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

6

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

7

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 .

8

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

14

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 W1 c li gii nh sau:t W1 t f 2 t(x) = t W1 + 2 1) Khi 1 = th t W1 t f 2 q= 1 + 2 t t t(x) = t W1 W1 f 2 x t W1 t f 2 2) Khi 1 = 2 = th q=

15

2.4.3. Vch c thay i theo nhit

Phng trnh cn bng nhit trong vch c (t) ph thuc t s c dngq ( x ) = ( t ) dt . Khi , c th tm t(x) theo phng trnh tch phn dx

(t )dt = q(x )dxKhi cho php tnh gn ng, c th dng cc cng thc tnh t v q nu trn, trong coi l mt hng s, bng tr trung bnh tch phn trong khong nhit [t1, t2] ca vch, l =1 ( t )dt t 2 t1 t1t2

V d, khi (t) c dng bc 1 v 2 tht1 + t 2 1 = 1 (a + bt )dt = a + b 2 t 2 t1 tt2 t +t t 2 + t1t 2 + t 2 1 2 (a + bt + ct 2 )dt = a + b 1 2 + c 1 = t 2 t1 2 3 t1 t2

2.4.4. Vch phng n lp 2.4.4.1. Pht biu bi ton

Cho vch phng n lp, mi lp i c i , i khng i, hai mt ngoi tip xc cht lng nng c tf1, 1 v cht lng lnh c tf2, 2 khng i. Tm dng nhit q qua vch, nhit cc mt tip xc ti v phn b nhit ti(x) trong mi lp.2.4.4.2. Xc nh q, ti, v ti(x). Hnh 7. Vch phng n lp

Khi n nh, dng nhit q qua cc lp l bt bin, do c h phng trnh: q= 1(tf1 t0) =t i t i +1 , (i = 1 n ) = ( t n t f 2 ) i / i

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y l h (n+2) phng trnh bc 1 ca n q v (n+1) n s ti, i=1n. Bng cch kh cc ti s tm c q, sau tnh ti v xc nh ti(x) nh vch 1 lp vi 2 bin loi 1, ta c: q tn = tf2 + , t i = t i +1 + i q, i = (n 1) 0 i 2 t f1 t f 2 , [W/m 2 ] q = 1 n i 1 + + 1 i =1 i 2 t i ( x ) = t i ( t i t i +1 ) x / i , i = 1 n

Phn b nhit trong vch phng nhiu lp c dng cc on thng gy khc, ging nh bin loi 4 Khi vch c bin loi 1 hoc ph thuc t, c th thay tw = tf, 1/ = 0 hoc = = const vo cc cng thc trn.

2.5. DN NHIT QUA VCH TR V VCH CU 2.5.1. Vch tr 1 lp c 2 bin w3 2.5.1.1. Pht biu bi ton

Cho mt ng tr ng cht di v cng, bn knh r2/r1, h s dn nhit khng i, mt r1 tip xc cht lng nng c tf1, 1 , mt r2 tip xc cht lng ngui hn c tf2, 2 . Tm phn b nhit t(r) trong vch v lng nhit qua vch. M t hnh hc trong to tr c dng Hnh 8. Trng t(r) trong ng tr c nh Hnh 82W3

Pht biu ton hc ca bi ny l gii h phng trnh sau:

17

d 2 t 1 dt =0 (1) 2 + r dr dr ( t ) 1 [ t f 1 t (r1 )] = t r (r1 ) (2) t (r ) = [ t (r ) t ] (3) r 2 2 2 f2

2.5.1. Tm trng nhit t(r)

1) Tch phn phng trnh (1) theo cc bc sau: i bin u =rdu + udr d(ur ) du u dt Phng trnh (1) c dng = =0 + = 0 rdr rdr dr r drC1 dt dr = t (r ) = C1 = C1 ln r + C 2 r r dr

d(ur)=0ur=C1 u =

2) Xc nh C1, C2 theo h phng trnh (2), (3): (t f 1 t f 2 ) C 1 C1 = , [K ] 1 [ t f 1 C1 ln r1 C 2 ] = r2 + ln + r1 1 r1 r1 2 r2 C1 = 2 [C1 ln r2 + C 2 t f 2 ] ln r1 ), [K ] C 2 = t f 1 + C1 ( r2 1 r1

Phn b nhit trong ng tr lt (r ) = t f 1 t f1 t f 2 r + ln 2 + r1 2 r2 1 r1 r ln + r r 1 1 1

th t(r) c dng logarit, tip tuyn ti r1 qua im R1(r1-/1, tf1), tip tuyn ti r2 qua im R2(r2+/2, tf2).2.5.1.3. Tnh nhit qua vch tr

1. Dng nhit qua 1m2 mt tr ng nhit bn knh r l q(r) = -tr(r) = -C1/r , [W/m2] q(r) l hm gim khi r tng, khng c trng cho vch tr. 2) Lng nhit truyn qua 1 m di ng tr, k hiu q l , nh ngha l: ql=

lng nhit qua mt tr bn knh r di l / chiu di l , [W/m]C q (r ).2rl = 1 .2r = 2C1 = const , r l r

ql =

Thay C1bigi tr trn, s thu c:18

ql =

t f1 t f 2 , [W/m]. r2 1 1 1 + ln + 2r11 2 r1 2r2 2

V q l = const, r, nn q l c dng c trng cho dn nhit qua vch tr.d 1 1 1 i lng R = + ln 2 + l d 2 d d 2 2 1 1 1 , [mK / W ] c gi l nhit tr dn

nhit ca 1m ng tr.2.5.2. Vch tr c bin hn hp

Khi th thay tw = tf v 1/ = 0 vo trn c li gii cho bi ton vch tr 2 bin hn hp (W1+ W3) hock 2 bin W1 nh sau:t w1 t f 2 r ln t ( r ) = t w1 r r1 ln 2 + r1 2 r2 t w1 t f 2 q l = r 1 1 ln 2 + 2 r1 2r2 2

1) Khi 1 = th

t t r t (r ) = t w1 w1 W 2 ln r r1 ln 2 r1 2) Khi 1 = 1 = th t t q l = w1 f 2 r 1 ln 2 2 r1

2.5.3. Vch tr n lp 2.5.3.1. Pht biu bi ton

Cho ng tr n lp, mi lp i c ri / ri+1 v i khng i, mt r0 tip xc vi cht lng nng c tf1, 1, mt rn tip xc vi cht lng lnh c tf2, 2 kh ng i Tm lng nhit q l , nhit ti ti ccHnh 9. Trng t(r) trong ng tr n l

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mt v phn b ti(n) trong mi lp i, i = 1 n2.5.3.2. Xc nh q l , ti v ti(r)

Khi n nh, phng trnh cn bng nhit cho 1m ng tr l : q l = 1[tf1 t0]2r1 =t i t i +1 , (i = 1 n ) = 2 ( t n t f 2 )2rn ri +1 1 ln 2 i ri

y l h (n+2) phng trnh bc 1 ca 1 n q l v (n+1) n ti. Bng cch kh cc ti tnh q l , sau tm ti theo q l v xc nh ti(r) nh vch c 2W1, s thu c: t f1 t f 2 ql = n r 1 1 1 + ln i +1 + 2r11 i =1 2 i ri 2rn 2 ql q r ; t i = t i 1 l ln i , i = 1 n t 0 = t f 1 2r11 2 i ri 1 t i t i +1 r ln , i = 1 n t i (r) = t i ri +1 ri ln ri

2.5.4. Dn nhit qua vch cu 2.5.4.1. Pht biu bi ton

Cho vch cu ng cht, bn knh r2/r1 c h s dn nhit khng i, mt r1 tip xc cht lng nng c tf1, 1 mt r2 tip xc cht lng lnh c tf2, 2 khng i. Tm phn b nhit t(r) v lng nhit Q qua vch. Trong to cu, trng t(r) c xc nh bi h phng trnh (t) sau:Hnh 10. Phn b t(r) trong vch cu

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d 2 t 2 dt (1) =0 2 + r dr dr ( t ) 1 [ t f 1 t (r1 )] = t r (r1 ) (2) t (r ) = [ t (r t )] (3) r 2 2 2 f2

2.5.4.2. Tm phn b t(r)

1) Tm nghim tng qut theo cc bc: i bin u = dng : u=

dt phng trnh (1) c dr

du u du dr +2 =0 + 2 = 0 tch phn ln 1 c lnu + 2lnr = ln(ur2) =lnC1 dr u r r

C1 dt = tch phn ln 2 c : t(x) = r 2 dr

r

C12

dr =

C1 + C2 r

2) Tm C1, C2 theo 2 iu kin bin (2) v (3):t f1 t f 2 [Km] C1 C1 C1 = 1 1 t f 1 + C 2 = 2 1 1 1 r1 r1 r2 + r2 + r r 2 2 11 1 2 C C 21 = 2 1 + C 2 t f 2 r C = t C + 1 [K ] r2 f1 1 1 2 1 r12 r1

Phn b nhit trong vch cu lt (r ) = t f 1 1 t f1 t f 2 1 1 + r r2 + r 1 1 1 1 1 1 + + + 1 r12 2 r22 r1 r2

th t(r) l ng hyperbol c tip tuyn ti bin qua 2 im R 1 r1 , t f 1 1

,tf2 v R 2 r2 + 2 2.5.4.3. Tnh cng sut nhit Q truyn qua v cuC1 4r2 = -4C1 = const, r 2 r

Q = q(r).(2r2)=-

Thay C1 bi gi tr nu trn, ta c:

21

Q=

t f1 t f 2 , [W] 1 1 1 1 1 1 + + 4 1 r12 2 r22 4 r1 r2 t W1 t W2 ,[W ] 1 1 1 4 r1 r2

Khi vch cu c 2 bin loi W1 th Q =

Khi vch cu c n lp vi 2 bin W3, sau khi gii h phng trnhQ = 1 ( t f 1 t f 0 )4r02 = ( t i t i +1 )4 i , (i = 1 n ) = 2 ( t n t f 2 )4 1 1 + ri ri +1

s tm c:Q= t f1 t f 2 1 1 1 n 1 + + 4 1 r12 2 r22 i =1 4 i 1 1 r r i +1 i ,[w ]

Nhit cc mt ti v trng ti(r) trong cc lp c xc nh nh trn.2.6.DN NHIT QUA THANH HOC CNH C TIT DIN KHNG I.

tng cng truyn nhit, ngi ta thng gn cc cnh ln mt ta nhit. Nhit qua gc cnh c dn qua chiu di x ca cnh, ri to ra mt xung quanh, lm tng lng nhit truyn qua gc. Nhit trong cnh t(x) gim dn theo chiu di x, cn ti mi tit din nhit c coi l phn b u.2.6.1. Pht biu bi ton

Ch mt thanh tr hoc cnh di l, tit din f = const c chu vi la U, mt xung quanh ta nhit ra cht lng nhit tf vi h s ta nhit ; nhit trn mi tit din c coi l phn b u, ti gc l t0 > tf , mt x = l ta nhit ra cng cht lng nhit tf vi h s ta nhit 2. Tm phn b nhit t(x) trong cnhHnh 11. Bi ton t(x) trong thanh tr v cnh phng c tit din khng i 22

v lng nhit Q0 qua gc cnh.2.6.2. Lp phng trnh cn bng nhit tm t(x) 2.6.2.1. Lp phng trnh cn bng nhit tm t(x)

V nhit bn trong thanh ng nht vi nhit bin W3 tc khng c im trong, nn phng trnh t = a 2 cn c thay bng phng trnh cn bng t nhit cho phn t thay dV = fdx, khi n nh c dng: Hiu cc lng nhit dn (vo ra) dV = nhit ta ra mt Udx Nu gi (x) = t(x) t f th phng trnh trn c dng: d d d f + + f = Udx dx dx dx

U d 2 Suy ra f 2 U = 0 . t m = , [m 1 ] f dxd 2 m 2 = 0 (1) Th phng trnh cn bng nhit tm (x) l 2 dx

Nghim tng qut ca (1) l (x) = C1.e mx + C 2 .e mx2.6.2.2 Tm (x) v Q0 cho thanh di hu hn

1)Cc hng s C1, C2 s c tm theo cc iu kin bin W1 ti x=0 v W3 ti x=l(0) = C1 + C 2 = t 0 + t f = 0 ml ml ml ml x (l) = 2(l) (m.C1.e m.C2 .e ) = 2 (C1.e + C 2 .e )

Gii h phng trnh bc nht tm c C1,C2 ri thay vo nghim tng qut v a v dng hm hyperbol shx = (ex + ex)/2 v chx = (ex + ex)/2, thx = shx/chx, s thu c(x) = 0 ch [ m(l x) ] + 2 sh [ m(l x) ] m. ch(ml) + 2 .sh(ml) m.

Trong tnh ton k thut,khi f