fundamentals of heat and mass transfer solutions manual phần 1 docx
fundamentals of heat and mass transfer solutions manual phần 1 docx
... The heat flux and heat rate when the outside wall temperature is T
2
= -15 °C
are
(
)
2
12
x
25 C 15 C
dT T T
q k k 1W m K 13 3.3W m
dx L 0.30m
−−
−
′′
=− = = ⋅ =
. (1)
22
xx
q q A 13 3.3W ... 7.85m 10 W/m K 15 0 25 K 0.8 5.67 10 W/m K 423 298 K
−
=⋅−+××⋅−
()()
22
q 7.85m 1, 250 1, 095 w/m 9 813 8592 W 18 ,405 W=+=+=
<
(b) The annual energy loss...
fundamentals of heat and mass transfer solutions manual phần 2 docx
... lens:
()
t,wo
23
2-3
3 311 1
Rm
4 0.35 W/m K 10 .2 12 .7
10
12 W/m K4 10 .2 10 m
π
π
−
=+−
×⋅
⋅×
()
2
2-3
3
19 1.2 K/W +13 .2 K/W+246.7 K/W=4 51. 1 K/W
6 W/m K4 12 .7 10 m
π
+=
⋅×
With lens:
t,w
3
311 1
R 19 1.2 K/W +13 .2 ... With the insulation,
s ,1
2
w1 2 i2 3
3
11 1 11 1 1
TTq
4k r r 4k r r
4rh
ππ
π
∞
=+ −+ −+
()
()
4
s ,1
2
11 1 1...
fundamentals of heat and mass transfer solutions manual phần 4 docx
... 12 6 2 81. 8 299.5 310 .5 317 .9 3 21. 4 323.3 324.2
8 14 4 279.8 296.2 308.6 315 .8 320.4 322.5
9 16 2 276.7 294 .1 306.0 314 .3 319 .0
10 18 0 274.8 2 91. 3 304 .1 312 .4
Hence, find
( ) ( )
10 10
03
T0, 18 0sT275C ... solution?
PROBLEM 5 .11 2 (Cont.)
() ()
()()
(
)
()
p +1 p +1 p +1 p +1
mm
m -1 m +1
p +1 p
p +1
mm
omp
TT TT
kd1 kd1
xx
TT
q x 1 2h x 1 T T x d 1 c...
fundamentals of heat and mass transfer solutions manual phần 6 docx
... condition), and the following results were obtained.
0.05 0.09 0 .13 0 .17 0. 21 0.25
Mass flowrate, mdot(kg/s)
10 00
14 00
18 00
2200
2600
3000
Heat rate, q(W)
0.05 0.09 0 .13 0 .17 0. 21 0.25
Mass flowrate, ... from Eqs. 8.20 and 8.22,
(
)
1/ 4
1/ 43
D
f0. 316 Re0. 316 9 .14 310 0.0323
−
−
==×=
2
m
u
pfL
2D
ρ
∆=
( )( )
2
3
2
0.7740 kg/m200 /10 1.36 m/s2 m
p0.0323 71. 1 N/...
fundamentals of heat and mass transfer solutions manual phần 3 pdf
... 75 10 0 12 5 15 0 17 5 200 225 250 275 300
0 18 0.7 18 0.2 17 8.4 17 5.4 17 1 .1 165.3 15 8 .1 149.6 14 0 .1 129.9 11 9.4 10 8.7 98.0
25 204.2 203.6 2 01. 6 19 8.2 19 3.3 18 6.7 17 8.3 16 8.4 15 7.4 14 5.6 13 3.4 12 1.0
50 ... isotherms.
- 10 0 10 0 10 0 10 0 10 0 -
50 86.0 10 5.6 11 9 13 1.7 15 1.6 200
50 88.2 11 7.4 13 8.7 15 6 .1 174.6 200
50 99.6 13...
fundamentals of heat and mass transfer solutions manual phần 5 ppsx
... flow
(
)
( )
L
1/ 2
1/ 3
1/ 21/ 35
L
Nu0.664 Re Pr0.66 41. 3 710 0. 712 19.2
==×=
( )
(
)
L
2
ss
k0.02 51 W/mK
qNuATT 219 .22m5C55 W.
L1m
∞
⋅
=−==
o
<
(b) The Reynolds number for the roof and collector of length ... K
D
==
⋅
=⋅
0625
0 01
347 21 690
2
.
.
,
′
=
qWm20 438,
Fluid: Engine Oil
()
D
62
5m s 0.01m
VD
Re 14 7
340 10 m s
ν
−
== =
×
()( )
()
4/5
1/ 2 1/ 3
5/8
D
1/ 4...
fundamentals of heat and mass transfer solutions manual phần 7 pot
... 8 0 0033 273 015
15 89 10 22 5 10
308 10
and from Eq. 9.34,
()
()
()
2
2
1/ 6
5
1/ 6
s,i
D,i
o
8/27 8/27
9 /16 9 /16
i
0.387 3.08 10 T T
0.387Ra
k 0.0263
h 0.60 0.60
D0 .15
1 0.559 Pr 1 0.559 0.707
∞
×−
=+ ... )
D
2
1/ 6
8/27
9 /16
0.38 710 , 410
Nu0.604.40
10 .559/0.690
=+=
+
D
2
3
k0.0338W/mK
hNu4.4 011 .9W/mK.
D
12 . 510 m
−
⋅
==×=⋅
×
The...
fundamentals of heat and mass transfer solutions manual phần 8 ppsx
... and Eqs. 11 .22 and 11 .36a yield
()
c,o c,i
h,i c,i
TT
1 exp NTU
TT
ε
−
==−−
−
(1)
From Eq. 11 .1,
() ()
t
oo
ch
R
11 1
UA hA L hA
ηη
′
=++
(2)
With
()
72
Di
Re 4m/ D 0 .12 kg /s/ 0.05m 19 6.4 10 ... = 4 .16
×
10
4
W/K( 410 K) = 1. 71
×
10
7
W. Hence,
ε
= (q/q
max
) = (1. 254
×
10
7
W /1. 71
×
10
7
W) =
0.735. With C
min
mixed and C
max
unmixed,...
fundamentals of heat and mass transfer solutions manual phần 9 ppt
... )
,bb
00
Gd/G1E ,10 00Kd/E1000K
λλλλ
ααλρλλ
∞∞
==−
∫∫
(
)
( )
(
)
( )
11
,1, 2
00
1F11F.
λλ
λλ
αρρ
→→
=−+−−
Using Table 12 .1 for λ
1
T
f
= 4 × 10 00 = 4000 µm⋅K, F
(0-λ)
= 0.4 91 giving
(
)
(
)
(
)
10 .20.4 911 0. 810 .4 910 .49.α
=−×+−×−=
<
(b) ... 8244
s
3
4 0.53 5.67 10 W m K 13 00K 15 .1W m K 10 00K 0.8 5.67 10 W m K 300 K
dT dt
8933kg m 385J kg K 0.01m
−−
××...
fundamentals of heat and mass transfer solutions manual phần 10 doc
... rates,
()
()
b1 1
in cv ,1 1 1 1
11 1
EJ
qq qhATT
1/ A
εε
∞
−
=+= −+
−
(2)
and the radiosity J
1
can be determined from the radiation energy balance, Eq. 13 . 21,
()
b1 1 1 312
11 1 11 2 11 3
EJ JJ
JJ
1 / A 1/ A ... A
2
and A
3
,
respectively,
A
1
:
()
b1 1 1 312 14
1 1 11 2 11 3 11 4
EJ JJJJ JJ
1 /A 1/ A F 1/ A F 1/ A F
ε
−−−−
=++
−
(3)
A
2
:
()
b2 1...
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