Lecture - 4 ( Impact of jet on stationary curved plates )


Let's continue after lecture 3

Jet strikes at the centre of stationary symmetric curved plate

Let us consider a stationary flat plate on the centre of which water Jet striking perpendicularly, denote this centre point of plate as point 1, which is entry point of water jet. After striking the plate water will split equally in two parts. Let's consider waterjet is coming out of nozzle with mass flow rate 'm' and with velocity 'V1'. After splitting into equal parts, water will exit from top and bottom ends of the plate. Denote both of these end points of plate with point 2, which is exit point of water jet. The mass flow rate exiting from top end of plate is 'm/2' and with exit velocity V2. Similarly the mass flow rate exiting from bottom end of plate is 'm/2' and with exit velocity V2. As water is traveling over the curved plate there is no change of pressure on the entry and exit point of plate so P1 = P2 = Patm .
At exit point 2, water jet is exiting with angle 'φ' and velocity 'V2' as shown in figure. So just resolve the component of this velocity in horizontal and vertical direction. So horizontal component of velocity at top and bottom end of plate is '-V2*cosφ' and vertical component of velocity is '±V2*sinφ'. Always remember sign convention will come with vector quantities. Sign conventions are also shown in figure below. Even if you flip this sign convention you can get same results.


We are considering the surface of plate to be frictionless so V1 = V2
'φ' denotes the vane angle at exit 
'δ' denotes the angle of deflection 
δ = 180 - φ 
Here mass flow rate entry (m) = ρ*A*V1
As we know 'A' denotes cross section area of water jet and 'ρ' denotes density of water.

Fx = FN = linear momentum of Jet at entry point 1 in horizontal direction - linear momentum of jet at exit point 2 in horizontal direction = m*V1 - [(m/2)*(-V2*cosφ) + (m/2)*(-V2*cosφ)]
Fx = m*V1( 1+cosφ )
Fy = Ft = linear momentum of jet in vertical direction at entry point 1 - linear momentum of jet in vertical direction at exit point 2 = m*V1 - [(m/2)*(V2*sinφ) + (m/2)*(-V2*sinφ)]
Fy = Ft = m*0 - [0] = 0

Jet strikes at the tip of stationary curved vane

In this case water jet will enter at the entry tip "point 1" making angle 'θ' with the horizontal axis. Waterjet will flow all the way over the surface of vane and exit from another tip named as exit tip and represented by "point 2" making angle 'φ' with horizontal axis. we are considering surface of plate to be frictionless so velocity of water Jet at exit and entry will be same. 
'δ' represents angle of deflection of water jet
δ = 180 - (θ+φ)


m = ρ*a*V1
Fx = Fn = m*V1*cosθ - m( -V2cosφ )
     = m*V1( cosθ + cosφ)
Fy = Ft = m*V1*sinθ - m*V2sinφ
             = m*V1(sinθ - sinφ)
For symmetrical vane
θ = φ
δ = 180 - 2θ = 180 - 2φ

Continue to lecture 5

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