Dan Nhiet on Dinh

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CHNG 2

DN NHIT N NH2.1. NH LUT FOURIER V H S DN NHIT 2.1.1. Thit lp nh lut Fourier v dn nhit nh lut Fourier l nh lut c bn ca dn nhit, n xc lp quan h gia 2 vect q v gr adt . thit lp nh lut ny ta s tnh nhit lng 2 Q dn qua mt dS nm gia 2 lp phn t kh c nhit T1 > T2, cch dS mt on x bng qung ng t do trung bnh cc phn t, trong thi gian d , nh hnh H2.Hnh 2. tm dng nhit q

V T1 v T2 sai khc b, nn coi mt phn t n0 v vn tc trung bnh ca cc phn t trong 2 lp l nh nhau , v bng:d 2n = i n 0 dSd 6 i 1 kT1 n 0 dSd v 2 61 1 kT2 n 0 dSd , 2 6 = 8314 = 1,3806.10 23 J / K l hng s Boltzmann, NA l s 6,02217

Lng nng lng qua dS t T1 n T2 ld 2 E 1 = E 1d 2 n =d 2 E 2 = E 2d 2 n =

trong k =

R NA

phn t trong 1 kmol cht kh (s Avogadro), I l s bc t do cu phn t cht kh. Tr 2 ng thc cho nhau, s thu c lng nhit trao i qua dS, bng: 2 Q = ( E 1 E 2 )d 2 n = i 1 k (T1 T2 ) n 0 dSd 2 6

V T1 T2 =

T 2 x v x

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R 1 i i i R n 0k = n 0 = n0 6 6 N A 3 N A 2

1 = C v nn c: 3

1 1 T 2 Q = C v x dSd , ddawtj = C v x 3 3 x

th c

2Q T = q x = . Ss x

y l dng nhit theo phng x. Khi dS c v tr bt k, th vct dng nhit qua T T T + j + k = gr adT dS l q = i x y z

2.1.2. Pht biu v h qu ca nh lut Fourier nh lut Fourier pht biu, rng vect dng nhit q t l thun vi vc t gradien nhit . Biu thc dng vect l q = gr adt , dng v hng lq = gradt = t n (M) . Du (-) v 2 vect ngc chiu nhau.

Nh nh lut Fourier, khi bit trng nhit t(x, y, z,), c th tnh c cng sut nhit Q[W] dn qua mt S [m2] theo cng thc Q = S gradt.dS v tm c lng nhit Q [J] dn qua S sau thi gian [s] theo cng thcQ = 0

S

gradtdSd , [J].

2.1.3. H s dn nhit H s dn nhit l h s ca nh lut Fourier:= q q , [W/mK] = t gradt n

V t l vi q nn c trng cho cng dn nhit ca vt liu. Vi cht kh, theo chng minh trn, c1 1 p 8kT kT 2C v = = C v x = C v m 2d 2 p 3Rd 2 3 3 RT k 2T 3 m

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H s dn nhit ca kh l tng khng ph thuc vo p sut p, tng khi tng nhit hoc tng CV, v gim khi tng hng s cht kh, R = ng knh d hoc tng khi lng m ca phn t cht kh. Vi cc vt liu khc tng theo nhit , c xc nh bng thc nghim v cho bng hoc cng thc thc nghim trong cc ti liu tham kho. V d, tr trung bnh ca h s ca mt s vt liu thng gp c nu ti bng 2. Vt liu Bc ng Vng Nhm Thp Cacbon Yhp CrNi [W/mK] 419 390 313 209 45 17 Vt liu Thu tinh Gch kh Nha PVC Bng thu tinh Polyurethan Khng kh [W/mK] 0,74 0,70 0,13 0,055 0,035 0,026R

, tng

Bng 2. H s dn nhit trung bnh ca cc vt liu thng dng 2.2. PHNG TRNH VI PHN DN NHIT 2.2.1. Ni dung v ngha ca PTVPDN PTVPDN l phng trnh cn bng nhit cho 1 vi phn th tch dV nm hon ton bn trong vt V dn nhit. PTVPDN l phng trnh c bn tm trng nhit t(M, ) trong V, bng cch tnh phng trnh ny. 2.2.2. Thitt lp PTVPDN Xt cn bng nhit cho vi phn th tch dV Hnh 3. Cn bng nhit cho dV bao quanh im M(x,y,z) bt k bn trong vt V, c khi lng ring , nhit dung ring Cp, h s dn nhit , cng sut sinh nhit qv , dng nhit qua M l q .

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nh lut bo ton nng lng cho dV pht biu rng: [ tng enthalpy ca dV] = [hiu s nhit lng (vo - ra)dV]+ [lng nhit sinh ra trong dV]. Trong thi gian 1 giy, phng trnh ny c dng :dVC p t = divq.dV + q v dV hay

t 1 = (q v divq ) C p

Theo nh lut Fourier q = gr adt , khi = const ta c t t t divq = div(gr adt ) = + + = 2 t x x y y z z

2t 2t 2t 2 + 2 + 2 (Trong tao vung goc (xyz)) z y x 2 2t 2t t 1 t 2 vi t = 2 + Trong toa tru (r, , z) + + r r r 2 2 z 2 r2 cos t 2t 2t t 2 t , trong toa cu (r, , ) + + 2 2 + 2 + 2 r 2 r r r r sin r sin 2 2

gi l ton t Laplace ca hm t(M) PTVPDN l phng trnh kt hp 2 nh lut ni trn, c dng:t qv q [m2/s] gi l h s khuch = + 2 t = a v + 2 t , vi a = C p C p

tn nhit, c trng cho mc tiu tn nhit trong vt. 2.2.3. Cc dng c bit ca PTVPDN Phung trnh VPDN tng qut1 T = q V div(gr adt ) s c dng n c P

[

]

gin hn, khi cn p ng cc iu kin c bit sau y: 1) Vt V khng c ngun nhit, qv = 0, th 2) Vi = const, M(x,y,z) V, th1 t = div gr adt C p

(

)

t = a 2 t

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3) Nu nhit n nh trong V,

t = 0 MV, th 2 t = 0

4) Khi trng t(M) l n nh 1 chiu th : t(x) trong to vung gc tm theo t(r) trong to tr tm theo t(r) trong to cu tm theo 2.3. CC IU KIN N TR Phng trnh vi phn dn nhit l phng trnh o hm ring cp 2, cha n l hm phn b nhit t(x,y,z,). Nghim tng qut thu c bng cch tch phn phng trnh ny lun cha mt s hng s tu chn. xc nh duy nht nghim ring ca PTVPDN, cn cho trc mt s iu kin, c gi chung l cc iu kin n tr. iu kin n tr l tp hp cc iu kin cho trc , xc nh duy nht nghim ca mt h phng trnh. 2.3.1. Phn loi cc iu kin n tr Theo ni dung, cc iu kin n tr c phn ra 4 loi sau 1) iu kin hnh hc: Cho bit mi thng s hnh hc xc nh hnh dng, kch thc v tr ca h vt V. 2) iu kin vt l: Cho bit lut xc nh cc thng s vt l ti mi im M V, tc l cho bit (, , a, qv, )= f(MV, t). 3) iu kin u: Cho bit lut phn b nhit ti thi im u = 0 ti mi im MV, tc l cho bit t(M V, = 0) = t(x, y, z). 4) iu kin bin: cho bit lut phn b nhit hoc lut cn bng nhit ti mi im M trn bin W ca vt V ti mi thi im kho st. Nu k hiu dng nhit dn trong vt V n M W l q = cc iu kin bin c dng:t = t n (M) th m t ton hc ca n d2t =0 dx 2

d 2 t 1 dt + =0 dr 2 r dr d 2 t 2 dt + =0 dr 2 r dr

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t w = t (M, ) hoc M W V q = t n (M ) = q (M, , t (M )) xet.

iu kin hnh hc, iu kin vt l v iu kin bin cn phi cho trc trong mi bi ton. Ring iu kin u ch cn cho trong bi ton khng n nh, c cha bin thi gian . 2.3.2. Cc loi iu kin bin. Trn cc bin Wi ca vt V, tu theo phng thc trao i nhit vi cc mi trng m V tip xc, ngi ta c th cho trc 7 loi iu kin bin khc nhau. Bng 3 sau y s tm tt ngha vt l v ton hc, minh ho hnh hc v cc trng hp c bit ca 7 loi iu kin bin quanh vt V bt k. Bng 3. Cc loi iu kin bin. Loi KB ngha vt l hay thng s cho trc M t ton hc hay pt CBN m t hnh hc hay th (t-x) Trng hp c bit

Cho nhit

tw1 = tf khi W1 tw1 = t(M1, ) tip xc cht lng c ln

1

tW1 ti M1W1V

q = const Cho dng 2 nhit q qua M2 W2V =const -tn(M2) = q(M2, ) q=0 W2 l mt i xng hoc cch nhit

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= 0 W3 l Cho mt W3 to nhit ra 3 cht lng nhit tf vi h s -tn(M3)= (t(M3),tf) cch nhit hoc i xng = t(M3) =tf W3 bin thnh W1. Khi (,,tf) = const R c nh Cho W4 tip 4 xc vt V2 ng yn, c 2 , t2 t2 = constW4 t n ( M 4 ) = -

bin thnh W1 (1, 2 )=constg c =const

2t2n(M4)

Cho W5 ho 5 rn t pha

t n ( M 5 ) =

W5 di ng vi tc ho rn bngx 5

x rc 5 f t fn (M 5 ) lng c thng

s (, rc, f, tf)

Mt bao chn cho W6 tip 6 xc chn khng khng c nhit -Tn(M6)= 0T4(M6) Tc. Tn(M6) = 0[T2(M6)Tc2]

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Cho W7 tip 7 xc cht kh c thng s (Tk, )

Quy ra trao i -Tn(M7)=[T(M7)- Tk]+

nhit phc hp -Tn(M7)=ph[t(M7)- Tk]

0[T4(M7) - T4k ]

M t ton hc cho mi loi iu kin bin l phng trnh cn bng cc dng nhit ra vo im M bt k trn bin. Phng trnh m t cc iu kin bin loi 2, 3, 4, 5 l cc phng trnh vi phn tuyn tnh cp 1 i vi t v tn . Phng trnh m t iu kin bin loi 6 v 7 l nhng phng trnh phi tuyn, cha T4 cha bit. 2.3.3. M hnh bi ton dn nhit dng tng qut, bi ton dn nhit c th c m t bi h phng trnh vi phn (t) gm phng trnh vi phn dn nhit v cc phng trnh m t cc iu kin n tr nh nu ti mc 2.3., c dng t 2 qv = a t + , M V Min xac nh va thng s vt ly cua M V t = t (M , ), M W 1 1 1 W1 t n (M 2 ) = q(M 2 , ), M 2 W2 t n (M 3 ) = [ t (M 3 ) t f ], M 3 W3 Hnh 4. M hnh tng qut t n (M 4 ) = 2 t n 2 (M 4 ), M 4 W4 bi ton dn nhit t(x,y,z,) dx 5 t n (M 5 ) = n t n (M 5 ) + rc d , M 5 W5 4 t n (M 6 ) = 0 T (M 6 ), M 6 W6 t n (M 7 ) = [ t (M 7 ) t k ] + 0 [T 4 (M 7 ) Tk4 ], M 7 W7

Gii bi ton dn nhit l tm hm phn b nhit t(M(x,y,z),) tho mn mi phng trnh ca h (t) ni trn. Vic ny gm c 2 bc chnh l tch phn phng trnh vi phn dn nhit tm nghim tng qut, sau xc nh cc hng s theo cc phng trnh m t cc iu kin n tr.

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2.4. DN NHIT QUA VCH PHNG Dn nhit n nh qua vch phng l bi ton n gin nht ca truyn nhit. Tu theo kt cu vch v iu kin bin, bi ton dn nhit s c phn ra cc loi sau y. 2.4.1. Vch phng 1 lp c 2 bin loi 3 2.4.1.1. Pht biu bi ton Cho 1 vch phng dy rng v hn, lm bng vt liu ng cht c h s dn nhit khng i, 2 mt bn tip xc vi 2 cht lng c nhit khc nhau tf1 > tf2 , vi h s to nhit vo ra vch l 1, 2. Tm phn b nhit t(x) trong vch v dng nhit q(x) qua vch. Theo ton hc, pht biu trn tng ng vi vic tm hm t(x), x[0, ] nh l nghim ca h phng trnh (t) sau y. d2t 2 =0 dx (t) 1 [ t f 1 t (0)] = t x (0) t () = [ t ( ) t ] x 2 f2 (1) ( 2) (3)

Hnh 6. Trng t(x) trong vch phng c 2W3

2.4.1.2. Tm phn b nhit t(x). 1) Tm nghim tng qut bng cch tch phn phng trnh (1), ta c :t ( x ) = dx 2 = C1 x + C 2

2) Xc nh C1 , C2 theo 2 iu kin bin (2) v (3) (t f 1 t f 2 ) , [ K / m] C1 = 1 [ t f 1 C 2 ] = C1 ++ 1 2 C1 = 2 [C1 + C 2 t f 2 ] C1 , [ K ] C 2 = t f 1 + 2

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Phn b nhit trong vch l t(x)= tf1 -

++ 1 2

t f1 t f 2

(x +

) 1

Bng cch thay x bng 0 hoc , ta d dng tm c nhit ti 2 mt vch. th t(x) l mmt on thng i qua 2 im nh hng R1(-/1, tf1) v R1( + /2, tf2) nh hnh H 2.4.1.3. Tm dng nhit q(x): theo nh lut Fourier c q(x) = -gradt(x) = -C1 = const, x hay q =t f1 t f 2 , [W/m2] 1 1 + + 1 2

Nu gi R =q=

1 1 + + , [m2K/W], l nhit tr dn nhit ca vch phng, th c 1 2

V V2 t f 1 t f 2 , tng t nh cng thc tnh dng in I = 1 . R R Bin loi 1 l trng hp c bit ca bin loi 3, khi mt vch tip xc vi

2.4.2. Vch phng c bin loi 1.

mt cht lng thc c h s to nhit rt ln. Theo phng trnh cn bng nhit cho bin loi 3, (tw-tf) = -tn ,v q = -tn l hu hn,nn khi th (tw-tf) 0, tc l tW = tf. khi ch cn thay tw = tf v 1/ =0 vo cc kt qu nu trn, ta c th tm t(x) v q(x) cho bi ton bin loi 1. V d: bi ton bin hn hp (W1 + W3) v bi ton 2 bin W