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Impact of jets

Force of jet impinging on a plate

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Contents

Force Of Jet Impinging Normally On A Fixed Plate
Force Of Jet Impinging On An Inclined Fixed Plate
Force Of Jet Impinging On A Moving Plate
Force Of Jet Impinging On A Moving Curved Vane
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Force Of Jet Impinging Normally On A Fixed Plate

Consider a jet of water impinging normally on a fixed plate as shown in fig-1.

Let,

v = Velocity of jet
a = Cross sectional area of the jet
w = Specific weight of water

$\therefore$ Mass of water flowing/s = $\frac{waV}{g}$ kg

We know that the velocity of jet, in its original direction, is reduced to zero after the impact (as the plate is fixed). Therefore force exerted by the jet on the plane.

$F = Mass\;of\;water\;flowing/s\times Change\;of\;velocity$

$\Rightarrow F = \frac{waV}{g}\times (V - 0) = \frac{waV^2}{g}\;KN$

Force Of Jet Impinging On An Inclined Fixed Plate

Consider a jet of water impinging normally on a fixed plate as shown in fig-2.

Let,

$\theta$ = Angle at which the plate is inclined with the jet

Force exerted by the jet on the plane = $\frac{waV^2}{g}$ KN

$\therefore$ Force exerted by the jet in a direction normal to the plate,

$F = \frac{waV^2\sin\theta}{g}$

and the force exerted by the jet in the direction of flow,

$F_x = F\sin\theta = \frac{waV^2\sin\theta}{g}\times \sin\theta = \frac{waV^2\sin^2\theta}{g}$

Similarly. force exerted by the jet in a direction normal to flow,

$F_y = F\cos\theta = \frac{waV^2\sin\theta}{g}\times \cos\theta$

$\therefore F_y = \frac{waV^2\sin 2\theta}{2g}$

Force Of Jet Impinging On A Moving Plate

Consider a jet of water imping normally on a plate. As a result of the impact of the jet, let the plate move in the direction of the jet as shown in fig-3.

Let,

= Velocity of the plate, as a result of the impact of jet

A little conversation will show that the relative velocity of the jet with respect to the plate equal to (V-v)m/s.

For analysis purposes, it will be assumed that the plate is fixed and the jet is moving with a velocity of (V-v)m/s. Therefore force exerted by the jet,

$F = Mass\;of\;water\;flowing\;per\;second\times Change\;of\;velocity$

$\Rightarrow F = \frac{wa(V-v)}{g}\times [(V-v)-0]$

$\Rightarrow F = \frac{wa(V-v)^2}{g}KN$

Force Of Jet Impinging On A Moving Curved Vane

Consider a jet of water entering and leaving a moving curved vane as shown in fig-4.

Let,

V = Velocity of the jet (AC), while entering the vane,
V₁ = Velocity of the jet (EG), while leaving the vane,
v₁, v₂ = Velocity of the vane (AB, FG)
$\alpha$ = Angle with the direction of motion of the vane, at which the jet enters the vane,
$\beta$ = Angle with the direction of motion of the vane, at which the jet leaves the vane,
V_r = Relative velocity of the jet and the vane (BC) at entrance (it is the vertical difference between V and v)
V_r1 = Relative velocity of the jet and the vane (EF) at exit (it is the vertical difference between v₁ and v₂)
$\theta$ = Angle, which V_r makes with the direction of motion of the vane at inlet (known as vane angle at inlet),
$\phi$ = Angle, which V_r1 makes with the direction of motion of the vane at outlet (known as vane angle at outlet),
V_w = Horizontal component of V (AD, equal to $V\cos\alpha$ ). It is a component parallel to the direction of motion of the vane (known as velocity of whirl at inlet),
V_w1 = Horizontal component of V₁ (HG, equal to $V_1\cos\beta$ ). It is a component parallel to the direction of motion of the vane (known as velocity of whirl at outlet),
V_f = Vertical component of V (DC, equal to $V\sin\alpha$ ). It is a component at right angles to the direction of motion of the vane (known as velocity of flow at inlet),
V_f1 = Vertical component of V₁ (EH, equal to $V_1\sin\beta$ ). It is a component at right angles to the direction of motion of the vane (known as velocity of flow at outlet),
a = Cross sectional area of the jet.

As the jet of water enters and leaves the vanes tangentially, therefore shape of the vanes will be such that V_r and V_r1 will be along with tangents to the vanes at inlet and outlet.

The relations between the inlet and outlet triangles (untill and unless given) are:

(i)

, and

(ii)

We know that the force of jet, in the direction of motion of the vane,

$F_x = Mass\;of\;water\;flowing\;per\;second\times Change\;of\;velocity\;of\;whirl$

$\Rightarrow F_x = \frac{waV}{g}(V_w - V_{w1})$

Example:

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Example - Force of jet impinging normally on a fixed plat

Problem

A jet of water of 100 mm diameter impinges normally on a fixed plate with a velocity of 30 m/s. Find the force exerted on the plate.

Workings

The cross sectional area of the jet,

$a = \frac{\pi}{4}\times d^2 = \frac{\pi}{4}\times 0.1^2 = 7.854\times 10^{-3}\;m^2$

Force exerted by the jet on the plane.

$F = \frac{waV^2}{g} = \frac{9.81\times (7.854\times 10^{-3})\times 30^2}{9.81}\;KN$

$\therefore F = 7.07\;KN$

Solution

Force exerted by the jet = 7.07 KN