Thermodynamics HW Solutions 143

# Thermodynamics HW Solutions 143 - Chapter 2 Heat Conduction...

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Chapter 2 Heat Conduction Equation 2-121 A long rectangular bar is initially at a uniform temperature of T i . The surfaces of the bar at x = 0 and y = 0 are insulated while heat is lost from the other two surfaces by convection. The mathematical formulation of this heat conduction problem is to be expressed for transient two-dimensional heat transfer with no heat generation. Assumptions 1 Heat transfer is transient and two-dimensional. 2 Thermal conductivity is constant. 3 There is no heat generation. Analysis The differential equation and the boundary conditions for this heat conduction problem can be expressed as α 2 2 2 2 1 T x T y T t += h , T Insulated a b h , T Tx t x Ty t x (,,) (, ,) 0 0 0 0 = = −= k Tayt x hTa yt T k Txbt x hT xbt T [(,,) ] ] Tx y T i 0 = 2-122 Heat is generated at a constant rate in a short cylinder. Heat is lost from the cylindrical surface at r = by convection to the surrounding medium at temperature with a heat transfer coefficient of h . The
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## This note was uploaded on 01/14/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.

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