(a) Explanation Given data: The diameter of the cylindrical canister is D = 75 mm . The length of cylindrical canister is L = 120 mm . The operating temperature is T f c = 800 °C . The convective heat transfer coefficient is h = 12 W / m 2 ⋅ K . The surface temperature is T s u r = 300 K . The ambient temperature is T ∞ = 298 K . Formula used: The expression for the temperature of canister surface is given as, T s = ( T b 4 ε s ) 1 4 Here, ε s is the emissivity of canister surface, σ is the Stefan Boltzmann’s Constant, and T b is the surface temperature. The expression for the radiation heat transfer coefficient is given as, h r = ε s σ ( T s + T s u r ) ( T s 2 + T s u r 2 ) Here, h r is the radiation heat transfer coefficient. The expression for the heat loss from the cylindrical wall is given as, q = q ∞ + q s u r T f c [ ln ( r 1 + t r 1 ) 2 π k L ] = T s − T ∞ [ 1 h 2 π ( r 1 + t ) L ] + T s − T s u r [ 1 h f 2 π ( r 1 + t ) L ] Here, k is the thermal conductivity, L is the length of the cylindrical canister, r 1 is the internal diameter of cylinder, t is the thickness of cylinder, h is the convective heat transfer coefficient, and T ∞ is the ambient temperature. Calculation: Calculations for the Case i: The outer surface is covered with the thin layer of dirt and insulating material is Calcium silicate with ε s = 0.90 and k = 0.09 W / m ⋅ K . The surface temperature can be calculated as, T s = ( T b 4 ε s ) 1 4 T s = ( ( 304 K ) 4 0.90 ) 1 4 T s = 312.1 K The radiation heat transfer coefficient can be calculated as, h r = ε s σ ( T s + T s u r ) ( T s 2 + T s u r 2 ) h r = ( 0.9 ) × ( 5.67 × 10 − 8 ) × ( 312.1 K + 300 K ) × [ ( 312.1 K ) 2 + ( 300 ) 2 ] h r = 5.853 W / m 2 ⋅ K The required thickness can be calculated as, T f c − T s [ ln ( r 1 + t r 1 ) 2 π k L ] = T s − T ∞ [ 1 h 2 π ( r 1 + t ) L ] + T s − T s u r [ 1 h f 2 π ( r 1 + t ) L ] { ( 800 °C + 273 ) K [ ln ( ( 37.5 mm × 1 m 10 3 mm ) + t ( 37.5 mm × 1 m 10 3 mm ) ) 2 π × ( 0.09 W / m ⋅ K ) × ( 120 mm × 1 m 10 3 mm ) ] = 312.1 K − 298 K [ 1 ( 12 W / m 2 ⋅ K ) 2 π ( ( 37.5 mm × 1 m 10 3 mm ) + t ) × ( 120 mm × 1 m 10 3 mm ) ] + 312.1 K − 300 K [ 1 ( 5.853 W / m 2 ⋅ K ) 2 π ( ( 37.5 mm × 1 m 10 3 mm ) + t ) × ( 120 mm × 1 m 10 3 mm ) ] } 51.633 ln ( 37.5 × 10 − 3 m + t 37.5 × 10 − 3 m ) = 180.968 × ( 37.5 × 10 − 3 m + t ) t = 0.1436 m Calculations for the Case ii: The outer surface is covered with the thin layer of dirt and insulating material is Calcium silicate with ε s = 0.08 and k = 0.09 W / m ⋅ K . The surface temperature can be calculated as, T s = ( T b 4 ε s ) 1 4 T s = ( ( 304 K ) 4 0.08 ) 1 4 T s = 571.61 K The radiation heat transfer coefficient can be calculated as, h r = ε s σ ( T s + T s u r ) ( T s 2 + T s u r 2 ) h r = ( 0.08 ) × ( 5.67 × 10 − 8 W / m 2 ⋅ K 4 ) × ( 571.61 K + 300 K ) × [ ( 571.61 K ) 2 + ( 300 ) 2 ] h r = 1.647 W / m 2 ⋅ K The required thickness can be calculated as, T f c − T s [ ln ( r 1 + t r 1 ) 2 π k L ] = T s − T ∞ [ 1 h 2 π ( r 1 + t ) L ] + T s − T s u r [ 1 h f 2 π ( r 1 + t ) L ] { ( 800 °C + 273 ) K − 571.16 K [ ln ( ( 37.5 mm × 1 m 10 3 mm ) + t ( 37.5 mm × 1 m 10 3 mm ) ) 2 π × ( 0.09 W / m ⋅ K ) × ( 120 mm × 1 m 10 3 mm ) ] = 571.61 K − 298 K [ 1 ( 12 W / m 2 ⋅ K ) 2 π ( ( 37.5 mm × 1 m 10 3 mm ) + t ) × ( 120 mm × 1 m 10 3 mm ) ] + 571.61 K − 300 K [ 1 ( 1.647 W / m 2 ⋅ K ) 2 π ( ( 37

6th Edition

ISBN: 9780470501962

Author: Frank P. Incropera, David P. DeWitt, Theodore L. Bergman, Adrienne S. Lavine

Publisher: Wiley, John & Sons, Incorporated

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Chapter 3, Problem 3.23P

(a)

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The thickness of insulation in case i.

The thickness of insulation in case ii.

The thickness of insulation in case iii.

The thickness of insulation in case iv.

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Introduction to Heat Transfer

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The thickness of insulation in case i. The thickness of insulation in case ii. The thickness of insulation in case iii. The thickness of insulation in case iv.

Solution Summary: The author explains the thickness of insulation in case i, and the convective heat transfer coefficient for the cylindrical canister.

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The thickness of insulation in case i. The thickness of insulation in case ii. The thickness of insulation in case iii. The thickness of insulation in case iv.

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Chapter 3 Solutions

The thickness of insulation in case i.

The thickness of insulation in case ii.

The thickness of insulation in case iii.

The thickness of insulation in case iv.