Fluid Mechanics Benno 7

8/3/2019 Fluid Mechanics Benno 7

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TL2101

Mekanika Fluida I

Benno Rahardyan

Pertemuan 7



Mg Topik Sub Topik Tujuan Instruksional (TIK)

1 Pengantar Definisi dan sifat-sifat fluida,

berbagai jenis fluida yang

berhubungan dengan bidang TL

Memahami berbagai

kegunaan mekflu

dalam bidang TL

Pengaruh tekanan Tekanan dalam fluida, tekananhidrostatik Mengerti prinsip-2tekanan statitka

2 Pengenalan jenis

aliran fluida Aliran laminar dan turbulen,

pengembangan persamaan untuk

penentuan jenis aliran: bilangan

reynolds, freud, dll

Mengerti, dapat

menghitung dan

menggunakan prinsip

dasar aliran staedy state

3 Prinsip kekekalan

energi dalam

aliran

Prinsip kontinuitas aliran,

komponen energi dalam aliran

fluida, penerapan persamaan

Bernoulli dalam perpipaan

Mengerti, dapat

menggunakan dan

menghitung sistem prinsi

hukum kontinuitas

4 Idem Idem + gaya pada bidang terendam Idem

5 Aplikasi kekekalan

energi Aplikasi kekekalan energi dalam

aplikasi di bidang TL Latihan menggunakan

prinsip kekekalan

eneri khususnya dalam

bidang air minum

Aplikasi kekekalan

energi

Darcy-Weisbach, headloss, major

losses dan minor lossesKeseimbangan bidang terapung



FLUID DYNAMICS

THE BERNOULLI EQUATION

The laws of Statics that we have learned cannot solveDynamic Problems. There is no way to solve for the flowrate, or Q. Therefore, we need a new dynamic approach

to Fluid Mechanics.



The Bernoulli Equation

By assuming that fluid motion is governed only by pressure and

gravity forces, applying Newton’s second law, F = ma, leads us tothe Bernoulli Equation.

P/ g + V 2 /2g + z = constant along a streamline (P=pressure g =specific weight V=velocity g=gravity z=elevation)

A streamline is the path of one particle of water. Therefore, at any twopoints along a streamline, the Bernoulli equation can be applied and,using a set of engineering assumptions, unknown flows and pressurescan easily be solved for.



Bernoulli Example Problem : Free Jets 2

R=1’ R=1’

4’

R=.5’ R=.5’ 2’

γH20=62.4 lbs/ft3

Q? Q?

Case 1 Case 2

A small cylindrical tank is filled with water, and thenemptied through a small orifice at the bottom.

Case 1

What is the flowrate Q exitingthrough the

hole when thetank is full?

Case 2

What is the flowrate Q exiting

through the hole

when the tank ishalf full?

-Hint-

TheContinuityEquation is

needed

Assumptions

Psurf = Pout = 0Because it’s a small tank,

Vsurf ≠ 0



Free Jets 2

R=1’ R=1’

4’

R=.5’ R=.5’ 2’

γH20=62.4 lbs/ft3

Q? Q?

Case 1 Case 2

Case 1

Apply Bernoulli’s Equation at the Surface and at the Outlet:

0 + Vsurf 2 /2g + 4 = 0 + Vout

2 /2g + 0

With two unknowns, we need another equation : The Continuity Equation

A surf V surf =A out V out

p(1)2 x Vsurf = p(.5)2 x Vout V surf =.25V out

Substituting back into theBernoulli Equation

(.25Vout)2 /2g + 4 = Vout2 /2g

Solving for Vout if g = 32.2 ft/s2

V out = .257 ft/s

Qout = AV = .202 ft3/s (cfs)



Bernoulli Example Problem : Free Jets 2

R=1’ R=1’

4’

R=.5’ R=.5’ 2’

γH20=62.4 lbs/ft3

Q? Q?

Case 1 Case 2

Case 2

Bernoulli’s Equation at the Surface and at the Outlet is changed:

0 + Vsurf 2 /2g + 2 = 0 + Vout

2 /2g + 0

Continuity eqn remains the same.

Substituting back into theBernoulli Equation

(.25Vout)2 /2g + 2 = Vout

2 /2g

Solving for Vout if g = 32.2 ft/s2

V out = .182 ft/s

Qout = AV = .143 cfs

Note that velocity is less in Case 2



Free Jets

The velocity of a jet of water is clearly related to the depth of waterabove the hole. The greater the depth, the higher the velocity. Similarbehavior can be seen as water flows at a very high velocity from the

reservoir behind the Glen Canyon Dam in Colorado



The Energy Line and the Hydraulic Grade Line

Looking at the Bernoulli equation again:

P/γ + V2 /2g + z = constant on a streamlineThis constant is called the total head (energy), H

Because energy is assumed to be conserved, at any point alongthe streamline, the total head is always constant

Each term in the Bernoulli equation is a type of head.

P/γ = Pressure Head

V2 /2g = Velocity Head

Z = elevation head

These three heads, summed together, will always equal H

Next we will look at this graphically…



The Energy Line and the Hydraulic Grade LineLets first understand this drawing:

Q

Measures theStatic Pressure

Measures theTotal Head

1 2

Z

P/γ

V 2/2gEL

HGL

12

1: Static Pressure Tap

Measures the sum of theelevation head and the

pressure Head.

2: Pilot Tube

Measures the Total Head

EL : Energy Line

Total Head along a systemHGL : Hydraulic Grade line

Sum of the elevation andthe pressure heads along a

system



The Energy Line and the Hydraulic Grade Line

Q

Z

P/γ

V 2/2g

EL

HGL

Understanding the graphical approach of

Energy Line and the Hydraulic Grade line iskey to understanding what forces aresupplying the energy that water holds.

V 2/2g

P/γ

Z

1

2

Point 1:

Majority of energystored in the water is in

the Pressure Head

Point 2:Majority of energy

stored in the water is inthe elevation head

If the tube wassymmetrical, then the

velocity would beconstant, and the HGL

would be level



The Complete ExampleSolve for the Pressure Head, Velocity Head, and Elevation Head at each

point, and then plot the Energy Line and the Hydraulic Grade Line

1

23 4

1’

4’

γH2O= 62.4 lbs/ft3

Assumptions and Hints:

P1 and P4 = 0 --- V3 = V4 same diameter tube

We must work backwards to solve this problem

R = .5’

R = .25’



1

23 4

1’

4’

γH2O= 62.4 lbs/ft3

Point 1:

Pressure Head : Only atmospheric P1/γ = 0 Velocity Head : In a large tank, V 1 = 0 V 1

2/2g = 0

Elevation Head : Z1 = 4’

R = .5’ R = .25’



1

23 4

1’

4’

γH2O= 62.4 lbs/ft3

Point 4:

Apply the Bernoulli equation between 1 and 40 + 0 + 4 = 0 + V 4

2/2(32.2) + 1

V 4 = 13.9 ft/s

Pressure Head : Only atmospheric P4/γ = 0

Velocity Head : V 42/2g = 3’


R = .5’ R = .25’



1

23 4

1’

4’

γH2O= 62.4 lbs/ft3

Point 3:

Apply the Bernoulli equation between 3 and 4 (V 3=V 4)P3/62.4 + 3 + 1 = 0 + 3 + 1

P3 = 0

Pressure Head : P3/γ = 0

Velocity Head : V 32/2g = 3’


R = .5’ R = .25’



1

23 4

1’

4’

γH2O= 62.4 lbs/ft3

Point 2:

Apply the Bernoulli equation between 2 and 3P2 /62.4 + V2

2 /2(32.2) + 1 = 0 + 3 + 1

Apply the Continuity Equation

(Π.52)V2 = (Π.252)x13.9 V2 = 3.475 ft/s

P2 /62.4 + 3.4752 /2(32.2) + 1 = 4 P2 = 175.5 lbs/ft2

R = .5’ R = .25’

Pressure Head :P2/γ = 2.81’

Velocity Head : V 2

2/2g = .19’

Elevation Head :Z2 = 1’



Plotting the EL and HGLEnergy Line = Sum of the Pressure, Velocity and Elevation heads

Hydraulic Grade Line = Sum of the Pressure and Velocity heads

EL

HGL

Z=1’ Z=1’ Z=1’

V2 /2g=3’ V2 /2g=3’

Z=4’

P/γ =2.81’

V2 /2g=.19’



Pipe Flow and the Energy EquationFor pipe flow, the Bernoulli equation alone is not sufficient. Friction loss

along the pipe, and momentum loss through diameter changes andcorners take head (energy) out of a system that theoretically conserves

energy. Therefore, to correctly calculate the flow and pressures in pipesystems, the Bernoulli Equation must be modified.

P1/γ + V 12/2g + z1 = P2/γ + V 2

2/2g + z2 + Hmaj + Hmin

Major losses: Hmaj Major losses occur over the entire pipe, as the friction of the fluid overthe pipe walls removes energy from the system. Each type of pipe as a

friction factor, f, associated with it.

Hmaj Energy line with no losses

Energy line with major losses

1 2



Pipe Flow and the Energy Equation

Minor Losses : Hmin

Momentum losses in Pipe diameter changes and in pipe bends are calledminor losses. Unlike major losses, minor losses do not occur over thelength of the pipe, but only at points of momentum loss. Since Minorlosses occur at unique points along a pipe, to find the total minor lossthroughout a pipe, sum all of the minor losses along the pipe. Each

type of bend, or narrowing has a loss coefficient, K L to go with it.

MinorLosses



Major and Minor Losses

Major Losses:

Hmaj = f x (L/D)(V2/2g)

f = friction factor L = pipe length D = pipe diameter V = Velocity g = gravity

Minor Losses:Hmin = K L(V 2/2g)

K l = sum of loss coefficients V = Velocity g = gravity

When solving problems, the loss terms are added to the system at thesecond point

P1/γ + V 12/2g + z1 = P2/γ + V 2

2/2g + z2 + Hmaj + Hmin



Loss Coefficients

Use this table to find loss coefficients:

i l l



Pipe Flow Example

1

2Z2 = 130 m

130 m

7 m

60 m

r/D = 2

Z1

= ?γoil= 8.82 kN/m3

f = .035

If oil flows from the upper to lower reservoir at a velocity of 1.58 m/s in the 15 cm diameter smooth pipe, what is the

elevation of the oil surface in the upper reservoir?

Include major losses along the pipe, and the minor lossesassociated with the entrance, the two bends, and the outlet.

K out=1

r/D = 0

i l l



Pipe Flow Example

1

2Z2 = 130 m

130 m

7 m

60 m

r/D = 2

Z1

= ?γoil= 8.82 kN/m3

f = .035

K out=1

r/D = 0

Apply Bernoulli’s equation between points 1 and 2: Assumptions: P1 = P2 = Atmospheric = 0 V1 = V2 = 0 (large tank)

0 + 0 + Z1 = 0 + 0 + 130m + Hmaj + Hmin

Hmaj = (fxLxV2) /(Dx2g)=(.035 x 197m x (1.58m/s)2)/(.15 x 2 x 9.8m/s2)

Hmaj= 5.85m

Pi Fl E l



Pipe Flow Example

1

2Z2 = 130 m

130 m

7 m

60 m

r/D = 2

Z1

= ?γoil= 8.82 kN/m3

f = .035

K out=1

r/D = 0

0 + 0 + Z1 = 0 + 0 + 130m + 5.85m + Hmin

Hmin= 2K bend V2 /2g + K ent V

2 /2g + K out V2 /2g

From Loss Coefficient table: K bend = 0.19 K ent = 0.5 K out = 1

Hmin = (0.19x2 + 0.5 + 1) x (1.582 /2x9.8)

Hmin = 0.24 m

Pi Fl E l



Pipe Flow Example

1

2Z2 = 130 m

130 m

7 m

60 m

r/D = 2

Z1

= ?γoil= 8.82 kN/m3

f = .035

K out=1

r/D = 0

0 + 0 + Z1 = 0 + 0 + 130m + Hmaj + Hmin

0 + 0 + Z1 = 0 + 0 + 130m + 5.85m + 0.24m

Z1 = 136.09 meters

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Fluid Mechanics Benno 7