Fluid Mechanics Benno 7

TL2101 Mekanika Fluida I Benno Rahardyan Pertemuan 7

Mg 1

Topik

Sub Topik

Tujuan Instruksional (TIK)

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, tekanan hidrostatik

Mengerti prinsip-2 tekanan 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 Mengerti, dapat energi dalam aliran fluida, menggunakan dan penerapan persamaan Bernoulli menghitung sistem prinsi dalam perpipaan hukum kontinuitas

4

Idem

Idem + gaya pada bidang terendam Idem

5

Aplikasi kekekalan energi

Aplikasi kekekalan energi dalam aplikasi di bidang TL

Aplikasi kekekalan energi

Darcy-Weisbach, headloss, major losses dan minor losses Keseimbangan bidang terapung

Latihan menggunakan prinsip kekekalan eneri khususnya dalam bidang air minum

FLUID DYNAMICS THE BERNOULLI EQUATION

The laws of Statics that we have learned cannot solve Dynamic Problems. There is no way to solve for the flow rate, 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 to the Bernoulli Equation.

P/γ + V2/2g + z = constant along a streamline (P=pressure γ =specific weight V=velocity g=gravity z=elevation)

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

Bernoulli Example Problem : Free Jets 2 A small cylindrical tank is filled with water, and then emptied through a small orifice at the bottom. Case 1 Case 2 -HintWhat is the What is the The flow rate Q flow rate Q Continuity exiting exiting through Equation is through the the hole when needed hole when the the tank is half tank full? R=1’ is full? R=1’ Assumptions Psurf = Pout = 0

γH20 =62.4 lbs/ft3

4 ’

R=.5’

Case 1

Q ?

2 ’

R=.5’

Case 2

Q ?

Because it’s a small tank, Vsurf ≠ 0

Free Jets 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 Asurf Vsurf =Aout Vout

π (1)2 x Vsurf = π (.5)2 x Vout  Vsurf =.25Vout R=1’

R=1’

Substituting back into the Bernoulli Equation 

γH20 =62.4 lbs/ft3

4 ’

R=.5’

Case

Q ?

(.25Vout )2/2g + 4 = Vout 2/2g 2 ’

R=.5’

Case 2

Q ?

Solving for Vout if g = 32.2 ft/s2 Vout = .257 ft/s Qout = AV = .202 ft3/s (cfs)

Bernoulli Example Problem : Free Jets2 2 Case 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 the Bernoulli Equation  R=1’

R=1’

(.25Vout )2/2g + 2 = Vout 2/2g γH20 =62.4 lbs/ft3

4 ’

R=.5’

Case

Q ?

Solving for Vout if g = 32.2 ft/s2 2 ’

R=.5’

Case 2

Q ?

Vout = .182 ft/s Qout = AV = .143 cfs Note that velocity is less in

Free Jets

The velocity of a jet of water is clearly related to the depth of water above the hole. The greater the depth, the higher the velocity. Similar behavior 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 equation again: Looking at the Bernoulli P/γ + V2/2g + z = constant on a streamline This constant is called the total head (energy), H Because energy is assumed to be conserved, at any point along the 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 Lets first understand Line this drawing: Measures the Static Pressure

Measures the Total Head

12

12

V2/2g

EL HGL

Q

P/γ

1: Static Pressure Tap Measures the sum of the elevation head and the pressure Head. 2: Pilot Tube Measures the Total Head EL : Energy Line Total Head along a system

Z

HGL : Hydraulic Grade line

The Energy Line and the Hydraulic Grade Line Understanding the graphical approach of Energy Line and the Hydraulic Grade line is key to understanding what forces are supplying the energy that water holds. V2/2g

EL

Q

P/γ

V2/2g

HGL

2

P/γ

Z

1

Z

Point 1:

Majority of energy stored in the water is in the Pressure Head Point 2: Majority of energy stored in the water is in the elevation head If the tube was symmetrical, then the velocity would be constant, and the HGL would be level

The Complete Example Solve for the Pressure Head, Velocity Head, and Elevation Head at each point, and then plot the Energy Line and the Hydraulic Grade Line Assumptions and Hints: P1 and P4 = 0 --- V3 = V4 same diameter tube We must work backwards to solve this problem

1 γH2O = 62.4 lbs/ft3

4’

R=. 5’

2

R = .25’

3

4 1’

Point 1: Pressure Head : Only atmospheric  P1/γ = 0 Velocity Head : In a large tank, V1 = 0  V12/2g = 0 Elevation Head : Z1 = 4’

1 4 ’

γH2O = 62.4 lbs/ft3 R=. R = .25’ 5’

2

3

4 1’

Point 4: Apply the Bernoulli equation between 1 and 4 0 + 0 + 4 = 0 + V42/2(32.2) + 1 V4 = 13.9 ft/s Pressure Head : Only atmospheric  P4/γ = 0 Velocity Head : V42/2g = 3’ Elevation Head : Z4 = 1’

1 4 ’

γH2O = 62.4 lbs/ft3 R=. R = .25’ 5’

2

3

4 1’

Point 3: Apply the Bernoulli equation between 3 and 4 (V3=V4) P3/62.4 + 3 + 1 = 0 + 3 + 1 P3 = 0 Pressure Head : P3/γ = 0 Velocity Head : V32/2g = 3’ Elevation Head : Z3 = 1’

1 4 ’

γH2O = 62.4 lbs/ft3 R=. R = .25’ 5’

2

3

4 1’

Point 2: Apply the Bernoulli equation between 2 and 3 P2/62.4 + V22/2(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 Pressure Head : P2/γ = 2.81’

1 4 ’

γH2O = 62.4 lbs/ft3 R=. R = .25’ 5’

2

3

Velocity Head : V22/2g = .19’

4 1’

Elevation Head : Z2 = 1’

Plotting the EL and HGL Energy Line = Sum of the Pressure, Velocity and Elevation heads Hydraulic Grade Line = Sum of the Pressure and Velocity heads V2/2g=.19’

EL P/γ =2.81’

V2/2g=3’ V2/2g=3’

Z=4’

HGL Z=1’

Z=1’

Z=1’

Pipe Flow and the Energy For pipe flow, the Bernoulli equation alone is not sufficient. Equation Friction loss along the pipe, and momentum loss through diameter changes and corners take head (energy) out of a system that theoretically conserves energy. Therefore, to correctly calculate the flow and pressures in pipe systems, the Bernoulli Equation must be modified.

P1/γ + V12/2g + z1 = P2/γ + V22/2g + z2 + Hmaj + Hmin Major losses: Hmaj Major losses occur over the entire pipe, as the friction of the fluid over the pipe walls removes energy from the system. Each Energy linefactor, with no f, associated with it. type of pipe as a friction losses

Hmaj

Energy line with major losses 1

2

Pipe Flow and the Energy Equation Minor Losses : H min

Momentum losses in Pipe diameter changes and in pipe bends are called minor losses. Unlike major losses, minor losses do not occur over the length of the pipe, but only at points of momentum loss. Since Minor losses occur at unique points along a pipe, to find the total minor loss throughout a pipe, sum all of the minor losses along the pipe. Each type of bend, or narrowing has a loss coefficient, KL to go with it.

Minor Losses

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 = KL(V2/2g) Kl = sum of loss coefficients V = Velocity g = gravity When solving problems, the loss terms are added to the system at the second point

P1/γ + V12/2g + z1 = P2/γ + V22/2g + z2 + Hmaj + Hmin

Loss Coefficients

Use this table to find loss coefficients:

Pipe Flow Example 1

Z1 = ?

γoil = 8.82 kN/m3 f = .035

60 m 7m r/D = 0

Z2 = 130 m Kout =1

2

130 m r/D = 2

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 losses associated with the entrance, the two bends, and the outlet.

Pipe Flow Example 1

γoil = 8.82 kN/m3

Z1 = ?

f = .035

60 m 7m r/D = 0

Z2 = 130 m Kout =1

2

130 m r/D = 2

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) H

= 5.85m

Pipe Flow Example 1

Z1 = ?

γoil = 8.82 kN/m3 f = .035

60 m 7m r/D = 0

Z2 = 130 m Kout =1

2

130 m r/D = 2

0 + 0 + Z1 = 0 + 0 + 130m + 5.85m + Hmin Hmin = 2Kbend V2/2g + Kent V2/2g + Kout V2/2g From Loss Coefficient table: Kbend = 0.19 Kent = 0.5 Hmin = (0.19x2 + 0.5 + 1) x (1.582/2x9.8) Hmin = 0.24 m

Kout = 1

Pipe Flow Example 1

Z1 = ?

γoil = 8.82 kN/m3 f = .035

60 m 7m r/D = 0

Z2 = 130 m Kout =1

2

130 m r/D = 2

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

0 + 0 + Z1 = 0 + 0 + 130m + 5.85m + 0.24m Z1 = 136.09 meters