cooling_the_aluminum_cylinder(1) -...

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View Full Document Right Arrow Icon Due April 16, 2010 at 11:50 AM Project: Cooling the Aluminum Cylinder Background: A solid aluminum cylinder treated as a lumped-mass 1 system is immersed in a bath of iced water. Let us develop the mathematical model for the problem. When the cylinder is placed in the iced water bath, the cylinder loses heat to its surroundings by convection. Rate of heat lost due to convection = ( 29 ( 29 a A h θ - . (1) where = ) ( t temperature of cylinder as a function of time, o C ) ( h = the convective cooling coefficient, W/(m 2 - o C) = A surface area, m 2 = a ambient temperature of iced water, o C The energy stored in the mass is given by Energy stored by mass = mC (2) where m = mass of the cylinder, kg C = specific heat of the cylinder, J/(kg-K) From an energy balance, Rate at which heat is gained ─ Rate at which heat is lost = Rate at which heat is stored gives ( 29 dt d mC hA a = - - (3) The ordinary differential equation is subjected to 0 ) 0 ( = where = 0 initial temperature of cylinder, o C Assuming the convective cooling coefficient, h to be a constant function of temperature, the exact solution to the differential equation (3) is 1 It implies that the internal conduction in the trunnion is large enough that the temperature throughout the trunnion is uniform. This allows us to assume that the temperature is only a function of time and not of the location in the trunnion. This means that if a differential equation governs this physical problem, it would be an ordinary differential equation for a lumped system and a partial differential equation for a non-
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This note was uploaded on 06/12/2011 for the course EML 3041 taught by Professor Kaw,a during the Spring '08 term at University of South Florida - Tampa.

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cooling_the_aluminum_cylinder(1) -...

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