... a layer of insulation
between it and the concrete.
PROBLEM 1 .43 (Cont.)
(
)
(
)
44 4 44 4
w,ss w,ss
0.65 1500 T K 0.65 330 T K
σσ
−+ −
() ()
22
w,ss w,ss
8W / m K T 700 K 4W / m ...
()
3
4 4 0.0181 m /s
V 4. 7 m/s
o
22
A
c
D
0.07 m
π
π
∀∀×
== = =
<
(b) Heat transfer from the casing is by convection and radiation, and from Eq. (1.10)
()
(...
... mdot(kg/s)
5000
10000
15000
20000
25000
30000
Heat rate, q(W)
0.1 0.2 0.3 0 .4 0.5
Mass flowrate, mdot(kg/s)
42
43
44
45
46
Outlet temperature, Tmo(C)
Although q increases with increasing
m
due to the attendant increase in Re
D
, and ... )
4/ 5
0 .4
4/50 .43
D
Nu0.023RePr0.0239. 143 100.68229.12
==×=
2
i
hNuk/D29.120.0373 W/mK/0.025 m43 .4 W/mK.
=⋅=×⋅=⋅
Hence, the outle...
... (a) Heat transfer rate by free convection to the room, per unit length of the tube; effect on
quality, x, at outlet of 30 m length of tube; (b) Effect of radiation on heat transfer and quality of ... emissivity, and power dissipation of cylindrical heater. Temperature of ambient
air and surroundings.
FIND:
Steady-state temperature of heater and time required t...
... K
3000W m K 279 W m K
=++
⋅
⋅⋅
(
)
45 32 32
h
1
4. 44 10 9.59 10 3.58 10 m K W 4. 12 10 m K W
U
−−− −
=×+×+× ⋅=× ⋅
2
h
U 243 WmK
=⋅
.
With C
h
= C
min
= 4. 160
×
10
4
W/K and C
c
= C
max
= 2.09
×
... NTU = 1. 54 and
()
()
42 2
hminh
A NTU C U 1. 54 4.160 10 W K 243 W m K 264m
==× ⋅=
.
Hence,
()
()()
2
h
L
2
oT
A 264m
N70
DWN
0.02 2 30 m
π
π
== =
<
(b) Using the IHT
C...
... or
()()
()
44
12
c2w
ss cc
TT
TT
Lk Lk
εσ
−
=−
+
(
)
8 244 44
2
2
1500K T
0.8 5.67 10 W m K T 90 K
0.008m 0.005m
25W m K 60W m K
−
−
=× × ⋅ −
+
⋅⋅
6 84
22
3.72 10 247 9T 4. 54 10 T 2.98
−
×− = × −
84 6
22
4. 54 ... 616
1
(4)
and G
w
is equal to the blackbody emissive power at T
ch
. Using Eqs. (3) and (4) and substituting
numerical values, find
σεσ εσ
TT T
bp
4
ww
4
w
ch...
... of heat transfer to tube without ash deposit, (b) Rate of heat transfer with an ash deposit
of diameter D
d
= 0.06 m, (c) Effect of deposit diameter and convection coefficient on heat rate and
contributions ... 8170 K / W R 6 944 K / W R 747 1 K / W.
===
The equivalent resistance is then
1
eq
111
R 249 8 K / W
8170 6 944 747 1
−
=++ =
and the total resistance...