4.2-2. Heat Removal of a Cooling Coil. A coòling coil of 1.0 ft of 304 stainless steel tubing having an inside diameter of 0.25 in. and an outside diameter of 0.40 in. is being used to remove heat from a bath. The temperature at the inside surface of the tube is 40°F and 80°F on the outside. The thermal conductivity of 304 stainless steel is a function of temperature. k = 7.75 + 7.78 × 10-3 T where k is in btu/h ft °F and T is in °F. Calculate the heat removal in btu/s and watts. Ans. 1.225 btu/s, 1292 W 4.2-3. Removal of Heat from a Bath. Repeat Problem 4.2-2 but for a cooling coil made of 308 stainless steel having an ayerage thermal conductivity of 15.23 W/m K.

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
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Answer 4.2-3

4.2-2. Heat Removal of a Cooling Coil. A coòling coil of 1.0 ft of 304 stainless steel
tubing having an inside diameter of 0.25 in. and an outside diameter of 0.40 in. is
being used to remove heat from a bath. The temperature at the inside surface of
the tube is 40°F and 80°F on the outside. The thermal conductivity of 304
stainless steel is a function of temperature.
k = 7.75 + 7.78 × 10-3 T
where k is in btu/h ft °F and T is in °F. Calculate the heat removal in btu/s
and watts.
Ans. 1.225 btu/s, 1292 W
4.2-3. Removal of Heat from a Bath. Repeat Problem 4.2-2 but for a cooling coil made
of 308 stainless steel having an average thermal conductivity of 15.23 W/m · K.
Transcribed Image Text:4.2-2. Heat Removal of a Cooling Coil. A coòling coil of 1.0 ft of 304 stainless steel tubing having an inside diameter of 0.25 in. and an outside diameter of 0.40 in. is being used to remove heat from a bath. The temperature at the inside surface of the tube is 40°F and 80°F on the outside. The thermal conductivity of 304 stainless steel is a function of temperature. k = 7.75 + 7.78 × 10-3 T where k is in btu/h ft °F and T is in °F. Calculate the heat removal in btu/s and watts. Ans. 1.225 btu/s, 1292 W 4.2-3. Removal of Heat from a Bath. Repeat Problem 4.2-2 but for a cooling coil made of 308 stainless steel having an average thermal conductivity of 15.23 W/m · K.
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