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Estimate the convective heat-transfer coefficient for natural convection from a horizontal steam pipe. The outside surface temperature of the insulated pipe is 80°C. The surrounding air temperature is 25°C. The outside diameter of the insulated pipe is 10 cm. please help me if my answer is wrong

Estimate the convective heat-transfer coefficient for natural convection from a horizontal steam pipe. The outside surface temperature of the insulated pipe is 80°C. The surrounding air temperature is 25°C. The outside diameter of the insulated pipe is 10 cm. please help me if my answer is wrong

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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Estimate the convective heat-transfer coefficient for natural convection from a horizontal
steam pipe. The outside surface temperature of the insulated pipe is 80°C. The
surrounding air temperature is 25°C. The outside diameter of the insulated pipe is 10 cm. please help me if my answer is wrong. 

Prodtem 4.16: horreomtal cylinder: for V, 0.317 Ne D- 0,1m NNu: 0.6 + (0.550 3,08 = 1- air Free Cmveetin alir Properties a Te Mar Too + To 20 + 86 ( 3. lox103) (m.sIE X10-)e P= 1.057 らん ト- 3.JDX16 K" They-K 4 - 1016 5,351,333 Nor Npr pea k 6.0272 M= 19. 515 x10" Pa.s - 5 1 g13 (0.a1) Mer - 6.71 3799 446 י רV 6.397 (3799 446)" 6.6 + יר ס
Transcribed Image Text:Prodtem 4.16: horreomtal cylinder: for V, 0.317 Ne D- 0,1m NNu: 0.6 + (0.550 3,08 = 1- air Free Cmveetin alir Properties a Te Mar Too + To 20 + 86 ( 3. lox103) (m.sIE X10-)e P= 1.057 らん ト- 3.JDX16 K" They-K 4 - 1016 5,351,333 Nor Npr pea k 6.0272 M= 19. 515 x10" Pa.s - 5 1 g13 (0.a1) Mer - 6.71 3799 446 י רV 6.397 (3799 446)" 6.6 + יר ס
0.387 (379466^(Y^) - (1 +( 0. 559÷0.7) 4.613 NNn = hd Nam 'k h= 4. 413(0.6272) = 1.254 6.1 m² -K Q -h VT - - Th D DT - -T(1, 254) (01)(20 – 80) = 23.64 Wm
Transcribed Image Text:0.387 (379466^(Y^) - (1 +( 0. 559÷0.7) 4.613 NNn = hd Nam 'k h= 4. 413(0.6272) = 1.254 6.1 m² -K Q -h VT - - Th D DT - -T(1, 254) (01)(20 – 80) = 23.64 Wm
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