Energy Equation

Energy equation The Lagrangian form of the energy equation is dE (1) dt Q-W total energy E Qrate of heat transfer; + when added to system W rate of work; + when done by system Here, E is our extensive property. The associated intensive property will be made up of two terms: e internal energy per unit mass 1 2 kinetic energy per unit mass 11 Thus, the intensive property is ple+u²) energy/volume, where de codT, c, specific heat at constant volume. c, is the ratio of the amount of heat, Q, it takes to raise a mass, m, by an amount AT. c,Q/MAT. RTT says that dE D IP D dV=Q-W (2) Recall this? When an intensive property is a product of p, the constant equation wipes out the ple+fu²) u term. Now consider Q, the heat transfer bit. We want it in integral form, and it's the heat transfer across the boundaries of our blob, so it will be a surface integral. (3) What's up with the sign? Q is positive when work is added to the system, but something fluxing into a control volume has a negative sign since da is positive outward. Now for the W term. Work is force times distance, WF-Ax. So the work rate will be force times distance divided by time, WF-, or force times velocity W Fu (4) Recall from our derivation of the momentum equations, that the forces acting on our blob of fluid are body and surface forces: (5) F = FR+Fs This ignores the line force, which is surface tension. Thus F-u+Fu 1 (6) From our work on the momentum equations, the only body force that we are considering so far, is gravity. It acts on all the mass in the blob, so we have the volume integral Fau (7) Note that W0 when work is done by the system, and since g is on the system, Fau < 0. Hence the minus sign. Again, from our work on momentum, we went through a lot of pain to convince ourselves that the surface force was given by the stress tensor. It's force on the surface of the blob, so it will be an area integral Fg-u = -ff (ru) dA (8) W>0 when work is done by the system, but here we have work being done on the system, so we need the minus sign. We now have all the terms for the energy equation (1) dE dt (-)------adv ---A dV q-dA Dt D (+) = If pg-udv+ff(r-u)-da-q-dA A (9) Now we need to convert the two area integrals on the RHS into volume integrals using the divergence theorem: (ru) dA --dA = -(ru)dV = --adv (10) (11) Inserting these results in the energy equation (9) yields D DI 2 (+)=u+V (ru) -V-qdV V (12) Since this integral must hold for all volumes (13) = (+) This is the energy equation. Note rising motion leads to an increase in potential energy as g-u < 0, as it should since it is on the RHS. The terms: pgu changes in potential energy changes due to heat flux V(ru) has a kinetic energy part and an internal heating part 2 Let's look into the term containing stress a bit more deeply as it is clearly the most painful of the terms: V(ru)u(r) +: Vu (14) The first term, u-(V-r), is the rate of work done by the surface forces. Imbalances in stress accelerate the fluid blob and change its kinetic energy. The second term, : Vu, is the work done by deformation. Not all stress moves the blob some of it deforms and causes heat change. Note the double contraction, A: B⇒ scalar. The u (Vr) terms is the easiest of the two since we came up with an expression for Vr when we looked at momentum conservation. VT = -(2---) = -p+²+(-) (r) = (-p+²+(-)) (15) We're advecting 7 through our fixed blob of fluid. It is bringing in kinetic energy associated with the stress on the blob. Now the second term, T: Vu double contraction. Notice, there is no between the V and the u. Vu is the gradient of the velocity... The velocity gradient tensor! (16) Vu = G Thus, this term looks like (17) But Ge+r, (1+0) (18) Recall that ris an antisymmetric tensor, and is a symmetric tensor. The double dot product of asymmetric and antisymmetric matrix is zero (19) 33 A:B = ΣΣΑΒ In 2D (0)+(-)+(2)+(0) [mm]:[ 711 712 731 722 = 712 0 (20) This last line is due to the fact that since r is symmetric, Thus, you can get rid of the r term and end up with rie (21) 3Substitute in our expression for (2e-p+-1):e 2 ul (22) Let's see what (if anything) we can learn by performing these double contractions. We'll start with an easy one ple θυ = = Vu (23) That is, change in energy due to compression or expansion. The last term is also pretty easy θυ θα + (u) () (24) Finally, the first term 2µε: ε 611 12 13 Cai C32 C116LL+ 021612+ €12022 22+23632+ #11 12 €13 €25 €22 ele = (25) Note = 12621 + 2 = ()) 18αθυ 1 (θυ = (26) etc. (27) Consider just the enen term. Remember, 2με: ε Ju (28) The u bit is associated with normal stress, and with the velocity gradient. Thus, this term has something to do with the work it takes to pull apart molecules in a blob of fluid. The other