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3_101_103

# 3_101_103 - PROBLEM 3.101 KNOWN Dimensions of a plate...

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PROBLEM 3.101 KNOWN: Dimensions of a plate insulated on its bottom and thermally joined to heat sinks at its ends. Net heat flux at top surface. FIND: (a) Differential equation which determines temperature distribution in plate, (b) Temperature distribution and heat loss to heat sinks. SCHEMATIC: ASSUMPTIONS: (1) Steady-state, (2) One-dimensional conduction in x (W,L>>t), (3) Constant properties, (4) Uniform surface heat flux, (5) Adiabatic bottom, (6) Negligible contact resistance. ANALYSIS: (a) Applying conservation of energy to the differential control volume, q x + dq = q x +dx , where q x+dx = q x + (dq x /dx) dx and ( ) o dq=q W dx . ʹஒʹஒ Hence, ( ) x o dq / dx q W=0. ʹஒʹஒ From Fourier’s law, ( ) x q k t W dT/dx. = Hence, the differential equation for the temperature distribution is 2 o o 2 q d dT d T ktW q W=0 0. dx dx kt dx ʹஒʹஒ Ρਟ Τ੏ ʹஒʹஒ + = ΢ਯ Υ੟ Σਿ Φ੯ < (b) Integrating twice, the general solution is, ( ) 2 o 1 2 q T x x C x +C 2kt ʹஒʹஒ = + and appropriate boundary conditions are T(0) = T o , and T(L) = T o . Hence, T o = C 2 , and 2 o o o 1 2 1 q q L T L C L+C and C . 2kt 2kt ʹஒʹஒ ʹஒʹஒ = + = Hence, the temperature distribution is ( ) ( ) 2 o o q L T x x Lx T . 2kt ʹஒʹஒ = + < Applying Fourier s law at x = 0, and at x = L, ( ) ( ) x=0 o o x=0 q q WL L q 0 k Wt dT/dx) kWt x kt 2 2 ʹஒʹஒ ʹஒʹஒ Ρਟ Τ੏ Ρਟ Τ੏ = = = ΢ਯ Υ੟ ΢ਯ Υ੟ Σਿ Φ੯ Σਿ Φ੯ ( ) ( ) x=L o o x=L q q WL L q L k Wt dT/dx) kWt x kt 2 2 ʹஒʹஒ ʹஒʹஒ Ρਟ Τ੏ Ρਟ Τ੏ = = = + ΢ਯ Υ੟ ΢ਯ Υ੟ Σਿ Φ੯ Σਿ Φ੯ Hence the heat loss from the plates is ( ) o o q=2 q WL/2 q WL. ʹஒʹஒ ʹஒʹஒ = < COMMENTS: (1) Note signs associated with q(0) and q(L). (2) Note symmetry about x = L/2. Alternative boundary conditions are T(0) = T o and dT/dx) x=L/2 =0.

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PROBLEM 3.102 KNOWN: Dimensions and surface conditions of a plate thermally joined at its ends to heat sinks at different temperatures.
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3_101_103 - PROBLEM 3.101 KNOWN Dimensions of a plate...

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