Energy CEEN 598D: Fluid Mechanics for Hydro Systems. Lindsay Bearup Berthoud Hall 121

Similar documents
Chapter 5: Mass, Bernoulli, and

Chapter 7 The Energy Equation

Where does Bernoulli's Equation come from?

Chapter 5: Mass, Bernoulli, and Energy Equations

Objectives. Conservation of mass principle: Mass Equation The Bernoulli equation Conservation of energy principle: Energy equation

vector H. If O is the point about which moments are desired, the angular moment about O is given:

SYSTEMS VS. CONTROL VOLUMES. Control volume CV (open system): Arbitrary geometric space, surrounded by control surfaces (CS)

where = rate of change of total energy of the system, = rate of heat added to the system, = rate of work done by the system

3.8 The First Law of Thermodynamics and the Energy Equation

Angular momentum equation

Chapter 2: Basic Governing Equations

Lecture 4. Differential Analysis of Fluid Flow Navier-Stockes equation

Thermodynamics ENGR360-MEP112 LECTURE 7

CEE 3310 Control Volume Analysis, Oct. 10, = dt. sys

4.1 LAWS OF MECHANICS - Review

Chapter Four fluid flow mass, energy, Bernoulli and momentum

Lesson 6 Review of fundamentals: Fluid flow

Chapter 6: Momentum Analysis

Conservation of Angular Momentum

Conservation of Momentum using Control Volumes

Section 2: Lecture 1 Integral Form of the Conservation Equations for Compressible Flow

ME 3560 Fluid Mechanics

MASS, MOMENTUM, AND ENERGY EQUATIONS

Shell/Integral Balances (SIB)

MOMENTUM PRINCIPLE. Review: Last time, we derived the Reynolds Transport Theorem: Chapter 6. where B is any extensive property (proportional to mass),

Chapter 7 Energy Principle

CEE 3310 Control Volume Analysis, Oct. 7, D Steady State Head Form of the Energy Equation P. P 2g + z h f + h p h s.

MAE 224 Notes #4a Elements of Thermodynamics and Fluid Mechanics

ENERGY TRANSFER BETWEEN FLUID AND ROTOR. Dr. Ir. Harinaldi, M.Eng Mechanical Engineering Department Faculty of Engineering University of Indonesia

Chapter 4 Continuity Equation and Reynolds Transport Theorem

In this section, mathematical description of the motion of fluid elements moving in a flow field is

ROAD MAP... D-0: Reynolds Transport Theorem D-1: Conservation of Mass D-2: Conservation of Momentum D-3: Conservation of Energy

Applied Thermodynamics for Marine Systems Prof. P. K. Das Department of Mechanical Engineering Indian Institute of Technology, Kharagpur

V (r,t) = i ˆ u( x, y,z,t) + ˆ j v( x, y,z,t) + k ˆ w( x, y, z,t)

In steady flow the velocity of the fluid particles at any point is constant as time passes.

Lecture 1: Kinematics, ideal mechanical systems and Bernoulli s equation

Transport processes. 7. Semester Chemical Engineering Civil Engineering

2.29 Numerical Fluid Mechanics Fall 2011 Lecture 5

Chapter 6: Momentum Analysis of Flow Systems

BERNOULLI EQUATION. The motion of a fluid is usually extremely complex.

Introduction to Turbomachinery

3.25 Pressure form of Bernoulli Equation

Chapter 5. Mass and Energy Analysis of Control Volumes

Chapter 5 Control Volume Approach and Continuity Equation

FORMULA SHEET. General formulas:

ME 2322 Thermodynamics I PRE-LECTURE Lesson 23 Complete the items below Name:

Conservation of Energy for a Closed System. First Law of Thermodynamics. First Law of Thermodynamics for a Change in State

Applied Thermodynamics for Marine Systems Prof. P. K. Das Department of Mechanical Engineering Indian Institute of Technology, Kharagpur

Lecture 3 The energy equation

Chapter 3 Bernoulli Equation

CLASS Fourth Units (Second part)

Aerodynamics. Basic Aerodynamics. Continuity equation (mass conserved) Some thermodynamics. Energy equation (energy conserved)

6. Laminar and turbulent boundary layers

M E 320 Professor John M. Cimbala Lecture 10. The Reynolds Transport Theorem (RTT) (Section 4-6)

Turbomachinery & Turbulence. Lecture 2: One dimensional thermodynamics.

Turbomachinery. Hasan Ozcan Assistant Professor. Mechanical Engineering Department Faculty of Engineering Karabuk University

Fundamentals of compressible and viscous flow analysis - Part II

The Impulse-Momentum Principle

Basic Fluid Mechanics

Detailed Outline, M E 320 Fluid Flow, Spring Semester 2015

Control Volume Revisited

FLOW IN PIPES. Mark J McCready University of Notre Dame July 24, chemeprof.com

The First Law of Thermodynamics. By: Yidnekachew Messele

CHAPTER 2 INVISCID FLOW

Basic Fluid Mechanics

ACCOUNTING FOR FRICTION IN THE BERNOULLI EQUATION FOR FLOW THROUGH PIPES

(Refer Slide Time 1:25)

COURSE NUMBER: ME 321 Fluid Mechanics I 3 credit hour. Basic Equations in fluid Dynamics

UNIT II CONVECTION HEAT TRANSFER

FLUID MECHANICS. Chapter 3 Elementary Fluid Dynamics - The Bernoulli Equation

9. Pumps (compressors & turbines) Partly based on Chapter 10 of the De Nevers textbook.

Isentropic Efficiency in Engineering Thermodynamics

Chapter Two. Basic Thermodynamics, Fluid Mechanics: Definitions of Efficiency. Laith Batarseh

T098. c Dr. Md. Zahurul Haq (BUET) First Law of Thermodynamics ME 201 (2012) 2 / 26

MAE 3130: Fluid Mechanics Lecture 7: Differential Analysis/Part 1 Spring Dr. Jason Roney Mechanical and Aerospace Engineering

Hydraulics for Urban Storm Drainage

NPTEL Course Developer for Fluid Mechanics DYMAMICS OF FLUID FLOW

Answers to questions in each section should be tied together and handed in separately.

Math Review Night: Work and the Dot Product

Eng Thermodynamics I conservation of mass; 2. conservation of energy (1st Law of Thermodynamics); and 3. the 2nd Law of Thermodynamics.

Unit C-1: List of Subjects

Heat and Mass Transfer Prof. S.P. Sukhatme Department of Mechanical Engineering Indian Institute of Technology, Bombay

Introduction to Fluid Mechanics

General Physics I (aka PHYS 2013)

Chemical Engineering 374

Fluids, Continuity, and Bernouli

ENT 254: Applied Thermodynamics

Lecture 2 Flow classifications and continuity

LECTURE 9. Hydraulic machines III and EM machines 2002 MIT PSDAM LAB

Chapter 5: The First Law of Thermodynamics: Closed Systems

Thermodynamics I Spring 1432/1433H (2011/2012H) Saturday, Wednesday 8:00am - 10:00am & Monday 8:00am - 9:00am MEP 261 Class ZA

Mechanical Engineering Programme of Study

Math S1201 Calculus 3 Chapters , 14.1

Chapter 4 DYNAMICS OF FLUID FLOW

piston control surface

Mass of fluid leaving per unit time

1 st Law Analysis of Control Volume (open system) Chapter 6

Physics 3 Summer 1990 Lab 7 - Hydrodynamics

ME332 FLUID MECHANICS LABORATORY (PART II)

Module 2 : Lecture 1 GOVERNING EQUATIONS OF FLUID MOTION (Fundamental Aspects)

Transcription:

Energy CEEN 598D: Fluid Mechanics for Hydro Systems Lindsay Bearup lbearup@mines.edu Berthoud Hall 11 GEGN 498A Fall 013

For the record: Material DerivaIves db sys dt Rate of change of property B of system and DB SYS Dt Total change in system Material Deriva@ve (or substanial derivaive) These two expressions do not mean the same thing! GEGN 498A Fall 013

Material DerivaIve In general, the material derivaive is represented by the expression: * can hold a variety of parameters (e.g., V, B sys, ρ, etc.). D( ) () () () () = u v w Dt t x y z () = ( V )( ) t gradient operator GEGN 498A Fall 013

Gradient Operator D() () = ( V )( ) Dt t gradient operator () ˆ () ˆ () () = i j kˆ x y z GEGN 498A Fall 013

For clarity: EGL & HGL review GEGN 498A Fall 013

For clarity: EGL & HGL review (4) P 4 /γ = V 4 /g = z 4 = EGL z=0 H (3) P 3 /γ = V 3 /g = z 3 = P /γ = V /g = z = () h (1) (0) P 1 /γ = V 1 /g = z 1 = P 0 /γ = V 0 /g = z 0 = GEGN 498A Fall 013

Bernoulli s EquaIon & the EGL Bernoulli s equaion assumes: 1) ) 3) 4) To capture all the energy in a system, what other terms must we include?? GEGN 498A Fall 013

Energy Energy: scalar physical quanity that describes the amount of work that can be performed by a force. Work: is a force acing through a distance when the force is parallel to the direcion of moion. Work = force distance = torque angular displacement GEGN 498A Fall 013

Power Expresses rate of work or energy: P quantity of work (or energy) interval of time ΔW = lim Δt 0 Δt = W If we let the amount of work be defined as the product of force and displacement: ΔW = FΔx P = lim Δt 0 FΔx Δt = FV GEGN 498A Fall 013

Torque Or, for a rotaing shac work is given as the product of the torque and angular displacement: ΔW = TΔθ So: TΔθ P = lim = Tω Δt 0 Δt P = FV = Tω GEGN 498A Fall 013

GEGN 351a - Maxwell - Fluid Mechanics

First Law of Thermodynamics Ime rate of increase in total stored energy of the system net Ime rate of energy addiion by heat transfer into the system = net Ime rate of energy addiion by work transfer into the system D Dt eρdv = ( Q in Q ) out ( W in sys W ) out sys sys D Dt $ eρdv = & % sys Q net in W net in ' ) ( sys GEGN 498A Fall 013

Energy DefiniIons e is the total stored energy per unit mass for each paricle in the system: u is the internal energy per unit mass V / is the kine0c energy per unit mass gz is the poten0al energy per unit mass Such that: e = u V gz GEGN 498A Fall 013

GEGN 351a - Maxwell - Fluid Mechanics

ConservaIon of Energy The First Law of Thermodynamics (FLT) is that the energy gained by a system is the difference between heat transferred TO the system minus the work done BY the system on the surroundings in some Ime interval: ΔE = Q W The energy may take many forms. We ll break it into readily observable (kine@c, poten@al) and not so obvious undiffereniated internal (temperature, chemical potenial, pressure, etc.) E = E E k p E u GEGN 498A Fall 013

ConservaIon of Energy The rate that the energy changes is directly related to the rate at which heat is received and work is done by/from the system: de dq dw = Q! W! dt dt dt = (FLT) In the RTT, the extensive property is the total energy E, so B sys = E. There is no trick where b = something cool (like for mass, b = 1), so we will just call the intensive energy b = e. de dt = d dt CV eρ dv CS eρv nda GEGN 498A Fall 013

Combine the FLT and the RTT: d Q! W! = eρdv dt eρv In a likewise fashion, define the intensive energies e CV = ek ep u What is the intensive kineic energy? The total kineic energy of a packet of mass divided my the mass: MV 1 e k = = M V How about the potenial energy of a packet of stuff? ρgvz 1 e p = = 1 ρv gz CS GEGN 498A Fall 013 nda

SubsItute into the RTT: Q! W! d dt = ( gz u) ρ dv ( gz u)ρv CV V Let s look at some of the terms. First, what is W dot? Work done by the system on the surroundings. Change the sign and it s work done by the surroundings on the fluid system. This is: W = W s W f SHAFT WORK (W s ): work by pumps and other machines FLOW WORK (W f ): work that the pipes and surrounding fluid does to change the fluid system via pressure CS V nda GEGN 498A Fall 013

Flow Work Assume the following system: GEGN 498A Fall 013

Flow Work Remember, work is the force applied Imes the length (W=Fd). So, for a moving system, the flow work, W f is calculated as: ΔW dw dt f f = = F d pa VΔt VpA = W! f GEGN 498A Fall 013

Work The system does work on the surroundings simply by pushing on them. Let s look at all the surfaces using an analogy the skateboard. V weight When the skateboard is cruising on level ground with a weight on it, how much work is being done? GEGN 498A Fall 013

Work Now push the board with the weight uphill. How much work has been done? The force F Imes the distance, or the component of the weight in the direc0on of travel. ΔW = FvΔt = Weight vδt F weight GEGN 498A Fall 013 This works in the previous flat case, since Weight v =0

Work Pressure always acts normal to a surface, so the force pa is normal to the surface. The work performed is then the component of that force in the direcion of the fluid velocity. GEGN 498A Fall 013

For an arbitrary surface, break it down into n Iny areas da and add them all up. The total flow work is then: dw f = v n( pda) = pv nda dt CS CS Using the shorthand notaion of a Ime derivaive and muliply and divide by density for future convenience: dw dt f = W! f = CS v n( pda) = CS p ρv ρ nda GEGN 498A Fall 013

Plug this back into the Energy form of the RTT: Q! W! d dt CV s V ( CS p ρv nda ρ gz u) ρdv = CS V ( gz u) ρv nda Note the similarity between the last term on the right hand side and the last term on the lec hand side. Put those together on the right hand side GEGN 498A Fall 013

Energy Principle Q! W! d dt = ( gz u) ρ dv ( gz u ) ρv CV s V CS V p ρ nda This is the Energy Principle, which complements our coninuity and momentum equaions. GEGN 498A Fall 013

Specific Enthalpy The combined term u p/ρ is actually called the specific enthalpy (h) of the fluid (yes, another h!). So the above equaion is someimes shortened a Iny bit: Q! W! d dt = ( gz u) ρ dv ( gz h)ρv CV s V CS V nda GEGN 498A Fall 013

Enthalpy Enthalpy: The internal heat, which most commonly changes with temperature and pressure, but can also change with state (liquid, gas) and chemical reacions. Consider steady flow of an incompressible fluid in a pipe system. GEGN 498A Fall 013

Steady- state means that the energy in the system does not change. Furthermore, the only places where V n is non- zero are the pipe inlets and outlets, leaving Q! W! = 1 gz1 u ) ρvda ( gz u 1 s V p V p 1 ( ) ρvda ρ ρ At surfaces 1 and, the quanity p/ρ z is constant and comes out of the integral. Also, let the internal energy be constant at a surface (it may change between the surfaces). GEGN 498A Fall 013

Write in a fudge factor called the kineic- energy correcion factor (α): V V ρ 1 da = αρ 1 3 3 A For turbulent flow, the velocity is preqy constant across the secion (WHY?), and α 1. For laminar flow, the velocity profile is parabolic and the integraion gives α =. GEGN 498A Fall 013

Let s recap We started from the FLT and RTT: de dq dw = = Q! W! dt dt dt de dt = d dt CV eρdv eρv nda Combined and broke e into different energy forms (kin/pot/int) and broke work into components of shac and flow: p Q! W! s ρv nda = ρ d dt CV V ( CS gz u) ρdv CS CS V ( (FLT) (RTT) gz u) ρv nda GEGN 498A Fall 013

Let s recap For steady- state flow we then divide all terms by gρq (i.e. assume flow in = flow out): Q! W! gρq s u1 p1 V1 u p z1 = z g γ g g γ V g We then look at the units of this equaion:!w t gρq = ML T L T L T M L 3 L 3 T = ML T 3 ML T 3 =L and realize that it is all in terms of length units, and GEGN 498A Fall 013

GEGN 498A Fall 013 Let s recap we realize that we can call those terms head gains or losses through pumps and/or turbines h p and h t. g V p g u z g V p g u z h h Q g Q p t 1 1 1 1 = γ γ ρ!

GEGN 498A Fall 013 There are several terms describing the loss or gain of heat: u 1, u, and Q dot. If we put those together, then The first term in the brackets is the internal energy (temperature) loss in the direcion of flow. The second term is the direct heat loss of the fluid system. In reality it s hard to differeniate between the two. We only know that the fluid system loses energy from heat transfer as fluid flows. This term can never be zero. = Q g Q g u u g V p z h h g V p z p t ρ γ γ! 1 1 1 1 Heat Terms

Energy EquaIon We lump these energy terms together and call it head loss h L and we get the Energy Equa@on: p α V p α 1 1 1 V z 1 hp = z hl γ g γ g This gives us the following definiions: Pump Head Turbine Head W h p p m g W h t t m g GEGN 498A Fall 013 Work/Ime done by pump on flow Weight/Ime of flowing fluid Work/Ime done by flow on turbine Weight/Ime of flowing fluid h t

So what does this mean? p α V p α V z 1 1 1 h = z h 1 p L γ g γ g Head carried into CV by flow Head added by pumps Head carried out of CV by flow Head lost to visc h t Head lost to turb. IN = OUT GEGN 498A Fall 013

h p h t z p γ α V g z p 1 1 γ α 1V 1 g h L GEGN 498A Fall 013

Example Given: The pump shown adds 10 hp to the water as it pumps from the lower lake to the upper lake. The elevaion difference between the lake surfaces is 30c and the head loss from fricion is 15c. Find: (a) the flowrate (b) the power loss associated with the flow GEGN 498A Fall 013

GEGN 498A Fall 013

Energy vs. Bernoulli EquaIons The Energy equaion looks a lot like the Bernoulli equaion, they are however somewhat different: Recall Bernoulli was derived by applying Newton s second law to a paricle and then integraing along a streamline Energy equaion derived using first law of thermodynamics and the Reynolds transport theorem Bernoulli only has mechanical energy Energy equaion has both mechanical and thermal energy GEGN 498A Fall 013

GEGN 498A Fall 013 Energy vs. Bernoulli EquaIons Energy EquaIon: Bernoulli EquaIon t L p h h g V p z h g V p z = 1 1 1 1 α γ α γ z 1 p 1 γ V 1 = z p γ V

Energy vs. Bernoulli EquaIons Most importantly, even though they look the same, the energy equaion is a control volume equaion and relates inflow and ouslow. The Bernoulli equaion is not a control volume equaion. It equates the mechanical energy of two points in a flow. The energy equaion applies to steady, viscous, incompressible flow in a pipe with energy added or lost. The Bernoulli equaion applies to steady, inviscid, incompressible flow. GEGN 498A Fall 013