Energy Transfer

ME2121 Engineering ThermodynamicsENERGY TRANSFERFirst Law Of ThermodynamicsA concise reviewChapter 3 of C&BASMRefer to text for details© Copyright 2005 Prof. Arun S. Mujumdar.

• This approach is also applied to solids and liquids -approximated as incompressible substances.• Energy can be neither created nor destroyed; it canonly change forms. Energy can cross the boundary ofa closed system in two distinct forms: heat and work.• Heat is defined as the form of energy that istransferred between two systems (or a system andits surroundings) by virtue of a temperaturedifference.• Heat is energy in transition. It is recognized only as itcrosses the boundary of a system. A process duringwhich there is no heat transfer is called an adiabaticprocess.Energy Transfer 3© Copyright 2005 Prof. Arun S. Mujumdar.

Heat and work are energy transfer mechanismsbetween a system and its surroundings, there aremany similarities between them:• Both are recognized at the boundaries of a system asthey cross them. Both heat and work are boundaryphenomena.• Systems possess energy, but not heat and work.• Both are associated with a process, not a state.Unlike thermodynamic properties, heat or work hasno meaning at a state.• Both are path functions (i.e. their magnitudes dependon the path followed during a process as well as theend states).These are basic concepts useful in further deliberations.Be clear about their precise meaning.Energy Transfer 5© Copyright 2005 Prof. Arun S. Mujumdar.

ENERGY TRANSFER BY WORKSome definitions/terminology• Work, like heat, is an energy interaction betweena system and its surroundings• If the energy crossing the boundary of a closedsystem is not heat, it must be work• Work is the energy transfer associated with aforce acting through a distance• Work done per unit mass of a system is denotedby w and is expressed asw Wm(kJ / kg)Energy Transfer 6© Copyright 2005 Prof. Arun S. Mujumdar.

ENERGY TRANSFER BY WORK (Cont’d)• Work done per unit time is called power and is denotedW (Figure 3-8). The unit of power is kJ/s or kW• Heat and work are directional quantities; requires thespecification of both the magnitude and direction; signconvention• Heat transfer to a system and work done by a systemare positive; heat transfer from a system and work doneon a system are negative• Use this intuitive approach in this book as it eliminatesthe need to adopt a formal sign conventionEnergy Transfer 7© Copyright 2005 Prof. Arun S. Mujumdar.

Similarities between Heat andWork - Summary• Both heat and work are boundaryphenomena• Systems possess energy but not heat orwork• Both are associated with a process, not astate –important to note!• Both are path functions (magnitudedepend on the path followed)Energy Transfer 8© Copyright 2005 Prof. Arun S. Mujumdar.

• Path functions have inexact differentialsdesignated by the symbol δ, e.g. δQ orδW• Properties are point functions , i.e. theydepend on the state only. They have exactdifferentials designated by the symbol d.Energy Transfer 9© Copyright 2005 Prof. Arun S. Mujumdar.

MECHANICAL FORMS OF WORK• The work done by a constant force F on a bodydisplaced a distance s in the direction of the forceis given byW = F s (kJ)• If the force F is not constant i.e. it is a function ofdistance, s ,thenW21F ds(kJ)Energy Transfer 10© Copyright 2005 Prof. Arun S. Mujumdar.

MECHANICAL FORMS OF WORK (cont’d)• Two requirements exist for a work interactionbetween a system and its surroundings• There must be a force acting on the boundary• The boundary must move• Displacement of the boundary without any forceto oppose or drive this motion (such as theexpansion of a gas into an evacuated space) isnot a work interaction since no energy istransferred• Mechanical work is associated with themovement of the boundary of a system or withthe movement of the entire system as a wholeEnergy Transfer 11© Copyright 2005 Prof. Arun S. Mujumdar.

MOVING BOUNDARY WORK• Moving boundary work is the primary form ofwork involved in automobile engines• We analyze the moving boundary work for aquasi-equilibrium process, a process duringwhich the system remains in equilibrium at alltimes• Consider the gas enclosed in the piston-cylinderdevice shown in Fig. 3-19, C &B. The initialpressure of the gas is P, the total volume is Vand the cross sectional area of the piston is A. Ifthe piston is allowed to move a distance ds, thedifferential work done during this process isδW b = F ds = PA ds = P dVEnergy Transfer 12© Copyright 2005 Prof. Arun S. Mujumdar.

MOVING BOUNDARY WORK (cont’d)• Note P is the absolute pressure, which is alwayspositive. However, the volume change dV ispositive during an expansion process (volumeincreasing) and negative during a compressionprocess (volume decreasing).• Thus, the boundary work is positive during anexpansion process and negative during acompression process. Therefore the aboveequation can be viewed as an expression forboundary work output, W b,out . A negative resultindicates boundary work input (compression)Energy Transfer 13© Copyright 2005 Prof. Arun S. Mujumdar.

MOVING BOUNDARY WORK (cont’d)• Moving boundary work is sometimes called theP dV work• The total boundary work done during the entireprocess as the piston moves is obtained byadding all the differential works from the initialstate to the final stateW2b P dV (kJ)1Energy Transfer 14© Copyright 2005 Prof. Arun S. Mujumdar.

MOVING BOUNDARY WORK (cont’d)• P = f(v) should be available. Note that P = f(v)is simply the equation of the process path on aP-V diagram• The total area A under the process curve 1-2 (to go from state 1 to state 2) is obtained byadding these differential areas:2 2Area A1dA1P dVEnergy Transfer 15© Copyright 2005 Prof. Arun S. Mujumdar.

MOVING BOUNDARY WORK (cont’d)• A gas can follow several different paths as it(say) expands from state 1 to state 2. Ingeneral, each path will have a different areaunderneath it, and since this area represents themagnitude of work, the work done will bedifferent for each process (See Fig 3-21).• Note: work is a path function (i.e., it depends onthe path followed as well as the end states).• If work were not a path function, no cyclicdevices (car engines, power plants) couldoperate as work producing devices.Energy Transfer 16© Copyright 2005 Prof. Arun S. Mujumdar.

Boundary Work during a Constant-Volume Process (Example 3-5)A rigid tank contains air at 500 kPa and 150ºC. Asa result of heat transfer to the surroundings,the temperature and pressure inside the tankdrop to 65ºC and 400 kPa, respectively.Determine the boundary work done during thisprocess. No volume change. Hence,W2b P dV 010Energy Transfer 17© Copyright 2005 Prof. Arun S. Mujumdar.

Boundary Work for a Constant-PressureProcess (Example 3-6)• A frictionless piston-cylinder device contains 10lbm of water vapor at 60 psia and 320ºF. Heat isnow transferred to the steam until thetemperature reaches 400ºF. If the piston is notattached to a shaft and its mass is constant,determine the work done by the steam duringthis process.• Not in SI units but illustrates a point!Note this is a constant pressure process.Energy Transfer 18© Copyright 2005 Prof. Arun S. Mujumdar.

Boundary Work for a Constant-PressureNote:Process (Example 3-6)2 P dV P0 dV P0(V2Wb V1)1W b mP0(V2 V 1 )Work = mass x pressure x change of volume ofsystem21Energy Transfer 19© Copyright 2005 Prof. Arun S. Mujumdar.

Isothermal Compression of anIdeal Gas (Example 3-7)• A piston-cylinder device initially contains0.4m 3 of air at 100 kPa and 80ºC. The airis now compressed to 0.1m 3 in such a waythat the temperature inside the cylinderremains constant. Determine the workdone during this process.• Assume ideal gas.Energy Transfer 20© Copyright 2005 Prof. Arun S. Mujumdar.

Isothermal Compression of anIdeal Gas (Analysis - Example 3-7)• For an ideal gas at constant temperature T 0 ,Then,Wb21P dVPV mRT 0 C or P 21CdVVC21dVVC lnVV21P1VCV1lnVV21Insert values given in problem statements to find W bEnergy Transfer 21© Copyright 2005 Prof. Arun S. Mujumdar.

POLYTROPIC PROCESS• During actual expansion and compressionprocesses of gases, pressure and volume areoften related by PV n = C, where n and C areconstants. A process of this kind is called apolytropic processP = CV -nThen,Wb21P dV21CVndVVCn12 V n 1n11P2V2 P1n1V1Energy Transfer 22© Copyright 2005 Prof. Arun S. Mujumdar.

Gravitational Work• Gravitational work is defined as the work done by oragainst a gravitational force field.F = mg2Wg z1)(kJ) F dZ mg dZ mg (z2121Energy Transfer 23© Copyright 2005 Prof. Arun S. Mujumdar.

Accelerational Work• The work associated with a change in velocity of asystem is called accelerational work. Theaccelerational work required to accelerate a body ofmass m from the definition of acceleration andNewton’s Second Law:F ma dVa dtFmdVdtEnergy Transfer 24© Copyright 2005 Prof. Arun S. Mujumdar.

dsV ds VdtdtWa21F ds2 dV m(V dt ) dt m21V dV12m(V22V21)Interesting Observation:The work done to accelerate a body is independent ofthe path followed and is equivalent to the change inthe kinetic energy of the body.Energy Transfer 25© Copyright 2005 Prof. Arun S. Mujumdar.

Shaft Work:• Energy transmission with a rotating shaft is verycommon in engineering practice. Often the torque applied to the shaft is constant, which means thatthe force F applied is also constant.• F acting through a moment arm r generates a torque ofFrFrThis force acts through a distance s, which is related tothe radius r bys2(2r)nEnergy Transfer 26© Copyright 2005 Prof. Arun S. Mujumdar.

Then the shaft work is given by:W shFs r Spring Work:2rn 2nTo determine the spring work, we need a frelationship between force F and displacement x.For linear elastic springs, the displacement x isproportional to the force applied,orF = kxEnergy Transfer 27© Copyright 2005 Prof. Arun S. Mujumdar.

Here k is the spring constant and has the units kN/m.The displacement x is measured from theundisturbed position of the spring (that is x = 0 whenF = 0). Substituting and integrating,Wspring12k(x22x21)(kJ)Here x1 and x2 are the initial and the finaldisplacements of the spring, measured from theundisturbed position of the spring.Energy Transfer 28© Copyright 2005 Prof. Arun S. Mujumdar.

Non-mechanical Forms of Work• Some examples of non-mechanical work modes areelectrical work, where the generalized force is thevoltage (the electrical potential) and the generalizeddisplacement is the electrical charge, and magneticwork, where the generalized force is the magneticfield strength and the generalized displacement is thetotal magnetic dipole moment; and electricalpolarization work.• These forms of work will not be considered inME2121. However you should be aware of these.Energy Transfer 29© Copyright 2005 Prof. Arun S. Mujumdar.

Total Energy – An Outcome of the First Law• A major consequence of the first law is the existenceand the definition of the property total energy E. thenet work is the same for all adiabatic processes of aclosed system between two specified states, thevalue of the net work must depend on the end statesof the system only, and thus it must correspond to achange in a property of the system.• This property is the total energy. The first law statesthat the change in the total energy during anadiabatic process must be equal to the net workdone.• Therefore, any convenient arbitrary value can beassigned to total energy at a specified state to serveas a reference point.Energy Transfer 30© Copyright 2005 Prof. Arun S. Mujumdar.

Energy Balance:The net change (increase or decrease) in thetotal energy of the system during a process isequal to the difference between the totalenergy entering and the total energy leavingthe system during that process. That is,during a process,TotalenergyTotalenergyChangein thetotal enteringthe systemleavingthe systemenergy of the system or, E in - E out = E systemEnergy Transfer 31© Copyright 2005 Prof. Arun S. Mujumdar.

Energy Change of a System, Esystem• Energy change = Energy at final state –Energy at initial stateE system = E final – E initial = E 2 – E 1• Energy can exist in numerous forms such asinternal (sensible, latent, chemical andnuclear), kinetic, potential, electrical, andmagnetic and their sum constitutes the totalenergy E of a system. In the absence ofelectric, magnetic and surface tension effectsE = U + KE PEEnergy Transfer 32© Copyright 2005 Prof. Arun S. Mujumdar.

whereUm(u2 u1 2KE m( V2V2PE mg (z z )21)121)Systems that do not involve any changes in theirvelocity or elevation during a process are stationarysystems.Changes in kinetic and potential energies are zero(that is KE = PE = 0), and E = U for stationarysystems.Energy Transfer 33© Copyright 2005 Prof. Arun S. Mujumdar.

First Law (Cont’d)Mechanisms of Energy Transfer, E in and E out• Heat Transfer, Q• Work, W• Mass Flow, m• When mass enters an open system, the energy of thesystem increases because mass carries energy withit.• The first law or energy balance relation for a closedsystem becomesQ – W = EEnergy Transfer 34© Copyright 2005 Prof. Arun S. Mujumdar.

First Law: Statement• Heat input to a system minus work output by thesystem is equal to the change in the energy of thesystem. A negative quantity for Q or W simply meansthat the assumed direction for that quantity is wrong.• Various forms of this “traditional” first law relation forclosed systems are given as1. General Q – W = E2. Stationary systems Q – W = U3. Per unit mass q – w = e4. Differential form q - w = deEnergy Transfer 35© Copyright 2005 Prof. Arun S. Mujumdar.

• The specific heat is defined as the energy required toraise the temperature of a unit mass of a substanceby one degree.• In thermodynamics, we are interested in two kinds ofspecific heats: specific heat at constant volume C vand specific heat at constant pressure C p .• The specific heat at constant volume C v is the energyrequired to raise the temperature of the unit mass ofa substance by one degree as the volume ismaintained constant.C vuTEnergy Transfer 36© Copyright 2005 Prof. Arun S. Mujumdar.

• An expression for the specific heat at constantpressure Cp can be obtained by considering aconstant-pressure process• (wb + u = h). It yieldsCphTp• The specific heats of a substance depend on thestate that, in general, is specified by twoindependent, intensive properties.• The energy required to raise the temperature of asubstance by one degree will be different at differenttemperatures and pressures. But this difference isusually not very large.Energy Transfer 37© Copyright 2005 Prof. Arun S. Mujumdar.

• These equations are property relations and as suchare independent of the type of processes. They arevalid for any substance undergoing any process.• C v is related to the changes in internal energy and C pto the changes in enthalpy. C v is a measure of thevariation of internal energy of a substance withtemperature, and C p is a measure of the variation ofenthalpy of a substance with temperature.• Since u and h depend only on temperature for anideal gas, the specific heats C v and C p also depend ontemperature only. Therefore, at a given temperature,u, h, C v and C p of an ideal gas will have fixed valuesregardless of the specific volume or pressure. Forideal gases, the partial derivatives can bereplaced by ordinary derivatives.Energy Transfer 38© Copyright 2005 Prof. Arun S. Mujumdar.

du = Cv(T) dT andHence,andudh = Cp (T) dT2 u2 u1 Cv(T)dT(kJ / kg)1h2h2 h1 Cp(T)dT (kJ / kg)1Energy Transfer 39© Copyright 2005 Prof. Arun S. Mujumdar.

• To carry out these integrations, we need to haverelations for C v and C p as functions of temperature.• The specific heats of gases with complex molecules(molecules with two or more atoms) are higher andincrease with temperature. The variation of specificheats with temperature is smooth and may beapproximated as linear over small temperatureintervals (a few hundred degrees or less).Henceu 2 – u 1 = C v,av (T 2 – T 1 )andh 2 – h 1 = C p,av (T 2 – T 1 )(kJ/kg)(kJ/kg)Energy Transfer 40© Copyright 2005 Prof. Arun S. Mujumdar.

The ideal gas specific heats of monatomic gases suchas argon, neon and helium remain constant over theentire temperature range.Specific-Heat Relations of Ideal GasesA special relationship between C p and C p for idealgases can be obtained by differentiating the relationh = u + RT, which yieldsdh = du + R dTReplacing dh by C p dT and du by C p dT and dividingthe resulting expression by dT we obtainC p = C p + R [kJ/(kg . K)]Energy Transfer 41© Copyright 2005 Prof. Arun S. Mujumdar.

Another ideal-gas property called the specific heat ratiok, defined asCpk CvThe specific heat ratio also varies mildly withtemperature. For monatomic gases, its value isessentially constant at 1.667. Many diatomic gases,including air, have a specific heat ratio of about 1.4at room temperature.Cv 0.718kJ /(kg.K) CR 0.287 kJ /(kg.K) orCRvu 20.80 kJ /(kmol .K) C 8.314 kJ /(kmol .K) p 1.005 kJ /(kg.K)p 29.114 kJ /(kmol.K)Energy Transfer 42© Copyright 2005 Prof. Arun S. Mujumdar.

Conservation of Mass PrincipleSection 3-5 C&BStatement: Net mass transfer to or from a systemduring a process is equal to the net change oftotal mass of the system during that process.Note: The change may be an increase or adecrease.This is the same principle as conservation ofmoney in your bank account, viz.,Amount entering the system – Amount leaving thesystem = Net change within the systemEnergy Transfer 43© Copyright 2005 Prof. Arun S. Mujumdar.

Mass Balance for Steady FlowProcesses (Figure 3-43 C&B)• In a steady process total mass in a control volume (CV)does not change with time. Hence mass entering a CVmust be equal to that leaving• There can be multiple entries and exits to a controlvolume• For steady flow, single streamm1m2or 1V1A1 2V2A2 V is average velocity over cross section A.Energy Transfer 44© Copyright 2005 Prof. Arun S. Mujumdar.

ExamplesFor incompressible fluid (not flow) density isconstant, hence V 1 A 1 = V 2 A 2• Example 3-12 – This is a simple probleminvolving a constant flow rate through a gardenhose nozzle.• Example 3-13 – This is an interesting example tocompute discharge time for water in a tank.Note that the discharge velocity is a function ofthe level of water in the tank which falls withtime as water is discharged. Clearly this is aproblem of mass conservation in rate form, i.e. itis a unsteady state problem. Pls review carefully.Energy Transfer 45© Copyright 2005 Prof. Arun S. Mujumdar.

Flow Work / Energy of a Flowing Fluid• Open systems or control volumes involve fluidflowing into or out of the CV. This involves socalled flow work or flow energy.• Flow work is needed to maintain fluid flowthrough CV (See figure 3-48)• If P is fluid pressure and the cross sectional areafor fluid flow is A then work required to makethe fluid element move a distance L isW flow = FL = PAL = PV (kJ)• Flow work per unit mass isw flow =Pv (kJ/kg)Energy Transfer 46© Copyright 2005 Prof. Arun S. Mujumdar.

Total Energy of a Flowing Fluid• It is sum of internal, kinetic, and potential energy of acompressible system, i.e.eukepeugz(kJ / kg)• For a flowing system on a per unit mass basis we mustadd flow energy to obtain the total energy of a flowingfluid, viz. = Pv + e = Pv + (u + ke + pe) or = h + ke + pe where h = Pv + uVh is called specific enthalpy of the fluid. Using h instead ofu allows us to account for flow work in control volumeproblems.2/2Energy Transfer 47© Copyright 2005 Prof. Arun S. Mujumdar.

Energy Transport by Mass• In open systems (control volumes) note thatenergy transfer can also occur due to mass flowin or out over control surfaces• Amount of energy transported by mass, E mass =m (kJ)• This equation can also be written in rate form bysimply taking time derivative of the aboveequation, i.e. simply place a dot over E and mStudy Example 3-14 (estimation of mass andenergy leaving a pressure cooker in a given timeby application of the rate form of conservationof mass to a CV (figure 3-53))Energy Transfer 48© Copyright 2005 Prof. Arun S. Mujumdar.

Closing Remarks – Chapter 3• This chapter defines various forms ofenergy and how energy transfer can occurin closed and open systems• Reviews definitions of work, heat, internalenergy, enthalpy, etc.• In chapter 4 these concepts are appliedalong with the First Law ofThermodynamics to a number of simplepractical systems involving gases andliquidsEnergy Transfer 49© Copyright 2005 Prof. Arun S. Mujumdar.

Some Quotable Quotes• “Thermodynamics is the only physicaltheory of universal content which, withinthe framework of the applicability of itsbasic concepts, I am convinced will neverbe overthrown” .– Albert Einstein• “Make it simple but not simpler”.- Albert EinsteinEnergy Transfer 50© Copyright 2005 Prof. Arun S. Mujumdar.