Bai 1 (nhö 10.2) Töôøng loø nung: h = 2,5 m; F = 39 m2; tw= 90
oC ; khoâng khí xung quanh tf= 30 oC.
Tính αvaø toån thaát nhieät Q.
KTXÑ: l = h = 2,5 m
NÑXÑ: 2
tt
twf
m
+
=C60
2
3090 o
=
+
=Tm= 333 oK
Cheânh leäch NÑ:
fw ttt −=Δ C603090 o
=−=
TSVL taïi tm= 60 oC ( phuï luïc Baûng 22):
KmW 109,2 2
m⋅×=λ −sm1097,18 2
6
m−
×=ν
696,0Prm=
()
=⋅ m
PrGr 1010 1034,5696,01067,7 ×=××
Giaûi
2
3
m
m
thg
Gr
ν
β
Δ
== 7,67 x 1010
Chaûy roái 135,0C =3
1
n=
PTTC:
(
)
31
m
mPrGr135,0Nu ⋅=
(
)
508
10
1034,5135,0 31
=×=
HSTN:
h
Nu mmλ
=α KmW 9,5
5,2
109,2508 2
2
=
××
=
−
Nhieät löôïng toån thaát:
()
fw ttFQ −α= kW 13,8 W1380606399,5
=
=
×
×
=
Bai 2 (nhö 10.7)
OÁng daãn khí noùng naèm ngang : d = 0,5 m; tw= 470 oC ; khoâng khí
xung quanh tf= 30 oC.
Tính αvaø toån thaát nhieät ql.
Neáu d* = d/2 thì α* = ?
KTXÑ:
NÑXÑ:
2
tt
twf
m
+
=C250
2
30470 o
=
+
=Tm= 523 oK
Cheânh leäch NÑ: fw ttt −=Δ C
o
44030470 =−=
TSVL taïi tm= 250 oC ( phuï luïc Baûng 22):
KmW 1027,4 2
m⋅×=λ −sm1061,40 2
6
m−
×=ν
68,0Prm=2
m
3
m
tdg
Gr ν
Δβ
=
()
8
12
2
3
10255,6
1061,40 523.
4405,081,9 ×=
×
××
=−
()
m
PrGr ⋅88 10253,468,010255,6 ×=××=
Giaûi
l = d = 0,5 m
Chaûy roái, C = 0,135 3
1
n=
PTTC:
(
)
31
1350 m
mPrGr,Nu ⋅=
(
)
5101
8
1025341350 31 ,,, =×=
HSTN:
d
Nu mmλ
=α KmW67,8
5,0
1027,45,101 2
2
=
××
=
−
Nhieät löôïng toån thaát treân 1m oáng:
()
fw ttFq −α=
=
×
×
×= 0445,014,367,8 kW/m 6 W/m5989,2
=
Khi d* = d/2 Gr giaûm 23= 8 laàn, töùc Ra* = (Gr*.Pr) = 5,3 x 107.
Vaãn chaûy roái. Cheá ñoä automodel : αkhoâng ñoåi
Bai 3: (10.12)
Tìm λtñ vaø q qua lôùp khoâng khí trong moät khe heïp δ= 20 mm, tw1 = 200 oC,
tw2 = 80 oC.
Neáu beà daøy giaûm 2 laàn thì λtñ seõ theá naøo?
2
21 ww
f
tt
t+
=C140
2
80200 o
=
+
=
Theo phuï luïc thoâng soá vaät lyù cuûa khoâng khí ta ñöôïc:
Km.W 0349,0
f=λ sm108,27 2
6
f−
×=ν
684,0Prf=
Tieâu chuaån Gr:
21 ww ttt −=Δ C
o
12080200 =−=
2
f
3
f
tlg
Gr
ν
Δβ
=
4
122
3
10952
1082741
120020819
×=
××
××
=−,
,3
,,
Giaûi