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KNOWN: Diameter and temperature of horizontal cylinder or wire. FIND: Heat flux due to natural convection and comparison with values shown in Figure 10.4 SCHEMATIC: D= 7 mm, 7μm Water, 1 atm а AT = 5°C ASSUMPTIONS: (1) Constant properties, (2) No bubble nucleation. PROPERTIES: Table A-6, Water (Saturated liquid, T₁ = (Tsat + Ts)/2 = 102.5°C = 375.5 K): k= 0.681 W/mK, v=2.85 × 107 m²/s, p=956.6 kg/m³, cp = 4221 J/kg∙K, Pr= 1.69, ß=765 × 106 K-¹.

KNOWN: Diameter and temperature of horizontal cylinder or wire. FIND: Heat flux due to natural convection and comparison with values shown in Figure 10.4 SCHEMATIC: D= 7 mm, 7μm Water, 1 atm а AT = 5°C ASSUMPTIONS: (1) Constant properties, (2) No bubble nucleation. PROPERTIES: Table A-6, Water (Saturated liquid, T₁ = (Tsat + Ts)/2 = 102.5°C = 375.5 K): k= 0.681 W/mK, v=2.85 × 107 m²/s, p=956.6 kg/m³, cp = 4221 J/kg∙K, Pr= 1.69, ß=765 × 106 K-¹.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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KNOWN: Diameter and temperature of horizontal cylinder or wire. FIND: Heat flux due to natural convection and comparison with values shown in Figure 10.4 SCHEMATIC: D= 7 mm, 7μm Water, 1 atm а AT = 5°C ASSUMPTIONS: (1) Constant properties, (2) No bubble nucleation. PROPERTIES: Table A-6, Water (Saturated liquid, T₁ = (Tsat + Ts)/2 = 102.5°C = 375.5 K): k= 0.681 W/mK, v=2.85 × 107 m²/s, p=956.6 kg/m³, cp = 4221 J/kg∙K, Pr= 1.69, ß=765 × 106 K-¹.
Transcribed Image Text:KNOWN: Diameter and temperature of horizontal cylinder or wire. FIND: Heat flux due to natural convection and comparison with values shown in Figure 10.4 SCHEMATIC: D= 7 mm, 7μm Water, 1 atm а AT = 5°C ASSUMPTIONS: (1) Constant properties, (2) No bubble nucleation. PROPERTIES: Table A-6, Water (Saturated liquid, T₁ = (Tsat + Ts)/2 = 102.5°C = 375.5 K): k= 0.681 W/mK, v=2.85 × 107 m²/s, p=956.6 kg/m³, cp = 4221 J/kg∙K, Pr= 1.69, ß=765 × 106 K-¹.
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