Explanation Given: The pipe emissivity is ε P = 0.6 . The temperature of surrounding is T ∞ = − 5 ° C . The temperature of hot water is T = 50 ° C . The outer diameter of steel pipe is D o = 100 mm . The inner diameter of steel pipe is D i = 84 mm . The thermal conductivity is k p = 60 W/m ⋅ K . Formula used: The expression for conduction heat transfer rate is given as, q cond . ' = q conv . ' + q rad . ' T − T s , o R cond . = h ¯ π D o ( T s , o − T ∞ ) + ε p π D o σ T s , o 4 ............ (1) Here q conv . ' is the convection heat transfer rate and q rad . ' is the radiation heat transfer rate, T s , o is the outer surface temperature, h ¯ is the convection coefficient, R cond . is the resistance in conduction, and σ is the Stefan Boltzmann constant. The expression for the conduction resistance is given as, R cond . = ln ( D o D i ) 2 π k p The expression of the Reynolds number is given as, R a D = g β ( T s , o − T ∞ ) D o 3 α v Here, g is the acceleration due to gravity, β is the volumetric thermal expansion, v is the kinematic viscosity and α is the thermal diffusitivity. The expression for Churchill and chu correlation is given as, N u ¯ D = { 0.60 + 0.387 Ra D 1 6 [ 1 + ( 0.559 Pr ) 9 16 ] 8 27 } 2 Here, N u D is the Nusselt number and Pr is the Prandtl number. The expression for convection coefficient is given as, h ¯ = N u D k a D o Here, h ¯ is the convection coefficient and k a is the thermal conductivity. The expression for convection heat transfer rate is given as, q conv . ' = h ¯ π D o ( T s , o − T ∞ ) Here, h ¯ is the convection coefficient and T s , o is the temperature of outer surface. The expression for radiation heat transfer rate is given as, q rad . ' = ε p π D o 4 σ T s , o 4 Calculation: The condition resistance can be calculated as, R cond . = ln ( D o D i ) 2 π k p R cond . = ln ( 100 mm 84 mm ) 2 π × 60 W / m ⋅ K R cond . = 4.62 × 10 − 4 m ⋅ K / W The Reynolds number can be calculated as, R a D = g β ( T s , o − T ∞ ) D o 3 α v R a D = 9.8 m / s 2 × 3.39 × 10 − 3 K − 1 ( T s , o − [ − 5 ( ° C + 273 ) K ] ) ( 0.1 m ) 3 ( 21.8 × 10 − 6 m 2 / s ) ( 15.45 × 10 − 6 m 2 / s ) R a D = 3.322 × 10 − 5 m 4 / s 2 × K − 1 ( T s , o − 268 K ) ( 21.8 × 15.45 × 10 − 12 m 4 / s 2 ) R a D = 98631 K − 1 ( T s , o − 268 K ) ............ (2) The nusselt number from Churchill and chu correlation can be calculated as, N u ¯ D = { 0.60 + 0.387 Ra D 1 6 [ 1 + ( 0.559 Pr ) 9 16 ] 8 27 } 2 N u ¯ D = { 0.60 + 0.387 × ( 98637 K − 1 ) 1 6 ( T s . o − 268 K ) 1 6 [ 1 + ( 0.559 0.708 ) 9 16 ] 8 27 } 2 N u ¯ D = { 0

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ISBN: 9780470501979

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

Publisher: Wiley, John & Sons, Incorporated

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Chapter 9, Problem 9.64P

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The daily cost of heat loss from an uninsulated pipe per unit length.

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Chapter 9, Problem 9.63P

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Fundamentals of Heat and Mass Transfer

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