2013 lect6 forces on pipe bends.ppt

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


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.


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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2013 lect6 forces on pipe bends.ppt

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