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Wednesday, 20 February 2019

February 20, 2019

FORCE EXERTED BY A JET ON A CURVED PLATE WHEN THE PLATE IS MOVING IN THE DIRECTION OF JET

We have already seen the derivation and fundamentals of stagnation properties i.e. stagnation pressure, stagnation temperature and stagnation density for compressible fluid flow in our previous posts. 


Now we will see here the derivation of expression of force exerted by a jet on a curved plate when the plate is moving in the direction of jet 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 a curved plate when the plate is moving in the direction of jet. 

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. 

Force exerted by a jet on a curved plate when the plate is moving in the direction of jet

Let us consider a jet of water striking a curved plate at the center of the plate which is moving with a uniform velocity in the direction of jet as displayed here in following figure. 
Let us consider the following terms from above figure
V = Absolute velocity of jet
a = Area of jet
u = Velocity of the plate in the direction of jet 

Relative velocity of water jet or velocity with which the water jet strikes the curved plate = (V-u) 

Let us consider that plate is smooth and loss of energy due to the impact of jet is zero. We will have the similar velocity with which the jet will be leaving the curved plate i.e. (V-u) 

Velocity will be resolved into two components. One component will be in the direction of jet and second component will be in the direction perpendicular to the jet. 

Component of the velocity in the direction of jet = - (V- u) Cos θ 

We have taken negative sign as at the outlet, the component will be in the opposite direction of jet. 

Component of the velocity in the direction perpendicular to the jet = (V- u) Sin θ 

Mass of the water striking the plate = ρ a (V-u) 

Force exerted by the jet of water on the curved plate in the direction of jet 


Work done by the jet per second on the plate will be given by following equation 


Above equation, derived above, force exerted by the jet of water on the curved plate in the direction of jet and works done by the jet per second on the plate when water jet strikes the curved plate when the plate is moving in the direction of jet.  

Do you have any suggestions? Please write in comment box. 

Reference: 
Fluid mechanics, By R. K. Bansal 
Image courtesy: Google  

Also read  

Friday, 8 February 2019

February 08, 2019

FORCE EXERTED BY JET ON HINGED PLATE

We have already seen the derivation of continuity equationBernoulli’s equationmomentum equationvelocity of sound in an isothermal processvelocity 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. 


Now we will see here the derivation of expression of force exerted by a jet on a hinged 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 hinged 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. 

Force exerted by a jet on a hinged plate 

Let us consider a jet of water striking a vertical plate at the center which is hinged at point O as displayed here in following figure. 

When jet of water will strike the plate, the plate will swing through an angle about the hinge as displayed here in following figure. 

Let us consider the following terms from above figure.
x = Distance of center of jet from hinge O
θ = Angle of swing about hinge
W = Weight of plate acting at C.G of the plate 

Initial position of the vertical plate is displayed above in figure by dotted lines. When water jet will strike the plate, the plate will swing through an angle θ. After the jet strikes the plate, point A will be now at point A’ as displayed in above figure. 

Distance OA = Distance OA’ = x 

Weight of plate will be acted through the point A’ after water jet strikes the plate. 

There will be two forces acting on the plate as mentioned here. 
  1. Force due to jet of water normal to the plate, F
  2. Weight of the plate, W 

Force due to jet of water normal to the plate

Fn = ρ a V2 Sin (90 – θ)  

Weight of the plate, W 

Once the jet strikes the plate, for equilibrium condition we will have following equation as mentioned here. 

Moment of force Fn about hinge = Moment of weight W about hinge 

Moment of force Fn about hinge = Fn x OB = ρ a V2 Sin (90 – θ)   x OB
Moment of force Fn about hinge = ρ a V2 Cos θ   x OB
Moment of force Fn about hinge = ρ a V2 Cos θ   x (OA / Cos θ)
Moment of force Fn about hinge = ρ a V2 x OA
Moment of force Fn about hinge = ρ a V2

Now we will determine the moment of weight W about hinge 

Moment of weight W about the hinge = W x OA’ Sin θ = W x Sin θ 

Once the jet strikes the plate, for equilibrium condition, we will have 

Moment of force Fn about hinge = Moment of weight W about hinge
ρ a V2 x = W x Sin θ

Sin θ = ρ a V2 /W 


Above equation, derived here, provides the angle of swing of plate when water jet strikes the vertical plate which is fixed at hinge.  

Do you have any suggestions? Please write in comment box. 

Reference: 

Fluid mechanics, By R. K. Bansal 
Image courtesy: Google  

Also read  

Friday, 1 February 2019

February 01, 2019

FORCE EXERTED BY A JET ON STATIONARY CURVED PLATE

We have already seen the derivation of continuity equationBernoulli’s equationmomentum equationvelocity 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.
  1. Jet strikes the curved plate at the center
  2. Jet strikes the curved plate at one end tangentially when the plate is symmetrical
  3. 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  

Also read  

Wednesday, 30 January 2019

January 30, 2019

FORCE EXERTED BY JET ON STATIONARY INCLINED FLAT PLATE

We have already seen the derivation of continuity equationBernoulli’s equationmomentum equationvelocity 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. 

Now we will start here the basics and derivation of expression of force exerted by a jet on stationary inclined flat 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 inclined flat 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 inclined flat plate

We have already seen the expression for force exerted by jet on vertical flat plate. Let us derive here the expression of force exerted by jet on stationary inclined flat plate. 

Let us consider a jet of water, which is coming from the outlet of nozzle fitted at the pipe, strikes a flat inclined flat plate as displayed here in following figure. 
V = Velocity of the jet in x direction 
d = Diameter of the jet 
a = Area of cross-section of the jet = (π/4) x d
θ = Angle between the jet and the plate 

Let us determine the mass of water per second striking the plate 
Mass of water per second striking the plate = ρ a V 

Let us assume that inclined plate is smooth and there is no loss of energy due to the impact of jet. Considering this assumption, we can say that jet will move over the surface of inclined plate after striking the plate with a velocity equivalent to the initial velocity i.e. V. 

Now we will determine the force Fn exerted by the jet on the plate in a direction normal to the plate. 

Fn = Mass of jet striking per second x [Initial velocity of jet before striking in the direction of n - final velocity of jet after striking in the direction of n] 

Fn = ρ a V x [V Sin θ - 0] 

Fn = ρ a V2 Sin θ 

Above force will be resolved in two components i.e. one component in the direction of jet and second component in the direction perpendicular to the jet. 


Above equation, derived here, provides the components of force exerted by the liquid jet on the inclined plate in the direction of jet and in the direction perpendicular to the jet.  

Now we will discuss the force exerted by a jet on stationary curved plate in our next post. 

Do you have any suggestions? Please write in comment box. 

Reference: 

Fluid mechanics, By R. K. Bansal 
Image courtesy: Google 

Also read  

January 30, 2019

FORCE EXERTED BY JET ON STATIONARY VERTICAL PLATE

We have already seen the derivation of continuity equationBernoulli’s equationmomentum equationvelocity of sound in an isothermal process, velocity of sound in an adiabatic process and also the basic fundamentals of stagnation properties i.e. stagnation pressure,stagnation temperature and stagnation density for compressible fluid flow in our previous posts. 

Now we will start here the basics and derivation of expression of force exerted by jet on stationary vertical 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 jet on stationary vertical 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 vertical plate 

Let us consider a jet of water, which is coming from the outlet of nozzle fitted at the pipe, strikes a flat vertical flat plate as displayed here in following figure. 
V = Velocity of the jet
d = Diameter of the jet
a = Area of cross-section of the jet = (π/4) x d

Liquid jet will move along the vertical plate after striking the plate. As we can see from the figure, vertical plate is located at right angle to the direction of liquid jet and hence liquid jet will be deflected at right angle i.e. 900 after striking the vertical plate. 

The component of velocity of liquid jet, in the direction of jet, after striking will be zero. 

Force exerted by liquid jet on the plate in the direction of jet will be determined by using the concept of impulse momentum equation. 

Above equation, derived here, provides the force exerted by the liquid jet on the plate in the direction of jet. 

Now we will discuss the force exerted by the liquid jet on the stationary inclined flat plate in our next post. 

Do you have any suggestions? Please write in comment box. 

Reference: 

Fluid mechanics, By R. K. Bansal 
Image courtesy: Google  

Also read  

Monday, 28 January 2019

January 28, 2019

AREA VELOCITY RELATIONSHIP FOR COMPRESSIBLE FLOW

Till now we were discussing the various concepts and equations such as continuity equationEuler equationBernoulli’s equation and momentum equation for incompressible fluid flow. In same way we have also discussed above equations for compressible fluid flow. 

We have already seen the derivation of continuity equationBernoulli’s equationmomentum equationvelocity of sound in an isothermal process, expression for velocity of sound in an adiabatic process and also the basic fundamentals of stagnation properties i.e. stagnation pressure,stagnation temperature and stagnation density for compressible fluid flow in our previous posts. 

Now we will discuss the area velocity relationship for compressible flow with the help of this post. 

Area velocity relationship for compressible flow 
As we know that for incompressible fluid, continuity equation provides the relationship between area and velocity and it is as mentioned here.
Area x velocity = constant
A x V = Constant

We can conclude from above equation that velocity will be decreased with increase in area. If we consider for compressible fluid flow, above equation will not be valid. 

Continuity equation for compressible fluid will be given by following equation as mentioned here. 

ρ x A x V = Constant

We can conclude from above equation that velocity and density both parameters will be affected with variation in the value of area. 

Let us differentiate the continuity equation of compressible fluid and we will have following equation as mentioned here. 

Let us recall the Euler’s equation as mentioned here.
Now we will neglect the term z and we will have following equation
Now we will recall the expression for the velocity of sound wave in a fluid as mentioned here.
Considering the above equation of velocity of sound wave in a fluid, we will have following equation as mentioned here.
Above equation shows the variation of area with variation of velocity for different mach numbers. 

We will discuss now force exerted by jet on stationary vertical plate in our next post. 

Do you have any suggestions? Please write in comment box. 

Reference: 

Fluid mechanics, By R. K. Bansal 
Image courtesy: Google  

Also read