Explanation Given information: The thickness of metal plate is 3 c m , the thermal conductivity of metal plate is 15 W / m ⋅ K , the uniform heat flux at the bottom of the plate is 5 k W / m 2 , the ambient air temperature is 30 ° C , the convection heat transfer coefficient of air is 10 W / m 2 ⋅ K and the maximum allowable temperature of SA-193 bolts is 260 ° C . The below figure represent the flow of heat through metal plate. Figure-(1) Write the expression for fourier-biot equation. ∂ 2 T ∂ 2 x + q g k = ( 1 α ) ∂ T ∂ t ...... (I) Here, the double differential of temperature with respect to length of plate is ∂ 2 T ∂ 2 x , the internal heat generation per unit volume is q g , the thermal conductivity of a material is k , the thermal diffusivity is α , the length of plate is x and the rate of change to temperature is ∂ T ∂ t . Substitute 0 for q g ,and 0 for ∂ T ∂ t in Equation (I). ∂ 2 T ∂ 2 x + 0 k = ( 1 α ) ( 0 ) ∂ 2 T ∂ 2 x = 0 ...... (II) Double integrate Equation (II). T = C 1 x + C 2 ...... (III) Here, the constants are C 1 and C 2 . Substitute 0 for x and T 1 for T in Equation (III). T 1 = C 1 ( 0 ) + C 2 = C 2 Substitute L for x , T 2 for T and T 1 for C 2 in Equation (III). T 2 = C 1 L + T 1 C 1 L = T 2 − T 1 C 1 = T 2 − T 1 L Substitute T 1 for C 2 and T 2 − T 1 L for C 1 in Equation (III). T = T 2 − T 1 L ⋅ x + T 1 ...... (IV) Here, the temperature variation is T , the above temperature of plate is T 2 ,the bottom temperature of plate is T 1 , the length of plate is L and the variation of length of plate is x . Write the expression forconvective heat flux. Q C o n v A = h ( T 2 − T ∞ ) ...... (V) Here, the convective heat flux is Q C o n v A , the convective heat transfer coefficient is h , the surface area of plate is A and the ambient temperature is T ∞ . Write the expression for conductive heat flux. Q C o n d A = k | T 1 − T 2 L | ...... (VI) Here,the conductive heat flux is Q C o n d A . Write the formula for temperature conversion from degree celcius to Kelvin. T K = T ° C + 273.15 ...... (VII) Here the temperature in Kelvin is T K and temperature in degree celcius is T ° C . Calculation: Substitute 30 ° C for T ° C in Equation (VII). T K = 30 ° C + 273.15 = 303.15 K Substitute 5 k W / m 2 for Q C o n v A , 303.15 K for T ∞ and 10 W / m 2 ⋅ K for h in Equation (V). 5 k W / m 2 = 10 W / m 2 ⋅ K ( T 2 − 303.15 K ) 5 k W / m 2 × 1000 W 1 k W = 10 W / m 2 ⋅ K ( T 2 − 303.15 K ) 500 K = T 2 − 303.15 K T 2 = 803

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Chapter 2, Problem 151P

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The equation of temperature variation in metal plate and temperature at x=0 cm , 1.5 cm and 3.0 cm and whether SA-193 bolts comply with ASME code.

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Chapter 2, Problem 150P

Chapter 2, Problem 152P

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Chapter 2 Solutions

HEAT+MASS TRANSFER:FUND.+APPL.(LOOSE)

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The equation of temperature variation in metal plate and temperature at x = 0 c m , 1.5 c m and 3.0 c m and whether SA-193 bolts comply with ASME code.

The equation of temperature variation in metal plate and temperature at x=0 cm , 1.5 cm and 3.0 cm and whether SA-193 bolts comply with ASME code.

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Chapter 2 Solutions