Heat flux and total heat
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1 Het flux nd totl het John McCun Mrch 14, Introduction Yesterdy (if I remember correctly) Ms. Prsd sked me question bout the condition of insulted boundry for the 1D het eqution, nd (bsed on glnce t the derivtion in the textbook) I gve her n incorrect explntion. Thinking more bout the question lter, I decided to try to clrify the question in my mind nd (mong other things) relized tht the book derivtion seems to be not quite correct. Fortuntely, I think my derivtion, given twice in clss is t lest somewht better nd leds to reltively resonble explntion of the insulted boundry condition. I will review the derivtion here, critique Zill s derivtion, nd discuss the boundry condition. 2 Derivtion My derivtion, just s Zill s, is bsed on two min ssumptions. The first is tht the het (energy) density q, given s function of position nd time within the medium, is proportionl to the temperture. Thus, the totl het energy in given volume within the medium is given by q = γ u. (1) It follows tht the rte of chnge of the totl het energy within is given by d q = γ u t. (2) dt 1
2 The second ssumption is tht the het energy flux density is proportionl to (nd in the opposite direction of) the grdient of the temperture: v = K u. (3) It will be noted tht this is vector field given s function of position nd time within the medium. In prticulr, when we sy this is flux vector field, we men tht the rte of chnge of het energy within ny volume cn be computed by determining the rte t which het enters through the boundry: v ( n) = divv = K u. (4) Here, of course, we hve used the divergence theorem nd n is the outwrd unit norml to. It should be noted tht I hve clled the reltion (3) Fourier s lw of het conduction, nd I think tht s correct. Equting the lst expressions in (2) nd (4), we get γ u t = K u, (5) or (u t Kγ ) u = 0. (6) The fundmentl lemm of vrition (of the domin ) then implies tht the het eqution holds: u t = K γ u. 3 The 1D cse In one spce dimension, i.e., for the model of thin rod, the flux condition becomes v = Ku x e 1, (7) the domin is n intervl, sy (, b), nd evlution of the flux integrl over the boundry the boundry is simply given by evlution of v ( n), tht is, Ku x (b, t) Ku x (, t) = K 2 u xx (ξ, t) dξ (8)
3 becuse the outwrd norml t x = b is e 1 nd the outwrd norml t x = is e 1. Specilizing to n intervl (, b) contining x the rest of the derivtion goes through in essentilly the sme wy s the higher dimensionl cse: so γ u t (ξ, t) dξ = K 4 Something to note u t = (K/γ)u xx. u xx (ξ, t) dξ, The sense in which the word density is used bove in regrd to (1) nd (3) is different in ech cse. This my be the source of the pprent confusion in Zill s book t discussed below. When we tlk bout the het density q we re tlking bout quntity which cn be integrted over set of full sptil dimension within the medium to get the totl het energy. Thus, this density is energy per mss, or equivlently energy per volume. Denoting the dimension of this volume, sy the volume of region in the medium, by n, the dimension of the boundry of is n 1. Thus, when we tlk bout the het energy flux density v ppering in (3), we re considering density with respect to the geometric mesure of, i.e., boundry density of dimension n 1. 5 Zill s derivtion Zill gives his two ssumptions s nd Q = γmu (9) Q t = KAu x. (10) Notice, in prticulr, tht Fourier s lw hs become sclr reltion, which is one problem, but not the only one. The big problem is tht these reltions must be interpreted some wy, nd probbly the only resonble wy to interpret them is tht there is quntity Q depending on spce nd time for which given length of rod between x nd x + h Q(x, t) = γρhu(x, t) (11) 3
4 nd Q t (x, t) = KAu x (x, t). (12) In this cse, it s resonble to ssume γ, ρ, h, K, nd A re ll constnts or t lest cn be constnts nd one obtins directly the reltion u t = KA γm u x. This is clerly not wht Zill hd in mind, but ccording to his nottion, which strongly suggests Q t = dq dt, (13) this conclusion seems to be pretty much unvoidble. The problem, which is hopefully pretty cler by now, is tht the rte of het flow (which Zill denotes Q t ) through the cross-section of the rod is not the derivtive with respect to time of the quntity of het (which Zill denotes Q) contined in prticulr (smll) length of rod from x to x + h. One wy to see the obvious identifiction (13) cnnot be correct is to imgine length of rod in which there is uniform distribution of het energy but het flows from one end to the other t uniform rte. Tht is, s het (energy) enters one end of the length of rod, het (energy) exits the other end t the sme rte. In this scenrio, the mount of hest in tht length of rod remins constnt s does the temperture, so dq dt = 0 nd u t = 0. On the other hnd, the het flux is nonzero, so we should hve v = Ku x e 1 0. Thus, one sees tht t no level is it correct to ssume Q t = dq dt = KAu x. This, it seems to me, is not correct sttement of Fourier s lw. 4
5 6 The boundry condition Finlly, let us note tht under my version of Fourier s lw (3), the interprettion of the bsense of het flux, or the flow of het energy, in prticulr direction t prticulr point, is cler. Tht is, if w is given direction t point x, then the bsence of het flow is quntified by v w = 0 tht is u(x, t) w = 0. In one spce dimension, this becomes the condition u x (x, t) = 0. Applying these conditions to the boundry, nd insulted boundry with norml w = n is one for which u(x, t) n = 0 where x is point on the boundry of the medium where the norml is n. For the 1D cse, we get simply u x (0, t) = 0 or u x (L, t) = 0 t the boundry points x = 0 nd x = L of the rod respectively. 5
Consequently, the temperature must be the same at each point in the cross section at x. Let:
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