Example: Saturated steam at 267°F is flowing inside a %-in steel pipe having an ID of 0.824 in. and an OD of 1.050 in. The pipe is insulated with 1.5 in of insulation on the outside. The convective coefficient for the inside steam surface of the pipe is estimated as h; = 1000 Btu/hr-ft? °F, and the convective coefficient on the outside of the lagging is estimated as ho = 2 Btu/hr-ft20°F. The mean thermal conductivity of the metal is 26 Btu/hr-ft °F and 0.037 Btu/hr-ft °F for the insulation a. Calculate for the heat loss for 1 ft of pipe using resistances if the surrounding air is at 80°F b. Repeat using the overall U; based on the area Ai Given: To= 80°F Ti = 267°F Dipipe = 0.824 in →r; = 0.412 in Do,pipe = 1.050 in→r¸ = 0.525 in Tins= 2.025" insulation r,-0.525" rins = 1.5 in + 0.525 in = 2.025 = ro %3D h, = 1000 Btu/hr-ft2 °F -1.5"- r=D0.42 T = 267 F ho = 2 Btu/hr-ft²°F km = 26 Btu/hr-ft °F Kins = 0.037 Btu/hr-ft °F %3D %3D
Example: Saturated steam at 267°F is flowing inside a %-in steel pipe having an ID of 0.824 in. and an OD of 1.050 in. The pipe is insulated with 1.5 in of insulation on the outside. The convective coefficient for the inside steam surface of the pipe is estimated as h; = 1000 Btu/hr-ft? °F, and the convective coefficient on the outside of the lagging is estimated as ho = 2 Btu/hr-ft20°F. The mean thermal conductivity of the metal is 26 Btu/hr-ft °F and 0.037 Btu/hr-ft °F for the insulation a. Calculate for the heat loss for 1 ft of pipe using resistances if the surrounding air is at 80°F b. Repeat using the overall U; based on the area Ai Given: To= 80°F Ti = 267°F Dipipe = 0.824 in →r; = 0.412 in Do,pipe = 1.050 in→r¸ = 0.525 in Tins= 2.025" insulation r,-0.525" rins = 1.5 in + 0.525 in = 2.025 = ro %3D h, = 1000 Btu/hr-ft2 °F -1.5"- r=D0.42 T = 267 F ho = 2 Btu/hr-ft²°F km = 26 Btu/hr-ft °F Kins = 0.037 Btu/hr-ft °F %3D %3D
Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)
8th Edition
ISBN:9781305387102
Author:Kreith, Frank; Manglik, Raj M.
Publisher:Kreith, Frank; Manglik, Raj M.
Chapter8: Natural Convection
Section: Chapter Questions
Problem 8.28P
Related questions
Question
![Example:
Saturated steam at 267°F is flowing inside a %-in steel pipe having an ID of 0.824 in. and an OD of
1.050 in. The pipe is insulated with 1.5 in of insulation on the outside. The convective coefficient for the
inside steam surface of the pipe is estimated as h; = 1000 Btu/hr-ft? °F, and the convective coefficient on
the outside of the lagging is estimated as ho = 2 Btu/hr-ft20°F. The mean thermal conductivity of the metal
is 26 Btu/hr-ft °F and 0.037 Btu/hr-ft °F for the insulation
a. Calculate for the heat loss for 1 ft of pipe using resistances if the surrounding air is at 80°F
b. Repeat using the overall U; based on the area Ai
Given:
To= 80°F
Ti = 267°F
Dipipe = 0.824 in →r; = 0.412 in
Do,pipe = 1.050 in→r¸ = 0.525 in
Tins= 2.025"
insulation
r,-0.525"
rins = 1.5 in + 0.525 in = 2.025 = ro
%3D
h, = 1000 Btu/hr-ft2 °F
-1.5"-
r=D0.42
T = 267 F
ho = 2 Btu/hr-ft²°F
km = 26 Btu/hr-ft °F
Kins = 0.037 Btu/hr-ft °F
%3D
%3D](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F426042b7-854b-4316-8bab-10794fa80326%2F521cb375-acfe-42a7-8c7b-671edfd3410f%2Frnbqyn_processed.png&w=3840&q=75)
Transcribed Image Text:Example:
Saturated steam at 267°F is flowing inside a %-in steel pipe having an ID of 0.824 in. and an OD of
1.050 in. The pipe is insulated with 1.5 in of insulation on the outside. The convective coefficient for the
inside steam surface of the pipe is estimated as h; = 1000 Btu/hr-ft? °F, and the convective coefficient on
the outside of the lagging is estimated as ho = 2 Btu/hr-ft20°F. The mean thermal conductivity of the metal
is 26 Btu/hr-ft °F and 0.037 Btu/hr-ft °F for the insulation
a. Calculate for the heat loss for 1 ft of pipe using resistances if the surrounding air is at 80°F
b. Repeat using the overall U; based on the area Ai
Given:
To= 80°F
Ti = 267°F
Dipipe = 0.824 in →r; = 0.412 in
Do,pipe = 1.050 in→r¸ = 0.525 in
Tins= 2.025"
insulation
r,-0.525"
rins = 1.5 in + 0.525 in = 2.025 = ro
%3D
h, = 1000 Btu/hr-ft2 °F
-1.5"-
r=D0.42
T = 267 F
ho = 2 Btu/hr-ft²°F
km = 26 Btu/hr-ft °F
Kins = 0.037 Btu/hr-ft °F
%3D
%3D
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