Thermodynamics HW Solutions 474

Thermodynamics HW Solutions 474 - the Row"...

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Chapter 5 Numerical Methods in Heat Conduction 5-85 "!PROBLEM 5-85" "GIVEN" L=0.08 "[m]" k=28 "[W/m-C]" alpha=12.5E-6 "[m^2/s]" T_i=100 "[C]" g_dot=1E6 "[W/m^3]" T_infinity=20 "[C]" h=35 "[W/m^2-C]" DELTAx=0.02 "[m]" "time=300 [s], parameter to be varied" "ANALYSIS" M=L/DELTAx+1 "Number of nodes" DELTAt=15 "[s]" tau=(alpha*DELTAt)/DELTAx^2 "The technique is to store the temperatures in the parametric table and recover them (as old temperatures) using the variable ROW. The first row contains the initial values so Solve Table must begin at row 2. Use the DUPLICATE statement to reduce the number of equations that need to be typed. Column 1 contains the time, column 2 the value of T[1], column 3, the value of T[2], etc., and column 7
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Unformatted text preview: the Row." Time=TableValue(Row-1,#Time)+DELTAt Duplicate i=1,5 T_old[i]=TableValue(Row-1,#T[i]) end "Using the explicit finite difference approach, the six equations for the six unknown temperatures are determined to be" T[1]=tau*(T_old[2]+T_old[2])+(1-2*tau)*T_old[1]+tau*(g_dot*DELTAx^2)/k "Node 1, insulated" T[2]=tau*(T_old[1]+T_old[3])+(1-2*tau)*T_old[2]+tau*(g_dot*DELTAx^2)/k "Node 2" T[3]=tau*(T_old[2]+T_old[4])+(1-2*tau)*T_old[3]+tau*(g_dot*DELTAx^2)/k "Node 3" T[4]=tau*(T_old[3]+T_old[5])+(1-2*tau)*T_old[4]+tau*(g_dot*DELTAx^2)/k "Node 4" T[5]=(1-2*tau-2*tau*(h*DELTAx)/k)*T_old[5]+2*tau*T_old[4]+2*tau*(h*DELTAx)/k*T_infinity+tau*(g_dot*DE LTAx^2)/k "Node 4, convection" 5-77...
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This note was uploaded on 01/22/2012 for the course PHY 4803 taught by Professor Dr.danielarenas during the Fall '10 term at UNF.

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