Pressure Volume Work

6 Key excerpts on "Pressure Volume Work"

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  Physical Chemistry

  Thermodynamics

  • Horia Metiu(Author)
  • 2006(Publication Date)
  • Taylor & Francis

    (Publisher)

  This is the case in most practical situations (the container does not leak). As we know from Chapter 2, pressure is related to molar volume and temperature through the equation of state p = f ( T , v ), which is independent of the number of moles in the system. Because of this, the work (given by δ W = − npdv = − nf ( T , v ) dv ) is proportional to the number of moles in the system. This means that work is an extensive quantity (see Chapter 1, §12, for the definition of extensive and intensive quantities). If I perform a transformation on a mole of substance and the work required is W , then the same transformation performed on 10 moles of the same substance will require 10 times as much work (i.e., the work is 10 W ). §6. No Change in Volume, No Work. The equation δ W = − F dx , which was our starting point in §2, indicates that to perform work I must displace the piston ( dx = 0). You are laboring very hard when you hold a big weight on your back, but mechanics says that, as long as the weight stands still, you are doing no work. It is not fair, but that’s the way it is. In thermodynamics, Eq. 5.3 says that no work is exerted on or by the system, unless the volume of the system changes. If I perform a chemical reaction in a vessel with very sturdy walls, the work performed by the reaction is zero. There may be tremendous changes of pressure inside the reactor How to Calculate the Work in a Finite Transformation 73 as the reaction proceeds, but the walls resist it and do not move; the volume does not change, and there is no work. Like the man with a weight on his back, the walls are straining hard to resist the pressure but no work is exchanged. §7. Work is Performed Against an Opposing Force. Work is always performed by a force or against a force. If a gas pushes a piston up, but the piston is weightless and there is no weight and no external pressure on it, the system is expanding its volume without performing work.

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  Thermodynamics

  Processes and Applications

  • Jr. Logan, Jr., Earl Logan, Earl Logan Jr.(Authors)
  • 1999(Publication Date)
  • CRC Press

    (Publisher)

  Like the compression process 1-2 this sweeping process is negative work, but it is different because the properties of the air do not change during the sweeping process. Since the pressure is constant, the work done by the piston in process 2-3 is/^ (^ r ^ 3 )- In flows through a pipe there occurs flow work across any arbitrarily chosen cross section. Choosing a parcel of flowing fluid, say a cylindrical volume of diameter D and length L which occupies the section of the pipe just upstream of a given section. Fluid upstream of the parcel acts like the piston in Example Problem 2.1, i.e., the upstream fluid pushes the parcel across the given section. The work to accomplish this action is the pressure times cross sectional area times displacement; thus, the flow work to move this parcel through the section is pnD2L/4 or simply pressure times volume, pV. Flow work per unit mass is then pressure times specific volume, pv. The concept of flow work is particularly useful in engineering analysis of thermodynamics of practical flow machines and Work 87 devices, e.g., compressors, turbines, pumps, fans, valves, etc. For analysis of these devices the control-volume method will be introduced in Chapter 5 and utilized in subsequent chapters. The control-volume method it utilized to account for the flow of some property across its boundaries. Applied to thermodynamics the method accounts for energy flow across the boundaries of the control volume. Flow work is treated as an energy flow associated with the fluid flow, as indeed it has been shown to be. Specific enthalpy h, defined by (2.33) as u + pv, allows the combination of two forms of energy contained in flows into or out of control volumes. Even though flow work is, in fact, work and not conceptually a property, it is the product of two properties and can be lumped together with specific internal energy to form the highly useful property enthalpy, which measures two forms of energy occurring in flowing systems.

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  The Mechanical Universe

  Mechanics and Heat, Advanced Edition

  • Steven C. Frautschi, Richard P. Olenick, Tom M. Apostol, David L. Goodstein(Authors)
  • 2008(Publication Date)
  • Cambridge University Press

    (Publisher)

  When work is performed by a gas, P and V are the natural variables to define the state, which can be represented by a point (V,P) on a pressure-volume diagram. Figure 20.4 illustrates this idea. As the gas is expanded from a smaller volume V } to a larger volume V 2 , it defines a set of points (V,P) which we call a PV curve. Since 510 THE ENGINE OF NATURE Pressure Volume Figure 20.4 The state of a gas is represented by a point on a pressure-volume graph and the work done by the expanding gas is equal to the area of the region under the curve. the work done by the gas is an integral, it is equal to the area of the region under the curve and over the interval [V, ,V 2 ]. The actual shape of the PV curve will depend on how P and V are related to each other and to the temperature T. Different dependencies will produce different curves, or paths, that represent how the pressure and volume were related during the expansion or compression process. Three such paths are illustrated in Fig. 20.5 . The work done by the expanding gas is different along each of these paths because the corresponding areas are different. So the work done in going from one state (V lf Pi) to another (V 2 ,P 2 ) depends not only on the initial and final states, but also on the path taken between the states. Pressure Volume Figure 20.5 The work done by an expanding gas is equal to the area of the region under the curve in a PV diagram and depends on the path taken. Example 1 A fricttonless piston compresses a gas in a chamber that keeps the pressure constant, a process known as an isobaric compression. If (V^P]) specifies the initial state of the 20.2 WORK AND THE PRESSURE-VOLUME DIAGRAM 511 gas and (V 2 ,P t ) the final state, find the work done by the gas in this compression if P { = 3 atm, V ] = 5, and V 2 = I (in liters). Pressure (atm) 3.0 expansion 1.0 5.0 Volume (L) As the gas is compressed it gains energy and does a negative amount of work.

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  Thermodynamics and Heat Power

  • Irving Granet, Maurice Bluestein(Authors)
  • 2014(Publication Date)
  • CRC Press

    (Publisher)

  Even though all systems have pressure and specific volume, the pv term represents flow work only in a flow system. For a nonflow, quasi-static, frictionless process, the work done is in the area under the pv curve. Key Terms Terms used for the first time in this chapter are as follows: closed system: See nonflow system . energy: The capacity to do work. equilibrium state: A state in which no finite rate of change can occur without a finite change, temporary or permanent, in the state of the environment. flow work: The product of the pressure and specific volume of a fluid in a given state in a flow process. heat: A form of energy in transition to or from a system due to the fact that temperature differences exist. internal energy: The energy possessed by a body by virtue of the motion of the mol-ecules of the body and the internal attractive and repulsive forces between molecules. kinetic energy: The energy possessed by a body due to its motion. mechanical equivalent of heat: In the English system of units, 1 Btu = 778 ft.·lb f . This is termed the mechanical equivalent of heat . nonflow system: A system in which there is no mass crossing the boundaries. potential energy: The energy possessed by a body due to its location with respect to an arbitrary reference plane. quasi-static process: A frictionless process carried out infinitely slowly so that it is in equilibrium at all times. work: The product of force and distance in which the distance is measured in the direction of the force; a form of energy in transition that is not stored in a system. 82 Thermodynamics and Heat Power Equations Developed in This Chapter Internal energy mu = U (2.1) Internal energy u U m = (2.2) Potential energy P.E. ft lb f = ⋅ mg g Z c (2.4) Potential energy (SI) P.E. = mgZ joules (2.6) Kinetic energy K. E. ft lb /lb f m = ⋅ mV g c 2 2 (2.12) Kinetic energy (SI) K.

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  Fundamentals of Chemical Engineering Thermodynamics, SI Edition

  • Kevin Dahm, Donald Visco(Authors)
  • 2014(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  Consequently the work done at the pipe outlet is: W · 5 PV · 5 P 1 m · H9267 2 5 s 0.195 MPa d 1 10 6 Pa MPa 2 1 1 N m 2 Pa 2 1 2 2 kg s 961.6 kg m 3 2 5 2 405.6 J s (1.27) An important observation about Example 1-5 is that the expression for flow work turned out to be independent of pipe area, or any other aspect of the geometry of the system. Consequently, Equation 1.28 is applicable to any material stream. SHAFT WORK Shaft work is characterized by rotary, rather than linear, motion. Work is as always equal to the distance travelled times the force opposing the motion, but “distance” is measured by revolutions, as indicated in Equation 1.3. Many chemical processes involve shaft work being either added to, or produced by, a moving fluid; the pump and turbine in the Rankine cycle are prime examples. The moving parts in pumps and compressors are powered by shaft work. Conversely, the kinetic energy in mov-ing liquids or gases can be converted into shaft work by devices like windmills and turbines. Shaft work can immediately be recognized as equal to zero in any system that has no moving parts. Equations 1.22 and 1.28 relate expansion work and flow work to pressure and volume. There is an analogous equation for shaft work but its derivation is more complex and requires the concept of energy balances. This derivation is shown in Example 3-8. Pressure times vol-ume is a quantity that occurs frequently in thermodynamics. Vol-ume can be expressed in absolute terms (e.g., m 3 ), a molar basis (m 3 /mol) or a specific basis (m 3 /kg), as discussed further in Section 2.2.3. Chapter 3 introduces an energy balance equation that we will use extensively. In this equation, flow work is accounted for using the enthalpy ( H ), which is formally defined in Section 2.3.1. Throughout this book, the hat signifies a property expressed on a mass basis (e.g., specific volume) and the underline signifies a property expressed on a molar basis (e.g., molar volume V ).

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  Thermodynamics for Engineers

  • Kaufui Vincent Wong(Author)
  • 2011(Publication Date)
  • CRC Press

    (Publisher)

  Concepts, Definitions, and the Laws of Thermodynamics 1 -7 Thus, the barometric pressure is P 13 6 kg/m 758 m 9 81m/s 1 11 1 N/m 1 11 atmos 3 2 3 2 = = × = ( , )( . )( . ) . . 00 0 0 0 0 kPa The absolute pressure in the pipeline is P 2 bar 1 11kPa 2 kPa 1 11kPa 3 11kPa abs = + = + = 0 00 0 0 . . . 1.4 Forms of Work 1.4.1 Mechanical Forms of Work The.work.done.by.a.constant.force.F.on.a.body.which.is.moved.a.distance.s.in.the.direc-tion.of.the.force.is.given.by . W Fs = . (1 .2) In.general,.if.the.force.is.not.constant,.the.work.is.obtained.by.integration, . W = Fd 1 2 ≡ s . (1 .3) There.are.two.requisites.for.a.work.interaction.between.a.system.and.its.surroundings . . There.must.be.(1).a.force.acting.on.the.boundary.and.(2).the.boundary.must.move . .If. there.are.forces.on.the.boundary,.but.there.is.no.displacement.of.the.boundary,.there.is. no.work.interaction . .Similarly,.if.there.is.a.displacement.of.the.boundary.without.a.force. to.oppose.or.drive.this.movement,.there.is.no.work.interaction . Consider.the.piston-cylinder.device.as.shown.in.Figure.1 .3. .Select.the.system.to.be.the. gas,.with.the.system.boundary.as.shown . .It.is.often.convenient.to.use.physical.bound-aries.to.be.system.boundaries . .Assume.that.the.gravity.force.is.negligible . . The. gas. is. assumed.to.undergo.a.series.of.equilibrium.states,.also.known.as.a.quasiequilibrium. Gas Gas ds FIGURE 1.3 Work.as.energy.transferred.by.a.moving.boundary . 1 -8 Thermodynamics for Engineers process.or.quasistatic.process . .The.gas.moves.the.piston.upward.a.small.distance.ds . .The. total.force.on.the.piston.is.the.pressure.multiplied.by.the.area.of.the.piston, . F PA = . (1 .4) The.differential.work.done.by.the.gas.on.the.piston.is . δ W PA ds = . (1 .5) Since.the.volume.element.is.dV.=.A.ds, . δ W P dV = . (1 .6) If.the.piston.moves.from.position.s 1 .to.position.s 2 ,.the.total.work.done.may.be.found.by. integration, . W PA ds = P dV 1 2 s s V V 1 2 -= ∫ ∫ 1 2 .

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