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8/11/2019 2013 lect6 FORCES ON PIPE BENDS.ppt
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THE FORCE DUE THE FLOW
AROUND A PIPE BEND
Consider a pipe bend
with a constant cross section
lying in the horizontal plane and
turning through an angle of .
Because the fluid changes direction, a force will act in the bend.If the bend is not fixed it will move and eventually break at the joints.
We need to know how much force a support (thrust block) must withstand.
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FORCES ON PIPE BENDS
In order to properly size thrust blocks, hangars, or other devices to hold a pipe
in place, the momentum equation is used to compute the necessary resistive
force to hold the pipe stationary.
Forces in a pipe bend in the horizontal plane are caused by the fluid's
momentum and pressure.
If the pipe undergoes a bend in the vertical plane, where the entrance to the
bend is above the exit (or vice-versa), then the weight of the liquid and pipe
material within the bend will contribute to the force.
Since computing the volume of fluid and pipe material within a bend requires
considerably more input, we kept our calculation relatively simple by keeping
it in the horizontal plane.
The forces Fxand Fycomputed by the calculation are the x and y components
of the total force F.
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Fy= P2A2sin(b) + d V2Q sin(b) F = (Fx2+ Fy
2)1/2
Q=VA A= D2/ 4
P2= P1+ d (V12- V2
2) / 2
Fx= -P1A1- P2A2cos(b) - d Q [V1+ V2cos(b)]
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THEORTICAL BASIC
Momentum Analysis of Flow Systems
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Objectives
Identify the various kinds of forces and
moments acting on a control volume.
Use control volume analysis to determinethe forces associated with fluid flow.
Use control volume analysis to determine
the moments caused by fluid flow and the
torque transmitted.
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Newtons Laws
Newtons laws are relations between motions of
bodies and the forces acting on them.
First law: a body at rest remains at rest, and a body in
motion remains in motion at the same velocity in a straightpath when the net force acting on it is zero.
Second law: the acceleration of a body is proportional to the
net force acting on it and is inversely proportional to its
mass.
Third law: when a body exerts a force on a second body,
the second body exerts an equal and opposite force on the
first.
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NEWTONS EQUATION OF MOTION
FOR RIGID BODY
dtdVmmaF
FdtmVd sys )(
maFManipulation into momentum
Resultant of vektor
F
MOMENTUM
IMPULS
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MOMENTUM BALANCEFOR FLUID FLOW
xoutoutxininxsysx
FmVmVdt
mVd
)(
inV
FdtdmVdmVmVd outoutininsys)(
FmVmVdt
mVdoutoutinin
sys
)(
SISTEM
Component-x
outV
Manipulation
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Force on the Control Volume
( )0
x sysd mV
dt
Externalpressure
Gravity
Steady Flow : 0 xin in xout out xV m V m F Identify forces
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PEOCEDURE OF CALCULATION
Draw a control volume
Decide on co-ordinate axis systemCalculate the totalforce
Calculate the pressureforce
Calculate the bodyforce
Calculate the resultantforce
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Draw a control volume
Decide on co-ordinate axis system
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Calculate the totalforce
In the x-direction: In the y-direction:
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Calculate the pressure force
Calculate the bodyforce
There are no body forces in the x or y directions. The only
body force is that exerted by gravity (which acts into thepaper in this example - a direction we do not need to
consider).
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Example : pipe nozzle
totalforce
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Calculate the body force
The only body force is the weight due to gravity in they-direction - but we need not consider this as the only
forces we are considering are in the x-direction.
Calculate the resultantforce
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MENGHITUNG GAYA YANG BEKERJA PADA BAUT
PENGIKAT FLANGES
xoutoutxininxsysx
FmVmVdt
mVd
)(
A nozzle is attached to a fire hose by a bolted flange. What is the force tending to tear apart thatflange when the valve is opened. When the valve is opened the fluid steadily flow out at a velocity
of 100 ft/s, the area of nozzle 1 in2. the pressure still 100 lbf/in2 gauge What is the force tending to
tear apart that flange now.
nozzle
Pin=100 lbf/in2
Aout=1 in2
Ain=10 in2
Vout=100 ft/s
SYSTEM
Flanges
Fbolt=?
Pout=Patmosfir
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xoutoutxininx FmVmV 0
110
10
outin xout xin
in
A ftV V V
A s
outxoutoutin
VAmm
( )x in atmosfir in x unknownF P P A F
nozzle
Pin=100 psig
Aout=1 in2
Ain=10 in2
Vout=100 ft/s
SYSTEM
Flanges
Fbolt=?
Pout=Patm=0psig
( ) ( )x xin xout in atmosfir inF m V V P P A
3362
ft
lbm.water
22
2 31 100 62.5 43.3
144
ft ft lbm lbmm in
in s ft s
2
2
243.3 (10 100 ) (100 10 0) 121 1000 879
32.2
f mx unknown f f f
m
lb s lblbm ft ft F in lb lb lb
s s s lb ft in
MENGHITUNG GAYA YANG BEKERJA PADA BAUT
PENGIKAT FLANGES
879 ( kiri)x unknown fF lb arah tarikan bolt=x unknown boltF gaya F
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APLIKASI NERACA MOMENTUM PADA STEDI FLOW
Soal 7--13
xoutoutxininx FmVmV 0Qmm
outin
( )in atmosfir in x unk F P P A F
nozzle
Pin=40 lbf/in2
Aout=3 in2
Ain=12 in2
Q=1200in3/s
SYSTEM
Flanges
Fbolt=?
Pout=0psig=Patmosfir
2 1 1( ) ( )x unk in inin out
F Q P AA A
3362
ft
lbm.water
2 ( ) ( )out inx unk in inin out
A AF Q P A
A A
2 4 42 2
2 4 4 3 2 2
9 .1200 62.4 40 12 33.6 480 446.4
12 36 32.2 .x unk
in in ft lbm lbf s lbf F in
s in ft in lbm ft in
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Soal 7.12
A1=1 in2
V=200ft/s
P1=30psig
F = ?
How is the force Transmitted
)( __ xoutxinx FVVm 0
P2=0psigA2=A1
10 ( ) ( 0) x unkownm V V P A F
1x unknownF P A
Via friction on pipe wall
kiri =x unknown boltF arah F
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Impact of a Jet on a Plane
We will first consider a jet hitting a flat plate (a plane) at an angl
of 90, as shown in the figure below.
We want to find the reaction force of the plate i.e. the force the
plate will have to apply to stay in the same position.
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Calculate the totalforce
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Calculate the pressure force.
The pressure force is zero as the pressure at both the inlet and the
outlets to the control volume are atmospheric.
5 Calculate the body force
As the control volume is small we can ignore the body force due to the
weight of gravity.
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Force due to a jet hitting an inclined
plane
We do not know the
velocities of flow in each
direction. To find these we
can apply Bernoulli
equation
The height differences are negligible i.e.z1 = z2 = z3and the pressures are all
atmospheric = 0. So
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Q1 = Q2 + Q3
u1A1 = u2A2 + u3A3
soA1 = A2 + A3
Q1 = A1u
Q2 = A2u
Q3 = (A1 - A2)u
u1 = u2 = u3 = u
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Calculate the totalforce in the x-direction.
Remember that the co-ordinate system is normal to the plate.
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2. Calculate the pressureforce
All zero as the pressure is everywhere atmospheric.
Calculate the bodyforce
As the control volume is small, hence the weight of
fluid is small, we can ignore the body forces.
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Fy=??
V=100ft/s
Fx=??
Soal 7.8 hal
1
2
P1=50psig
P2=40psig
sistim
A=0.5 ft2
xxoutxin FVVm )(0
10 ( 0) ( )x unk atmm V F P P A 1( )x unkF mV P A
yyoutyin FVVm )(0
20 (0 ( ) ( )y unk atmm V F P P A
2( )y unkF mV P A
Fx
Fy
Momentum
Masuk +, Keluar -Kecepatan & GayaKekanan, keatas +Kekiri, kebawah -
FR
Arah X (horizontal)
Arah Y (vertikal)
& adalah gaya dorongx unk y unkF F
Patm=0psig
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BESAR DAN ARAH GAYA-GAYA PADA ELBOW
FmVmV outoutinin 0
x
y
F
F1tan
2 2 1 1( sin sin )yF m V V
Fx
Fy
V2 2
V1
1
2
y
2
x FFF F
2 2 1 1( cos cos )xF m V V
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Free Jet Flows
Fx= m(Uxout
Uxin) = m(U2cos2
U1cos1)Fy= m(UyoutUyin) = m(U2sin2U1sin1)
Deflector plate
Fx
Fy
U
U2control
volume
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Free Jet Flows
y
x
Fy
V
V 2
Thrust reverser
Fx
Fx= mV(cos21) = AV2(cos21)
Fy= mV(sin2) = AV2(sin2)
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