Explanation Given Information: Diameter of the copper wire D = 0.8 mm = 0.8 × 10 − 3 m Length of the copper wire L = 5 cm = 0.05 m Temperature of air T ∞ = ( 10 + 14 t ) ℃ t = Time in seconds Initial temperature of the wire is T 0 = 10 ℃ Convective heat transfer coefficient (h)= 40 W / m 2 K Explanation: For properties of copper, refer to Appendix-2 and Table 12. At 20 ℃ , density of copper ( ρ ) = 8933 kg/m 3 Specific heat ( c ) = 383 J / kgK Thermal conductivity ( k ) = 399 W / mK Initially (at time t = 0), the temperature of wire and the temperature of air are the same. As time passes, the temperature of air increases and it causes an increase in the temperature of the copper wire. Biot number ( Bi ) = hL k L = Characteristic dimension = Volume Surface Area L = π 4 D 2 H πDH = D 4 Bi = hD 4 k = 40 × 0.8 × 10 − 3 4 × 399 = 2 × 10 − 5 < 0.1 Since Bi < 0.1 , internal resistance of the copper wire is negligible. For copper wire energy balance equation is written as follows: [ Rate of change of internal energry of the copper wire ] = [ Rate of convective heat transfer from air to wire ] mc dT dt = hA s ( T ∞ − T ) m c dT dt = hA s ( 10 + 14 t − T ) dT dt = [ hA s m c ] ( 10 + 14 t − T ) Let λ = [ hA s mc ] dT dt = λ ( 10 + 14 t − T ) dT dt + λT = λ ( 10 + 14 t ) … .. Equation ① Solving the above Differential Equation, we get: T = [ Complementary solution ] + [ Particular solution ] T = [ C 1 e − λt ] + [ C − 2 + C 3 t ] … . Equation ② dT dt = − λC 1 e − λt + C 3 … … .. Equation ③ Substitute Equation ② in Equation ① dT dt + λ [ C 1 e − λt + C 2 + C 3 t ] = λ [ 10 + 14 t ] … .

Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)

8th Edition

ISBN: 9781305387102

Author: Kreith, Frank; Manglik, Raj M.

Publisher: Cengage Learning

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Principles of Heat Transfer (Activate Learning with these NEW titles from Engineering!)

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