btktn3(1 bai)
TRANSCRIPT
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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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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.