(a) Explanation Given: The thickness of a plate is t = 1 mm . The vertical length of the plate is L = 250 mm . The horizontal breadth of the plate is b = 100 mm . The temperature of the surface of plate is T s = 90 ° C . Formula used: The expression for total heat transfer is given as, q = h ¯ A ( T s a t − T s ) Here, h ¯ is the average convection coefficient, A is area of the plate, T s a t is the saturation temperature, and T s is the surface temperature. The expression for Jacob number is calculated as, J a = c p . l ( T s a t − T s ) h f g Here, h f g is the specific enthalpy of gas and fluid, and c p . l is the specific heat of liquid. The expression for the specific enthalpy is given as, h ′ f g = h f g ( 1 + 0.68 J a ) The expression for dimensionless parameter P is given as, P = k l L ( T s a t − T s ) μ l h ′ f g ( ν l 2 g ) 1 3 Here, L is the length of the plate, g is the acceleration due to gravity, k l is the thermal conductivity of liquid, μ l is the dynamic viscosity of liquid, ν l is the kinematic viscosity of the liquid, and T s a t is the saturation temperature. The expression for average convection coefficient is given as, h ¯ l = k l ( ν l 2 g ) 1 3 × 1 P ( 0.68 P + 0.89 ) 0.82 The expression for film temperature is given as, T f = T s + T s a t 2 Calculation: For 2024 -T 6 aluminum at temperature 300 K , the thermal conductivity of aluminum is, k A l = 177 W / m ⋅ K The value of film temperature is calculated as, T f = T s + T s a t 2 = 90 ° C + 100 ° C 2 = 95 ° C Obtain the value from the table A.4 water vapor at temperature T ∞ = 100 ° C “Thermophysical Properties of Gases at Atmospheric Pressure” as, P s a t = 1.008 bar h f g = 2.257 J / kg Obtain the value from the table A.4 water liquid at film temperature T f = 95 ° C “Thermophysical Properties of Gases at Atmospheric Pressure” as, μ l = 0.00029858 N ⋅ s / m 2 c p . l = 4212 J / kg ⋅ K k l = 0.6782 W / m ⋅ K ν l = 3.076 × 10 − 7 m 2 / s Applying energy balance for the control surface at T s , q = h ¯ A ( T s a t − T s ) = k m A t ( T s − T c ) Here, k m is the thermal conductivity of metal plate. So, the cold temperature is given as, T c = T s − t h ¯ k m ( T s a t − T s ) Here, t is the thickness of the plate (b) To determine The cold temperature for AISI 302 stainless steel.

7th Edition

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 10, Problem 10.71P

(a)

To determine

The cold temperature for 2024-T6 aluminum.

(b)

To determine

The cold temperature for AISI 302 stainless steel.

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Chapter 10, Problem 10.70P

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The cold temperature for 2024 -T 6 aluminum.

Solution Summary: The author explains how the temperature of the surface of a plate is T_s=90°C.