48
Impact of Jet Lecture slides by Sachin Kansal NATIONAL INSTITUTE OF TECHNOLOGY KURUKSHETRA
Impact of Jet
Lecture slides by
Sachin Kansal
NATIONAL INSTITUTE OF TECHNOLOGY
KURUKSHETRA
2
Objectives
• Have an understanding of the various effects produce
by a jet on stationary and moving plates
• Estimate the force and work done associated with
impact of jet on series of vane
• Understand the effects jet in the propulsion of ships
3
4
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 Newton’s 2nd law ofmotion or from the Impulse – Momentum equation.
5
Impulse – Momentum Principle:Newton’s 2nd law of motion states that “The rate of change of
momentum is equal to the force applied and takes place in the
direction of the force.”
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
𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 𝑚𝑎𝑠𝑠 × 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡y = 𝑚 × 𝑑𝑣
and 𝑅𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 𝑚 ×𝑑𝑣/𝑑𝑡
According to Newton’s 2nd law of motion,
𝐹𝑜𝑟𝑐𝑒 = 𝑅𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚, ∴𝐹 =𝑚×𝑑𝑣/𝑑𝑡
∴ 𝐹𝑑𝑡 = 𝑚𝑑𝑣…………………….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.
6
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
7
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
8
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 90° and then jet Fig. – 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.
9
Let, V = velocity of the jet
d = diameter of the jet
a = area of cross-section of the jet =𝜋d2/4
ρ = density of fluid
Q = volume flow rate of fluid
�� = mass flow rate of fluid = 𝜌𝑄 = 𝜌𝑎𝑉
The force exerted by the jet on the plate in the direction of jet,
𝐹𝑥 = 𝑅𝑎𝑡𝑒 𝑜𝑓𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑓𝑜𝑟𝑐𝑒
=
𝐹𝑥 = × 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑗𝑒𝑡
𝐹𝑥 = × [𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑗𝑒𝑡 𝑏𝑒𝑓𝑜𝑟𝑒 𝑠𝑡𝑟𝑖𝑘𝑖𝑛𝑔 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑗𝑒𝑡 −𝐹𝑖𝑛𝑎𝑙 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑗𝑒𝑡 𝑎𝑓𝑡𝑒𝑟 𝑠𝑡𝑟𝑖𝑘𝑖𝑛𝑔 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑗𝑒𝑡 ]
𝐹𝑥 = 𝑚 × ∆𝑉
𝐹𝑥 = 𝜌𝑎𝑉[𝑉 − 0]
𝐹𝑥 = 𝜌𝑎𝑉2
Time
tumFinalMomenentumInitialMom
Time
Mass
Time
Mass
[Note: If the force exerted on the jet is to be calculated then (Final – Initial)
velocity should be taken]
10
(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
θ = angle between the
jet and plate
d = diameter of the jet
a = area of cross-section
=𝜋d2/4
ρ = density of fluid
Q = volume flow rate of fluid Fig. – Jet striking a fixed inclined Plate
�� = Mass of water striking
the plate per sec,
�� = 𝜌𝑎𝑉
11
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,
𝐹n = 𝑅𝑎𝑡𝑒 𝑜𝑓𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛normal to the plate
=
𝐹n = × 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛normal to the plate
𝐹n= × [𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑗𝑒𝑡 𝑏𝑒𝑓𝑜𝑟𝑒 𝑠𝑡𝑟𝑖𝑘𝑖𝑛𝑔 𝑖𝑛𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 normal to plate −𝐹𝑖𝑛𝑎𝑙 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑗𝑒𝑡 𝑎𝑓𝑡𝑒𝑟𝑠𝑡𝑟𝑖𝑘𝑖𝑛𝑔 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 normal to plate ]
𝐹n = �� × ∆𝑉𝐹n = 𝜌𝑎𝑉[𝑉Sin 𝜃 − 0]
𝐹n = 𝜌𝑎𝑉2Sin 𝜃
Time
tumFinalMomenentumInitialMom
Time
Mass
Time
Mass
12
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)
∴ 𝐹𝑥 = 𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝐹𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑓𝑙𝑜𝑤
∴ 𝐹𝑥 = 𝐹𝑛 cos(90 − 𝜃)
∴ 𝐹𝑥 = 𝐹𝑛 sin 𝜃
∴ 𝐹𝑥 = 𝜌𝑎𝑉2 sin 2𝜃
and
𝐹𝑦 = 𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝐹𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑓𝑙𝑜𝑤
∴ 𝐹𝑦 = 𝐹𝑛 sin(90 − 𝜃)
∴ 𝐹𝑦 = 𝐹𝑛 cos 𝜃
∴ 𝐹𝑦 = 𝜌𝑎𝑉2 sin 𝜃 cos 𝜃
(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
θ = Vane angle= angle between
the jet and plate
d = diameter of the jet
a = area of cross-section
ρ = density of fluid
Q = volume flow rate of fluid
𝑚 = Mass of water striking the
plate per sec = 𝜌𝑎𝑉
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Fig. – 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)
14
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 = − 𝑉 cos 𝜃 (−ve 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 = 𝑉 sin 𝜃
The forces exerted by the jet on the plate in the direction
of X and Y are,
15
𝐹𝑥 =(𝑀𝑎𝑠𝑠× 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑖𝑛 𝑋 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛)/ 𝑇𝑖𝑚𝑒
𝐹𝑥 = 𝑚 [𝑉 − (−𝑉 cos 𝜃)]
𝐹𝑥 = 𝜌𝑎𝑉[𝑉 + 𝑉 cos 𝜃]
𝐹𝑥 = 𝜌𝑎𝑉2 [1 + 𝑐𝑜𝑠 𝜃]
Due to symmetry of the plate, net force acting in
perpendicular direction= 𝐹𝑦 =0
So, F= 𝐹𝑥 = 𝜌𝑎𝑉2 [1 + 𝑐𝑜𝑠 𝜃]
If θ= 900, 𝐹= 𝜌𝑎𝑉2 (Case of flat plate)
If θ= 00, 𝐹= 2𝜌𝑎𝑉2 (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,𝑉 =Velocity of the jet of water
𝜃1 = Angle made by tangent at
inlet tip with X-axis
𝜃2 = Angle made by tangent at
outlet tip with X-axis
16
Fig. – Jet striking a fixed
curved Plate at one end of
unsymmetrical plate
𝜃1
𝜃2
𝜃1
𝜃2
17
The forces exerted by the jet of water on the plate in
the direction of X and Y are,
𝐹𝑥 =(𝑀𝑎𝑠𝑠× 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑖𝑛 𝑋 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛)/ 𝑇𝑖𝑚𝑒
𝐹𝑥 = �� [𝑉 cos 𝜃1 − (−𝑉 cos 𝜃2)]
𝐹𝑥 = 𝜌𝑎𝑉[𝑉 cos 𝜃1 + 𝑉 cos 𝜃2]
𝐹𝑥 = 𝜌𝑎𝑉2 (cos 𝜃1 + cos 𝜃2)
Similarly,𝐹𝑦 = �� [𝑉 sin 𝜃1 − 𝑉 sin 𝜃2]
𝐹𝑦 = 𝜌𝑎𝑉2 (sin 𝜃1 − sin 𝜃2)
Resultant force =
Direction of Resultant Force=
Total angle of Deflection= 180-(𝜃1 + 𝜃2)
22yxr FFF
direction- with xtan 1
Fx
Fy
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For symmetrical vane, 𝜃1 = 𝜃 2
𝐹𝑥 = 2𝜌𝑎𝑉2𝑐𝑜𝑠 𝜃
𝐹𝑦 = 𝑚 [𝑉 sin 𝜃 − 𝑉 sin 𝜃] = 0
If inlet is parallel to x-axis, i.e. 𝜃1 =0
𝐹𝑥 = 𝜌𝑎𝑉2(1+c𝑜𝑠 𝜃 2 )
𝐹y = 𝜌𝑎𝑉2sin 𝜃 2
For semi-circular vane, 𝜃1 = 𝜃 2 =0
𝐹𝑥 = 2𝜌𝑎𝑉2
𝐹𝑦 = 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,𝑉=Velocity of the jet (absolute)
𝑢 = 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).
• The relative velocity of the jet to
plate = (𝑉 − 𝑢)19
Fig. – Jet striking a moving
vertical plate
20
𝑚 = Mass of water striking the plate per sec,
= 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 × 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑗𝑒𝑡 × 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑤𝑖𝑡ℎ 𝑤ℎ𝑖𝑐ℎ 𝑗𝑒𝑡 𝑠𝑡𝑟𝑖𝑘𝑒𝑠𝑡ℎ𝑒 𝑝𝑙𝑎𝑡𝑒= 𝜌𝑎(𝑉 − 𝑢)
The force exerted by the jet on the plate in the direction of jet,
𝐹𝑥 = 𝑅𝑎𝑡𝑒 𝑜𝑓𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑓𝑜𝑟𝑐𝑒
=
𝐹𝑥 = × 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑗𝑒𝑡
𝐹𝑥 = m × [𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑗𝑒𝑡 𝑏𝑒𝑓𝑜𝑟𝑒 𝑠𝑡𝑟𝑖𝑘𝑖𝑛𝑔 𝑖𝑛 𝑡ℎ𝑒𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑗𝑒𝑡 −𝐹𝑖𝑛𝑎𝑙 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑗𝑒𝑡 𝑎𝑓𝑡𝑒𝑟 𝑠𝑡𝑟𝑖𝑘𝑖𝑛𝑔 𝑖𝑛 𝑡ℎ𝑒𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑗𝑒𝑡 ]
𝐹𝑥 = 𝜌𝑎(𝑉 − 𝑢) × [(𝑉 − 𝑢) − 0]
𝐹𝑥 = 𝜌𝑎(𝑉 − 𝑢)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)
21
Work done per second by the jet on the plate=
𝑊 = 𝐹𝑜𝑟𝑐𝑒 × (𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑓𝑜𝑟𝑐𝑒 )/
𝑡𝑖𝑚𝑒
= 𝐹𝑥×𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑤𝑖𝑡ℎ 𝑤ℎ𝑖𝑐ℎ 𝑝𝑙𝑎𝑡𝑒 𝑚𝑜𝑣𝑒𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓𝑓𝑜𝑟𝑐𝑒
𝑊 = 𝐹𝑥 × 𝑢
𝑊 = 𝜌𝑎(𝑉 − 𝑢)2 × 𝑢
𝑊 = 𝜌𝑎𝑢(𝑉 − 𝑢)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
𝑢 = Velocity of the flat plate
𝑎 = Cross-section area of jet
𝜃 = Angle between jet and plate
The relative velocity of the jet
of water = (𝑉 − 𝑢)
Mass of water striking the plate
per second=𝑚 = 𝜌𝑎(𝑉 − 𝑢)
Fig. – Jet striking a moving
inclined Plate
22
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 – u)
23
Force exerted by the jet of water on the plate in the
direction normal to the plate
𝐹n = 𝑅𝑎𝑡𝑒 𝑜𝑓𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛normal to the plate
=
𝐹n = × 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛normal to the plate
𝐹n= × [𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑗𝑒𝑡 𝑏𝑒𝑓𝑜𝑟𝑒 𝑠𝑡𝑟𝑖𝑘𝑖𝑛𝑔𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 normal to plate −𝐹𝑖𝑛𝑎𝑙 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑗𝑒𝑡 𝑎𝑓𝑡𝑒𝑟 𝑠𝑡𝑟𝑖𝑘𝑖𝑛𝑔 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 normal to plate ]
𝐹n = 𝑚 × ∆𝑉
𝐹n = 𝜌𝑎(𝑉- 𝑢) [(𝑉- 𝑢)Sin 𝜃 − 0]
𝐹n = 𝜌𝑎(𝑉- 𝑢)2Sin 𝜃
Time
tumFinalMomenentumInitialMom
Time
Mass
Time
Mass
24
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)
∴ 𝐹𝑥 = 𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝐹𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑓𝑙𝑜𝑤
∴ 𝐹𝑥 = 𝐹𝑛 cos(90 − 𝜃)
∴ 𝐹𝑥 = 𝐹𝑛 sin 𝜃
∴ 𝐹𝑥 = 𝜌𝑎(𝑉- 𝑢)2 sin 2𝜃
and
𝐹𝑦 = 𝐶𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝐹𝑛 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑓𝑙𝑜𝑤
∴ 𝐹𝑦 = 𝐹𝑛 sin(90 − 𝜃)
∴ 𝐹𝑦 = 𝐹𝑛 cos 𝜃
∴ 𝐹𝑦 = 𝜌𝑎(𝑉- 𝑢)2 sin 𝜃 cos 𝜃
25
Work done per second by the jet on the plate
𝑊 = 𝐹𝑥 × 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑤𝑖𝑡ℎ 𝑤ℎ𝑖𝑐ℎ 𝑝𝑙𝑎𝑡𝑒 𝑚𝑜𝑣𝑒𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑋 −
𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
𝑊 = 𝐹𝑥 × 𝑢
𝑊 = 𝜌𝑎(𝑉 − 𝑢)2 sin2 𝜃 × 𝑢
𝑊 = 𝜌𝑎𝑢(𝑉 − 𝑢)2 sin2 𝜃
(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, 𝑉 = Absolute velocity
of the jet of water
𝑢 = Velocity of the flat plate
in the direction of the jet
𝑎 = Cross-section area of jet
(𝑉 − 𝑢) =The relative velocity
of the jet of water or the
velocity with which jet
strikes the curved plate
26
Fig. – 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 – u)
27
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 = − 𝑉 cos 𝜃 (−ve 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 = 𝑉 sin 𝜃
The forces exerted by the jet on the plate in the direction
of X and Y are,
28
𝐹𝑥 =(𝑀𝑎𝑠𝑠× 𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑖𝑛 𝑋 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛)/ 𝑇𝑖𝑚𝑒
𝐹𝑥 = 𝑚 [(𝑉- 𝑢) − {−(𝑉- 𝑢) cos 𝜃}]
𝐹𝑥 = 𝜌𝑎(𝑉- 𝑢)[(𝑉- 𝑢) + (𝑉- 𝑢) cos 𝜃]
𝐹𝑥 = 𝜌𝑎(𝑉- 𝑢 )2 [1+ 𝑐𝑜𝑠 𝜃]
𝐹y = 0
Work done per second by the jet on the plate=
𝑊 = 𝐹𝑥 × 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑤𝑖𝑡ℎ 𝑤ℎ𝑖𝑐ℎ 𝑝𝑙𝑎𝑡𝑒 𝑚𝑜𝑣𝑒𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑋 − 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛
𝑊 = 𝐹𝑥 × 𝑢
𝑊 = 𝜌𝑎(𝑉 − 𝑢)2 [1 + cos 𝜃] × 𝑢
𝑊 = 𝜌𝑎𝑢(𝑉 − 𝑢)2 [1 + cos 𝜃]
The kinetic energy of the jet per second,
𝐾𝐸 = ½ 𝑚 𝑉2 = ½ (𝜌𝑎𝑉)𝑉2 = ½ 𝜌𝑎𝑉3
29
The efficiency of the wheel,𝜂=
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,𝑉1 = Absolute velocity of the jet at the inlet
𝑉2 = Absolute velocity of the jet at the outlet
𝑉𝑟1 = Relative velocity of the jet
and plate at inlet
𝑉𝑟2 = Relative velocity of the jet
and plate at outlet
𝑢1 = Velocity of the vane at the
inlet
𝑢2 = Velocity of the vane at the
outlet
𝛼 = Angle between the direction
of the jet and direction of
motion of the plate at inlet=
Guide blade angle
𝜃 = Angle made by the relative
velocity 𝑉𝑟1 , with the
direction of motion of the
vane at the inlet= Vane/blade
angle at inlet31
Fig. – Jet striking a moving
curved Plate at one end of
unsymmetrical plate
𝑉𝑤1 𝑎𝑛𝑑 𝑉𝑓1 = The components
of the velocity of the jet 𝑉1 ,
in the direction of motion and
perpendicular to the direction
of motion of the vane
respectively.
𝑉𝑤1 = Velocity of whirl at the
inlet
𝑉𝑓1 = Velocity of flow at the inlet
𝛽 = Angle made by the velocity
𝑉2 with the direction of
motion of the vane at the
outlet
32
Fig. – Jet striking a moving
curved Plate at one end of
unsymmetrical plate
𝜑 = Angle made by the relative
velocity 𝑉𝑟2 , with the
direction of motion of the
vane at the outlet
= Vane/blade angle at the
outlet
𝑉𝑤2 𝑎𝑛𝑑 𝑉𝑓2 = The components
of the velocity 𝑉2 , in the
direction of motion of vane
and perpendicular to the
direction of motion of the
vane at outlet respectively
𝑉𝑤2 = Velocity of whirl at the
outlet
𝑉𝑓2 = Velocity of flow at the
outlet33
Fig. – Jet striking a moving
curved Plate at one end of
unsymmetrical plate
34
The triangles ABD and B’C’D’ 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,𝑢1=𝑢2 = 𝑢 =
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑣𝑎𝑛𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑚𝑜𝑡𝑖𝑜𝑛 and 𝑉𝑟1 = 𝑉𝑟2
Mass of water striking the vane per second=𝑚 = 𝜌𝑎𝑉𝑟1
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 –Final velocity ofthe jet in the direction of motion]
The initial velocity with which jet strikes the vane = 𝑉𝑟1
and, The component of this velocity in the direction ofmotion = 𝑉𝑟1 cos 𝜃 = (𝑉𝑤1 − 𝑢1 )
Similarly, The component of the relative velocity at theoutlet in the direction of motion = −𝑉𝑟2 cos 𝜑 = −[𝑢2 + 𝑉𝑤2 ]
35
So,∴ 𝐹𝑥 = 𝑚 × [𝑉𝑟1 cos 𝜃 − (−𝑉𝑟2 cos 𝜑)]
𝐹𝑥 = 𝜌𝑎𝑉𝑟1 × [(𝑉𝑤1 − 𝑢1 ) + (𝑢2 + 𝑉𝑤2 )]
As we know 𝑢1 = 𝑢2
𝐹𝑥 = 𝜌𝑎𝑉𝑟1 × [𝑉𝑤1 + 𝑉𝑤2 ]
It is true only when angle 𝛽 shown in Fig is acute angle (<90°)
• If 𝛽 = 90°, then 𝑉𝑤2 = 0 and Eq. becomes, 𝐹𝑥 = 𝜌𝑎𝑉𝑟1 𝑉𝑤1
• If 𝛽 is an obtuse angle (> 90°), the expression for 𝐹𝑥 willbecome, 𝐹𝑥 = 𝜌𝑎𝑉𝑟1 × [𝑉𝑤1 − 𝑉𝑤2 ]
In general,𝐹𝑥 = 𝜌𝑎𝑉𝑟1 × [𝑉𝑤1 ± 𝑉𝑤2 ]
Work done per second on the vane by the jet=𝑊
= 𝐹𝑜𝑟𝑐𝑒 × 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑 𝑝𝑒𝑟 sec 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑓𝑜𝑟𝑐𝑒
W = 𝐹𝑥 × 𝑢
𝑊 = 𝜌𝑎𝑢𝑉𝑟1 × [𝑉𝑤1 ± 𝑉𝑤2 ]
36
Work done per second per unit weight of fluid striking per
second,=𝜌𝑎𝑢𝑉𝑟1 × [𝑉𝑤1 ± 𝑉𝑤2 ] /{(𝜌𝑎𝑉𝑟1 ) × 𝑔}
= [𝑉𝑤1 ± 𝑉𝑤2 ] × 𝑢
Work done per second per unit mass of fluid striking per
second,=𝜌𝑎𝑢𝑉𝑟1 × [𝑉𝑤1 ± 𝑉𝑤2 ] / (𝜌𝑎𝑉𝑟1 )
= 𝑢 × [𝑉𝑤1 ± 𝑉𝑤2 ]
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. – 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,𝑉 = Velocity of jet
𝑑 = Diameter of jet
𝑢 = 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 = 𝜌𝑎𝑉
also, The jet strikes a plate with velocity = (𝑉 − 𝑢)
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,
𝐹𝑥 = 𝜌𝑎𝑉[(𝑉 − 𝑢) − 0]
𝐹𝑥 = 𝜌𝑎𝑉(𝑉 − 𝑢)
39
Work done by the jet on the series of plates per second,
𝑊 =𝐹𝑜𝑟𝑐𝑒 × 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑 𝑝𝑒𝑟 sec 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛𝑜𝑓𝑓𝑜𝑟𝑐𝑒
∴𝑊 = 𝐹𝑥 × 𝑢
∴ 𝑊 = 𝜌𝑎𝑉(𝑉 − 𝑢) × 𝑢
∴ 𝑊 = 𝜌𝑎𝑉𝑢(𝑉 − 𝑢)
The kinetic energy of the jet per second,
𝐾𝐸 = ½ 𝑚 𝑉2
∴ 𝐾𝐸 = ½(𝜌𝑎𝑉)𝑉2 = ½ 𝜌𝑎𝑉3
The efficiency of the wheel,𝜂=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,𝑅1 = Radius of the
wheel at the inlet of the
vane
𝑅2 = Radius of the
wheel at the outlet of
the vane
41Fig. – Jet striking a series of radial curved
vanes mounted on a wheel
42
𝜔 = Angular speed of the wheel
Then,𝑢1 = 𝜔𝑅1
𝑎𝑛𝑑 𝑢2 = 𝜔𝑅2
The mass of water striking per second for a series of vanes= The mass of water coming out from nozzle per sec = 𝜌𝑎𝑉1
Where,𝑎 = Area of the jet, and 𝑉1 = 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
∴ 𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑎𝑡 inlet 𝑝𝑒𝑟 𝑠𝑒𝑐 = 𝜌𝑎𝑉1 × (𝑉1 cos 𝛼)
∴ 𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑎𝑡 𝑖𝑛𝑙𝑒𝑡 𝑝𝑒𝑟 𝑠𝑒𝑐 = 𝜌𝑎𝑉1 × 𝑉𝑤1
(∵ 𝑉𝑤1 = 𝑉1 cos 𝛼)
Similarly, Momentum of water at outlet per sec = 𝜌𝑎𝑉1 Xcomponent of V2 in the tangential direction
∴ 𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑎𝑡 𝑜𝑢𝑡𝑙𝑒𝑡 𝑝𝑒𝑟 𝑠𝑒𝑐 = 𝜌𝑎𝑉1 × (−𝑉2 cos 𝛽)
∴ 𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑎𝑡 𝑜𝑢𝑡𝑙𝑒𝑡 𝑝𝑒𝑟 𝑠𝑒𝑐 = −𝜌𝑎𝑉1 × 𝑉𝑤2
(∵ 𝑉𝑤2 = 𝑉2 cos 𝛽)
43
Now angular momentum,
𝐴𝑛𝑔𝑢𝑙𝑎𝑟 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑝𝑒𝑟 sec 𝑎𝑡 𝑖𝑛𝑙𝑒𝑡 = 𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑎𝑡 𝑖𝑛𝑙𝑒𝑡
× 𝑅𝑎𝑑𝑖𝑢𝑠 𝑎𝑡 𝑖𝑛𝑙𝑒𝑡= 𝜌𝑎𝑉1 × 𝑉𝑤1 × 𝑅1
𝐴𝑛𝑔𝑢𝑙𝑎𝑟 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑝𝑒𝑟 sec 𝑎𝑡 𝑜𝑢𝑡𝑙𝑒𝑡 = 𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑎𝑡 𝑜𝑢𝑡𝑙𝑒𝑡× 𝑅𝑎𝑑𝑖𝑢𝑠 𝑎𝑡 𝑜𝑢𝑡𝑙𝑒𝑡= −𝜌𝑎𝑉1 × 𝑉𝑤2 × 𝑅2
Torque exerted by the water on the wheel,
𝑇 = 𝑅𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚
𝑇 = [𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑝𝑒𝑟 𝑠𝑒𝑐 − 𝐹𝑖𝑛𝑎𝑙 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑝𝑒𝑟 𝑠𝑒𝑐]
∴ 𝑇 = [𝜌𝑎𝑉1 × 𝑉𝑤1 𝑅1 − (−𝜌𝑎𝑉1 × 𝑉𝑤2 𝑅2 )]
∴ 𝑇 = 𝜌𝑎𝑉1 [𝑉𝑤1 𝑅1 + 𝑉𝑤2 𝑅2 ]
Work done per sec on the wheel,
𝑊𝐷/𝑠𝑒𝑐 = 𝑇𝑜𝑟𝑞𝑢𝑒 × 𝐴𝑛𝑔𝑢𝑙𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
∴ 𝑊𝐷/𝑠𝑒𝑐 = 𝑇 × 𝜔
∴ 𝑊𝐷/𝑠𝑒𝑐 = 𝜌𝑎𝑉1 [𝑉𝑤1 𝑅1 + 𝑉𝑤2 𝑅2 ] × 𝜔
44
∴ 𝑊𝐷/𝑠𝑒𝑐 = 𝜌𝑎𝑉1[𝑉𝑤1 𝑅1 𝜔 + 𝑉𝑤2 𝑅2 𝜔]
∴ 𝑊𝐷/𝑠𝑒𝑐 = 𝜌𝑎𝑉1[𝑉𝑤1 𝑢1 + 𝑉𝑤2 𝑢2 ], It is valid only when,𝛽 < 90
• If the angle 𝛽 is an obtuse angle (𝛽 > 90) then,
𝑊𝐷/𝑠𝑒𝑐 = 𝜌𝑎𝑉1 [𝑉𝑤1 𝑢1 − 𝑉𝑤2 𝑢2 ]
In general,𝑊𝐷/𝑠𝑒𝑐 = 𝜌𝑎𝑉1 [𝑉𝑤1 𝑢1 ± 𝑉𝑤2 𝑢2 ]
If the discharge is radial at the outlet then, 𝛽 = 90° and
hence 𝑉𝑤2 = 0,
∴ 𝑊𝐷/𝑠𝑒𝑐 = 𝜌𝑎𝑉1 [𝑉𝑤1 𝑢1 ]
Efficiency of the radial curved vanes,
𝜂=𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 𝑝𝑒𝑟 𝑠𝑒𝑐𝑜𝑛𝑑 / 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝑒𝑛𝑒𝑟𝑔𝑦 𝑝𝑒𝑟 𝑠𝑒𝑐𝑜𝑛𝑑
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
∴ 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= 𝜌𝑎𝑉r(𝑉r - 𝑢 )
𝑊𝐷/𝑠𝑒𝑐 = 𝜌𝑎𝑉r (𝑉r − 𝑢) x 𝑢
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 = 𝑉r
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
uuVaV
3
2
2
1
0).(2
1
r
rr
aV
VaV
5.02
).2(2
2 020)(
0])(2
[du
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 = 𝑉r
Energy Supplied= K.E at outlet- K.E. at Inlet
uV
u
uVuV
uuV
uVaV
uuVaV
r
rr
r
rr
rr
2
))((
)(2
)(2
1
)(
secondper suppliedEnergy
secondper doneWork
22
)(2
1
).(2
1).(
2
1
22
22
uVaV
uaVVaV
rr
rrr
