2. Losses by Natural Convection from diameter and 121.9 mm Cylinder. A vertical cylinder 76.2 mm in high is maintained at 397.1 K at its surface. It loses heat by natural convection to air at 294.3 K. Heat is lost from the cylindrical side and the flat circular end at the top. Calculate the heat loss, neglecting radiation losses. Use the simplified equations of Table 15.5-2 and those equations for the lowest range of NGr Npr. The equivalent L to use for the top flat surface is 0.9 times the diameter. Ans. q 26.0 W a 11 able 15.5-2. Simplified Equations for Natural Convection from Various Surfaces Equation h= W/m2. K h btu/h ft. °F L = m, AT = K °F L -f, ΔT= L ft, AT Physical Geometry D=m D=ft NGNpr Ref. Air at 101.32 kPa (1 atm) abs pressure h = 0.28(ATIL)4 h 1.37(AT/L)4 h 1.24 AT/3 h =0.27(ATID)/4 h 1.32(AT/D)4 h = 1.24 AT1/3 Vertical planes and cylinders 1/4 104-10 >10 h0.18(AT) 103- 10 >109 for (P1) odasrO(P1) (M1) (M1) PM 1/3 m Horizontal cylinders h = 0.18(AT)3 Horizontal plates h = 0.27(AT/L)14 h = 0.32(AT/L)1/4 0.22(AT)s 105-2 x 10 Heated plate facing upward or cooled 2 x 107-3 x 1010 h plate facing downward (M1) 9(M1) h 0.52(AT)13 3x 105 x 3 - 1010 h 0.12(ATIL)14 h= 0.59(AT/L) 1/4 Heated plate facing downward or (M1) d AI cooled plate facing upward M Water at 70°F (294 K) bn 1loo (P1) h = 26(AT/L)4 h=127(AT/L)4 104-10° 1/4 Vertical planes and cylinders 1/4 Organic liquids at 70°F (294 K) 104-10 h = 12(AT/L)4 h= 59(AT/L)4 1/4 1/4 Vertical planes and cylinders (P1) EXAMPLE 15.5-2. Natural Convection and the Simplified Equation Repeat Example 15.5-1 but use the simplified equation. o The film temperature of 408.2 K is in the range 255-533 K. Also, Solution: L'AT = (0.305m) (194,4K) = 5.5m2 K This is slightly greater than the value of 4.7 given as the approximate maximum for use of the simplified equation. However, in Example 15.5-1 the value of NNpr is below 10°, so the simplified equation from Table 15.5-2 will be used: 1/4 \1/4 AT 1.37 194.4 0.305 h=1.37 L = 6.88 W/m2 K (1.21 btu/h ft2.F)
2. Losses by Natural Convection from diameter and 121.9 mm Cylinder. A vertical cylinder 76.2 mm in high is maintained at 397.1 K at its surface. It loses heat by natural convection to air at 294.3 K. Heat is lost from the cylindrical side and the flat circular end at the top. Calculate the heat loss, neglecting radiation losses. Use the simplified equations of Table 15.5-2 and those equations for the lowest range of NGr Npr. The equivalent L to use for the top flat surface is 0.9 times the diameter. Ans. q 26.0 W a 11 able 15.5-2. Simplified Equations for Natural Convection from Various Surfaces Equation h= W/m2. K h btu/h ft. °F L = m, AT = K °F L -f, ΔT= L ft, AT Physical Geometry D=m D=ft NGNpr Ref. Air at 101.32 kPa (1 atm) abs pressure h = 0.28(ATIL)4 h 1.37(AT/L)4 h 1.24 AT/3 h =0.27(ATID)/4 h 1.32(AT/D)4 h = 1.24 AT1/3 Vertical planes and cylinders 1/4 104-10 >10 h0.18(AT) 103- 10 >109 for (P1) odasrO(P1) (M1) (M1) PM 1/3 m Horizontal cylinders h = 0.18(AT)3 Horizontal plates h = 0.27(AT/L)14 h = 0.32(AT/L)1/4 0.22(AT)s 105-2 x 10 Heated plate facing upward or cooled 2 x 107-3 x 1010 h plate facing downward (M1) 9(M1) h 0.52(AT)13 3x 105 x 3 - 1010 h 0.12(ATIL)14 h= 0.59(AT/L) 1/4 Heated plate facing downward or (M1) d AI cooled plate facing upward M Water at 70°F (294 K) bn 1loo (P1) h = 26(AT/L)4 h=127(AT/L)4 104-10° 1/4 Vertical planes and cylinders 1/4 Organic liquids at 70°F (294 K) 104-10 h = 12(AT/L)4 h= 59(AT/L)4 1/4 1/4 Vertical planes and cylinders (P1) EXAMPLE 15.5-2. Natural Convection and the Simplified Equation Repeat Example 15.5-1 but use the simplified equation. o The film temperature of 408.2 K is in the range 255-533 K. Also, Solution: L'AT = (0.305m) (194,4K) = 5.5m2 K This is slightly greater than the value of 4.7 given as the approximate maximum for use of the simplified equation. However, in Example 15.5-1 the value of NNpr is below 10°, so the simplified equation from Table 15.5-2 will be used: 1/4 \1/4 AT 1.37 194.4 0.305 h=1.37 L = 6.88 W/m2 K (1.21 btu/h ft2.F)
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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Question
Transcribed Image Text:2. Losses by Natural Convection from
diameter and 121.9 mm
Cylinder. A vertical cylinder 76.2 mm in
high is maintained at 397.1 K at its surface. It loses heat by
natural convection to air at 294.3 K. Heat is lost from the cylindrical side and the flat
circular end at the top. Calculate the heat loss, neglecting radiation losses. Use the
simplified equations of Table 15.5-2 and those equations for the lowest range of
NGr Npr. The equivalent L to use for the top flat surface is 0.9 times the diameter.
Ans. q 26.0 W
a
11
Transcribed Image Text:able 15.5-2. Simplified Equations for Natural Convection from Various Surfaces
Equation
h= W/m2. K
h btu/h ft. °F
L = m, AT = K
°F
L -f, ΔT=
L ft, AT
Physical Geometry
D=m
D=ft
NGNpr
Ref.
Air at 101.32 kPa (1 atm) abs pressure
h = 0.28(ATIL)4 h 1.37(AT/L)4
h 1.24 AT/3
h =0.27(ATID)/4 h 1.32(AT/D)4
h = 1.24 AT1/3
Vertical planes and
cylinders
1/4
104-10
>10 h0.18(AT)
103- 10
>109
for
(P1)
odasrO(P1)
(M1)
(M1)
PM
1/3
m
Horizontal cylinders
h = 0.18(AT)3
Horizontal plates
h = 0.27(AT/L)14 h = 0.32(AT/L)1/4
0.22(AT)s
105-2 x 10
Heated plate facing
upward or cooled 2 x 107-3 x 1010 h
plate facing
downward
(M1)
9(M1)
h 0.52(AT)13
3x 105 x 3 - 1010 h 0.12(ATIL)14 h= 0.59(AT/L)
1/4
Heated plate facing
downward or
(M1)
d
AI
cooled plate
facing upward
M
Water at 70°F (294 K)
bn 1loo
(P1)
h = 26(AT/L)4 h=127(AT/L)4
104-10°
1/4
Vertical planes and
cylinders
1/4
Organic liquids at 70°F (294 K)
104-10
h = 12(AT/L)4 h= 59(AT/L)4
1/4
1/4
Vertical planes and
cylinders
(P1)
EXAMPLE 15.5-2. Natural Convection and the Simplified Equation
Repeat Example 15.5-1 but use the simplified equation.
o
The film temperature of 408.2 K is in the range 255-533 K. Also,
Solution:
L'AT = (0.305m) (194,4K) = 5.5m2 K
This is slightly greater than the value of 4.7 given as the approximate maximum
for use of the simplified equation. However, in Example 15.5-1 the value of
NNpr is below 10°, so the simplified equation from Table 15.5-2 will be used:
1/4
\1/4
AT
1.37 194.4
0.305
h=1.37
L
= 6.88 W/m2 K (1.21 btu/h ft2.F)
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