HW2+09F+solution

# HW2+09F+solution - ME487 HW#2 Solutions 1 Set a control...

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ME487 HW#2 Solutions 1. Set a control volume as shown below on a fixed coordinate system. The rate of change of internal energy for the element x: ˙ Q v = ρcA Δ x Δ T Δ t (1) Heat ﬂux across the upper face of the element is: ˙ Q u = kA dT dx x That across the lower face is: ˙ Q l = kA dT dx x x Thus net heat ﬂux for the element becomes: ˙ Q = ˙ Q u ˙ Q l = kA dT dx x x kA dT dx x (2) - 1 -

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Internal heat source due to Joule heating is: ˙ Q s = I 2 R = I 2 ρ r Δ x A (3) From first principle, ˙ Q v = ˙ Q + ˙ Q s (4) Thus plug Eqs. (1), (2) and (3) into (4), ρcA Δ x Δ T Δ t = kA dT dx x x dT dx x + I 2 ρ r Δ x A (5) Divide Eq. (5) by A x ρc Δ T Δ t = k ( dT dx x x dT dx x ) Δ x + I 2 ρ r A 2 Taking the limit as x 0 and t 0 becomes: ρc dT dt = k 2 T ∂x 2 + I 2 ρ r A 2 (6) But dT dt = ∂T ∂t + ∂T ∂x dx dt (7) and dx dt = V w (8) Plug in Eqs. (7) & (8) into (6) ρc ∂T ∂t = k 2 T ∂x 2 ρcV w ∂T ∂x + I 2 ρ r A 2 - 2 -
2. Approximated value: q a = ( θ m + 273) 2 300000 = 2 . 9 J/mm 3 Theoretic value: q t = ρ [ c ( θ m θ 0 ) + L ] = 2700 · [1 . 028 · 630 + 398] = 2 . 823 J/mm 3 Thus, the error can be obtained as: %error = 2 . 9 2 . 823 2 . 823 · 100 = 2 . 79 % 3. (a) r = ξ 2 + y 2 = ( 8) 2 + 5 2 = 9 . 43 mm κ = k ρc = 0 . 05 8 × 10 6 · 850 = 7

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