We have already seen the derivation of continuity equation, Bernoulli’s equation, momentum equation, velocity of sound in an isothermal process,
velocity of sound in an adiabatic process,
fundamentals of stagnation properties i.e. stagnation pressure, stagnation temperature and stagnation density for compressible fluid flow in our previous
posts.
We have also discussed the fundamentals of impact of
jets, force exerted by a jet on vertical flat plate and force exerted by a jet on stationary inclined flat plate in our recent post.
Now we will see here the derivation of expression of
force exerted by a jet on stationary curved plate with the help of this
post. Let us first brief here the basic concept of impact of jets and
after that we will derive the expression of force exerted by a jet on
stationary curved plate.
Impact of jets
Let us consider that we have one pipe through which
liquid is flowing under pressure. Let us assume that a nozzle is fitted at
outlet of pipe. Liquid which will come through the outlet of nozzle will be in
the form of jet.
If a plate, which may be moving or fixed, is placed
in the path of jet, there will be one force which will be exerted by the jet
over the surface of plate. The force which will be exerted by the jet over the
surface of plate, which might be moving or fixed, will be termed as impact of
jet.
In order to determine the expression of force
exerted by the jet over the surface of plate i.e. impact of jet, we will use
the basic concept of Newton’s second law of motion and impulse-momentum
equation.
Force exerted by jet on stationary curved plate
We will see here three conditions as mentioned here.
- Jet strikes the curved plate at the center
- Jet strikes the curved plate at one end tangentially when the plate is symmetrical
- Jet strikes the curved plate at one end tangentially when the plate is symmetrical
First we will see the case of Jet striking the
curved plate at the center and we will secure the expression for force exerted
by jet.
Jet strikes the curved plate at the center
Let us consider that a jet of water strikes a fixed
stationary curved plate at its center as displayed here in following figure.
Let us assume the following data from above figure.
V = Velocity of the jet
d = Diameter of the jet
a = Area of cross-section of the jet = (π/4) x d2
θ = Angle made by jet with x- axis
Let us assume that the plate is smooth and there is
no loss of energy due to impact of water jet. Water jet, after striking the
stationary curved plate, will come with similar velocity in a direction
tangentially to the curved plate.
We will resolve the velocity at the outlet of curved
plate in its two components i.e. in the direction of jet and in a direction
perpendicular to the jet.
Component of velocity of water jet in the direction
of jet = - V Cos θ
We have taken negative sign because velocity at the
outlet is in the opposite direction of the water jet coming out from nozzle.
Component of velocity of water jet perpendicular to
the jet = V Sin θ
Force exerted by the water jet in the direction of
jet
FX = Mass per second x [V1x –V2x]
Where,
V1x = Initial velocity in the direction
of jet = V
V2x = Final velocity in the direction of
jet = - V Cos θ
FX = ρ a V x [V + V Cos θ]
FX = ρ a V2 x [1+ Cos θ]
Force exerted by the water jet in the direction
perpendicular to the jet
FY = Mass per second x [V1y –V2y]
Where,
V1y = Initial velocity in Y direction = 0
V2y = Final velocity in Y direction = V Sin
θ
FY = ρ a V x [0 - V Sin θ]
FY = - ρ a V2 Sin θ
Jet strikes the curved plate at one end tangentially when the plate is symmetrical
Let us consider that a jet of water strikes a fixed
stationary curved plate at its one end tangentially, when the plate is
symmetrical, as displayed here in following figure.
Let us assume the following data from above figure.
V = Velocity of jet
θ = Angle made by jet with x - axis at inlet tip of
the curved plate
Let us assume that the plate is smooth and there is
no loss of energy due to impact of water jet. Water jet, after striking the
stationary curved plate, will come at the outlet tip of the curved plate with
similar velocity i.e. V in a direction tangentially to the curved plate.
Force exerted by the water jet in the x - direction
FX = Mass per second x [V1x –V2x]
Where,
V1x = Initial velocity in the x direction
= V Cos θ
V2x = Final velocity in the x direction =
- V Cos θ
FX = ρ a V x [V Cos θ + V Cos θ]
FX = 2 ρ a V2 Cos θ
Force exerted by the water jet in the Y- direction
FY = Mass per second x [V1y –V2y]
Where,
V1y = Initial velocity in Y direction = V
Sin θ
V2y = Final velocity in Y direction = V
Sin θ
FY = ρ a V x [V Sin θ - V Sin θ]
FY = 0
Jet strikes the curved plate at one end tangentially when the plate is unsymmetrical
Let us consider that a jet of water strikes a fixed
stationary curved plate at its one end tangentially, when the plate is unsymmetrical,
as displayed here in following figure.
V = Velocity of jet
θ = Angle made by tangent with x - axis at inlet tip
of the curved plate
ϕ = Angle made by tangent with x - axis at outlet tip
of the curved plate
Let us assume that the plate is smooth and there is
no loss of energy due to impact of water jet. Water jet, after striking the
stationary curved plate, will come at the outlet tip of the curved plate with
similar velocity i.e. V in a direction tangentially to the curved plate.
Force exerted by the water jet in the x - direction
FX = Mass per second x [V1x –V2x]
Where,
V1x = Initial velocity in the x direction
= V Cos θ
V2x = Final velocity in the x direction =
- V Cos ϕ
FX = ρ a V x [V Cos θ + V Cos ϕ]
FX = ρ a V2 [Cos θ + Cos ϕ]
Force exerted by the water jet in the Y- direction
FY = Mass per second x [V1y –V2y]
Where,
V1y = Initial velocity in Y direction = V
Sin θ
V2y = Final velocity in Y direction = V
Sin ϕ
FY = ρ a V x [V Sin θ - V Sin ϕ]
FY = ρ a V2 x [Sin θ - Sin ϕ]
Above equation, derived here, provides the components
of force exerted by the liquid jet on the stationary curved plate for above
three cases.
Do you have any suggestions? Please write in comment
box.
Reference:
Fluid mechanics, By R. K. Bansal
Image courtesy: Google
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