A light oil flows through a copper tube of 2.6-cm-ID and 3.2-cm-OD. Air flows perpendicular over the exterior of the tube as shown in the following sketch. The convection heat transfer coefficient for the oil is 120 W/(m2 K) and for the air is 35 W/(m2 K). Calculate the overall heat transfer coefficient based on the outside area of the tube (a) considering the thermal resistance of the tube, (b) neglecting the resistance of the tube but assume that fouling factors of 0.0009 (m2 K)/W and 0.0004 (m2K)/W have developed on the inside and on the outside, respectively.

GIVEN

• Air flow over a copper tube with oil flow within the tube

• Tube diameters: Inside (Di) = 2.6 cm = 0.026 m

Outside (Do) = 3.2 cm = 0.032 m

• Convective heat transfer coefficients: Oil h i= 120 W/(m2 K) Air h o= 35 W/(m2 K)

• Fouling factors: Inside (Rdi) = 0.0009 (m2 K)/W

Outside (Rdo) = 0.0004 (m2 K)/W

FIND

• The overall heat transfer coefficient based on the outside tube area (Uo),

(a) considering the thermal resistance of the tube, and

(b) neglecting the tube resistance

ASSUMPTIONS

• Uniform heat transfer coefficients

• Variation of thermal properties is negligible

SKETCH



PROPERTIES AND CONSTANTS

the thermal conductivity of copper (k) = 392 W/(m K) (at 127°C)


From the solution with or without tube wall resistance



(a) The overall heat transfer with fouling can be calculated by rearranging



Based on the outside tube area



(b) The tube wall resistance is negligible as shown in the solution

COMMENTS

The given fouling factors lead to a 4% decrease in the overall heat transfer coefficient based on the outer tube wall area.

Physics & Space Science

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