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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 = dQrev /T (but not dWrev/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 V1 and V2 at the same temperature that, respectively, contain nitrogen and oxygen. If the partition between them is removed, then N2 and O2 gases have a new set of quantum states due to extension of volume from V1 and V2 to V1+ V2 . 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 6.023x1026 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 H2O 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 6x1026 molecules) of water in the mixture ^uH2O, is different from that of a kmole of pure water ¯uH2O. The heat at the top of ¯uH2O indicates property when the component is inside the mixture. Here, ^uH2O 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 ^uH2O = ¯uH2O, for an ideal gas mixture and µk=µk since attractive forces do not influence the property. However, even then, ^sH2O would not equal ¯sH2O, 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 CO2 0.033 0.050 N2 78.084 75.521 O2 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(lx1, lx2,...) = lmF(x1,x2,...), and (79a) x1,new = lx1, x2,new = lx2,.... Differentiating Eq. (80) with respect to l (and treating it as a vari-able), (¶F/¶(lx1,new))(¶x1,new/¶l)+(¶F/¶(lx2,new))(¶x2,new/¶l)+… = mlm–1F(x1,x2,...). (81b) Since ¶x1,new/¶l = x1, ¶x2,new/¶l = x2, …, Eq. (81b) assumes the form (¶F/¶(lx1,new))x1 + (¶F/¶(lx2,new))x2 + … = mlm–1F(x1,x2,...). Multiplying both sides of the above equation by l, and noting that lmF(x1,x2,...) = F(x1,new, x2,new,...), we have the relation (¶F/¶(lx1,new))x1,new + (¶F/¶(lx2,new))x2,new + … = mF(x1,new, x2,new,...). (80) If m = 1, åK 0 xk (¶F / ¶xk ) = mF . (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 by a power boat that applies a force of F on the board. The work done is given as dW = F×ds = F cosq ds, where q denotes the angle between the force and the elemental path. ii. Work Done to Move an Electrical Charge r Similarly if an electrical charge of strength Q is located at an origin, the force F ex- erted by it on another charge of strength q situated at distance r removed from the origin is F = (eqQ r)/|r3 |, where e 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 dW = F×dr. b. Vector or Cross Product The area A due to a vector product A = x ´ y, (82) can be written in the form A = k | x || y |sinq, (83) r where k denotes the unit vector in a plane normal to that containing the vectors x and y, and q 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 F = i Fx + jFy = i Fcosq+ jFsinq, (84) where q denotes the angle between the force and an arbitrary x–wise coordinate at any instant, and i and j denote unit vectors in the x– and y– directions, respectively. The torque exerted about the center B = F´ r = (i Fcosq+ jFsinq)´ (i x+ jy) = k(yFcosq+ xFsinq), (85) where i ´ i = 0, i ´ j= k, and j´ i= −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 q y – Fsinq x) = 0 in Eq.(87), then there is no rotation around the z–axis. In general, a force has three spatial components, i.e., F = iFx + jFy + kFz , (86) and the torque is described by the relation B = F´ r = i(Fyz+ Fzy)+ j(Fzx + Fxz)+ k(Fxy+ Fyx), i.e., (87) r there are rotational components in the x– and y– directions also. If F and r are parallel to each other, e.g., F = iFx and r = ix, then B = F´ r = 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 ÑT (i¶ /=¶x j¶ /¶+y k¶ /¶+z)T, (88) which for the one–dimensional problem assumes the form ÑT = j¶T /¶y, (89) The x–z plane contains isotherms, since T¹T(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 T¹T(z). Assume an axisymmetric problem for which the iso- therms are circular in the x–y plane and form cylindrical surfaces. In this case, Ñ dT = (¶T/¶x)dx + (¶T/¶y)dy = ÑT·ds, (90) where , ÑT i¶T /=¶x j¶T /+¶y. Therefore, dT/ds = ÑT·ds/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 ÑT·ds = 0 according to Eq. (93), since ÑT and ds 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,ÑT represents a vector that lies normal to a scalar surface on which T is constant. d. Curl of a Vector Consider a vector Ñ F´ =i¶ /¶x j¶ /¶y k¶ /¶z) ´iFx j+Fy k+Fz) = i(¶Fy /¶z−¶Fz /¶y)+ j(¶Fz /¶x −¶Fx /¶z)+ k(¶Fx /¶y−¶Fy /¶x) (92) The LHS of Eq. (94) is a vector called curl F. If Ñ´ F= 0, then the two are parallel to each other, i.e., the vector field is irrotational. Assume that r r F = T. (93) Now assume that instead of a spatial coordinate system, x denotes pressure P, y denotes the specific volume v, and z represents x1 (i.e., the mole fraction of component 1 in a binary mix-ture), i.e., Ñ ´ ÑT = i(¶ /¶x1 (¶T /¶v)−¶ /¶v(¶T /¶x1)) + j(¶ /¶P(¶T /¶x1)−¶ /¶x1 (¶T /¶P)) (94) + k(¶ /¶v(¶T /¶P)−¶ /¶P(¶T /¶v)). The vector ÑT lies in a direction normal to the isothermal surface T, and Ñ´ÑT lies normal to the plane containing Ñ and ÑT. This implies that Ñ´ÑT is a vector that lies back on the isothermal scalar surface T, and, therefore, Ñ´ÑT = 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,x1) are known as properties, point functions, scalar functions, or scalar potentials. Terms in exact differential form, such as dT = ¶T/¶P dP + ¶T/¶v dv + ¶T/¶x1 dx1, are called Pfaffians. ... - tailieumienphi.vn
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