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Phn II

Truyn nhit

Chng 3Dn nhit v i lu

3.1 Dn nhit

3.1.1 Dn nhit n nh mt chiu khng c ngun nhit bn trong

3.1.1.1 Dn nhit qua vch phng

=

+

=

n

1i i

i

)1n(W1W ttq , W/m2 (3-1)

q mt dng nhit, W/m2

i - chiu dy ca lp th i, mi - h s dn nhit, W/m.K;tW1 nhit b mt trong,tW(n+1) nhit b mt ngoi ca lp th n.

Phn b nhit theo chiu dy vch c qui lut ng thng(khi I = const).

3.1.1.2 Dn nhit qua vch tr

,

d

dln

2

1tt

qn

1i i

1i

i

)1n(W1Wl

=

+

+

= , (W/m) (3-2)

q mt dng nhit trn mt mt chiu di, W/m

di - ng knh ca lp th i, mPhn b nhit theo chiu dy vch c qui lut ng cong logarit.

3.1.2 Dn nhit n nh mt chiu khi c ngun nhit bn trong

3.1.2.1 Tm phng c chiu dy 2)x(

2

qqtt 22vvf

+

+= (3-3)

Nhit b mt tm:

+= vfw

qtt (3-4)

Nhit ti tm ca tm:

2vvf0 2

qqtt

+

+= (3-5)

tf nhit moi trng xung quanh,

i - h s to nhit, W/m2.K;

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qv nng sut pht nhit ca ngun bn trong, W/m3

Phn b nhit theo chiu dy vch c qui lut ng cong parabol.

3.1.2.2. Thanh tr ng cht bn knh r0

)xr(4

qrqtt 220

v0vf

+

+= (3-6)

Nhit b mt thanh tr:

+=

2

rqtt 0vfw (3-7)

Nhit ti tm ca tm:2

0v0v

f0 r4

q

2

rqtt

+

+= (3-8)

Mt dng nhit ti b mt:

2

rqq 0vw = , W/m

2 (3-9)

Phn b nhit theo chiu dy vch c qui lut ng cong parabol.

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3.1.2 Dn nhit khng n nh

Vi tm phng rng 2Nhit ti tm ca tm:*x=0 = f1(Bi/Fo) tra th hnh 3.1

Nhit b mt tm:*x=1 = f2(Bi/Fo) tra th hnh 3.2

trong :

=Bi , l tiu chun Biot,

2

aFo

= , l tiu chun Fourier

= xX , kch thc khng th nguyn.

Phn b nhit theo chiu dy vch c qui lut ng cong parabol.

3.2 trao i nhit i lu

Khi tnh ton lng nhit trao i bng i lu ta dng cng thc Newton:],W[),tt(FQ fW =

trong :Q lng nhit trao i trong mt n v thi gianl mt giy, s.

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F din tch b mt trao i nhit, m2TW Nhit trung bnh ca b mt,Tf Nhit trung bnh ca mI trng (cht lng hoc kh). - h s ta nhit, W/m2.K.H s ta nhit c xc nh t cc phng trnh tiu chun. Cc

phng trnh tiu chun ny c xc nh t thc nghim.Nu = f(Re,Pr,Gr, . . . )

Trong :

- Nu =l

l tiu chun Nusselt,

aPr

= l tiu chun Prandtl,

=l

Re l tiu chun Reynolds,

2

3 tlgGr

=

l tiu chun Grashof,

vi - h s to nhit, W/m2.K; - h s dn nhit, W/m.K; - tc chuyn ng, m/s - nht ng hc, m2/s;a - h s dn nhit , m2/s;g - gia tc trng trng 9,81 m/s2

t = (tw - tf) - h s dn n th tch, (1/0K)vi cht lng ta tra bng;vi cht kh:

T

1= , 0K-1.

l kch thc xc nh.

3.2.1 Ta nhin t nhin3.2.1.1 Ta nhin t nhin trong khng gian v hn

i vi ng hoc tm t ng, khi (Grf.Prf ) > 109 :

25,0

W

f

Pr

Pr(15,0

= 0,33fff )PrGrNu (3-10)

i vi ng hoc tm t nm ngang, khi 103< (Grf.Prf ) < 109 :25,0

W

f

Pr

Pr(5,0

= 0,25fff )PrGrNu (3-11)

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Nhit xc nh l nhit tf, kch thc xc nh vi ng hoc tm tng l chiu cao h, vi ng t nm ngang l ng knh, vi tm t nmngang l chiu rng.

3.2.2 Ta nhit cng bc khi cht lng chuyn ngtrong ng

3.2.2.1 Ch chy tng25,0

W

f1,0f

43,0

Pr

PrGrRe15,0

= f

0,33

ff PrNu (3-12)

i vi khng kh:1,0

fGrRe13,00,33

ffNu = (3-13)

Cng thc trn p dng cho trng hp d

l

> 50

Nud

l< 50 th h s to nhit cn nhn thm h s hiu chnh.

3.2.1.2 Ta nhit khi cht lng chy ri

Rl

25,0

W

f43,0 ..Pr

PrRe021,0

= f

0,8

ff PrNu (3-14)

trng hp:

dl > 50 th 1 = 1

Nud

l< 50: 1 tra bng

3.2.2 Ta nhit khi cht lng chy ngang qua chm ng

3.2.3.1. i vi chm ng song song

sl

25,0

W

f33,0 ..PrPrRe026,0

= f0,65ff PrNu (3-15)

i - h s k n th t hng ng.i vi hng ng th nht 1 = 0,6, hng ng th hai 2 = 0,9, hng ng th ba tri 3 = 1.s - h s k n nh hng ca bc ng theo chiu su.

15,0

2s d

Ss

=

3.2.3.1. i vi chm ng so le

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sl

25,0

W

f33,0 ..Pr

PrRe41,0

= f

0,6

ff PrNu (3-16)

i - h s k n th t hng ng.i vi hng ng th nht 1 = 0,6, hng ng th hai 2 = 0,7, hng ng th ba tri 3 = 1.s - h s k n nh hng ca bc ng theo chiu su.

2

1

S

S< 2

15,0

2

1s S

S

=

2

1

S

S> 2 s = 1,12

S2 bc dc, S1 bc ngang,Trong cc cng thc trn, R

=103 105. Kch thc xc nh l ng

knh ngoi. Nhit xc nh l nhit trung bnh ca cht lng t f .

3.2.4 To nhit khi bin i pha

3.2.4.1. To nhit khi si

Khi nc si bt p sut p = 0,2 80 bar: = 0,46.t2,33.p0,5, W/m2.K

t = tw tstw - nhit b mt vch t nng,ts - nhit bo ho ng vi p sut si;p - p sut si;

3.2.4.1. To nhit khi ngng mng

Ngng trn b mt vch ng ng:

4

ws

3

d d)tt(

.g..r943,0

= , w/m2.K (3-18)

Ngng trn b mt ng nm ngang:

4

ws

3

n d)tt(

.g..r724,0

= , w/m2.K (3-18)

trong :g - Gia tc trng trng , 9,81 m/ss

- h s dn nhit cu cht lng, W/m.K;r - nhit n ho hI, J/kg; - khi lng ring ca cht lng ngng, kg/m3; - nht ng hc, m2/s;

h chiu cao ca vch hoc ng t ng, m;

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d - ng knh ngoI ca ng, m;tw - nhit b mt vch,

0C;ts - nhit bo ho ng vi p sut si;

Trong cc cng thc trn, nhit xc nh l tm = 0,5(tw + ts).

3.3 BI tp v dn nhit

Bi 3.1 Vch bung sy c xy bng hai lp gch c dy 250 mm, ch s dn nhit bng 0,7 W/mK; lp n bc ngoi c h s dn nhit bng 0,0465W/mK. Nhit mt tng bn trong bung sy bng 1100C. Nhit mt tngbn ngoi bng 250C. Xc nh chiu dy lp n tn tht nhit qua vch bungsy khng vt qu 110W/m2. Tnh nhit tip xc gia hai lp.

Li gii

Mt dng nhit qua vch bung sy:

2

2

1

1

2W1W ttq

+

= , W/m2,

21

12W1W2 .q

tt

=

0465,0.7,0

25,0

110

251102

= = 0.019 m.

Vy chiu dy lp n bng 0,019 m.Nhit tip xc gia hai lp da vo Iu kin dng nhit n nh:

==

q

ttqq 2W1W1 :

1

11W1 qtt

=

C7,707,0

25,0.110110t 01 ==

Bi 3.2 Vch phng hai lp c chnh nhit 105 0C, chiu dy dy v h sdn nhit tng ng ca hai lp: 1 = 100 mm, 2 = 50 mm, 1 = 0,5 W/mK, 2 =0,1 W/mK. Xc nh mt dng nhit qua vch

Li gii

Mt dng nhit qua vch phng hai lp theo (3-1) vi 1 = 100 mm =0,1 m; 2 = 50 mm = 0,05 m v t = tW1 tW2 = 105 0C:

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150

1.0

05,0

5,0

1,0105tt

q

2

2

1

1

2W1W =+

=

+

= , W/m2,

Bi 3.3 Bit dng nhit qua vch phng dy 20 cm, c h s dn nhit 0,6W/m.K l 150 W/m2. Xc nh chnh nhit gia hai mt vch.

Li giiTheo (3-1), mt dng nhit qua vch phng mt lp vi q = 150 W/m2,

= 20 cm = 0,2 m; t = tW1 tW2 :

=

q

ttq 2W1W ; t =

6,0

2,0.150.q =

= 50 0C.

Bi 3.4 Vch tr di 1 m, ng knh d2/d1 = 144/120 mm,c chnh nhit gia hai mt vch 60C0, h s dn nhit ca vch 0,4 W/m.K. Xc nh dngnhit dn qua vch.

Li gii

Dng nhit qua vch tr mt lp theo (3-2) vi l = 1 m; ; t = tW1 tW2 =60 0C:

W7,826120144ln

4,0.14,3.21

60.1

ddln

21

)tt.(l

q.lQ1

2

21

l ==

==

Bi 3.5 Mt ng dn hi bng thp ng knh d2/d1 = 110/100 mm, h s dnnhit 1 = 55 W/mK c bc mt lp cch nhit c 2 = 0,09 W/mK. Nhit mt trong ng tw1 = 200

0C, nhit mt ngai lp cch nhit tw3 = 500C.

Xc nh chiu dy v nhit tW2 tn tht nhit qua vch ng khngvt qu 300W/m.

Li giiDng nhit trn 1 m chiu di ng theo (3-2) vi vch 2 lp:

2

3

21

2

1

3W1Wl

d

dln

2

1

d

dln

2

1

)tt(q

+

=

21

2

1l

3W1W

2

3 2d

dln

2

1

q

)tt(

d

dln

=

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282,009,0.14,3.2100

110ln

55.14,3.2

1

300

50200

d

dln

2

3 =

=

282,0

2

3 e

d

d=

d3 = d2.e0,282 = 110. e0,282 = 146 mm.

Chiu dy cch nhit :

182

110146

2

dd 23 =

=

= mm.

tm nhit gia hai lp tW2 ta da vo Iu kin trng nhit nnh: q1 = q11 =q12 = const.

1

2

1

2W1W1ll

d

dln

2

1

)tt(qq

==

1

2

111W2W d

dln

2

1qtt

=

9,199100

110ln

55.14,3.2

1300200t 2W ==

0C.

Bi 3.6 Mt thit b sy bng In c ch to t cc dy hp kim niken-cromng knh d = 2 mm, di 10 m. Khng kh lnh thi vo thit b sy c nhit

200C. Tnh nhit lng to ra trn 1 m dy, nhit b mt v nhit tm cady. Nu dng in t nng c cng 25 A, in tr sut = 1,1 mm2/m,h s dn nhit = 17,5 W/mK, h s to nhit t b mtdy ti khng kh =46,5 W/m2.K.

Li giiin tr ca dy t nng:

5,31.14,3

10.1,1

S

lR

2=== ,

Nhit do dy to ra:Q = R.I2 = 3,5. 252 = 2187,5 W,

Nhit lng to ra trn 1 m dy:

m/W75,21810

5,2187

I

Qq l ===

Nng sut pht nhit:

3622

0

lv m/W10.7,69

001,0.14,3

75,218

r

qq ==

=

Nhit b mt dy:

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7695,46.2

10.1.10.7,6920

2

rqtt

360v

fw =+=+=

C0,

Nhit ti tm dy:

5,17.4

10.10.1.7,69

5,46.2

10.1.10.7,6920r

4

q

2

rqtt

66362

0v0v

f0

+=

+

+=

t0 = 770 C0.

Bi 3.7 Mt tm cao su dy = 2 mm, nhit ban u t0 = 1400C c lm

ngui trong mi trng khng kh c nhit tf = 1400C. Xc nh nhit b

mt v nhit tm ca tm cao su sau 20 ph. Bit h s dn nhit ca cao su = 0,175 W/mK, h s dn nhit a = 8,33.10-8 m2/s. H s to nhit t b mt

tm cao su n khng kh = 65 W/m2.K.

Li gii

71,3075,0

01,0.65Bi ==

= ,

101,0

60.20.10.33,8.aFo

2

8

2==

=

Cn c Bi = 3,71 v Fo = 1, t th hnh 3-2 v 3-1 ta c:038,0* 1X = = 26,0* 0X = =

Vy nhit b mt:tX= = tf+ *X=.(t0-tf)tX= = 15 + 0,038.(140 15) = 25,4 C

0,Nhit tai tm:

tX=0 = tf+ *X=0.(t0-tf)tX=0 = 15 + 0,26.(140 15) = 47,5 C

0,

Bi 3.8 Mt tng gch cao 5 m, rng 3m, dy 250 mm, h s dn nhit cagch = 0,6 W/mK. Nhit b mt tng pha trong l 70 0C v b mt tngpha ngoi l 20 0C. Tnh tn tht nhit qua tng.

Tr li Q = n1800W,

3.4. BI tp v to nhit i lu

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Bi 3.9 Bao hi ca l t nm ngang c ng knh d = 600 mm. Nhit mtngoi lp bo n tW = 60

0C, nhit khng kh xung quanh tf = 400C. Xc nh

lng nhit to t 1 m2 b mt ngoi ca bao hi ti khng kh xung quanh.

Li giiT nhit khng kh tf = 40

0C tra bng 6 trong phn ph lc ca khngkh ta c:

= 0,00276 W/m.K , = 16,69.01-6 [ ]sm /2 , Prf= 0,699,Cng t bng 6 vi tf = 40 C

0, ta c: PrW = 0,696. Ta nhn thy Prf PrW

nn 1Pr

Pr25,0

W

f =

,

Theo tiu chun Gr:2

3

f

tlg.Gr

=

..

y g = 9,81 m/ s2, 0032,027340

1

T

1

f

=+

== , t = tW tf= 200C.

826

3

10.87,4)10.69,16(

20.0,60,0032.9,81.==

fGr

Grf.Prf= 4,87.108.0,699 = 3,4.108

Ta dng cng thc (3-11):

Nuf= 0,5.(Grf.Prf)0,25 = 0,5.(3,4.108)0,25 = 68.

Nuf=

d.

Vy h s to nhit i lu:

6,0

027,0.68

d

.Nu =

=

Lng nhit to t 1 m2 b mt ngoi ca bao hi:Q = .t = 3,13.20 = 62,6 W/m2.

Bi 3.10 Tnh h s to nhit trung bnh ca du my bin p chy trong ng cng knh d = 8 mm, dI 1 m, nhit trung bnh ca du t f = 80

0C, nhit

trung bnh ca vchng tW = 20 0C. tc chy du trong ng = 0,6m/s.

Li giiKch thc xc nh : ng knh trong d = 8.10-3 m.Nhit xc nh: tf= 80

0C.Tra cc thng s ca du bin p theo t f= 80

0C, bng 8 ph lc: = 0,1056 W/m.K , = 3,66.10-6 sm /2 , = 7,2.10-40K-1, Prf= 59,3, PrW = 298 Tra theo tW = 20

0C,

131010.66,3

10.8.6,0lRe

6

3

==

=

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Ref < 2300 du chy tng, do :25,0

W

f1,0f

43,0

Pr

PrGrRe15,0

= f

0,33

ff PrNu

Tnh26

94

)10.66,3()2080.(01.8.10.2,7.81,9..

=

=2

3

f

tlg.Gr

16198=fGr

Nuf= 0,15.13100,33.161980,1.59,30,43 25,0298

3,59

Nuf= 16,3

Tnh 21510.8

1056,0.3,16

d

.Nu3

ff ==

=

W/m2.K

Bi 3.11 Bit phng trnh tiu chun trao i nhit i lu ca khng khchuyn ng trong ng Nu = 0,021Re0,5. Nu tc ca khng kh gim I 2 lncn cc Iu kin khc khng i, lc ny h s to nhit 2 s l bao nhiu sovi 1. Ngc li nu tc tng ln 2 ln th 2 bng bao nhiu?

Li gii

V Nu =l

;

=l

Re nn ta c:

Nu = 0,021.Re0,5,5,0d

021,0l

=

Ch khi c tc thay i, cc thng s khc khng i, ta c:0,5 ( t l vi 0,5)11

0,5 ; 220,5

1

5,0

1

2

1

2

2

1;

2

1==

=

2

Vy h s to nhit 2 gim i 2 ln so vi 1.

Ngc li, nu tc tng ln 2 ln th 2 tng ln 2 ln so vi 1. Ch nu tc gi khng i cn ng knh gim i 2 ln th 2 tng ln 2 ln,khi ng knh tng ln 2 ln th 2 gim i 2 ln so vi 1.

Bi 3.12 Khng kh nhit 27 C0 c nht ng hc 16.10-6 m2/s, trao inhit i lu t nhin vi ng tr nm ngang ng knh 80 mm vi nhit bmt 67 . Xc nh tiu chun ng dng.

Li gii

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Tiu chun ng dng Grf vi ng tr nm ngang c kch thc xc nhl =d:

2

3

f

tlg.Gr

=

..

y: g = 9,81 m/s2 ( gia tc trng trng),3001

272731

T1

f

=+

==

d = 80 mm = 0,08 m; t = tW tf= 67 27= 40 C0; = 16.10-6 m2/s.

626

3

10.616,2)10.16.(300

40.9,81.0,08.==

fGr .

Bi 3.13 Mt chm ng so le gm 10 dy. ng knh ngoI ca ng d = 38mm. Dng khng kh chuyn ng ngang qua chm ng c nhit trung bnh t f

= 500 C0. Tc dng khng kh l 12 m/s. Xc nh h s to nhit trung bnhca chm ng.

Li giiKch thc xc nh: d = 38.10-3 m,Nhit xc nh: tf= 500 C

0.Tra cc thng s vt l ca khng kh ng vi 500 C0 bng 6 ph lc, ta

c: = 5,74.10-2 W/m.K , = 79,38.10-6 sm /2 , Prf= 0,687.

Tnh: 6

3

10.38.79

10.38.12d.

Re

=

= Ref= 5745,

Tnh theo (3-16) vi hng ng th 3:33,0Re41,0 f

0,6

ff PrNu = (vi khng kh coi Prf = PrW v b qua nh

hng ca bc ng S = 1),33,05745.41,0 .0,687Nu 0,6f =

N = 65,2.

Tnh3

2

310.38

10.74,5.2,65

d

.Nu

=

=

2 = 98,5 W/m2.K,

H s to nhit trung bnh ca chm ng so le:

n

).2n( 321 ++=

6,9110

3,9

10

).210(.7,0.6,0 3333 =

=++

= W/m2.K.

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103

Bi 3.14 Xc nh h s to nhit v lng hi nhn c khi nc si trn bmt c din tch 5 m2. Bit nhit ca vch tW = 156

0C v p sut hi p = 4,5bar.

Li giiNhit si (nhit bo ho ) tng ng vi p = 4,5 bar l ts = 148

0C.Nhit n ho hi r = 2120,9 kJ/kg. (tra bng 4 ph lc):

t = tW ts = 156 148 = 80C,H s to nhit khi si bt theo (3-17):

= 46. t2,33.p0,5 = 46.82,33.4,50,5 = 12404 W/m2.K.

Nhit lng b mt vch truyn cho nc:Q = .F.( tW ts) = 12404.5.(156 148)Q = 496160 W,Lng hI nhn c sau 1 gi:

84210.9,2120

3600.496160G

3== kg/h.