equal (u1=u2)
Let,
R1 =radius of wheel at inlet of
vane
R2 = radius of wheel at outlet
of vane

V1 =velocity of jet at inlet of vane
V2 =velocity of jet at outlet of vane
ω = angular velocity of the wheel
U1 = tangential velocity of vane at inlet = ωR1
U2 = tangential velocity of vane at outlet = ωR2
In the series of vanes, the mass of water striking per
second is same as mass of water coming out from nozzle
per second, and ρAV1
momentum of water striking the vane in the tangential
direction at inlet/sec =(mass/sec)x component of velocity
V1 in tangential direction = ρAV1xV1cosα1
= ρAV1xVw1

Angular momentum/sec at inlet = (ρAV1xVw1)xR1
momentum of water at outlet/sec =(mass/sec)x component
of velocity V2 in tangential direction = ρAV1x(-V2cosα2)
= ρAV1xVw2
Angular momentum/sec at inlet = -(ρAV1xVw2)xR2
Torque exerted by the water on the wheel,
T = rate change of angular momentum
= (ρAV1xVw1)xR1 - (-ρAV1xVw2)xR2
= ρAV1(Vw1xR1+Vw2xR2)

Work done by water on wheel/sec
= torque x angular velocity
= T x ω
= ρAV1[Vw1ωR1+Vw2ωR2 ]
= ρAV1 (Vw1u1+Vw2u2 )
If α2 > 90° ,WD/sec= ρAV1 (Vw1u1 - Vw2u2 )
In general for α2 < 90° and α2 > 90°
WD/sec= ρAV1 (Vw1u1 ± Vw2u2 )
For α2 = 90°, Vw2 = 0
WD/sec= ρAV1 (Vw1u1)

Effficiency of system(vane)
η = work done/sec /kinetic energy/sec
= ρAV1 (Vw1u1 ± Vw2u2 )/ .5ρAV1 xV1²
= 2(Vw1u1 ± Vw2u2 )/ V1²
If there is no loss of energy when water is flowing over
vanes, the work done and effiency in terms of absolute
velocities.
WD/sec= (initial kinetic energy/sec –final kinetic energy/sec)
= 1/2 (ρAV1)V1² -1/2 (ρAV1)V2²
η = 1/2 (ρAV1)V1² -1/2 (ρAV1)V2²/ 1/2 (ρAV1)V1²
=V1² -V2²/ V1² = 1- V2²/ V1²

From above equation following points are observed :
(1) For given value of V1 , η increases with decrease V2 ,at
V2 = 0 , η = max.actual practice ,V2 ≠0 ,without outlet
velocity V2 ,incoming jet will not move out of the vane
(2) efficiency of the vane is max when α2 < 90° .hence ,
Vw2 added to Vw1 .also Vw2 is max when α2 = 0.but
actual practice α2 ≠0 .so efficiency is max when α2 is
minimum

Impact of jets

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