ADVANCED THERMODYNAMICS ENGINEERING phần 2 ppsx

component of the molecular velocity V that increases the “te”. The energy level of this group of molecules is raised as shown in Figure 32e. Thus, the total number of states do not change even though the energy level for each group has increased due to work input. Now consider the energy transfer due to heat (i.e. due to temperature difference) through solid walls into a gas with the solid being at a higher temperature. The molecules within a group of gas molecules impinging on the wall pick up the energy randomly and these can be placed at different energy levels as shown in Figure 32c. The energy transfer through heat results in an entropy increase while energy transfer through work does not. In Chapter 3 we will see that dS = δQ rev /T (but not δW rev /T or PdV/T). The entropy increases as two different species are mixed. This can be illustrated through the example of two adjacent adiabatic containers of volumes V 1 and V 2 at the same temperature that, respectively, contain nitrogen and oxygen. If the partition between them is removed, then N 2 and O 2 gases have a new set of quantum states due to extension of volume from V 1 and V 2 to V 1 + V 2 . This increases the entropy of each species. Hence mixing causes an increase in entropy, and, consequently the system entropy. In this instance, mixing causes the entropy to increase even though total energy of nitrogen and oxygen is unchanged due to mixing. 11. Properties in Mixtures – Partial Molal Property A kmole of any substance at standard conditions contains 6.023x10 26 molecules known as Avogadro number. The molecular energy is in the form of vibrational, rotational, and translational energy, and the molecules are influenced by the intermolecular potential en- ergy (ipe). At the standard state, the energy of pure water ¯u is 1892 kJ/kmole (the bar at the top indicates pure property on a kmole basis). If a kmole of water is mixed at the molecular level at standard conditions with 2 kmoles of ethanol, each H 2 O molecule is now surrounded by 2 molecules of ethanol. Since the temperature is unchanged, the intermolecular distance is virtually unaltered before and after mixing. The attractive forces due to the water-ethanol molecules are different from those between water-water molecules (this is true of non-ideal solutions and will be discussed in Chapter 8) and, consequently, the potential energy is differ- ent for the two cases. Therefore, the combined energy contribution to the mixture by a kmole (or 6x10 26 molecules) of water in the mixture ^u H2O, is different from that of a kmole of pure water ¯u H2O . The heat at the top of ¯u H2O indicates property when the component is inside the mixture. Here, ^u H2O denotes the partial molar internal energy. Similarly, the enthalpy and entropy of the water are different in the mixture from its unmixed condition. This is further discussed in Chapter 8. If the solution were ideal, i.e., if the ethanol-ethanol intermolecular attractive forces were the same as those for water-water molecules, the water-ethanol attractive forces would equal those in the pure states. In that case ^u H2O = ¯u H2O , for an ideal gas mixture and µ k =µ k since attractive forces do not influence the property. However, even then, ^s H2O would not equal ¯s H2O , since the water molecules would be spread over greater distances in the mixture with the result that the number of quantum states for water molecules would increase. E. SUMMARY We have briefly reviewed various systems (such as open, closed, and composite), mixtures of substances, exact and inexact differentials and their relation to thermodynamic variables, homogeneous functions and their relation to extensive and intensive variables, Tay- lor series, the LaGrange multiplier method for optimization, and the Gauss and Stokes theo- rems. The background material and mathematical concepts will be used through a quantitative language useful to engineers involved with the design and optimization of thermodynamic systems. We have also briefly covered the nature of intermolecular forces and potential, the physical meanings of energy, pressure; of temperature, and of thermal, mechanical, and species equilibrium; boiling and saturation relations; and, finally, entropy. These concepts are useful in the physical interpretation of various thermodynamic relations that are presented in later chap- ters. F. APPENDIX 1. Air Composition Species Mole % Mass % Ar 0.934 1.288 CO 2 0.033 0.050 N 2 78.084 75.521 O 2 20.946 23.139 Rare gases 0.003 0.002 Molecular Weight: 28.96 kg kmole –1 . 2. Proof of the Euler Equation Assume that our objective is to determine a system property F, where F(λx 1 , λx 2 , ) = λ m F(x 1 , x 2 , ), and (79a) x 1,new = λx 1 , x 2,new = λx 2 , Differentiating Eq. (80) with respect to λ (and treating it as a vari- able), (∂F/∂(λx 1,new ))(∂x 1,new /∂λ)+(∂F/∂(λx 2,new ))(∂x 2,new /∂λ)+… = mλ m–1 F(x 1 , x 2 , ). (81b) Since ∂x 1,new /∂λ = x 1 , ∂x 2,new /∂λ = x 2 , …, Eq. (81b) assumes the form (∂F/∂(λx 1,new ))x 1 + (∂F/∂(λx 2,new ))x 2 + … = mλ m–1 F(x 1 , x 2 , ). Multiplying both sides of the above equation by λ, and noting that λ m F(x 1 , x 2 , ) = F(x 1,new , x 2,new , ), we have the relation (∂F/∂(λx 1,new ))x 1,new + (∂F/∂(λx 2,new ))x 2,new + … = mF(x 1,new , x 2,new , ). (80) If m = 1, xFx mF k k K k = ∑ = 0 (/ )∂∂ . (81) 3. Brief Overview of Vector Calculus a. Scalar or Dot Product i. Work Done to Move an Object Consider a surfboard being dragged over water along an elemental path ds r by a power boat that applies a force of r F on the board. The work done is given as δθWFdsF ds=⋅ = r r cos , where θ denotes the angle between the force and the elemental path. ii. Work Done to Move an Electrical Charge Similarly if an electrical charge of strength Q is located at an origin, the force r F ex- erted by it on another charge of strength q situated at distance r r removed from the origin is r r FqQrr= ()/||ε 3 , where ε denotes the Coulomb constant. If the product (qQ) > 0 (i.e., the two are like charges), the force is repulsive. In case (qQ) < 0 (i.e., the charges are unlike) the force is one of attrac- tion. The work done to move charge q away from Q δWFdr=⋅ r r . b. Vector or Cross Product The area r A due to a vector product r rr Axy=× , (82) can be written in the form rr Akxy= | || | sinθ , (83) where r k denotes the unit vector in a plane normal to that containing the vectors r x and r y , and θ the angle between these two vectors. The vector product yields an area vector in a direction normal to the plane containing the two vectors. Consider the circular motion of an object around an origin in a plane. The force due to that object in the plane rr r r r FiF jF iF jF xy =+= +cos sinθθ , (84) where θ denotes the angle between the force and an arbitrary x–wise coordinate at any instant, and r i and r j denote unit vectors in the x– and y– directions, respectively. The torque exerted about the center rr r rr rrr BFr iF jF ixjy kyF xF=×= + ×+= +( cos sin ) ( ) ( cos sin )θθ θθ , (85) where rr ii× = 0, rr ij× = r k , and rr ji× = − r k . When a screw is loosened from a flat surface by rotating it in the counter clockwise direction, it emerges outward normal to the surface, say, in the z–direction. To place the screw back into the surface, it must be rotated in the clockwise direction, i.e., it may be visualized as moving towards the origin of the z–direction. The rotation is caused by an applied torque that is a vector. If the term (F cos θ y – Fsinθ x) = 0 in Eq.(87), then there is no rotation around the z–axis. In general, a force has three spatial components, i.e., rr r r FiF jFkF xyz =++ , (86) and the torque is described by the relation rr r rr r B F r iFz Fy jFx Fz kFy Fx yz z x xy =×=+++++()()( ) , i.e., (87) there are rotational components in the x– and y– directions also. If r F and r r are parallel to each other, e.g., rr FiF x = and r r rix= , then rr r BFr=×=0 . c. Gradient of a Scalar Consider a one–dimensional heat transfer problem in which the temperature T is only a function of one spatial coordinate, say, y, i.e., T = T(y). In this case T(y) is a point or scalar function of y, since its value is fixed once y is specified. In general, the gradient of T is defined as rrrr ∇=++Ti xjykzT(/ / /)∂∂ ∂∂ ∂∂ , (88) which for the one–dimensional problem assumes the form rr ∇=TjTy∂∂/ , (89) The x–z plane contains isotherms, since T≠T(x,z), and r ∇ T is a vector along normal to the isotherms in the y–direction. Consider, now, the temperature profile in an infinite cylindrical rod. Assume that the temperature is constant along the axial direction z, once a cross–sectional location (x,y) is specified, i.e., T=T(x,y), and T≠T(z). Assume an axisymmetric problem for which the iso- therms are circular in the x–y plane and form cylindrical surfaces. In this case, dT = (∂T/∂x)dx + (∂T/∂y)dy = r ∇ T· ds r , (90) where , rr r ∇=+TiTxjTy∂∂ ∂∂// . Therefore, dT/ds = r ∇ T· ds r /ds, i.e., (91) the gradient dT/ds varies, depending upon the direction of the gradient between any two iso- therms. Along any circular isotherm r ∇ T· ds r = 0 according to Eq. (93), since r ∇ T and ds r are normal to each other. In general, if T=T(x,y,z) then isotherms form surfaces that lie in all three (x,y,z) coor- dinates, and, at any location, r ∇ T represents a vector that lies normal to a scalar surface on which T is constant. d. Curl of a Vector Consider a vector rrrrr rrr ∇×= + + × + +Fi xj ykz iFjFkF xyz (/ / /)( )∂∂ ∂∂ ∂∂ =−+−+− rr r iF z F y jF x F z kF y F x yz zx xy (/ /)(/ /)(/ /)∂∂∂∂ ∂∂∂∂ ∂∂∂∂ (92) The LHS of Eq. (94) is a vector called curl r F . If rr ∇×F = 0, then the two are parallel to each other, i.e., the vector field is irrotational. Assume that r F = r ∇ T. (93) Now assume that instead of a spatial coordinate system, x denotes pressure P, y denotes the specific volume v, and z represents x 1 (i.e., the mole fraction of component 1 in a binary mix- ture), i.e., rr r r r ∇×∇ = − +− +− Ti xTv vTx jPTx xTP kvTP PTv (/ (/) /(/ )) (/(/ ) / (/)) (/(/) /(/)). ∂∂ ∂ ∂ ∂∂ ∂ ∂ ∂∂ ∂ ∂ ∂∂ ∂ ∂ ∂∂ ∂ ∂ ∂∂ ∂ ∂ 11 11 (94) The vector r ∇ T lies in a direction normal to the isothermal surface T, and r ∇ × r ∇ T lies normal to the plane containing r ∇ and r ∇ T. This implies that r ∇ × r ∇ T is a vector that lies back on the isothermal scalar surface T, and, therefore, r ∇ × r ∇ T = 0. Note that the terms in the brackets satisfy the criteria for exact differentials and the RHS of Eq. (96) equals zero. All thermody- namic properties satisfy the irrotationality condition. Functions such as T=T(P,v,x 1 ) are known r ∇ as properties, point functions, scalar functions, or scalar potentials. Terms in exact differential form, such as dT = ∂T/∂P dP + ∂T/∂v dv + ∂T/∂x 1 dx 1 , are called Pfaffians. Chapter 2 2. FIRST LAW OF THERMODYNAMICS A. INTRODUCTION Chapter 1 contains an introduction to thermodynamics, provides some basic defini- tions, a microscopic overview of thermodynamic properties and processes, and briefly reviews the necessary mathematics. We will use that material to formulate thermodynamic laws based either on a generalization of experimental observations, or in terms of four mathematical pos- tulates that are not necessarily based on these experimental results. The laws of thermody- namics are presented in Chapters 2 and 3, and the postulate concepts are addressed in Chapter 5. The thermodynamic laws are simply restrictions on the transformation of energy from one form into another. For examzple, If the thermal energy content of a given mass of steam is 100,000 kJ, it is impossible to obtain a work output of 150,000 kJ from it in the absence of another energy input. Here, the First Law of Thermodynamics provides a restriction. If that same mass of steam containing the same energy content exists at a temperature, it is impossible to obtain a work output of 90,000 kJ from steam at 1000 K. In this case, the re- striction is due to the second law of thermodynamics that constrains the degree of conver- sion of heat energy. In this chapter we will briefly discuss the zeroth and first laws that deal with energy conservation, examine problems involving reversible and irreversible, and transient and steady processes; and, finally, present the formulation of the conservation equations in differential form. The second law and its consequences will be considered in Chapter 3. 1. Zeroth Law The Zeroth law forms the basis for the concept of thermal state (or temperature). Con- sider the body temperature of two persons (systems P 1 and P 2 ) read using an oral thermometer (system T). If the systems P 2 and T are in thermal equilibrium, and so are systems T and P 1 , then systems P 2 and P 1 must exist at the same thermal state. Therefore, both persons will mani- fest the same body temperature. Similarly, if the hot gas inside an electric bulb is in thermal equilibrium both with the electrical filament and the glass wall of the bulb, the glass wall is necessarily in thermal equilibrium with the filament. 2. First Law for a Closed System We will present the First law of thermodynamics for a closed system, and illustrate applications pertaining to both reversible and irreversible processes. a. Mass Conservation For closed systems the mass conservation equation is simply that the mass m = Constant, (1) In the field of atomic physics, mass and energy E are considered convertible into each another and, taken together, are conserved through the well–known Einstein relation E = mc 2 , where c denotes the light speed. However, in the field of thermodynamics it is customary to assume that the conversion of mass and energy into each other is inconsequential and, therefore, either is separately conserved. b. Energy Conservation An informal statement regarding energy conservation is as follows: “Although energy assumes various forms, the total quantity of energy is constant, with the consequence that when energy disappears in one form, it appears simultaneously in others”. i. Elemental Process For a closed system undergoing an infinitesimally slow process, (Figure 1a) during which the only allowed interactions with its environment are those involving heat and work, the first law can be expressed quantitatively as follows δQ – δW = dE, (2) where δQ denotes the elemental (heat) energy transfer across the system boundaries due to temperature differences (Figure 1a), δW the elemental (work) energy in transit across the boundaries (e.g., the piston weight lifted due to the expansion of the system), and dE the en- ergy change in the system. The "E" includes internal energy U (=TE+VE+RE etc.) which re- sides in the matter, kinetic energy KE and potential energy PE. Note that Q and W are transi- tory forms of energy and their differentials are written in the inexact forms δQ and δW (see Chapter 1) while differential of resident energy E is written as an exact differential. Dividing Eq. (2) by m, δq – δw = de, (3a) where q denotes the heat transfer per unit mass Q/m, w the analogous work transfer W/m, and, likewise, e = E/m. It is customary to choose a sign convention for the work and heat transfer that follows common sense. In the absence of work transfer, i.e., δW = 0, addition of heat causes an in- crease in energy. Therefore, it is usual to accord a positive sign for heat transfer into a system. For an adiabatic system (δQ = 0), if the work done by the system is finite and conferred with a positive sign (W > 0), then, from Eq. (2), dE < 0. This is intuitively appropriate, since in order to perform work, the system must expend energy. On the other hand, if the system of Fig. 2 is adiabatically compressed, work is done on the system (so that W < 0), and the stored energy in the system increases (dE > 0). The system energy consists of the internal, potential, and kinetic energies. Equation (2) may be rewritten for a static system in the form δQ – δW = dU. (3b) ii. Internal Energy At a microscopic level the internal energy is due to the molecular energy which is the sum of the (1) molecular translational, vibrational and rotational energies (also called the ther- mal portion of the energy), (2) the molecular bond energy (also called the chemical energy), and the (3) intermolecular potential energy, ipe (cf. Chapter 1). At a given temperature the energy depends upon the nature of a substance and, hence, is known as an intrinsic form of energy. iii. Potential Energy The potential energy of a system is due to the work done on a system to adiabatically move its center of gravity through a force field. The potential energy of a system whose center of gravity is slowly raised vertically (so as not to impart a velocity to it) in the earth’s gravity field through a distance of dz increases by a value equal to mgdz. The first law δQ - δW = dU + d(PE) + d(KE), where PE and KE denote the potential and kinetic energies, can be applied after noting that for this case δQ = dU = d(KE) = 0, so that 0 – δW = 0 + d(PE) + 0. (4) Now, δW = –F dz. The negative sign arises since work is done on the system by a force F that lifts it through a distance dz. In raising the mass, the direction of the force is vertically upward. In the absence of any acceleration of the mass, this force is also called a body force. Using the relation F = mg for the force with g denoting the local gravitational acceleration, the work done W = –mg dz, and using Eq. (4) d(PE) = –δW = mg dz. Integrating this expression across a vertical displacement that extends from z 1 to z 2 , the potential energy change is given as ∆PE = mg(z 2 – z 1 ). The potential energy per unit mass due to the gravitational acceleration at a location z above a stipulated datum is also called the gravitational potential pe, i.e., pe = gz. (5) In SI units, pe can be expressed in J kg –1 or in units of m 2 s –2 , where pe (in units of kJ kg –1 ) = g(in units of m s –2 ) z(in units of m)/1000. In English units, pe can be expressed as BTU lb –1 = g(ft s –2 ) z(ft)÷(g c J), where g c = 32.174 (lb ft s –2 lbf –1 ) is the gravitational constant, and J denotes the work equivalent of heat of value 778.1(ft lbf BTU –1 ). iv. Kinetic Energy In order to move a mass along a level frictionless surface, a boundary or surface force must be exerted on it. Applying the first law, namely, δQ - δW = dU + d(PE) + d(KE), the adiabatic work due to these forces can be expressed as 0 – δW = 0 + 0 + d(KE). (6) The work performed in moving the center of gravity of a system through a distance dx is (– F dx), where the force F = m dV/dt, the velocity V = dx/dt, and t denotes time. In order to be con- sistent with the standard sign convention, the work done on the system is considered negative. Therefore, d(KE) = m (dV/dt)×(V dt) = mVdV. Upon integration, the kinetic energy change of the system as it changes its velocity from a value V 1 to V 2 is ∆KE = (1/2)m( VV 2 2 1 2 − ). The kinetic energy per unit mass ke is ke =1/2 V 2 .(7) In SI units, ke is expressed in J kg –1 , namely, ke(J kg –1 ) = (1/2)V 2 (m 2 s –2 ). Often, it is prefer- able to express ke as ke (kJ kg –1 ) = (1/2000)V 2 . In English units, ke(BTU lb –1 ) = V 2 (ft 2 s –2 )÷(g c J). The kinetic and potential energies are inde- pendent of the nature of the matter within a system, and are known as extrinsic forms of en- ergy. v. Integrated Form Integrating Eq. (2) between any two thermodynamic states (1) and (2) we have Q 12 – W 12 = E 2 – E 1 = ∆E. (8) The heat and work transfers are energy forms in transit and, hence, do not belong to the matter within the system with the implication that neither Q nor W is a property of matter. Therefore, while it is customary to write the energy change for a process E 12 = E 2 –E 1 , we cannot write Q 12 = (Q 2 –Q 1 ) or W 12 = (W 2 –W 1 ). Since for a cycle the initial and final states are identical, δ ∫∫ =QW =0. Writing Eq. (8) on a unit mass basis q 12 – w 12 = e 2 – e 1 = ∆e. (9) The application of the first law to systems require these to be classified as either cou- pled systems in which the transit energy modes, namely, Q and/or W, affect particular storage forms of energy, or as uncoupled systems if the heat and/or work transfer affect more than one mode of energy as illustrated below. vi. Uncoupled Systems Consider an automobile that is being towed uphill on a frictionless road during a sunny summer afternoon from initial conditions Z 1 = V 1 = U 1 = 0 to an elevation Z 2 , velocity V 2 and energy U 2 . Taking the automobile as a system, the heat transfer Q 12 from the ambient to the car is determined by applying Eq. (8), i.e., Q 12 – W 12 = E 2 – E 1 = ∆U + ∆PE + ∆KE, so that Q 12 = ∆U. Therefore, the heat transfer across the boundary increases the system internal energy by ∆U which changes the static state of the system. The work performed to tow the automobile is – W 12 = ∆ PE + ∆ KE, which influences the dynamic state of the system. a. Example 1 A car of mass 2000 kg is simultaneously accelerated from a velocity V = 0 to 55 mph (24.6 m s –1 ) and elevated to a height of 100 m. Determine the work required. Treat the problem as being uncoupled. Solution Q 12 – W 12 = = U 2 – U 1 + KE 2 – KE 1 + PE 2 – PE 1 . 0 – W 12 = (0+(2000÷2)(24.6 m s –1 ) 2 –0+2000×(9.81×100)–0)÷1000 = 2568 kJ. Remark All of the work can be recovered if the car is made to slide down on a frictionless road to ground level (i.e., to zero potential energy) so that the potential energy is completely converted into kinetic energy. Upon impact against a spring the vehicle kinetic energy is fur- ther transformed into the spring potential energy, thereby recovering the work. Hence, the process is uncoupled. vii. Coupled Systems In coupled systems two or more interactions across the system boundary (e.g., heat and work) influence the same energy mode. For example, if a tow truck pulls a car on a rough high-friction road, the work performed is higher than that for an uncoupled system, since addi- tional work is required in order to overcome the external friction. Frictional heating can cause the internal energy of the car tires to increase, and if the tires do not serve as good insulators, heat transfer to the road can occur. Therefore, the work is coupled with both internal energy and heat transfer. This is illustrated in the following example. b. Example 2 A car of mass 2000 kg that is simultaneously accelerated from a velocity V = 0 to 55 mph (24.6 m s –1 ) and elevated to a height of 100 m requires a work input of 3000 kJ. If the car is well insulated, what is the change in the internal energy of the car? Solution Q 12 – W 12 = = U 2 – U 1 + KE 2 – KE 1 + PE 2 – PE 1 . 0+3000=U 2 –U 1 +((2000÷2)(24.6ms –1 ) 2 –0+2000×(9.81×100)–0)÷1000=2568 kJ, i.e., U 2 – U 1 = 3000 – 2568 = 432 kJ. Remarks The work input is more than ∆KE and ∆PE. Thus additional work is used to over- come friction. Frictional work results in heating. If the tires (which are part of the car) are well insulated, their internal energy increases by 432 kJ. In this case the work is coupled to changes in the internal, kinetic, and potential energies of the system. Dividing the work into intrinsic and extrinsic contributions W 12 = W 12,int + W 12 , ext , we find that W 12,int = 432 kJ, which results in the change in U, and that W 12 , ext = 3000 – 432 = 2568 kJ, which results in a change in the kinetic and potential energies, (diathermic) and the tire remains at fixed temperature. Then there is no change in the internal energy. Hence heat must be lost from the tires, i.e., Q 12 – W 12 = ∆U + ∆PE + ∆KE = 0 + ∆PE + ∆KE, where Q 12 = –432 kJ, and W 12 = –3000 kJ. In this case the work is coupled with the heat transfer. The heat transfer affects the intrinsic energy by changing U. When the car moves at a high velocity, frictional drag due to the atmosphere can cause its body to heat, thereby increasing the internal energy. The work done on the car also increases its potential and kinetic energies, and the process becomes coupled. The work cannot be recovered, since the car will contain a higher internal energy even after impacting it against the spring, as illustrated in the previous example. The heating of matter offers another example involving a coupled system. Consider constant pressure heating that causes a system of gases to expand and lift a weight of 100 kg through a distance of 2 m, if Q 12 = 10 kJ, W 12 = 1.96 kJ. (Here we neglect any change in the center of gravity of the matter contained in the system.) If there is no change in the system kinetic energy, from Eq. (8) Q 12 – W 12 = U 2 – U 1 = ∆U. (10) In this case Q 12 ≠ ∆U. As a result of the work and heat interaction, ∆U = 10 – 1.96 = 8.04 kJ. If the system is confined to include only the moving boundary and the lifted weight, and these are considered adiabatic, then (–W 12 ) = ∆(PE), so that the work performed alters the system potential energy. The illustrations of coupled and uncoupled systems demonstrate that it is necessary to understand the nature of a problem prior to applying the mathematical equations. c. Systems with Internal Motion Consider a mass of warm water contained in a vessel. If it is stirred, the entire effort imparts kinetic energy to that mass in the absence of frictional forces, and the center of gravity of each elemental mass of water moves with a specific kinetic energy ke. If the kinetic energy distribution is uniform throughout the system, its total value equals m x ke. Such a situation exists in an automobile engine when fresh mixture is admitted or when the exhaust valve opens. Oftentimes, the kinetic energy is destroyed due to internal frictional forces between the system walls and moving matter, which converts the kinetic energy into internal energy, as in coupled systems. [...]... position, which requires extra work P2 A = Wt, or P2 = Wt /A (A) With Wt = 2 kN, P2= 2 ÷ 0.01 = 20 0 kPa Applying the First Law to the system, Q 12 – W 12 = 0 = E2 – E1, or E2 = E1 Neglecting the kinetic energy, E2 = U2 + Wt Z2, and E1 = U1 + Wt Z1 (B) Substituting Eq (A) in (B), since E2 = E1, U2 –U1 = Wt (Z2 – Z1) = Wt (V2 – V1) ÷ A = P2 (V1 – V2), or m (u2 – u1) = Wt (V1 – V2) ÷ A (C) Treating the air as... /M = 8.314 ÷ 28 .97 = 0 .28 7 kJ kg K , –1 –1 T2/T1 = (20 0÷100 + 0.7÷0 .28 7) ÷ (1 + 0.7 ÷ 0 .28 7) = 1 .29 , or T2= 387 K Substituting this result in Eq (E), V2/V1 = (1 + (cvo/R)(P1/P2))/(1 + cvo/R) (G) ∴ V2/V1 = (1+ 0.7 × 100 ÷ (0 .28 7 × 20 0)) ÷ (1+ 0.7 ÷ 0 .28 7) = 0.65, and V2 = 0.65 × 0.1 = 0.065 m3 Remarks The potential energy of the weight is converted into thermal energy in air Once P2 and T2 are known,... V1/A = 10 cm3 ÷ 10 cm2 = 1 cm, Z2 = 10 cm (B) Applying Eq (8) to system B, i.e., Q 12 – W 12 = E2 – E1 = ∆U + ∆PE + ∆KE, (C) where ∆PE = 50 kg×9.81 m s 2 (10–1)cm×0.01 m cm–1 ÷ (1000 J kJ–1) = 0.044 kJ, and ∆U = 0 Using this result and Eq (A) in Eq (C), 0 – (–0 .23 0) = 0 + ∆KE + 0.044, i.e., ∆KE = (1 /2) m(V 22 – V 12) = 0 .23 0 – 0.044 = 0.186 kJ Since the initial velocity V1 is zero, (1 /2) mV 22/ 1000 = 0.186 kJ,... written in the form m cvo (T2 – T1) = Wt (V1 – V2) ÷ A (D) The two unknowns in Eq (D) are T2, V2, so that an additional equation is required to solve the problem Invoking the ideal gas law for the fixed mass P1V1/RT1 = P2V2/RT2, (E) Equations (D) and (E) provide the solution for V2 and T2 Substituting for V2 from Eq (E) in (D), we obtain a solution for T2/T1, namely, T2/T1 = (P2/P1 + cvo/R)/(1 + cvo/R)... time t+δt to the RHS of Eq (29 ), mc.v,t+δt = mc.v,t + (dmc.v/dt) δt + (1 /2! )(d2mc.v/δ 2) t(δt )2 + …, (30b) where (dmc.v,/dt)t denotes the time rate of change of mass in the c.v at time t Substituting Eq (30b) in Eq (30a) mc.v,t + δmi = mc.v,t + (dmc.v/dt)tδt + (1 /2! )(d2mc.v/dt2)t(δt )2 + … + δme Simplifying, and dividing throughout by δt, δmi/δt= (dmc.v/dt)t + (1 /2! )(d2mc.v/dt2)t(δt) + … + δme/δt (31)... given by the relation Wu = ∫PdV – ∫PodV = ∫(P – Po)dV = ∫PdV – Po(V2 – V1), i.e., Wu = 0 .23 0 – 1 × 100 × 90 × 10–6 = 0 .22 1 kJ Therefore, the kinetic energy change is ∆KE = (0 .22 1 – 0.044) = 0.177 kJ, and V2 = (2 × 1000 × 0.177 ÷ 50)1 /2 = 2. 66 m s–1 g First Law in Enthalpy Form If the kinetic and potential energies are neglected, Eq (2) transforms into δQ – δW = dU The enthalpy can replace the internal... implies that h2 = h1 A process during which the enthalpy is unchanged (i.e., h2 = h1) is called a throttling process Furthermore, since v2 = v1= v, and u2 + P2v2 = u1 + P1v1 (as a consequence of h2 = h1), Slug 5 Slug 1 Inlet q w cv Slug 5 Exit Slug 1 he, Pe , T V Figure 12: First Law applied to a steady state open system in a Lagrangian reference frame c (T2 – T1) = –v(P2 – P1), i.e., T2 – T1 = (0.001... the first law (Eq (2) ) yields, ∫ δQ = ∫ δW , and Eq ( 12) implies that if ( 12) ∫ δQ ≠0, then ∫ δW ≠0 On a unit mass basis ∫ δq = ∫ δw Figure 1b illustrates the cyclical process in a steam power plant for which the heat transfer during the various processes is indicated Applying Eq ( 12) for all processes, i.e., 1 2, 2 3, 3–4, , and 8–1, ∫ δQ = Q 12 + Q 23 +L+ Q 81 = ∫ δW = W 12 + W23 + L + W81 , so that... subscript e refers to the exit conditions Using Eqs (2) , (37), and (39), δQc.m.s – δWc.m.s = Ec.v,t+δt + δmeee – Ec.v,t – δmiei (40) Expanding Ec.v,t+δt using a Taylor's series, Ec.v,t+δt = Ec.v,t + (dEc.v/dt)tδt +(1 /2) (d2Ec.v/dt2)t(δt )2 + …, and using this result in Eq (40), and simplifying δQc.m.s – δWc.m.s = (dEc.v/dt)tδt +(1 /2) (d2Ec.v/dt2)t(δt )2 + … + δmeee – δmiei (41) In general, the work interaction... temperature by integrating Eq (25 ), namely, dhp = cp dT (26 ) The ratio of the two specific heats k = cp/cv is an important thermodynamic parameter Typically the value of k is 1.6 for monatomic gases (such as Ar, He, and Ne), 1.4 for diatomic gases (such as CO, H2, N2, O2) and 1.3 for triatomic gases (CO2, SO2, H2O) g Example 7 Consider an electron gas, the enthalpy of which is h = 3CT6/P2 Obtain an expression . car? Solution Q 12 – W 12 = = U 2 – U 1 + KE 2 – KE 1 + PE 2 – PE 1 . 0+3000=U 2 –U 1 +( (20 00 2) (24 .6ms –1 ) 2 –0 +20 00×(9.81×100)–0)÷1000 =25 68 kJ, i.e., U 2 – U 1 = 3000 – 25 68 = 4 32 kJ. Remarks The. as being uncoupled. Solution Q 12 – W 12 = = U 2 – U 1 + KE 2 – KE 1 + PE 2 – PE 1 . 0 – W 12 = (0+ (20 00 2) (24 .6 m s –1 ) 2 –0 +20 00×(9.81×100)–0)÷1000 = 25 68 kJ. Remark All of the work. V 1 to V 2 is ∆KE = (1 /2) m( VV 2 2 1 2 − ). The kinetic energy per unit mass ke is ke =1 /2 V 2 .(7) In SI units, ke is expressed in J kg –1 , namely, ke(J kg –1 ) = (1 /2) V 2 (m 2 s 2 ). Often,