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The Energy Equation + APPENDIX A: Conservation of Energy: The Energy Equation 507 APPENDIX A CONSERVATION OF ENERGY: THE ENERGY EQUATION The derivation of energy equation (2.15) in y dz Section 2.6 is presented in detail. We dy consider the element dxdydz in Fig. A.1 dx and apply conservation of energy (first law of thermodynamics). We assume contin- x uum and neglect nuclear, electromagnetic z and radiation energy transfer. Our starting Fig. A.1 point is equation (2.14) [1]: A B Rate of change of Net rate of internal and kinetic internal and kinetic energy transport by convection energy of element (2.14) C D _ + Net rate of heat added Net rate of work done by by conduction element on surroundings Note that net rate in equation (2.14) refers to rate of energy added minus rate of energy removed. We will formulate expressions for each term in equation (2.14). (1) A = Rate of change of internal and kinetic energy of element The material inside the element has internal and kinetic energy. Let uˆ internal energy per unit mass V magnitude of velocity Thus w A >@U (uˆ V 2 / 2) dxdydz . (A-1) wt 508 APPENDIX A. Conservation of Energy: The Energy Equation (2) B = Net rate of internal and kinetic energy transport by convection w (uˆ V 2 / 2)U w dxdy ( uˆ V 2 / 2)U w@ dxdydz wz > (uˆ V 2 / 2)Uu dydz (uˆ V 2 / 2)Uu dydz dy w 2 (uˆ V / 2)Uu@dxdydz wx> dz dx (uˆ V 2 / 2)U w dxdy Fig. A.2 Mass flow through the element transports kinetic and thermal energy. Fig. A.2 shows energy convected in the x and y-directions only. Not shown is energy carried in the z-direction. To understand the components of energy transport shown in Fig. A.2, we examine the rate of energy entering the dydz surface. Mass flow rate through this area is U udydz . When this is 2 multiplied by internal and kinetic energy per unit mass, (u V / 2) , gives 2 the rate of energy entering dydz due to mass flow (u V / 2)U udydz. Similar expressions are obtained for the energy transported through all sides. Using the components shown in Fig. A.2 and including energy transfer in the z-direction (not shown) we obtain B = (uˆ V 2 / 2) U u dydz (uˆ V 2 / 2) v dxdz (uˆ V 2 / 2) U wdxdy w (uˆ V 2 / 2) U u dydz >@(uˆ V 2 / 2) U u dxdydz wx w (uˆ V 2 / 2) U v dxdz >@(uˆ V 2 / 2) U v dxdydz wy w (uˆ V 2 / 2) U wdxdy >@(uˆ V 2 / 2) U w dxdydz . wz Simplifying ­ w 2 w 2 w 2 ½ B ® >@(uˆ V / 2) U u >@(uˆ V / 2) U v >@(uˆ V / 2) U w ¾u dxdydz. ¯wx w y w z ¿ APPENDIX A. Conservation of Energy: The Energy Equation 509 Making use of the definition of divergence (1.19) the above becomes 2 B ^ x >(uˆ V / 2) UV @ `dxdydz . (A-2) (3) C = Net rate of heat addition by conduction Let wqcc (qcc y dy)dxdz y w y qcc heat flux = rate of heat conduc- wqxcc tion per unit area qccdydz (qcxc dx)dydz x dy w x Fig. A.3 shows the z-plane of the dx element dxdydz. Taking into consid- qccdxdz eration conduction in the z-direction, y the net energy conducted through the Fig. A.3 element is given by w qcxc C qcxcdydz qcycdxdz qczcdxdy (qcxc dx)dydz w x w qcyc w qczc (qcyc dy)dxdz (qczc dz)dxdy. w y w z Simplifying ªwqcxc wqcyc wqczc º C « »dxdydz . ¬ wx wy wz ¼ Introducing the definition of divergence C ( x qcc) dxdydz . (A-3) (4) D = Net rate of work done by the element on the surroundings Rate of work is defined as forceu velocity. Thus Rate of work = force u velocity. Work done by the element on the surroundings is negative because it represents energy loss. We thus examine all forces acting on the element and their corresponding velocities. As we have done previously in the formulation of the equations of motion, we consider body and surface forces. Thus 510 APPENDIX A. Conservation of Energy: The Energy Equation D Db Ds , (A-4) where Db = Net rate of work done by body forces on the surroundings Ds = Net rate of work done by surface forces on the surroundings ConsiderDb first. Let g x , g y and g z be the three components of gravitational acceleration. Thus Db is given by & V Db = U(g xu g y v g z w) dxdydz , or D = (V x g) (A-5) & b U U g To further clarify the negative sign consider a fluid particle Fig. A.4 being raised vertically, as shown Fig. A.4. If the particle is & & being raised, work is being done on it, and yet the dot product V g is negative; hence the negative sign being added to the work term to make the work positive. Next we formulate an equation for rate of work done by surface stresses Ds . Fig. A.5 shows an element with some of the surface stresses. For the purpose of clarity, only stresses on two faces are shown. Each stress is associated with a velocity component. The product of stress, surface area and velocity represents rate of work done. Summing all such products, we obtain wW W xz dx xz wx u W xy wu u dx V xx wx W xz dz dy dx y z wW xy W xy dx x wx Fig. A.5 APPENDIX A. Conservation of Energy: The Energy Equation 511 § wu ·§ wV xx · Ds ¨u dx¸¨V xx dx¸dydz u V xx dydz © wx ¹© wx ¹ § ww ·§ wIJ xz · ¨w dx¸¨IJ xz dx¸dydz w IJ xz dydz © wx ¹© wx ¹ § wv ·§ wW xy · dx ¨ dx¸dydz dydz ¨v ¸¨W xy ¸ v W xy © wx ¹© wx ¹ § ww ·§ wV zz · ¨w dz¸¨V zz dz¸dxdy w V zz dxdy © wz ¹© wz ¹ § wv ·§ wW zy · dz ¨ dz¸dxdy dxdy ¨v ¸¨W zy ¸ v W zy © wz ¹© wz ¹ § wv ·§ wV yy · dy ¨ dy¸dxdz dxdz ¨v ¸¨V yy ¸ v V yy © wy ¹© wy ¹ § wu ·§ wIJ yx · u dy ¨ dy¸dxdz u dxdz ¨ ¸¨IJ yx ¸ IJ yx © wy ¹© wy ¹ § ww ·§ wIJ yz · w dy ¨ dy¸dxdz w dxdz ¨ ¸¨ IJ yz ¸ IJ yz © wy ¹© wy ¹ § wu ·§ wIJ zx · ¨u dz¸¨IJ zx dz¸dxdy u IJ zx dxdy . © wz ¹© wz ¹ Note that negative sign indicates work is done by element on the surroundings. Neglecting higher order terms the above simplifies to °­ § wV wW yx wW · § wW xy wV yy wW zy · D u¨ xx zx ¸ ¨ ¸ s ® ¨ ¸ v¨ ¸ ¯° © wx wy wz ¹ © wx wy wz ¹ § wW wW yy wV · § wu wu wu · w¨ xz zz ¸ ¨ ¸ ¨V xx W yx W zx ¸ © wx wy wz ¹ © wx wy wz ¹ § wv wv wv · § ww ww ww ·½ ¨W xy V yy W zy ¸ ¨W xz W yz V zz ¸¾dxdydz. © wx wy wz ¹ © wx wy wz ¹¿ (A 6) Substituting (A-5) and (A-6) into (A-4) 512 APPENDIX A. Conservation of Energy: The Energy Equation & & ª w D U V g dxdydz « (uV xx vW xy wW xz ) ¬wx w w º (uW yx vV yy wW yz ) (uW zx vW zy wV zz )»dxdydz. (A - 7) wy wz ¼ Substituting (A-1), (A-2), (A-3) and (A-7) into (2.14) w ª § 1 2 ·º ª§ 1 2 · &º & & «U¨uˆ V ¸» «¨uˆ V ¸UV » qcc U V g wt ¬ © 2 ¹¼ ¬© 2 ¹ ¼ w w (uV xx vW xy wW xz ) (uW yx vV yy wW yz ) wx wy w (uW zx vW zy wV zz ). (A - 8) wz Note that equation (A-8) contains the nine normal and shearing stresses that appear in the formulation of the momentum equations (2.6). We will now use (2.6) to simplify (A-8). Multiplying equations (2.6a), (2.6b) and (2.6c) by the velocity components u, v and w, respectively, and adding the resulting three equations, we obtain § Du Dv Dw · U¨u v w ¸ U ug x vg y wg z © Dt Dt Dt ¹ § wV wW yx wW · § wW xy wV yy wW zy · § wW wW yz wV · u¨ xx zx ¸ ¨ ¸ w¨ xz zz ¸. ¨ ¸ v¨ ¸ ¨ ¸ © wx wy wz ¹ © wx wy wz ¹ © wx wy wz ¹ (A-9) However, 2 § Du Dv Dw· U DV U¨u v w ¸ , (A-10) © Dt Dt Dt ¹ 2 Dt and & & ug x vg y wg z V g (A-11) Substituting (A-10) and (A-11) into (A-9) APPENDIX A. Conservation of Energy: The Energy Equation 513 * 2 U DV & & § wV wW yx wW · § wW xy wV yy wW zy · V g u¨ xx zx ¸ ¨ ¸ U ¨ ¸ v¨ ¸ 2 Dt © wx wy wz ¹ © wx wy wz ¹ § wW wW yz wV · w¨ xz zz ¸ ¨ ¸ © wx wy wz ¹ . (A-12) Returning to (A-8), the first and second terms are rewritten as follows w ª § 1 2 ·º § 1 2 · wU w § 1 2 · «U¨uˆ V ¸» ¨uˆ V ¸ U ¨uˆ V ¸ , (A-13) wt ¬ © 2 ¹¼ © 2 ¹ wt wt © 2 ¹ ª§ 1 2 · &º § 1 2 · & & § 1 2 · «¨uˆ V ¸UV » ¨uˆ V ¸ UV UV ¨uˆ V ¸ . (A-14) ¬© 2 ¹ ¼ © 2 ¹ © 2 ¹ Substituting (A-12), (A-13) and (A-14) into (A-8) D § 1 2 · ¨uˆ V ¸ Dt 2 0 © ¹ § 1 2 ·§ wU &· ª w § 1 2 · & § 1 2 ·º ¨uˆ V ¸¨ UV ¸ U« ¨uˆ V ¸ V ¨uˆ V ¸» qcc © 2 ¹© wt ¹ ¬wt © 2 ¹ © 2 ¹¼ U DV 2 § wu wu wu · § wv wv wv · ¨V xx W yx W zx ¸ ¨W xy V yy W yz ¸ 2 Dt © wx wy wz ¹ © wx wy wz ¹ § ww ww ww · ¨W xz W yz V zz ¸ 0.
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