Impact of Jet - Weebly
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Impact of Jet Lecture slides by Sachin Kansal NATIONAL INSTITUTE OF TECHNOLOGY KURUKSHETRA
Transcript of Impact of Jet - Weebly
Impact of Jet
Lecture slides by
Sachin Kansal
NATIONAL INSTITUTE OF TECHNOLOGY
KURUKSHETRA
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Objectives
β’ Have an understanding of the various effects produce
by a jet on stationary and moving plates
β’ Estimate the force and work done associated with
impact of jet on series of vane
β’ Understand the effects jet in the propulsion of ships
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Introduction Analysis and design of Fluid machines are essentially
based on the knowledge of forces exerted on or bythe moving fluids.
The liquid comes out in the form of a jet from theoutlet of a nozzle with high velocity, which is fitted to apipe through which the liquid is flowing underpressure.
If some plate, which may be fixed or moving, is placedin the path of the jet, a force is exerted by the jet onthe plate.
This force is obtained from Newtonβs 2nd law ofmotion or from the Impulse β Momentum equation.
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Impulse β Momentum Principle:Newtonβs 2nd law of motion states that βThe rate of change of
momentum is equal to the force applied and takes place in the
direction of the force.β
If the mass of the fluid is m which flows with a velocity v, the
momentum = mv
Let the change in velocity in dt time interval is dv, then
πΆβππππ ππ ππππππ‘π’π = πππ π Γ πβππππ ππ π£ππππππ‘y = π Γ ππ£
and π ππ‘π ππ πβππππ ππ ππππππ‘π’π = π Γππ£/ππ‘
According to Newtonβs 2nd law of motion,
πΉππππ = π ππ‘π ππ πβππππ ππ ππππππ‘π’π, β΄πΉ =πΓππ£/ππ‘
β΄ πΉππ‘ = πππ£β¦β¦β¦β¦β¦β¦β¦β¦.Eq. (1)
Where F.dt is the impulse of the force and m.dv is the change
in momentum. Eq. (1) is known as the Impulse-Momentum
principle.
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Force Exerted by the Jet on the PlateForce Exerted by the Jet on the plate is discussed here for
the following cases:
Force Exerted by the Jet on Stationary Plate
A Flat plate is vertical to
the jet
A Flat plate is inclined to the jet
A Curved Plate
(i) Jet impacts at the center of the curved plate
(ii)Jet strikes at one end of the curved plate when the
plate is unsymmetrical
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Following assumptions are made in general for the
discussion of all the cases:
The plate is smooth and there is no loss of energy due
to fluid friction with the plate
No loss of energy due to impact of jet
Velocity is uniform throughout
Force Exerted by the Jet on Moving Plate
A Flat plate is vertical to
the jet
A Flat plate is inclined to
the jetA Curved Plate
(i) Jet impacts at the center of the curved plate
(ii)Jet strikes at one end of the curved plate when
the plate is unsymmetrical
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Force exerted by the jet on a stationary
Plate(a) A Flat Plate Vertical to the Jet
Consider a jet of water coming out from the nozzle,strikes a flat vertical plate as shown in Fig.
The plate is stationary
and does not deflect even
after the jet strikes on it.
The plate deflects the
jet by 90Β° and then jet Fig. β Jet striking a fixed vertical plate
leaves the plate tangentially.
Hence the component of the velocity of jet V, in thedirection of the jet, after striking will be zero.
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Let, V = velocity of the jet
d = diameter of the jet
a = area of cross-section of the jet =πd2/4
Ο = density of fluid
Q = volume flow rate of fluid
οΏ½οΏ½ = mass flow rate of fluid = ππ = πππ
The force exerted by the jet on the plate in the direction of jet,
πΉπ₯ = π ππ‘π πππβππππ ππ ππππππ‘π’π ππ π‘βπ ππππππ‘πππ ππ πππππ
=
πΉπ₯ = Γ πΆβππππ ππ π£ππππππ‘π¦ ππ π‘βπ ππππππ‘πππ ππ πππ‘
πΉπ₯ = Γ [πΌπππ‘πππ πππππππ‘π¦ ππ πππ‘ ππππππ π π‘ππππππ ππ π‘βπ ππππππ‘πππ ππ πππ‘ βπΉππππ πππππππ‘π¦ ππ πππ‘ πππ‘ππ π π‘ππππππ ππ π‘βπ ππππππ‘πππ ππ πππ‘ ]
πΉπ₯ = π Γ βπ
πΉπ₯ = πππ[π β 0]
πΉπ₯ = πππ2
Time
tumFinalMomenentumInitialMom
Time
Mass
Time
Mass
[Note: If the force exerted on the jet is to be calculated then (Final β Initial)
velocity should be taken]
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(b) A Flat Plate Inclined to the Jet Consider a jet of water coming out from the nozzle, strikes a
inclined plate as shown in Fig.
Let,V = velocity of the
jet in the direction of x
ΞΈ = angle between the
jet and plate
d = diameter of the jet
a = area of cross-section
=πd2/4
Ο = density of fluid
Q = volume flow rate of fluid Fig. β Jet striking a fixed inclined Plate
οΏ½οΏ½ = Mass of water striking
the plate per sec,
οΏ½οΏ½ = πππ
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The plate is very smooth and there is no loss of energydue to the impact of jet then, the jet will moveover the plateafter striking, with a velocity equals to initial velocity, i.e. V
The force exerted by the jet on the plate in the directionnormal to the plate,
πΉn = π ππ‘π πππβππππ ππ ππππππ‘π’π ππ π‘βπ ππππππ‘πππnormal to the plate
=
πΉn = Γ πΆβππππ ππ π£ππππππ‘π¦ ππ π‘βπ ππππππ‘πππnormal to the plate
πΉn= Γ [πΌπππ‘πππ πππππππ‘π¦ ππ πππ‘ ππππππ π π‘ππππππ πππ‘βπ ππππππ‘πππ normal to plate βπΉππππ πππππππ‘π¦ ππ πππ‘ πππ‘πππ π‘ππππππ ππ π‘βπ ππππππ‘πππ normal to plate ]
πΉn = οΏ½οΏ½ Γ βππΉn = πππ[πSin π β 0]
πΉn = πππ2Sin π
Time
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This force can be resolved into two components,
I. In the direction of the jet (Fx) and,
II. Perpendicular to the direction of flow (Fy)
β΄ πΉπ₯ = πΆππππππππ‘ ππ πΉπ ππ π‘βπ ππππππ‘πππ ππ ππππ€
β΄ πΉπ₯ = πΉπ cos(90 β π)
β΄ πΉπ₯ = πΉπ sin π
β΄ πΉπ₯ = πππ2 sin 2π
and
πΉπ¦ = πΆππππππππ‘ ππ πΉπ ππ π‘βπ ππππππ‘πππ ππππππππππ’πππ π‘π ππππ€
β΄ πΉπ¦ = πΉπ sin(90 β π)
β΄ πΉπ¦ = πΉπ cos π
β΄ πΉπ¦ = πππ2 sin π cos π
(c) A Curve Plate (i) Jet impiges at the center
Consider a jet of water coming out from the nozzle,
strikes a fixed curved plate at the center as shown in Fig.
At Inlet, Let,V = velocity of the jet
in the direction of x
ΞΈ = Vane angle= angle between
the jet and plate
d = diameter of the jet
a = area of cross-section
Ο = density of fluid
Q = volume flow rate of fluid
π = Mass of water striking the
plate per sec = πππ
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Fig. β Jet striking a fixed curved
Plate the center
At Outlet, The jet after striking the plate comes out with the
same velocity in the tangential direction of the curved plate
if the plate is smooth, and there is no loss of energy due to
the impact of a jet. (Assumption)
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Now the velocity at the outlet of the plate can be resolved
into two components:
i)In the direction of the jet and
ii)Perpendicular to the direction of the jet.
The component of velocity in the direction of jet i.e. in X-
direction = β π cos π (βve sign is taken as the velocity at
the outlet is in the opposite direction of the jet of water
coming out at the nozzle)
The component of velocity perpendicular to the direction
of the jet i.e. Y- direction = π sin π
The forces exerted by the jet on the plate in the direction
of X and Y are,
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πΉπ₯ =(πππ π Γ πΆβππππ ππ π£ππππππ‘π¦ ππ π ππππππ‘πππ)/ ππππ
πΉπ₯ = π [π β (βπ cos π)]
πΉπ₯ = πππ[π + π cos π]
πΉπ₯ = πππ2 [1 + πππ π]
Due to symmetry of the plate, net force acting in
perpendicular direction= πΉπ¦ =0
So, F= πΉπ₯ = πππ2 [1 + πππ π]
If ΞΈ= 900, πΉ= πππ2 (Case of flat plate)
If ΞΈ= 00, πΉ= 2πππ2 (Semi-circular vane jet deflected
back in incoming direction)
(c) A Curve Plate (ii) Jet strikes the curved plate at one end tangentially when the plate is unsymmetrical
Let the jet strikes the unsymmetrical curved fixed plate at one end tangentially as shown in Fig.
Let the curved plate is
unsymmetrical about X-axis,
then the angle made by tangents
drawn at the inlet and outlet tips
of the plate with the X-axis will
be different
Let,π =Velocity of the jet of water
π1 = Angle made by tangent at
inlet tip with X-axis
π2 = Angle made by tangent at
outlet tip with X-axis
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Fig. β Jet striking a fixed
curved Plate at one end of
unsymmetrical plate
π1
π2
π1
π2
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The forces exerted by the jet of water on the plate in
the direction of X and Y are,
πΉπ₯ =(πππ π Γ πΆβππππ ππ π£ππππππ‘π¦ ππ π ππππππ‘πππ)/ ππππ
πΉπ₯ = οΏ½οΏ½ [π cos π1 β (βπ cos π2)]
πΉπ₯ = πππ[π cos π1 + π cos π2]
πΉπ₯ = πππ2 (cos π1 + cos π2)
Similarly,πΉπ¦ = οΏ½οΏ½ [π sin π1 β π sin π2]
πΉπ¦ = πππ2 (sin π1 β sin π2)
Resultant force =
Direction of Resultant Force=
Total angle of Deflection= 180-(π1 + π2)
22yxr FFF
direction- with xtan 1
Fx
Fy
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For symmetrical vane, π1 = π 2
πΉπ₯ = 2πππ2πππ π
πΉπ¦ = π [π sin π β π sin π] = 0
If inlet is parallel to x-axis, i.e. π1 =0
πΉπ₯ = πππ2(1+cππ π 2 )
πΉy = πππ2sin π 2
For semi-circular vane, π1 = π 2 =0
πΉπ₯ = 2πππ2
πΉπ¦ = 0
Force exerted by the jet on a moving
Plate(a) A Flat Plate Vertical to the Jet
Consider jet of water striking a flat vertical plate moving with
a uniform velocity away from in direction of the jet as shown
in Fig.
Let,π=Velocity of the jet (absolute)
π’ = Velocity of the flat Plate, (u<V)
In this case, the jet does not strike
the plate with velocity V, but it
strikes with a relative velocity
(because the plate is not
stationary).
β’ The relative velocity of the jet to
plate = (π β π’)19
Fig. β Jet striking a moving
vertical plate
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π = Mass of water striking the plate per sec,
= ππππ ππ‘π¦ Γ ππππ ππ πππ‘ Γ π£ππππππ‘π¦ π€ππ‘β π€βππβ πππ‘ π π‘πππππ π‘βπ ππππ‘π= ππ(π β π’)
The force exerted by the jet on the plate in the direction of jet,
πΉπ₯ = π ππ‘π πππβππππ ππ ππππππ‘π’π ππ π‘βπ ππππππ‘πππ ππ πππππ
=
πΉπ₯ = Γ πΆβππππ ππ π£ππππππ‘π¦ ππ π‘βπ ππππππ‘πππ ππ πππ‘
πΉπ₯ = m Γ [πΌπππ‘πππ πππππππ‘π¦ ππ πππ‘ ππππππ π π‘ππππππ ππ π‘βπππππππ‘πππ ππ πππ‘ βπΉππππ πππππππ‘π¦ ππ πππ‘ πππ‘ππ π π‘ππππππ ππ π‘βπππππππ‘πππ ππ πππ‘ ]
πΉπ₯ = ππ(π β π’) Γ [(π β π’) β 0]
πΉπ₯ = ππ(π β π’)2
Time
tumFinalMomenentumInitialMom
Time
Mass
In this case, the work will be done by the jet on the plate, asthe plate is moving. (for the stationary plate, the work done iszero)
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Work done per second by the jet on the plate=
π = πΉππππ Γ (πππ π‘ππππ π‘πππ£πππππ ππ π‘βπ ππππππ‘πππ ππ πππππ )/
π‘πππ
= πΉπ₯Γπππππππ‘π¦ π€ππ‘β π€βππβ ππππ‘π πππ£ππ ππ π‘βπ ππππππ‘πππ πππππππ
π = πΉπ₯ Γ π’
π = ππ(π β π’)2 Γ π’
π = πππ’(π β π’)2
(Here SI unit of W is Watt because it is work done per sec,
i.e. Power)
(b) A Flat Plate Inclined to the Jet
Consider a jet of water strikes an inclined plate, which is
moving with a uniform velocity in the direction of the jet as
shown in Fig
Let,V = Absolute velocity of the
jet in the direction of x
π’ = Velocity of the flat plate
π = Cross-section area of jet
π = Angle between jet and plate
The relative velocity of the jet
of water = (π β π’)
Mass of water striking the plate
per second=π = ππ(π β π’)
Fig. β Jet striking a moving
inclined Plate
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If the plate is smooth and 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 equals to (V β u)
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Force exerted by the jet of water on the plate in the
direction normal to the plate
πΉn = π ππ‘π πππβππππ ππ ππππππ‘π’π ππ π‘βπ ππππππ‘πππnormal to the plate
=
πΉn = Γ πΆβππππ ππ π£ππππππ‘π¦ ππ π‘βπ ππππππ‘πππnormal to the plate
πΉn= Γ [πΌπππ‘πππ πππππππ‘π¦ ππ πππ‘ ππππππ π π‘ππππππππ π‘βπ ππππππ‘πππ normal to plate βπΉππππ πππππππ‘π¦ ππ πππ‘ πππ‘ππ π π‘ππππππ ππ π‘βπ ππππππ‘πππ normal to plate ]
πΉn = π Γ βπ
πΉn = ππ(π- π’) [(π- π’)Sin π β 0]
πΉn = ππ(π- π’)2Sin π
Time
tumFinalMomenentumInitialMom
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Mass
Time
Mass
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This force can be resolved into two components,
I. In the direction of the jet (Fx) and,
II. Perpendicular to the direction of flow (Fy)
β΄ πΉπ₯ = πΆππππππππ‘ ππ πΉπ ππ π‘βπ ππππππ‘πππ ππ ππππ€
β΄ πΉπ₯ = πΉπ cos(90 β π)
β΄ πΉπ₯ = πΉπ sin π
β΄ πΉπ₯ = ππ(π- π’)2 sin 2π
and
πΉπ¦ = πΆππππππππ‘ ππ πΉπ ππ π‘βπ ππππππ‘πππ ππππππππππ’πππ π‘π ππππ€
β΄ πΉπ¦ = πΉπ sin(90 β π)
β΄ πΉπ¦ = πΉπ cos π
β΄ πΉπ¦ = ππ(π- π’)2 sin π cos π
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Work done per second by the jet on the plate
π = πΉπ₯ Γ πππππππ‘π¦ π€ππ‘β π€βππβ ππππ‘π πππ£ππ ππ π‘βπ π β
ππππππ‘πππ
π = πΉπ₯ Γ π’
π = ππ(π β π’)2 sin2 π Γ π’
π = πππ’(π β π’)2 sin2 π
(c) A Curve Plate (i) Jet impiges at the center
Consider a jet of water strikes a curved plate at the center
of the plate which is moving with a uniform velocityin the
direction of the jet as shown in Fig
Let, π = Absolute velocity
of the jet of water
π’ = Velocity of the flat plate
in the direction of the jet
π = Cross-section area of jet
(π β π’) =The relative velocity
of the jet of water or the
velocity with which jet
strikes the curved plate
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Fig. β Jet striking a moving
curved Plate the center
If the plate is smooth and loss of energy due to the impact
of the jet is assumed zero, then the velocity with which the
jet will be leaving the curved vane equals to (V β u)
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Now the velocity at the outlet of the plate can be resolved
into two components:
i)In the direction of the jet and
ii)Perpendicular to the direction of the jet.
The component of velocity in the direction of jet i.e. in X-
direction = β π cos π (βve sign is taken as the velocity at
the outlet is in the opposite direction of the jet of water
coming out at the nozzle)
The component of velocity perpendicular to the direction
of the jet i.e. Y- direction = π sin π
The forces exerted by the jet on the plate in the direction
of X and Y are,
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πΉπ₯ =(πππ π Γ πΆβππππ ππ π£ππππππ‘π¦ ππ π ππππππ‘πππ)/ ππππ
πΉπ₯ = π [(π- π’) β {β(π- π’) cos π}]
πΉπ₯ = ππ(π- π’)[(π- π’) + (π- π’) cos π]
πΉπ₯ = ππ(π- π’ )2 [1+ πππ π]
πΉy = 0
Work done per second by the jet on the plate=
π = πΉπ₯ Γ πππππππ‘π¦ π€ππ‘β π€βππβ ππππ‘π πππ£ππ ππ π‘βπ π β ππππππ‘πππ
π = πΉπ₯ Γ π’
π = ππ(π β π’)2 [1 + cos π] Γ π’
π = πππ’(π β π’)2 [1 + cos π]
The kinetic energy of the jet per second,
πΎπΈ = Β½ π π2 = Β½ (πππ)π2 = Β½ πππ3
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The efficiency of the wheel,π=
For maximum efficiency,
If vane is semi circular,
secondper Energy Kinetic
secondper doneWork
3
3
)cos1()(2
2
1
)cos1()(
2
2
V
uVu
aV
uVau
3u V plate), thestrike liquid (nou V
0])(
[)cos1(2
0])cos1()(2
[ i.e. , 0
2
3
3
2
or
uVu
du
d
V
V
uVu
du
d
du
d
2cos
27
16..,
2cos2
27
8
0])3(
)cos1()3(2
2max
2max
3
2
max
ei
u
uuu
592.027
16 and 0 max
(c) A Curve Plate (ii) Jet strikes the curved plate at one
end tangentially when the plate is unsymmetrical
Consider a jet striking a moving curved plate/vane/blade
tangentially at one of its tips
As the jet strikes tangentially, the loss of energy due to the
impact of the jet will be zero.
In this case, as the plate is moving, the velocity with which
jet of water strikes is equal to the relative velocity of the jet to
the plate.
As the direction of jet velocity and vane velocity is not the
same, the relative velocity at the inlet will be vector difference
of the jet velocity and plate velocity at inlet.
Let,π1 = Absolute velocity of the jet at the inlet
π2 = Absolute velocity of the jet at the outlet
ππ1 = Relative velocity of the jet
and plate at inlet
ππ2 = Relative velocity of the jet
and plate at outlet
π’1 = Velocity of the vane at the
inlet
π’2 = Velocity of the vane at the
outlet
πΌ = Angle between the direction
of the jet and direction of
motion of the plate at inlet=
Guide blade angle
π = Angle made by the relative
velocity ππ1 , with the
direction of motion of the
vane at the inlet= Vane/blade
angle at inlet31
Fig. β Jet striking a moving
curved Plate at one end of
unsymmetrical plate
ππ€1 πππ ππ1 = The components
of the velocity of the jet π1 ,
in the direction of motion and
perpendicular to the direction
of motion of the vane
respectively.
ππ€1 = Velocity of whirl at the
inlet
ππ1 = Velocity of flow at the inlet
π½ = Angle made by the velocity
π2 with the direction of
motion of the vane at the
outlet
32
Fig. β Jet striking a moving
curved Plate at one end of
unsymmetrical plate
π = Angle made by the relative
velocity ππ2 , with the
direction of motion of the
vane at the outlet
= Vane/blade angle at the
outlet
ππ€2 πππ ππ2 = The components
of the velocity π2 , in the
direction of motion of vane
and perpendicular to the
direction of motion of the
vane at outlet respectively
ππ€2 = Velocity of whirl at the
outlet
ππ2 = Velocity of flow at the
outlet33
Fig. β Jet striking a moving
curved Plate at one end of
unsymmetrical plate
34
The triangles ABD and BβCβDβ are called the velocitytriangles at the inlet and outlet respectively.
If the vane is smooth and having velocity in the direction ofmotion at inlet and outlet equal then we have,π’1=π’2 = π’ =
π£ππππππ‘π¦ ππ π£πππ ππ π‘βπ ππππππ‘πππ ππ πππ‘πππ and ππ1 = ππ2
Mass of water striking the vane per second=π = ππππ1
Force exerted by the jet in the direction of motion=
Fx= mass of water striking per sec X [Initial velocity withwhich jet strikes in the direction of motion βFinal velocity ofthe jet in the direction of motion]
The initial velocity with which jet strikes the vane = ππ1
and, The component of this velocity in the direction ofmotion = ππ1 cos π = (ππ€1 β π’1 )
Similarly, The component of the relative velocity at theoutlet in the direction of motion = βππ2 cos π = β[π’2 + ππ€2 ]
35
So,β΄ πΉπ₯ = π Γ [ππ1 cos π β (βππ2 cos π)]
πΉπ₯ = ππππ1 Γ [(ππ€1 β π’1 ) + (π’2 + ππ€2 )]
As we know π’1 = π’2
πΉπ₯ = ππππ1 Γ [ππ€1 + ππ€2 ]
It is true only when angle π½ shown in Fig is acute angle (<90Β°)
β’ If π½ = 90Β°, then ππ€2 = 0 and Eq. becomes, πΉπ₯ = ππππ1 ππ€1
β’ If π½ is an obtuse angle (> 90Β°), the expression for πΉπ₯ willbecome, πΉπ₯ = ππππ1 Γ [ππ€1 β ππ€2 ]
In general,πΉπ₯ = ππππ1 Γ [ππ€1 Β± ππ€2 ]
Work done per second on the vane by the jet=π
= πΉππππ Γ πππ π‘ππππ π‘πππ£πππππ πππ sec ππ π‘βπ ππππππ‘πππ ππ πππππ
W = πΉπ₯ Γ π’
π = πππ’ππ1 Γ [ππ€1 Β± ππ€2 ]
36
Work done per second per unit weight of fluid striking per
second,=πππ’ππ1 Γ [ππ€1 Β± ππ€2 ] /{(ππππ1 ) Γ π}
= [ππ€1 Β± ππ€2 ] Γ π’
Work done per second per unit mass of fluid striking per
second,=πππ’ππ1 Γ [ππ€1 Β± ππ€2 ] / (ππππ1 )
= π’ Γ [ππ€1 Β± ππ€2 ]
g
1
Force exerted by a jet of water on
(a) Series of flat vanes The force exerted by a jet of water on a single moving plate
is not practically feasible. It's only a theoretical one.
In actual practice, a large number of plates/blades are
mounted on the circumference of a wheel at a fixed distance
apart as shown in Fig.
The jet strikes a plate and due to
the force exerted by the jet on the
plate, the wheel starts moving and
the 2nd plate mounted on the
wheel appears before the jet, which
again exerts the force on the 2nd
plate.
37
Fig. β Jet striking a series of flat
vanes mounted on a wheel
Thus each plate appears successively before the jet and jet
exerts a force one each plate and the wheel starts moving at
a constant speed
38
Let,π = Velocity of jet
π = Diameter of jet
π’ = Velocity of vane
In this case, the mass of water coming out from the nozzle
per second is always in contact with the plates, when all
the plates are considered.
Hence, the mass of water per sec striking the series of
plates = πππ
also, The jet strikes a plate with velocity = (π β π’)
After striking, the jet moves tangential to the plate and
hence the velocity component in the direction of motion of
plate is equal to zero.
Force exerted by the jet in the direction of motion of plate,
πΉπ₯ = πππ[(π β π’) β 0]
πΉπ₯ = πππ(π β π’)
39
Work done by the jet on the series of plates per second,
π =πΉππππ Γ π·ππ π‘ππππ π‘πππ£πππππ πππ sec ππ π‘βπ ππππππ‘ππππππππππ
β΄π = πΉπ₯ Γ π’
β΄ π = πππ(π β π’) Γ π’
β΄ π = ππππ’(π β π’)
The kinetic energy of the jet per second,
πΎπΈ = Β½ π π2
β΄ πΎπΈ = Β½(πππ)π2 = Β½ πππ3
The efficiency of the wheel,π=secondper Energy Kinetic
secondper doneWork
2
3
)(2
2
1
)(
V
uVu
aV
uVaVu
40
Condition for the maximum efficiency
For a given jet velocity V, the efficiency will be maximum
when,
Force exerted by a jet of water on
(b) Series of Radial Curved Vanes For a radial curved vane, the radius of the vane at inlet and
outlet is different and hence the tangential velocities of the
radial vane at inlet and outlet will not be equal
Consider a series of radial curved vanes mounted on a wheel
as shown in Fig.
The jet of water strikes
the vanes and the
wheel starts rotating at
constant angular speed
Let,π 1 = Radius of the
wheel at the inlet of the
vane
π 2 = Radius of the
wheel at the outlet of
the vane
41Fig. β Jet striking a series of radial curved
vanes mounted on a wheel
42
π = Angular speed of the wheel
Then,π’1 = ππ 1
πππ π’2 = ππ 2
The mass of water striking per second for a series of vanes= The mass of water coming out from nozzle per sec = πππ1
Where,π = Area of the jet, and π1 = Velocity of the jet
Momentum of water striking the vanes in the tangentialdirection per sec at inlet = mass of water striking per sec Xcomponent of V1 in the tangential direction
β΄ ππππππ‘π’π ππ π€ππ‘ππ ππ‘ inlet πππ π ππ = πππ1 Γ (π1 cos πΌ)
β΄ ππππππ‘π’π ππ π€ππ‘ππ ππ‘ πππππ‘ πππ π ππ = πππ1 Γ ππ€1
(β΅ ππ€1 = π1 cos πΌ)
Similarly, Momentum of water at outlet per sec = πππ1 Xcomponent of V2 in the tangential direction
β΄ ππππππ‘π’π ππ π€ππ‘ππ ππ‘ ππ’π‘πππ‘ πππ π ππ = πππ1 Γ (βπ2 cos π½)
β΄ ππππππ‘π’π ππ π€ππ‘ππ ππ‘ ππ’π‘πππ‘ πππ π ππ = βπππ1 Γ ππ€2
(β΅ ππ€2 = π2 cos π½)
43
Now angular momentum,
π΄πππ’πππ ππππππ‘π’π πππ sec ππ‘ πππππ‘ = ππππππ‘π’π ππ‘ πππππ‘
Γ π ππππ’π ππ‘ πππππ‘= πππ1 Γ ππ€1 Γ π 1
π΄πππ’πππ ππππππ‘π’π πππ sec ππ‘ ππ’π‘πππ‘ = ππππππ‘π’π ππ‘ ππ’π‘πππ‘Γ π ππππ’π ππ‘ ππ’π‘πππ‘= βπππ1 Γ ππ€2 Γ π 2
Torque exerted by the water on the wheel,
π = π ππ‘π ππ πβππππ ππ ππππ’πππ ππππππ‘π’π
π = [πΌπππ‘πππ ππππ’πππ ππππππ‘π’π πππ π ππ β πΉππππ ππππ’πππ ππππππ‘π’π πππ π ππ]
β΄ π = [πππ1 Γ ππ€1 π 1 β (βπππ1 Γ ππ€2 π 2 )]
β΄ π = πππ1 [ππ€1 π 1 + ππ€2 π 2 ]
Work done per sec on the wheel,
ππ·/π ππ = πππππ’π Γ π΄πππ’πππ π£ππππππ‘π¦
β΄ ππ·/π ππ = π Γ π
β΄ ππ·/π ππ = πππ1 [ππ€1 π 1 + ππ€2 π 2 ] Γ π
44
β΄ ππ·/π ππ = πππ1[ππ€1 π 1 π + ππ€2 π 2 π]
β΄ ππ·/π ππ = πππ1[ππ€1 π’1 + ππ€2 π’2 ], It is valid only when,π½ < 90
β’ If the angle π½ is an obtuse angle (π½ > 90) then,
ππ·/π ππ = πππ1 [ππ€1 π’1 β ππ€2 π’2 ]
In general,ππ·/π ππ = πππ1 [ππ€1 π’1 Β± ππ€2 π’2 ]
If the discharge is radial at the outlet then, π½ = 90Β° and
hence ππ€2 = 0,
β΄ ππ·/π ππ = πππ1 [ππ€1 π’1 ]
Efficiency of the radial curved vanes,
π=ππππ ππππ πππ π πππππ / πΎππππ‘ππ ππππππ¦ πππ π πππππ
Jet Propulsion of Ships
Ship is steered through water due to reaction of force of the
jet which is issued out at the back of ship
Water for this is taken in through the inlet orifice by
centrifugal pump
Let, u = velocity of ship
V= Absolute velocity of
issuing jet (opp. To u)
a=area of cross-section
β΄ Vr= V+u =Issing velocity
of the jet in backward
direction
45
46
The final velocity of water is u relative to ship in backwarddirection
F= mass x change in velocity in direction of jet
F= πππr(πr - π’ )
ππ·/π ππ = πππr (πr β π’) x π’
Energy supplied depends upon the way in which the wateris supplied to the ship
(a) Inlet orifice at right the angle to the direction of motion ofship
(b) Inlet orifice faces the direction of motion of ship
secondper suppliedEnergy
secondper doneWork Efficiency
47
(a) Inlet orifice at right the angle to the direction of motion of
ship
Velocity at inlet in the direction of jet = 0
Velocity at outlet in the direction of jet = πr
Energy Supplied= K.E at outlet- K.E. at Inlet
2
3
)(2
2
1
)(
secondper suppliedEnergy
secondper doneWork
r
r
r
rr
V
uVu
aV
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3
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0 ,efficiency maximumFor
2max
2
2
u
uuu
VuoruVuV
du
dor
V
uVud
du
d
rrr
r
r
48
(b) Inlet orifice in the direction of motion of ship
Velocity at inlet in the direction of jet = u
Velocity at outlet in the direction of jet = πr
Energy Supplied= K.E at outlet- K.E. at Inlet
uV
u
uVuV
uuV
uVaV
uuVaV
r
rr
r
rr
rr
2
))((
)(2
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secondper suppliedEnergy
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