Explanation Given information: The outer diameter of tube is 5 cm , the temperature of the wall is 180 ° C , the thermal conductivity of the fin is 186 W / m ⋅ K , the outer diameter of circular fin is 6 cm , the thickness of the circular fin is 1 mm , the space between the fins is 3 mm , the number of fins per meter length of tube is 250 , the temperature of surrounding air is 25 ° C , the convective heat transfer coefficient is 40 W / m 2 ⋅ K . Write the expression for corrected outer radius. r 2 c = D f 2 + t f 2 ....... (I) Here, the outer diameter of circular fin is D o and thickness of the circular fin is t f . Write the expression for area of the circular fin. A f = 2 π ( r 2 c 2 − r o 2 ) ....... (II) Here, the corrected radius is r 2 c , the outer radius of tube is r o . Write the expression for the corrected fin length of the fin L c = L + t f 2 = D f − D o 2 + t f 2 ....... (III) Here, the diameter of fin is D f , the outer diameter of tube is D o , thickness of the circular fin is t f . Write the expression for profile area of fin. A P = L c t f ....... (IV) Write the expression for ξ ξ = L c 3 / 2 h k A P ....... (V) Here, the corrected length is L c , the heat transfer coefficient of outer surface of pipe is h , the thermal conductivity of fin is k , the profile area is A P Write the expression for the ratio r 2 c r 1 . r 2 c r 1 = r 2 c ( D o / 2 ) ....... (VI) Write the expression for total heat transfer from finned surface area of the tube Q f = η f N h A f ( T s − T ) ....... (VII) Here, the efficiency of the fin is η f Write the expression for unfinned surface area of tube. A u f = N π D o x ....... (VIII) Here, the number of fins per meter length of tube is N , the outer diameter of tube is D o , the space between the fins is x . Write the expression for heat transfer from unfinned surface area of tube. Q u f = h A u f ( T s − T ) ....... (IX) Here, the convective heat transfer coefficient is h , the unfinned surface area of tube is A u f , the temperature of tube's wall is T s , the temperature of surrounding air is T . Write the expression for total heat transfer with fins. Q = Q f + Q u f ....... (X) Write the expression for area of the tube A t u b e = π D o L ....... (XI) Write the expression for heat transfer from the tube without fins Q t u b e = h A t u b e ( T s − T ) ....... (XII) Write the expression for increase in heat transfer from the tube per unit length. Q i n c r e a s e = Q f − Q t u b e ....... (XIII) Calculation: Substitute 6 cm for D f and 1 mm for t f in the Equation (I). r 2 c = D f 2 + t f 2 = ( 6 cm × 1 m 100 cm ) 2 + ( 1 mm × 1 m 1000 mm ) 2 = 0.0305 m Substitute 0.0305 m for r 2 c , 5 cm 2 for r o in the Equation (II). A f = 2 π ( r 2 c 2 − r o 2 ) = 2 π [ ( 0.0305 m ) 2 − ( 5 cm 2 × 1 m 100 cm ) 2 ] = 0.00192 m 2 Substitute 6 cm for D f , 5 cm for D o , 1 mm for t f in the Equation (II). L c = D f − D o 2 + t f 2 = ( 6 cm × 1 m 100 cm − 5 cm × 1 m 100 cm ) 2 + ( 1 mm × 1 m 1000 mm ) 2 = 0.0055 m Substitute 0.0055 m for L c , 1 mm for t f in the Equation (IV). A P = L c t f = ( 0.0055 m ) ( 1 mm × 1 m 1000 mm ) = 0.0000055 m 2 Substitute 0.0000055 m 2 for A P , 0.0055 m for L c , 40 W / m 2 ⋅ K for h , 186 W / m ⋅ K for k in the Equation (V)

5th Edition

ISBN: 9781307573060

Author: CENGEL

Publisher: MCG/CREATE

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Question

Chapter 3, Problem 128P

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The increase in the heat transfer from the tube per meter of its length.

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Chapter 3, Problem 127P

Chapter 3, Problem 129P

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Chapter 3 Solutions

HEAT+MASS TRANSFER-SI ED. EBOOK >I<

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