Heat Transfer Solver 3.00 ... Problem-Solving Application #7 Heat Transfer Solver: a database of solutions. Heat Transfer Solver shows a group of Heat Transfer problems with solutions. It Solves Heat Transfer BVP, IVP, IP problems. Source code + Output files are included. The source codes are short; less then 100 lines. They were written to demonstrate Calculus (level) Programming. Calculus Programming makes solving partial differential equations, ordinary differential equations and algebraic equations simple. You state the equations, initial or boundary conditions, and any parameter values and leave the solving method to the compiler. Equations may be nonLinear, implicit, any order, any degree and/or constrained. An online Calculus Compiler is scheduled for release around mid-2009. So learn Calculus-level Programming for free using this front-end editor. We use a DOS version of Fortran Calculus to compile and execute those problems e-mailed to us and e-mail the output back, ALL for free. Do you have a Heat Transfer problem to solve? If so, download this program and use it to e-mail us your problem. In return, we ask that you share this tool and your results with others. For Example, Consider a tapered fin of trapezoidal profile: The governing equation for the temperature distribution in the fin is the non-dimensionalized second order differential equation where y = non-dimensional temperature = (T - T∞ ) / (Tw + T∞ ) x = non-dimensional distance = (Tw - T∞ ) where q is the uniform rate of internal heat generation per unit volume (via nuclear fission, electrical dissipation, chemical reaction, etc.) and k is the thermal conductivity. . . . where is the average convective heat transfer coefficient Based upon the known physical conditions for a fin, it is usually necessary to specify the value of the boundary conditions at two points, unless both the temperature and the temperature gradient at x=0 are known. For example, in a fin with an adiabatic end, one can specify the value of Y at x=0 and the value of dy/dx at x=1. It is then necessary to find a value of dy/dx at x=0 which allows the resulting temperature profile to satisfy the end condition at x=1.

This is accomplished in Fortran Calculus using the statement:

  FIND dydx0; in bound; by Ajax(acon); to match dydx1

which invokes the AJAX solver to operate on the model Bound. Bound contains an integration process employing the ISIS solver. The process is initiated via the statements

  x=0 : y=1 : dydx = dydx0 : dx=.05  ! initial conditions
  INITIATE Isis; for diff; equations d2ydx2/dydx, dydx/y;
     &      of x; step dx; to xf;

where the model DIFF contains the differential equation:

Model Diff
  d2ydx2=((2*p4*y/cos(a)+(p3-p2)*dydx)/(p2+(p3-p2)*(1-x)))-p1
End

The process is integrated to the boundary and the boundary condition dydx1 is computed, via the statements:

  do while (xf .le. 1)
    INTEGRATE diff; by Isis
    print '(3e14.5)', x, y, dydx
    xf = xf + xp
  end do
  dydx1 = dydx   !  boundary condition

This program solves the two point boundary value problem of the 2nd order differential equation that describes the one dimensional, steady state temperature distribution in a tapered Fin, where y is a non-dimensional temperature and x is a non-dimensional distance.

The Calculus-level Program for this problem contains 33 lines of text (Note: plots were removed for this demo), as follows: ...

Problem Fin   ! Heat Transfer in a tapered fin
  common dydx0,d2ydx2,dydx,y,x,dydx1
  common/params/p1,p2,p3,p4,a
  ! Fortran problem ... sub'ing P's 4 N's
  p1=.1 : p2=.03 : p3=.08 : p4=.01
  a=atan(.5*(p3-p2))
  dydx0=-.5 !  initial guess
  FIND dydx0; in bound; by Ajax(acon); to match dydx1
end
controller acon(ajax)
  damp=0
end
model bound
  common dydx0,d2ydx2,dydx,y,x,dydx1
  data it/0/
  xp=.1 : xf=xp : it=it+1
  x=0 : y=1 : dydx=dydx0 : dx=.05  ! initial conditions
  INITIATE Isis; for diff; equations d2ydx2/dydx,dydx/y;
&     of x; step dx; to xf;
  print '(a,i2/a)',' integration for iteration ',it,
&    '     distance           y         dydx   '
  do while (xf.le.1)
    INTEGRATE diff; by isis
    print '(3e14.5)',x,y,dydx
    xf=xf+xp
  end do
  dydx1=dydx   !  boundary condition
  terminate diff
end
model diff

                 common dydx0,d2ydx2,dydx,y,x,dydx1
  common/params/p1,p2,p3,p4,a
  d2ydx2=((2*p4*y/cos(a)+(p3-p2)*dydx)/(p2+(p3-p2)*(1-x)))-p1
End

A full discussion of this problem is given in Computer Aided Heat Transfer Analysis by J. A. Adams and D. F. Rogers, McGraw Hill, 1973, pp. 55-74. A BASIC program is given which contains 165 lines of text.

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