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BC X NHIT__________________________

1. nh ngha TNBX

Qu trnh trao i nhit c thc hin bng sng in t.

c im

- Mi vt c T > 00K c kh nng bc x nng lng, bin i ni nng do dao ng in t.

- Cc sng in t c cng bn cht v ch khc nhau vchiu di bc sng.

V d:

- Cc tia v tr v tia gama: m- Cc tia Rn nghen: m- Cc tia t ngoi : m- Cc tia nh sng: m- Cc tia hng ngoi: m- Cc sng v tuyn: mm

4 40,1.10 10.10 4 410 .1 0 200 .10

0 , 0 2 0 , 4 0 , 4 0 , 7 6 0 , 7 6 4 0 0

0,2

I.CC KHI NIM C BN

1. nh ngha TNBX

- Cc tia hng ngoi v nh sng trng c bc sng c hiu ng nhit cao cn gi l cc tia nhit.

- Bc x nhit: Qu trnh truyn cc tia nhit trong khng gian.

- Hp th bc x: Qu trnh hp th mt phn hay ton btia nhit bin thnh nhit nng.

- Cch trao i nhit:

Nhit nng Nng lng bc x Nhit nng

- Vt trng thi cn bng s c nng lng bc x bng nng lng hp th.

( 0,4 400 m) I.CC KHI NIM C BN

m 0,4 400

2. Vt en tuyt i, trng tuyt i, trong tuyt i

Gi s:C vt nh hnh v:

Dng bc x Q p ti vt s sinh ra 3 phn:

+ Phn b phn x QR,+ Phn c vt hp th QA+ Phn xuyn qua vt QD.

Theo nh lut bo ton nng lng, ta c:

( 0,4 400 m) I.CC KHI NIM C BN

A R DQ Q Q Q

QQR

QA

QD


- Vit li biu thc bo ton nng lng:

Trong : : Gi l h s hp th.

: Gi l h s phn x.

: Gi l h s xuyn qua.

- Biu thc dng cc h s: A + R + D = 1

A R DQ Q Q 1Q Q Q

AQ AQ

RQ RQ

DQ DQ

I.CC KHI NIM C BN


- Biu thc dng cc h s: A + R + D = 1

+ Nu A = 1 (R v D = 0): Vt c gi l en tuyt i - Vt c kh nng hp th ton b nng lng p ti n.

+ Nu R = 1 (A v D = 0): Vt c gi l trng tuyt i - vt c kh nng phn x li ton b nng lng p ti n.

+ Nu D = 1 (A v R = 0): Vt c gi l trong sut tuyt i- vt c kh nng cho ton b nng lng i qua.

+ Cc kh c s nguyn t trong phn t 2 c th xem l vt

trong sut tuyt i vi tia nhit, D = 1.

I.CC KHI NIM C BN


+ Vt c: Khng cho cc tia nhit xuyn qua. Cc vt rn v

cht lng c th coi D = 0. Khi A + R = 1 ngha l vt

hp th tt th phn x ti v ngc li.

+ Vt xm: L trng hp c bit ca vt c trong kh

nng hp th tt hn kh nng phn x.

Hoc: Vt m c ng cong nng sut bc x n sc

E() (ph thuc vo nhit v bc sng) c dng

ging nh ca vt en tuyt i, tc l:

+ Thc nghim cho thy phn ln cc vt liu dng trong k

thut u c th coi l vt xm.

,0

E constE

I.CC KHI NIM C BN

3. Dng bc x, nng sut BX, nng sut BX ring, nng sut BX hiu dng

3.1 Dng bc x

- Tng nng lng bc x t b mt c din tch F ca vt theo mi phng ca khng gian bn cu v mi bc sng ( = 0 ti ) trong mt n v thi gian.

- K hiu l: Q (W; J/s).

- Nu bc x tnh trong khong n (+d) th gi l bc x n sc dng bc x n sc Q.

I.CC KHI NIM C BN

3. Dng bc x, nng sut BX, nng sut BX ring,

nng sut BX hiu dng

3.2 Nng sut bc x (kh nng bc x)

- Dng bc x ton phn ng vi mt n v din tch ca bmt bc x.

- K hiu l: E (W/m2);

+ Vi bc x n sc: Nng sut bc x n sc.

(W/m3)

+ Nu ti mi im trn b mt, nng sut bc x c gitr khng i:

hay Q = EF3.3 Nng sut bc x ring

Nng sut bc x ca bn thn vt.

d QEd F

dEEd

QEF

I.CC KHI NIM C BN


3.4 Nng sut bc x hiu dng

Gi s :- Vt c (A + R = 1) c nhit T.

- H s hp th A, nng sut bc

x p ti l Et. - Nng sut bc x ring ca bn thn vt l E.

Khi vt s hp th mt phn : A.Et Phn cn li vt s phn x tr li : R.Et = (1-A).Et

I.CC KHI NIM C BN

E

Et

(1-A)Et

Ehd

T, A


3.4 Nng sut bc x hiu dng (tip)

- Thc t vt s pht i nng

sut bc x hiu dng Ehd :

Nng sut bc x hiu dng: Tng ca nng sut bc x

ring v nng sut bc x phn x .

- Tng t, dng bc x hiu dng c xc nh nh sau:

hd tE E 1 A E

hd R tQ Q Q Q 1 A Q

I.CC KHI NIM C BN

E

Et

(1-A)Et

Ehd

T, A

1. nh lut Stefan-Boltzmann

- Nng sut bc x ca vt en tuyt i t l bc 4 vi nhit tuyt i.

hay

Trong : 0: Hng s bc x ca vt en tuyt i

0 = 5,67.10-8 (W/m2.K4)C0: H s bc x ca vt en tuyt iC0 = 108 0 = 5,67 (W/m2.K4)

- nh lut ny cng ng cho vt xm:

Trong : C: H s bc x ca vt xm.

T: L nhit ca vt xm.

4

0 0 0E = T4

00 0

TE =C100

4TE = C100

II.CC NH LUT C BN V BC X

1. nh lut Stefan-Boltzmann

- en: T s gia nng sut bc x ca vt en tuyt i v vt xm cng nhit (T = To).

- Rt ra biu thc :

C = C0- Gi tr c xc nh bng thc nghim. en ph thuc

vo nhit v trng thi b mt , = 0 1

- Vit li nh lut Stefan-Boltmann cho vt xm :

(W/m2)

4

4

0 000

TCE C100

= = = E CTC

100

4

0

TE = C

100


2. nh lut Kirchhoff

Gi s:

+ Hai vt t song song vi nhau:

- Vt th nht l vt c :

T1, A1, E1- Vt th hai l vt en tuyt i :

T0, E0, A0 = 1

- V vt en tuyt i hp th ton b nng lng ti (A0=1) :Ehd 0 = E0

- Vi vt c (A + R = 1): Ehd 1 = E1 + (1 A1)E0

Gi s nhit 2 vt nh nhau T0 = T1: Ehd1 = Ehd0


E1

(1-A1)E0

Ehd

T1A1

E0

T0A0

4. nh lut Kirchhoff

- Bin i va rt gn:

- Thay vt c 1 bng vt c 2 c E2, A2, T2 v khi T2=T0:

- Biu thc tng qut :

Pht biu nh lut Kirchhoff:

T s gia nng sut bc x v h s hp th ca cc vt c nhit nh nhau v bng nhit ca vt en tuyt i th bng nhau v bng nng sut bc xca vt en tuyt i.

10

1

E = EA


20

2

E =EA

1 20 1 2 0

1 2

E E= = ...= E (T = T = ...= T )A A

4. nh lut Kirchhoff

- T L suy ra: (1)

- Mt khc: (2)

T (1) v (2) suy ra = A, ngha l en ca vt c = hs hp th ca n.

0

0

E E=E =AA E

0

EE


Lng nhit trao i ph thuc vo nhiu yu t nh:

+ Bn cht vt l ca cc vt,

+ Hnh dng, kch thc v trng thi b mt vt.

+ Nhit ,

+ V tr tng i gia cc vt.

+ Tnh cht ca mi trng gia cc vt.

Xem xt trng hp vt t trong mi trng trong sut.

III. TRIII. TRAO AO I NHII NHIT BT BC XC X GIGIA CA CC VC VT T T TRONGT TRONGMI TRMI TRNG TRONG SUNG TRONG SUTT

1. Trao i nhit bc x gia hai vch phng, rng

v hn t song song

1.1 Khi khng c mn chn gia hai vt

- Gi thit: Hai vt c (2 vch)

+ Vch 1 c: T1, A1, 1

+ Vch 2 c: T2, A2, 2

T1 > T2 v F1 = F2 = F


E1

(1-A1)Ehd2

Ehd1

T1A1E1

T2A2E2

E2

(1-A2)Ehd1

Ehd2


v hn t song song


- Lng nhit trao i:

Trong :

Gii h (1,2,3) :

12 hd1 hd 2 1 2q E E q q (1)

hd1 1 1 hd2

hd2 2 2 hd1

E E 1 A E (2)

E E 1 A E (3)

2 1 1 2 2 1 1 212

1 2 1 2 1 2 1 2

A E A E E Eq (*)A A A A



v hn t song song


- Lng nhit trao i:

Mt khc: v

Thay vo PT (*):

4

11 1 0

TE C100

4

22 2 0

TE C100

4 4

1 212 0

1 2

1 T Tq C1 1 100 1001


1. Trao i nhit bc x gia hai vch phng, rng v hn t song song


- Lng nhit trao i:

t

Vi qd : l en quy dn ca h

PT tnh nhit :

qd

1 2

11 1 1

4 4

21 212 qd 0

T Tq C ; (W /m )100 100



1.1 Khi c mn chn gia hai vt

- Gi thit thm:

Gia 2 b mt 1 v 2 t thm mt mn chn c en m,nhit mn Tm cha bit, T1>T2.

Qu trnh trao i nhit n nh & mt chiu: q12 = q1m = q2m

p dng:

4 4

1 m1m 0

1 m

4 4

m 2m 2 0

m 2

1 T Tq C1 1 100 1001

1 T Tq C1 1 100 1001




- Lng nhit trao i:

+ Trong trng hp c n mn chn c en nh nhau :


4 4

21 212 0

1 2 m

1 T Tq C ; (W / m )1 1 2 100 1001 1

4 4

21 212 0

1 2 m

1 T Tq C ;(W / m )100 1001 1 21 n 1



- Lng nhit trao i:

Nhn thy:

Cng c nhiu mn chn, nhit lng trao i cng gim.

Vi gi thit: 1 = 2 = m, suy ra:

Nh vy, lng nhit trao i gia hai b mt 1 v 2 khit n mn chn s gim i (n+1) ln so vi khi khng t

mn chn.


12

12 m

qqn 1

2.Trao i nhit bc x gia hai vt bc nhau

Gi thit: Hai vt c bc nhau:

+ Vt 1 li, c: F1, 1, A1, T1+ Vt 2 lm, c: F2, 2, A2, T2

Nhn thy:

- Ton b dng bc x hiu dng

ca vt 1 u c th n c vt 2.

- Ngc li, ch c mt phn dng

nhit bc x hiu dng ca vt 2 ri

trn vt 1.

- Phn cn li s ri ngay trn bn

thn vt 2.


1T1, 1

2 T2, 2

21.Qhd2

(1-21).Qhd2

2. Trao i nhit bc x gia hai vt bc nhau

- T s gia dng bc x ca vt 2 pht i p ti vt 1 (Q21)so vi ton b dng bc x ca vt 2 pht i (Q2) gi l

h s gc bc x ca vt 2 ti vt 1.

- Dng bc x hiu dng ca vt 1 ln vt 2:

- Dng bc x hiu dng ca vt 2 ln vt 1:

- Phn cn li (1 21) Qhd2 li p ngay vo bn thn vt 2.

2121

2

QQ


1 2 hd1 1 1 21 hd2Q Q Q 1 A Q

2 1 hd 2 21Q Q


- Lng nhit trao i gia b mt 1 v 2:

(*)

Vi

v


1 2 h d 1 h d 2Q Q Q

hd 1 1 1 21 hd 2

hd 2 2 2 hd 1 21 2 hd 2

Q Q 1 A Q

Q Q 1 A Q 1 1 A Q

4

11 1 1 1 0 1

4

22 2 2 2 0 2

TQ E .F C F100

TQ E .F C F100



+ Rt ra PT:

+ H s gc 21 (cn gi l h s hnh dng) c xc nh t iu kin cn bng T1 = T2. Khi Q12 = 0, t (*) rt ra:

+ Thay vo PT (*):


4 4

1 212 0 1 2 21

21

1 2

1 T TQ C F F. (*)100 1001 1 1

12 1

2

FF

4 4

1 212 0 1

1

1 2 2

1 T TQ C F (**)1 0 0 1 0 01 F 1 1

F



t ; qd : en quy dn ca h.

+ PT cui cng:

+ Trng hp c bit: Nu F2 ln hn F1 rt nhiu nn: qd 1:


qd

1

1 2 2

11 F 1 1

F

4 4

1 212 qd 0 1

T TQ C F (W)100 100

4 4

1 212 1 0 1

T TQ C F (W )(***)100 100



Cng thc (**) v (***) ch s dng khi b mt vt 1 li

hoc phng, khng c lm v vt 2 phi lm.

C th dng cng thc trn c khi vt li 1 v vt lm 2

to nn mt khng gian kn.


12

I. KHI NIM C BN

Trong thc t, ba dng trao i nhit gm: Dn nhit, i lu,

bc x thng xy ra ng thi v c nh hng ln

nhau. Khi xy ra ng thi cc dng trao i nhit c bn

trn se c trao i nhit hn hp (hay phc tp).

V d:

Qu trnh dn nhit ca vt liu xp lun km theo qu trnh

trao i nhit i lu v bc x trong cc l rng

Trong trng hp ny, ta c th chn mt dng trao i nhit

c bn tnh ton cn nh hng ca cc dng trao i

nhit khc c th xem xt bng cc h s hiu chnh.

1. Truyn nhit qua vch phng mt lp

Gi s:

+ Vch phng mt lp, c = const,

+ Chiu dy


Nhn thy:

Trong trng hp n nh nhit, nhit truyn t mi trng

nng ti mt tri ca vch (bng phng thc i lu)

bng lng nhit truyn qua vch (bng dn nhit) v

bng lng nhit truyn t b mt phi ca vch ti mi

trng lnh (bng i lu).

C h PT

II. TRUYN NHIT HN HP QUA VCH PHNG

f 1 w1

1 f 1 w1 1

w1 w 2 w1 w 2

2 w 2 f 2w 2 f 2

2

1t t qq t t

q t t t t q

1q t t t t q


Gii h PT, rt ra:

t

k : gi l h s truyn nhit, W/m2K

Khi : q = k(tf1-tf2)

Dng nhit Q : Q = F.q = kF(tf1-tf2)

f 1 f 2

1 2

t tq 1 1

1 2

1k 1 1



i lng nghch o ca h s truyn nhit gi l nhittr truyn nhit:

Khi xc nh cc nhit trn b mt vch:

w 1 f 1 w 2 f 2 w1

1 2

1 1t t q ; t t q t q (da sua )

1 2

1 1 1Rk


2. Truyn nhit qua vch phng nhiu lp

Gi s:

+ Vch phng ngn cch gia hai mi trng l vchphng nhiu lp,

+ Chng hn c n lp c h s dn nhit tng ng l: 1, 2, 3 ..

+ C chiu dy tng ng l: 1, 2, 3 Chng minh tng t, ta c:

q = k(tf1 tf2)Trong k l h s truyn nhit ca vch phng nhiu lp:

ni

i 11 i 2

1k 1 1


1. Vch tr mt lp

Gi s:

- Vch tr mt lp chiu di l, ng

knh trong l d1, ng knh ngoi l d2.

- Chiu di >> chiu dy rt nhiu

nhit trong vch ch thay i theo

hng bn knh.

- Vch tr ng cht v c h s dn

nhit: = const

III. TRUYN NHIT HN HP QUA VCH TR

tf11 tw1

tf22

tw2

t

r

=const

0r1

r2

lQq ; (W /m )l

1. Vch tr mt lp

- Pha trong tip xc vi mi trng nng c nhit tf1, h s

to nhit l 1.

- Pha ngoi tip xc vi mi trng lnh c nhit tf2, h s

to nhit l 2.

Gi tw1 v tw2 l nhit b mt tip xc vi cc mi trng.

+ Dng nhit ng vi mt n v chiu di:


1. Vch tr mt lp

ch nhit n nh, mt dng nhit truyn t mi trngnng ti vch bn trong (bng i lu) bng dng nhittruyn qua vch (bng dn nhit) v bng dng nhittruyn t mt ngoi ca vch ti mi trng lnh (bngi lu).

Do ta c:

f 1 w1 ll 1 1 f 1 w1

1 1

w1 w 2 2

l w1 w 2 l

2 1

1

w 2 f 2 l

l 2 2 w 2 f 2 2

1t t qq d t t dt t 1 dq t t q . ln1 d 2 dln

2 d 1t t q dasuaq d t t d


1. Vch tr mt lp

Gii h trn ta c:

t

kl: Gi l h s truyn nhit ca vch tr.

Do : ql = kl(tf1 tf2)

Nhit lng: Ql = l.ql = l.kl(tf1 tf2)

l f 1 f 22

1 1 1 2 2

1q t t1 1 d 1lnd 2 d d


l

2

1 1 1 2 2

1k 1 1 d 1lnd 2 d d

1. Vch tr mt lp

Nhit tr truyn nhit ca vch tr:

Nhit trn b mt vch l:

2

l

l 1 1 1 2 2

1 1 1 d 1R lnk d 2 d d

w1 f 1 l

1 1

w 2 f 2 l

2 2

1t t q .d

1t t q .d


2. Vch tr nhiu lp

Gi s:

- Vch tr nhiu lp (n lp) vi cc ng knh: d1, d2, d3dn+1,- Vt liu khc nhau c h s dn nhit tng ng: 1, 2 n.

Chng minh tng t, ta c: ql = kl(tf1 tf2)

Trong :

kl: Gi l h s truyn nhit ca vch tr nhiu lp

l ni 1

i 11 1 i i 2 n 1

1k 1 1 d 1lnd 2 d d


2. Vch tr nhiu lp

Nhit tr ca vch tr nhiu lp s bng:

Nhit ca b mt vch tr tip xc vi cc mi trngc th xc nh:

ni 1

li 1

l 1 1 i i 2 n 1

1 1 1 d 1R lnk d 2 d d

w 1 f 1 l f 2 lw n 1

1 1 2 n 1

1 1t t q . ; t t q .d d


CHUONG VI. TN_BUC XABC X NHIT__________________________I.CC KHI NIM C BNI.CC KHI NIM C BNI.CC KHI NIM C BNI.CC KHI NIM C BNI.CC KHI NIM C BNI.CC KHI NIM C BNI.CC KHI NIM C BNI.CC KHI NIM C BNI.CC KHI NIM C BNI.CC KHI NIM C BNII.CC NH LUT C BN V BC XII.CC NH LUT C BN V BC XII.CC NH LUT C BN V BC XII.CC NH LUT C BN V BC XII.CC NH LUT C BN V BC XIII. TRAO I NHIT BC X GIA CC VT T TRONG MI TRNG TRONG SUTIII. TRAO I NHIT BC X GIA CC VT T TRONG MI TRNG TRONG SUTIII. TRAO I NHIT BC X GIA CC VT T TRONG MI TRNG TRONG SUTIII. TRAO I NHIT BC X GIA CC VT T TRONG MI TRNG TRONG SUTIII. TRAO I NHIT BC X GIA CC VT T TRONG MI TRNG TRONG SUTIII. TRAO I NHIT BC X GIA CC VT T TRONG MI TRNG TRONG SUTIII. TRAO I NHIT BC X GIA CC VT T TRONG MI TRNG TRONG SUTIII. TRAO I NHIT BC X GIA CC VT T TRONG MI TRNG TRONG SUTIII. TRAO I NHIT BC X GIA CC VT T TRONG MI TRNG TRONG SUTIII. TRAO I NHIT BC X GIA CC VT T TRONG MI TRNG TRONG SUTIII. TRAO I NHIT BC X GIA CC VT T TRONG MI TRNG TRONG SUTIII. TRAO I NHIT BC X GIA CC VT T TRONG MI TRNG TRONG SUTIII. TRAO I NHIT BC X GIA CC VT T TRONG MI TRNG TRONG SUTIII. TRAO I NHIT BC X GIA CC VT T TRONG MI TRNG TRONG SUT

CHUONG VII. TN_ HON HOPI. KHI NIM C BNII. TRUYN NHIT HN HP QUA VCH PHNGII. TRUYN NHIT HN HP QUA VCH PHNGII. TRUYN NHIT HN HP QUA VCH PHNGII. TRUYN NHIT HN HP QUA VCH PHNGII. TRUYN NHIT HN HP QUA VCH PHNGIII. TRUYN NHIT HN HP QUA VCH TRIII. TRUYN NHIT HN HP QUA VCH TRIII. TRUYN NHIT HN HP QUA VCH TRIII. TRUYN NHIT HN HP QUA VCH TRIII. TRUYN NHIT HN HP QUA VCH TRIII. TRUYN NHIT HN HP QUA VCH TRIII. TRUYN NHIT HN HP QUA VCH TR

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