3.4 APPLICATION OF
MOMENTUM
Introduction
The study of forces resulting from the impact of fluid jets and
when fluids are diverted round pipe bends involves the application of newtons
second law in the form of F = m.a. The forces are determined by calculating the
change of momentum of the flowing fluids. In nature these forces
manifest themselves in the form of wind forces, and the impact forces of the
sea on the harbour walls. The operation of hydro-kinetic machines
such as turbines depends on forces developed through changing the momentum of
flowing fluids.
I have created a excelcalcs spreadsheet for convenient access to
all of the equations found on this page . This is located at ExcelCalcs.com calculation Fluid Jets
Symbols
|
α = jet angle (radian)
a = Acceleration (m/s 2) ρ = density (kg/m 3) ρ = density (kg/m 3) F = Force (N) m = mass (kg) V = fluid velocity(m/s) |
u 1 = initial velocity (m/s)
u 2 = final velocity velocity (m/s) P = Power (watts) Q = Volumetric Flow Rate (m 3/s) θ 1 vane inlet angle angle (radian) θ 2 vane outlet angle angle (radian) |
Momentum
Newtons Second Law can be stated as: The force acting on a body in
a fixed direction is equal to rate of increase of momentum of the body in that
direction. Force and momentum are vector quantities so the direction
is important. A fluid is essentially a collection of particles and the net
force, in a fixed direction, on a defined quantity of fluid equals the total
rate of momentum of that fluid quantity in that direction.
Consider a mass m which has an initial velocity u and is brought to rest. Its loss of momentum is m.u and if it stopped in a time interval t then the rate of change of momentum is m.u /t. The force F required to stop the moving mass is therefore F = m.u / t . Now if this is applied to a jet of fluid with a mass flow rate ( m / t ) which is equivalent to the volumetric flow rate times the density ( Qρ ) the equivalent force on a flowing fluid is F = Qρ u. Also in accordance with Newtons third law the resulting force of the fluid by a flowing fluid on its surroundings is (-F). Newtons third law states that for every force there is an equal and opposite force.
Consider a mass m which has an initial velocity u and is brought to rest. Its loss of momentum is m.u and if it stopped in a time interval t then the rate of change of momentum is m.u /t. The force F required to stop the moving mass is therefore F = m.u / t . Now if this is applied to a jet of fluid with a mass flow rate ( m / t ) which is equivalent to the volumetric flow rate times the density ( Qρ ) the equivalent force on a flowing fluid is F = Qρ u. Also in accordance with Newtons third law the resulting force of the fluid by a flowing fluid on its surroundings is (-F). Newtons third law states that for every force there is an equal and opposite force.
The figure below illustrates this principle at two locations.
The fluid flowing into the tank is brought to rest from a velocity u to zero velocity and the force on the jet is F = Q.ρ. u. . The reaction force on the tank contents is -F.
The fluid flowing in the pipe in the horizontal direction is forced to change direction at the bend such that its velocity in the orginal direction is zero. Therefore the force on the flowing fluid is F = Q. ρ u. The reaction force on the pipe is -F in the horizontal direction as shown.
In its simplest form, with steady flow conditions, the force on a
fluid flow in a set direction is equal to its mass flow rate times by the
change in velocity in the set direction. The fluid flow also exerts
an equal and opposite reaction force as a result of this change in momentum.
F = Qρ (u 1 - u 2). ..(
F and u are vector quantities)
The resultant force on a fluid in a particular direction is equal
to the rate of increase of momentum in that direction.
Jet Forces on Stationary Plates
Jet force on a flat plate
Considering only forces in a horizontal direction u 1 =
V and u 2 = 0 therefore
F = QρV = ρAV 2
Jet force on a flat plate at an angle θ
Considering only forces Normal to plate surface u 1 =
V sinθ and u 2 = 0 therefore
F = QρV sin θ = ρAV 2 sin θ
when θ =90o then F = ρAV 2 as
above
Jet force on an angled plate. (θ < 90 o)
Considering only forces in a horizontal direction u 1 =
V and u 2 = V cos θ therefore
F = QρV (1 - cosθ ) = ρAV 2(1 - cosθ )
Jet Force on an angled plate (θ > 90 o)
Considering only forces in a horizontal direction u 1 =
V and u 2 = V cos θ therefore
F = QρV (1 - cosθ ) = ρAV 2(1 - cosθ )
Jet Force on an angled plate (θ = 180 o)
u 1 = V and u 2 = V
cos θ therefore
F = QρV (1 - cos180 o ) = 2QρV
= 2ρAV 2
Jet force on a moving flat plate
Considering only forces in a horizontal direction
u 1 = V and u 2 = V p
and let r = V p / V therefore
F = Qρ( V - V p ) = ρAV(V -V p)
and F = ρAV 2( 1 - r
The power (P) generated by the force on the moving plate = P = F. V p
Jet force on an angled moving plate
Considering only forces in a horizontal direction
u 1 = V and u 2 = V p +
(V - V p) cos θ
and let r = V p / V therefore
F = ρA V( V - V p ) ( 1 - cosθ ) = ρA
V 2 ( 1 - r ) ( 1 - cosθ )
The power (P) generated by the force on the moving plate P
= F. V p
Note: The moving plate with an angle θ = 180o is the
typical of the rotating plate of the pelton wheel.
The ideal value for r resulting in the maximum power output is clearly 0,5
The ideal value for r resulting in the maximum power output is clearly 0,5
Jet Forces on Vanes......
Additional notes can be found on webpages
Fluid Machines - Francis Wheel
Steam Turbines - Impulse blades
Fluid Machines - Francis Wheel
Steam Turbines - Impulse blades
Force on fixed Vane.
In the x Direction: u 1x = V cos θ 1 ,
u 2x = -V cos θ 2
F x = QρV(cos θ 1 +
cos θ 2 ) = ρAV 2(cos θ 1 +
cos θ 2 )
In the y Direction u 1y = V sin θ 1 u 2y =
V sin θ 2
F y = QρV(sin θ 1 -
sin θ 2 ) = ρAV 2(sin θ 1 -
sin θ 2 )
Force on Moving Vane.
The notes below related to vanes as used in impulse
turbines. These turbines derive the mechanical energy mainly from
the change in momentum as the fluid passes through the vanes. The
conditions as shown when the vectorial sum of V v +
V r1 = V1 results in smooth entry with
efficient transfer of energy of the fluid to the vane. When this
does not occur there will be turbulent flow over the vane with significant
losses.
In the x
Direction
u 1x = V 1 cos α
u 2x = V v - V r2 cos θ 2 .... ( V r2 = V r1 = V 1 sin α /sin θ 1 )
F x = QρV 1 (cos α + [sin α /sin θ 1 ] cos θ 2 - r > )
r = V v / V 1
In the y Direction u 1y = V 1 sin α
u 2y = V r2 sin θ 2 .... ( V r2 = V r1 = V 1 sin α /sin θ 1 )
In the y direction F y = QρV 1sin α (1 - sin θ 2 / sin θ 1 )
If the vane is moving in the x direction the power developed by the vane P = F x.V V
u 1x = V 1 cos α
u 2x = V v - V r2 cos θ 2 .... ( V r2 = V r1 = V 1 sin α /sin θ 1 )
F x = QρV 1 (cos α + [sin α /sin θ 1 ] cos θ 2 - r > )
r = V v / V 1
In the y Direction u 1y = V 1 sin α
u 2y = V r2 sin θ 2 .... ( V r2 = V r1 = V 1 sin α /sin θ 1 )
In the y direction F y = QρV 1sin α (1 - sin θ 2 / sin θ 1 )
If the vane is moving in the x direction the power developed by the vane P = F x.V V
Force on Pipe Wall
The notes below related to the force on a pipe wall resulting from
the changes in fluid pressure and fluid momentum as the fluid flows round a
pipe bend . Gravity and friction effects are not considered. The
fluid is assumed to be flowing under steady state conditions.
In the x Direction
u 1 = V 1
u 2 = V 2
F x = p1.A1 - p2.A2 + ρ (A1V12 - A2V2)2
u 1 = V 1
u 2 = V 2
F x = p1.A1 - p2.A2 + ρ (A1V12 - A2V2)2
|
In the x Direction
u 1x = V 1 u 2x = V 2cos θ F x = p1.A1 - p2.A2cos θ + ρ (A1V12 - A2V22 cos θ ) |
In the y Direction
u 1y = 0 u 2y = V 2sin θ F y = - p2.A2sin θ - ρA2V22 sin θ |
The resultant reaction force on the pipe = Fr =√ (Fx2 +
Fx2 )
The angle α of the resultant force to the x axis = tan-1 ((Fy /(Fx)
The angle α of the resultant force to the x axis = tan-1 ((Fy /(Fx)
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