Used to estimate energy loss due to friction in pipe. D = internal diameter of pipe (feet) L = length of pipe (feet) Penn State-Harrisburg

Similar documents
HEADLOSS ESTIMATION. Mekanika Fluida 1 HST

When water (fluid) flows in a pipe, for example from point A to point B, pressure drop will occur due to the energy losses (major and minor losses).

Hydraulic validation of the LHC cold mass heat exchanger tube.

Slopes of Secant and!angent (ines - 2omework

Chapter 3 Water Flow in Pipes

Example 1. Examples for walls are available on our Web page: Columns

Reynolds, an engineering professor in early 1880 demonstrated two different types of flow through an experiment:

ENTROPY GENERATION IN RECTANGULAR DUCTS WITH NONUNIFORM TEMPERATURE ON THE CONTOUR

LECTURE 6- ENERGY LOSSES IN HYDRAULIC SYSTEMS SELF EVALUATION QUESTIONS AND ANSWERS

OE4625 Dredge Pumps and Slurry Transport. Vaclav Matousek October 13, 2004

Section 2.1 The Definition of the Derivative. We are interested in finding the slope of the tangent line at a specific point.

An Improved Expression for a Classical Type of Explicit Approximation of the Colebrook White Equation with Only One Internal Iteration

. Compute the following limits.

Taylor Series and the Mean Value Theorem of Derivatives

Midterm #1B. x 8 < < x 8 < 11 3 < x < x > x < 5 or 3 2x > 5 2x < 8 2x > 2

Pipe Flow/Friction Factor Calculations using Excel Spreadsheets

FLOW FRICTION CHARACTERISTICS OF CONCRETE PRESSURE PIPE

Simple Iterative Model for Adjusting Hazen-Williams Friction Coefficient for Drip Irrigation laterals.

MATH CALCULUS I 2.1: Derivatives and Rates of Change

Hydraulics and hydrology

Continuity and Differentiability

RESOLUTION MSC.362(92) (Adopted on 14 June 2013) REVISED RECOMMENDATION ON A STANDARD METHOD FOR EVALUATING CROSS-FLOODING ARRANGEMENTS

Engineers Edge, LLC PDH & Professional Training

Chapter 4: Numerical Methods for Common Mathematical Problems

= h. Geometrically this quantity represents the slope of the secant line connecting the points

Open Channel Hydraulics I - Uniform Flow

LQ 128(0.0012)(1.2 m)q πρgd π(789)(9.81)(0.002) Solve for Q 1.90E 6 m /s = m /h. Ans.

Water Circuit Lab. The pressure drop along a straight pipe segment can be calculated using the following set of equations:

UNIVERSITY OF SASKATCHEWAN Department of Physics and Engineering Physics

1. Which one of the following expressions is not equal to all the others? 1 C. 1 D. 25x. 2. Simplify this expression as much as possible.

Conductance from Transmission Probability

Chapter 2 Limits and Continuity. Section 2.1 Rates of Change and Limits (pp ) Section Quick Review 2.1

Honors Calculus Midterm Review Packet

Laminar and turbulent flows

5.1 The derivative or the gradient of a curve. Definition and finding the gradient from first principles

5. (a) Find the slope of the tangent line to the parabola y = x + 2x

12d Model. Civil and Surveying Software. Version 7. Drainage Analysis Module Hydraulics. Owen Thornton BE (Mech), 12d Model Programmer

Fluids and Buoyancy. 1. What will happen to the scale reading as the mass is lowered?

FLUID MECHANICS D203 SAE SOLUTIONS TUTORIAL 2 APPLICATIONS OF BERNOULLI SELF ASSESSMENT EXERCISE 1

ME 305 Fluid Mechanics I. Part 8 Viscous Flow in Pipes and Ducts. Flow in Pipes and Ducts. Flow in Pipes and Ducts (cont d)

FLUID FLOW. Note that the balance is per unit mass. In differential form:

Outline. MS121: IT Mathematics. Limits & Continuity Rates of Change & Tangents. Is there a limit to how fast a man can run?

Sedimentation. Treatment Processes. Sedimentation. Sedimentation. Sedimentation. Sedimentation. CIVL 1112 Water Treatment - Sedimentation 1/7

Pipe Flow. Lecture 17

Practice Problem Solutions: Exam 1

UNIT #6 EXPONENTS, EXPONENTS, AND MORE EXPONENTS REVIEW QUESTIONS

Chemical Engineering 374

Solving Continuous Linear Least-Squares Problems by Iterated Projection

ME 305 Fluid Mechanics I. Chapter 8 Viscous Flow in Pipes and Ducts

FACULTY OF CHEMICAL & ENERGY ENGINEERING FLUID MECHANICS LABORATORY TITLE OF EXPERIMENT: MINOR LOSSES IN PIPE (E4)

DEFINITION OF A DERIVATIVE

Transport Properties: Momentum Transport, Viscosity

MAT 1800 FINAL EXAM HOMEWORK

Section 15.6 Directional Derivatives and the Gradient Vector

12d Model. Civil and Surveying Software. Drainage Analysis Module Hydraulics. Owen Thornton BE (Mech), 12d Model Programmer.

Chapter 8 Flow in Pipes. Piping Systems and Pump Selection

Chapter 10 Flow in Conduits

ACE Engineering Academy. Hyderabad Delhi Bhopal Pune Bhubaneswar Lucknow Patna Bengaluru Chennai Vijayawada Vizag Tirupati Kukatpally Kolkata

Open Channel Flow I - The Manning Equation and Uniform Flow COURSE CONTENT

Major and Minor Losses

2, where dp is the constant, R is the radius of

Part 2: Introduction to Open-Channel Flow SPRING 2005

Ballistic Atmospheric Entry

Study fluid dynamics. Understanding Bernoulli s Equation.

Limits and Derivatives

Chapter 7 FLOW THROUGH PIPES

CEE 3310 Open Channel Flow,, Nov. 18,

5.1 We will begin this section with the definition of a rational expression. We

Chapter 8: Flow in Pipes

Path to static failure of machine components

Convective Mass Transfer

Chapter 6. Losses due to Fluid Friction

Introduction to Derivatives

1 2 x Solution. The function f x is only defined when x 0, so we will assume that x 0 for the remainder of the solution. f x. f x h f x.

SOIL MECHANICS

Piping Systems and Flow Analysis (Chapter 3)

Review for Exam IV MATH 1113 sections 51 & 52 Fall 2018

Krazy Katt, the mechanical cat

Problem Solving. Problem Solving Process

Prediction of Coating Thickness

The derivative function

A Model Answer for. Problem Set #7

1watt=1W=1kg m 2 /s 3

Key Concepts. Important Techniques. 1. Average rate of change slope of a secant line. You will need two points ( a, the formula: to find value

Chezy s Resistance Coefficient in a Circular Conduit

FE Exam Fluids Review October 23, Important Concepts

Lecture 4. Lab this week: Cartridge valves Flow divider Properties of Hydraulic Fluids. Lab 8 Sequencing circuit Lab 9 Flow divider

Journal of Applied Science and Agriculture. The Effects Of Corrugated Geometry On Flow And Heat Transfer In Corrugated Channel Using Nanofluid

Homework 1. Problem 1 Browse the 331 website to answer: When you should use data symbols on a graph. (Hint check out lab reports...

Computer Applications in Hydraulic Engineering

ACCOUNTING FOR FRICTION IN THE BERNOULLI EQUATION FOR FLOW THROUGH PIPES

Series Sytem of Pipes Parallel Sytem of Pipes

INTRODUCTION DEFINITION OF FLUID. U p F FLUID IS A SUBSTANCE THAT CAN NOT SUPPORT SHEAR FORCES OF ANY MAGNITUDE WITHOUT CONTINUOUS DEFORMATION

CONVECTIVE HEAT TRANSFER CHARACTERISTICS OF NANOFLUIDS. Convective heat transfer analysis of nanofluid flowing inside a

FLUID THEORY RELATING TO WATER

ρg 998(9.81) LV 50 V. d2g 0.062(9.81)

FLOW CHARACTERISTICS OF HFC-134a IN AN ADIABATIC HELICAL CAPILLARY TUBE

Bernoulli and Pipe Flow

MTH-112 Quiz 1 Name: # :

MVT and Rolle s Theorem

Transcription:

Module b: Flow in Pipes Darcy-Weisbac Robert Pitt University o Alabama and Sirley Clark Penn State-Harrisburg Darcy-Weisbac can be written or low (substitute V Q/A, were A (π/4)d in te above equation): 8 π D 5 g Q Darcy-Weisbac can be rewritten to solve or velocity: V g D Based upon teory. Used to estimate energy loss due to riction in pipe. V D g Were ead loss (eet) lengt o pipe (eet) riction actor D internal diameter o pipe (eet) V /g velocity ead (eet) Friction actor or : Based upon te ynolds number, N R or, and a dimensionless parameter called te relative rougness, e/d or ε/d (ε absolute rougness; d diameter). For laminar low: 64 N R 64 For turbulent low, riction actor must be read o a Moody diagram (or o a relative rougness vs. riction actor diagram or completely turbulent low). 1

V D g To use Darcy-Weisbac to calculate ead loss, need : A polymeric coagulant, undiluted, as an absolute viscosity o 0.48 kg/(m-sec) [0.01 lb-sec/t] and a speciic gravity o 1.15. Tis luid is to be pumped at te rate o.78 /min (1 gallon/min) troug 15.5 m (50 t) o ½-inc diameter scedule 40 pipe (ID 0.6 in 15.8 mm 0.0158 m). Wat is te ead loss due to riction? Example: Equivalent sand rougness diagram [(.78 /min)(1 m /1000 )(1 min/60 sec)] [(π / 4)(0.0158 m) ] V 0.1m / sec or V 0.m / sec V I te low is laminar, te riction actor can be calculated. Oterwise, it must be looked up o te cart. Need to determine velocity in order to calculate ynolds number (determine i low is laminar or turbulent). By Continuity Equation: Q VA or V Q/A Substituting: Moody Diagram

µ ρ Friction Slope Friction Slope: 6.7 m 1.75 175 % 15.5 m (5.7)(15.5 m)(0.1 m/sec) (0.0158 m)[()(9.806 m/sec )] 6.7m Substituting into Darcy-Weisbac equation: (0.1 m/sec)(0.0158 m)(1.15)(1000 kg/m ) (0.48 kg/m sec) 1.15 Deinition o Speciic Gravity Fluid Density/Water Density Density o Water ρho 1000 kg/m Substituting: Calculate te ynolds number to see i low is laminar or turbulent. ε/d 0.000 From te relative rougness diagram (assume similar to aspalted cast iron, read ε/d. A 4-inc class I ductile iron pipe (ID 0.6 m 4.79 in.) 90 m long wit a neat cement lining (aspalted cast iron) carries a low o water at 1.5 m/sec (5.9 t/sec). Wat is te riction loss in te pipe? Example: Darcy-Weisbac Equation 64 64 1.15 5.7 HAVE AMINAR FOW!! Tereore, te riction actor or Darcy-Weisbac is calculated as ollows:

Calculate ynolds number. Assuming tat te luid is water: ν 1.00 x 10-6 m /sec at 0 o C. Find velocity o low. Q 1.5 m / sec V A π ( 0.6m ) 4 V 4.81 m / sec Find te ynolds number..0x 10 ( 4.81m /sec)( 0.6m ) 6 ν ( 1.00x 10 m /sec) 6 Moody Diagram ading rom te Moody Diagram: (ε/d 0.000 &.0 x 10 6 ) 0.014 Substituting into Darcy-Weisbac: (0.014)(90 m)(4.81m /sec) (0.6 m)()(9.806 m /sec.6 m ) 4 Example: Determine te lowrate in a 500-m section o a 1-m diameter commercial steel pipe wen tere is a -m drop in te energy grade line over te section. Given: 500 m D 1 m Commercial steel pipe m Want to use Darcy-Weisbac equation wit low rate. 8 Q 5 π D g

1/ 1/ 4 π π 4 V.79m / sec.147m / sec Q VA V (1.0m ) ν (.79 m / sec)(1.0 m ) 1.19 x10 6 m / sec.40 x10 6 Calculate te ynolds number. (Note tat no temperature is given or te luid. Assume te luid is water and te temperature is 15oC). At 15oC, ν 1.19 x 10-6 m/sec Substituting: Must ceck assumption o ully turbulent low (i.e., was te selected rom te igure acceptable?). Using te continuity equation: (m)π (9.81m / sec )(1m) 5 Q 8(0.0105)(500m) Q.147m / sec Substituting : 8 π gd 5 8 Q π D 5 g π gd 5 Q 8 Q Solve Darcy-Weisbac or low rate: 0.0105 As a irst assumption about te low in te pipe, will assume ully turbulent low. Using Moody diagram or relative rougness in turbulent low: Need to ind te riction actor. Darcy-Weisbac Equation 5

1/ Are we done? (m)π (9.81m / sec )(1m) 5 Q 8(0.0115)(500m) Q.05m / sec π gd 5 Q 8 Substituting : 1/ Substituting back into Darcy-Weisbac or low: ε/d (0.000046 m)/ (1.0 m) 0.000046 In order to use te ull Moody diagram, need te relative rougness. For commercial steel, ε 0.000046 m. Calculate: Darcy-Weisbac Equation (.61m / sec)(1.0m) ν 1.19 x10 6 m / sec.x106 Calculate te ynolds number associated wit tis velocity: Q.05m / sec A π (1.0m) 4 V.61m / sec V To ceck tis low rate, repeat te process. By continuity: Using te ull Moody diagram, or.40 x 106 and ε/d 0.000046: 0.0115 Moody Diagram: Draw imaginary line interpolating at ε/d 0.000046 6

Need to solve or V and A to get Q, te low. Example: A 14-inc scedule 80 pipe (commercial steel) as an inside diameter o 1.5 in (17.5 mm). How muc low can tis pipe carry i te allowable ead loss is.5 m in a lengt o 00 m? Using te ull Moody diagram, or.0 x 106 and ε/d 0.000046: 0.0115 Moody Diagram: Draw imaginary line interpolating at ε/d 0.000046 ε/d 0.0046 mm/ 17.5 mm 0.000014 In order to use te ull Moody diagram, need te relative rougness. For commercial steel, ε 0.000046 m. Calculate: Darcy-Weisbac Equation Are we done? Yes. Friction actors approximately te same between last two iterations. Q.05 m/sec Friction actors approximately te same between last two iterations. Can use value rom previous iteration as Q. 7

er to Moody Diagram 8 For tis ε/d, ranges rom 0.0087 to 0.04, depending on te low (expressed as te ynolds number). Assume 0.01 (in range near low end). I 0.01, ten te ynolds number is approximately 4 x 10 6. Substituting tis into Darcy-Weisbac: [ V V V g D ] [ ] [(.5 m)()(9.806 m /sec )(0.175 m)] [(00 m)(0.01)].0m /sec Are we done? Now need to ceck te riction actor assumption: Calculate ynolds number or V.0 m/sec. ν.0m / sec(0.175m ) 6 1.00x 10 m /sec 6 1.04x 10 er to Moody Diagram From te Moody diagram, te or tis ynolds number is 0.01.

[ ] [ g D ] Calculate ε/d. D 46 cm 0.46 m ε 0.001 t (0.048 m/t) 0.00005 mm Tereore ε/d 0.000 m/0.46 m 0.00066 Example: Determine te ead loss in a 46-cm concrete pipe wit an average velocity o 1.0 m/sec and a lengt o 0 m. Darcy-Weisbac Equation V [(.5 m)()(9.806 m / sec )(0.175 m)] [(00 m)(0.01)] V.01m / sec V calculate using Darcy-Weisbac: ν Yes. ν (1 m / sec)(0.46 m) 1.00 x10 6 m / sec 4.58 x105 Calculate ynolds number: Are we done? π Q VA (.01m / sec ) (0.175 m ) 4 Q 0.8 m / sec COSE ENOUGH, So pipe will carry: (.01 m / sec)(0.175 m) 1.00x10 6 m / sec 0.95 x106 Calculate te ynolds number or V.01 m/sec. 9

er to Moody Diagram 10 From te Moody diagram, te or tis ynolds number is 0.019. Substituting into Darcy-Weisbac: D 0.06 V (0.019) m g 0 0.46 m m (1 m (9.806 / sec) m / sec ) Are we done? Yes.