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Free Powerpoint Templates Page 1 KULIAH VIII - KULIAH VIII - IX IX MEKANIKA FLUIDA II MEKANIKA FLUIDA II Nazaruddin Sinaga Nazaruddin Sinaga

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KULIAH VIII - IX. MEKANIKA FLUIDA II Nazaruddin Sinaga. Entrance Length. Shear stress and velocity distribution in pipe for laminar flow. Typical velocity and shear distributions in turbulent flow near a wall: (a) shear; (b) velocity. Solution of Pipe Flow Problems. Single Path - PowerPoint PPT Presentation

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Page 1: KULIAH VIII - IX

Free Powerpoint TemplatesPage 1

KULIAH VIII - IXKULIAH VIII - IX

MEKANIKA FLUIDA IIMEKANIKA FLUIDA II

Nazaruddin SinagaNazaruddin Sinaga

Page 2: KULIAH VIII - IX

Entrance LengthEntrance Length

2

Page 3: KULIAH VIII - IX

Shear stress and velocity distribution in pipe for laminar flow

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Typical velocity and shear distributions in turbulent flow near a wall: (a) shear; (b) velocity.

4

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Solution of Pipe Flow ProblemsSolution of Pipe Flow Problems

• Single Path– Find p for a given L, D, and Q

Use energy equation directly

– Find L for a given p, D, and Q Use energy equation directly

Page 6: KULIAH VIII - IX

Solution of Pipe Flow ProblemsSolution of Pipe Flow Problems

• Single Path (Continued)– Find Q for a given p, L, and D

1. Manually iterate energy equation and friction factor formula to find V (or Q), or

2. Directly solve, simultaneously, energy equation and friction factor formula using (for example) Excel

– Find D for a given p, L, and Q1. Manually iterate energy equation and friction factor

formula to find D, or2. Directly solve, simultaneously, energy equation and

friction factor formula using (for example) Excel

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7

Example 1Example 1 Water at 10C is flowing at a rate of 0.03 m3/s through a pipe. The

pipe has 150-mm diameter, 500 m long, and the surface roughness is estimated at 0.06 mm. Find the head loss and the pressure drop throughout the length of the pipe.

Solution: From Table 1.3 (for water): = 1000 kg/m3 and =1.30x10-3 N.s/m2

V = Q/A and A=R2

A = (0.15/2)2 = 0.01767 m2

V = Q/A =0.03/.0.01767 =1.7 m/sRe = (1000x1.7x0.15)/(1.30x10-3) = 1.96x105 > 2000 turbulent

flowTo find , use Moody Diagram with Re and relative roughness (k/D).

k/D = 0.06x10-3/0.15 = 4x10-4

From Moody diagram, 0.018The head loss may be computed using the Darcy-Weisbach equation.

The pressure drop along the pipe can be calculated using the relationship: ΔP=ghf = 1000 x 9.81 x 8.84ΔP = 8.67 x 104 Pa

.m84.881.9x2x15.0

7.1x500x018.0

g2

V

D

Lh

22

f

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8

Example 2Example 2 Determine the energy loss that will occur as 0.06 m3/s water

flows from a 40-mm pipe diameter into a 100-mm pipe diameter through a sudden expansion.

Solution: The head loss through a sudden enlargement is given by;

Da/Db = 40/100 = 0.4From Table 6.3: K = 0.70Thus, the head loss is

g2

VKh

2a

m

smA

QV

aa /58.3

)2/04.0(

06.02

m47.081.9x2

58.3x70.0h

2

Lm

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9

ExExample 3ample 3

Calculate the head added by the pump when the water system shown below carries a discharge of 0.27 m3/s. If the efficiency of the pump is 80%, calculate the power input required by the pump to maintain the flow.

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Solution:Applying Bernoulli equation between section 1 and 2

(1)

P1 = P2 = Patm = 0 (atm) and V1=V2 0 Thus equation (1) reduces to:

(2)

HL1-2 = hf + hentrance + hbend + hexit

From (2):

21L

22

22

p

21

11 H

g2

Vz

g

PH

g2

Vz

g

P

21L12p HzzH

g2

V4.39

14.05.04.0

1000x015.0

g2

VH

2

2

21L

81.9x2

V4.39200230H

2

p

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The velocity can be calculated using the continuity equation:

Thus, the head added by the pump: Hp = 39.3 m

Pin = 130.117 Watt ≈ 130 kW.

s/m15.2

2/4.0

27.0

A

QV

2

in

pp P

gQH

8.0

3.39x27.0x81.9x1000gQHP

p

pin

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EGL & HGL for a Pipe System

• Energy equation

• All terms are in dimension of length (head, or energy per unit weight)

• HGL – Hydraulic Grade Line

• EGL – Energy Grade Line

• EGL=HGL when V=0 (reservoir surface, etc.)

• EGL slopes in the direction of flow

22

22

211

21

1 22z

p

g

Vhz

p

g

VL

zp

HGL

g

VHGLz

p

g

VEGL

22

22

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EGL & HGL for a Pipe System

• A pump causes an abrupt rise in EGL (and HGL) since energy is introduced here

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EGL & HGL for a Pipe System

• A turbine causes an abrupt drop in EGL (and HGL) as energy is taken out

• Gradual expansion increases turbine efficiency

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EGL & HGL for a Pipe System

• When the flow passage changes diameter, the velocity changes so that the distance between the EGL and HGL changes

• When the pressure becomes 0, the HGL coincides with the system

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EGL & HGL for a Pipe System

• Abrupt expansion into reservoir causes a complete loss of kinetic energy there

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EGL & HGL for a Pipe System

• When HGL falls below the pipe the pressure is below atmospheric pressure

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FLOW MEASUREMENTFLOW MEASUREMENT• Direct Methods

– Examples: Accumulation in a Container; Positive Displacement Flowmeter

• Restriction Flow Meters for Internal Flows– Examples: Orifice Plate; Flow Nozzle; Venturi; Laminar

Flow Element

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Definisi tekanan pada aliran di sekitar sayap

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Flow Measurement

• Linear Flow Meters– Examples: Float Meter

(Rotameter); Turbine; Vortex; Electromagnetic; Magnetic; Ultrasonic

Float-type variable-area flow meter

Page 25: KULIAH VIII - IX

Flow Measurement• Linear Flow Meters

– Examples: Float Meter (Rotameter); Turbine; Vortex; Electromagnetic; Magnetic; Ultrasonic

Turbine flow meter

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Flow Measurement

• Traversing Methods– Examples: Pitot (or Pitot Static) Tube; Laser Doppler

Anemometer

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The measured stagnation pressure cannot of itself be used to determine the fluid velocity (airspeed in aviation).

However, Bernoulli's equation states:

Stagnation pressure = static pressure + dynamic pressure

Which can also be written

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Solving that for velocity we get:

Note: The above equation applies only to incompressible fluid.where:

V is fluid velocity;pt is stagnation or total pressure;ps is static pressure;and ρ is fluid density.

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The value for the pressure drop p2 – p1 or Δp to Δh, the reading on the manometer:

Δp = Δh(ρA-ρ)g

Where:ρA is the density of the fluid in the manometerΔh is the manometer reading

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EXTERNAL EXTERNAL INCOMPRESSIBLE INCOMPRESSIBLE VISCOUS FLOWVISCOUS FLOW

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Main TopicsMain Topics• The Boundary-Layer Concept• Boundary-Layer Thickness• Laminar Flat-Plate Boundary Layer: Exact Solution• Momentum Integral Equation• Use of the Momentum Equation for Flow with Zero

Pressure Gradient• Pressure Gradients in Boundary-Layer Flow• Drag• Lift

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The Boundary-Layer Concept

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The Boundary-Layer Concept

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Boundary Layer Thickness

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Boundary Layer Thickness

• Disturbance Thickness, where

Displacement Thickness, *

Momentum Thickness,

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Boundary Layer LawsBoundary Layer Laws

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Laminar Flat-PlateBoundary Layer: Exact Solution

• Governing Equations

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Laminar Flat-PlateBoundary Layer: Exact Solution

• Boundary Conditions

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Laminar Flat-PlateBoundary Layer: Exact Solution

• Equations are Coupled, Nonlinear, Partial Differential Equations

• Blassius Solution:– Transform to single, higher-order, nonlinear, ordinary

differential equation

Page 49: KULIAH VIII - IX

Laminar Flat-PlateBoundary Layer: Exact Solution

• Results of Numerical Analysis

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Momentum Integral Equation

• Provides Approximate Alternative to Exact (Blassius) Solution

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Momentum Integral Equation

Equation is used to estimate the boundary-layer thickness as a function of x:

1. Obtain a first approximation to the freestream velocity distribution, U(x). The pressure in the boundary layer is related to the freestream velocity, U(x), using the Bernoulli equation

2. Assume a reasonable velocity-profile shape inside the boundary layer

3. Derive an expression for w using the results obtained from item 2

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Use of the Momentum Equation for Flow with Zero Pressure Gradient

• Simplify Momentum Integral Equation(Item 1)

The Momentum Integral Equation becomes

Page 53: KULIAH VIII - IX

Use of the Momentum Equation for Flow with Zero Pressure Gradient

• Laminar Flow– Example: Assume a Polynomial Velocity Profile (Item 2)

• The wall shear stress w is then (Item 3)

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Use of the Momentum Equation for Flow with Zero Pressure Gradient

• Laminar Flow Results(Polynomial Velocity Profile)

Compare to Exact (Blassius) results!

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Use of the Momentum Equation for Flow with Zero Pressure Gradient

• Turbulent Flow– Example: 1/7-Power Law Profile (Item 2)

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Use of the Momentum Equation for Flow with Zero Pressure Gradient

• Turbulent Flow Results(1/7-Power Law Profile)

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Pressure Gradients in Boundary-Layer Flow

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Drag

• Drag Coefficient

with

or

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Drag

• Pure Friction Drag: Flat Plate Parallel to the Flow

• Pure Pressure Drag: Flat Plate Perpendicular to the Flow

• Friction and Pressure Drag: Flow over a Sphere and Cylinder

• Streamlining

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Drag• Flow over a Flat Plate Parallel to the Flow: Friction

Drag

Boundary Layer can be 100% laminar, partly laminar and partly turbulent, or essentially 100% turbulent; hence several different drag coefficients are available

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Drag• Flow over a Flat Plate Parallel to the Flow: Friction

Drag (Continued)

Laminar BL:

Turbulent BL:

… plus others for transitional flow

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Drag Coefficient

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Drag• Flow over a Flat Plate Perpendicular to the

Flow: Pressure Drag

Drag coefficients are usually obtained empirically

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Drag• Flow over a Flat Plate Perpendicular to the

Flow: Pressure Drag (Continued)

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Drag• Flow over a Sphere and Cylinder: Friction and

Pressure Drag

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Drag• Flow over a Sphere and Cylinder: Friction and

Pressure Drag (Continued)

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Streamlining• Used to Reduce Wake and Pressure Drag

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Lift• Mostly applies to Airfoils

Note: Based on planform area Ap

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Lift

• Examples: NACA 23015; NACA 662-215

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Lift• Induced Drag

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Lift• Induced Drag (Continued)

Reduction in Effective Angle of Attack:

Finite Wing Drag Coefficient:

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Lift

• Induced Drag (Continued)

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