(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)

To determine

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.