Floating point exception error

Dear all,

I am very confused with this error in the following code. I've searched through the forums but I could'n find an answer, even a documentation about that error. As in the title, it says:

model project I.F95(49) : error 637 - Internal compiler error - floating point exception

The code is as followed, and [color=red:8330a2d139]error is on the first DO statement[/color:8330a2d139]:

!*************************************
!*******ChE 352 Term Project 1********
!*****Dynamics of Shell and Tube******
!***********Heat Exchanger************
!****with 3-pass Shell 1-pass Tube****
!*************************************

PROGRAM TERM_PROJECT1
IMPLICIT NONE
DOUBLE PRECISION :: PI,NT,U,MT,DTUBE,A,DENS_T,CPT,CONST1,CONST2,DSHELL,DENS_S,CPS,CONSS1,CONSS2
DOUBLE PRECISION, DIMENSION(10,50000) :: T1,T2,T3,TS,ERR1,ERR2,ERR3,ERRS
DOUBLE PRECISION :: TIME,EPS,DT,DX,MS,A_TUBE,A_SHELL,A_S

INTEGER, DIMENSION(0:10) :: I
INTEGER, DIMENSION(0:50000) :: J

PRINT*, 'Enter DeltaT?'
READ*, DT
! OPEN (20,file='Temp Distribution.xls') ! Open xls file to obtain temperature distribution

PI=4.0*ATAN(1.0)
NT=300
U=840

EPS=0.005
DX=0.4

! Define tube side properties and constants of equation
MT=31.4
DTUBE=0.02
A=PI*(NT/3.0)*(DTUBE**2)/4.0
DENS_T=800
CPT=2000
CONST1=MT/(A_TUBE*DENS_T)
CONST2=NT*U*PI*DTUBE/(3.0*A*DENS_T*CPT)

! Define shel side properties and constants of equation
MS=31.4
DSHELL=0.6
A_S=PI*(DSHELL**2)/4.0-(PI*NT*DTUBE**2)/4.0
DENS_S=1500
CPS=2200
CONSS1=MS/(A_S*DENS_S)
CONSS2=NT*U*PI*DSHELL/(3.0*A_S*DENS_S*CPS)
   
   
! Set initial temperatures of tubes and shell side along the system

DO I=0,10,1
 T1(I,0)=295.0
 T2(I,0)=295.0
 T3(I,0)=295.0
 TS(I,0)=295.0
END DO

DO J=1,50000,1
 TS(10,J)=420.0
END DO

DO J=0,50000,1

! Calculate the temperature of pass 1
DO I=0,8,1
  T1(I,J+1)=(-1.0)*CONST1*(DT/(2*DX))*((-1.0)*3.0*(T1(I,J))+4.0*(T1(I+1,J))-T1(I+2,J))+&
  (DT*CONST2)*(TS(I,J)-T1(I,J))+T1(I,J)
END DO

DO I=9,10,1
  T1(I,J+1)=(-1.0)*CONST1*(DT/(2*DX))*(3.0*(T1(I,J))-4.0*(T1(I-1,J))+T1(I-2,J))+&
  (DT*CONST2)*(TS(I,J)-T1(I,J))+T1(I,J)
END DO

! Application of boundary condition 3
T2(10,J)=T1(10,J)

! Calculate the temperature of pass 2
DO I=10,2,-1
T2(I,J+1)=CONST1*(DT/(2*DX))*(3.0*(T2(I,J))-4.0*(T2(I-1,J))+T2(I-2,J))+&
  (DT*CONST2)*(TS(I,J)-T2(I,J))+T2(I,J)
END DO

DO I=2,0,-1
  T1(I,J+1)=CONST1*(DT/(2*DX))*((-1.0)*3.0*(T1(I,J))+4.0*(T1(I+1,J))-T1(I+2,J))+&
  (DT*CONST2)*(TS(I,J)-T2(I,J))+T2(I,J)
END DO

! Application of boundary condition 2
T3(0,J)=T2(0,J)

! Calculate the temperature of pass 3
DO I=0,8,1
  T3(I,J+1)=(-1.0)*CONST1*(DT/(2*DX))*((-1.0)*3.0*(T3(I,J))+4.0*(T3(I+1,J))-T3(I+2,J))+&
  (DT*CONST2)*(TS(I,J)-T3(I,J))+T3(I,J)
END DO

DO I=9,10,1
  T3(I,J+1)=(-1.0)*CONST1*(DT/(2*DX))*(3.0*(T3(I,J))-4.0*(T3(I-1,J))+T3(I-2,J))+&
  (DT*CONST2)*(TS(I,J)-T3(I,J))+T3(I,J)
END DO

cont'd

! Calculate the temperature of shell side
DO I=10,0,-1
TS(I,J+1)=CONSS1*(DT/(2*DX))*(3.0*(TS(I,J))-4.0*(TS(I-1,J))+TS(I-2,J))+&
  (DT*CONSS2)*(3*(TS(I,J))-T1(I,J)-T2(I,J)-T3(I,J))+TS(I,J)
END DO


! Define errors here and let the loop to stop if satisfied

DO I=0,10,1
  	ERR1(I,J)=ABS(T1(I,J+1)-T1(I,J))
	ERR2(I,J)=ABS(T2(I,J+1)-T2(I,J))
	ERR3(I,J)=ABS(T3(I,J+1)-T3(I,J))
	ERRS(I,J)=ABS(TS(I,J+1)-TS(I,J))
		
		IF(ERR1(I,J)>=EPS)THEN
        GO TO 10
        ELSEIF(ERR2(I,J)>=EPS)THEN
        GO TO 10
        ELSEIF(ERR3(I,J)>=EPS)THEN
        GO TO 10
        ELSEIF(ERRS(I,J)>=EPS)THEN
        GO TO 10
        ELSEIF GO TO 20
        ENDIF
END DO

10	END DO  !End the grand DO (deltime)

20	TIME=J*DT
PRINT*, 'Time for steady state=',TIME


END PROGRAM TERM_PROJECT1

You have declared I and J to be arrays. They must be integer scalars if you want to use them as DO loop indices. The compiler error is not helpful, but if you fix this you will find other errors you can fix.

The notation:

INTEGER, DIMENSION(0:10) :: I INTEGER, DIMENSION(0:50000) :: J

does NOT mean I goes from 0:10 and J goes from 0:50000

Change the lines to:

INTEGER :: I, J

There are numerous other errors too. Your array T1 has indices (1:10, 1:50000) but your first loop accesses elements (0,0).

Thanks davidb, I changed as you proposed but I just tried to define an array for i=0 to 10 and j=0 to 50000 but my definition might be wrong so how i can fix that?

Edit: Thanks to david, i just corrected the definition of dimensions to

DOUBLE PRECISION, DIMENSION(0:10,0:50000) :: T1,T2,T3,TS,ERR1,ERR2,ERR3,ERRS

and

INTEGER :: I
INTEGER :: J

but i still have some errors to be solved

Problems are solved. I will be posting the last final code at late midnight with no errors and physically meaningful one (for those who are interested in). Thanks all for your support, especially David 😉

Edit: The code is now working with zero errors.

  	PROGRAM SHELLTUBE
	DOUBLE PRECISION :: PI,NT,U,MT,DTUBE,DENS_T,CPT,CONST1,CONST2,DSHELL,DENS_S,CPS,CONSS1,CONSS2
    DOUBLE PRECISION :: DX, ERRO, EPS, MS, A_TUBE,A_S,DISTANCE
    DOUBLE PRECISION :: T1(11,0:100000),T2(11,0:100000),T3(11,0:100000),TS(11,0:100000)
  	PRINT*, 'INPUT TIME STEP'
	READ*, DT
  	PI=4.0*ATAN(1.0)
	NT=300.0
	U=840.0

	EPS=0.000001
	DX=0.4

! Define tube side properties and constants of equation
MT=31.4
DTUBE=0.02
A_TUBE=PI*(NT/3.0)*(DTUBE**2)/4.0
DENS_T=800.0
CPT=2000.0
CONST1=MT/(A_TUBE*DENS_T)
CONST2=NT*U*PI*DTUBE/(3.0*A_TUBE*DENS_T*CPT)

! Define shel side properties and constants of equation
MS=31.4
DSHELL=0.6
A_S=PI*(DSHELL**2)/4.0-(PI*NT*DTUBE**2)/4.0
DENS_S=1500.0
CPS=2200.0
CONSS1=MS/(A_S*DENS_S)
CONSS2=NT*U*PI*DSHELL/(3.0*A_S*DENS_S*CPS)

  	OPEN (8,FILE='STEADY STATE TEMPERATURE DISTRIBUTION.TXT',STATUS='REPLACE')
  	OPEN (9,FILE='TIME EVOLUTION OF TEMPERATURES.TXT',STATUS='REPLACE')
  
  	DO I=1,11
     T1(I,0)=295
     T2(I,0)=295
     T3(I,0)=295
     TS(I,0)=295
  	END DO

	TIMEZERO=0
    WRITE(9,50) 'TIME', 'T1', 'T2', 'T3', 'TSHELL'
  	WRITE(9,40) TIMEZERO, T1(11,0), T2(1,0), T3(11,0), TS(1,0) 

    DO J=0,100000
  	T1(1,J)=295.0
	END DO

	DO J=1,100000
  	TS(11,J)=420.0
	END DO
    
  	DO J=0,100000 !Grand DO LOOP for time evaluation
    
! First pass equations
	DO I=2,11
    T1(I,J+1)=(-CONST1*(DT/DX)*(T1(I,J)-T1(I-1,J)))+(CONST2*DT*(TS(I,J)-T1(I,J)))+T1(I,J)
   	END DO
    ! Apply boundary condition
    T2(11,J+1)=T1(11,J+1)

    
! Second pass equations  
	DO I=10,1,-1
	T2(I,J+1)=(CONST1*(DT/DX)*(T2(I+1,J)-T2(I,J)))+(CONST2*DT*(TS(I,J)-T2(I,J)))+T2(I,J)
	END DO
    ! Apply boundary condition
   	T3(1,J+1)=T2(1,J+1)
    	
! Third pass equations	  
	DO I=2,11
    T3(I,J+1)=(-CONST1*(DT/DX)*(T3(I,J)-T3(I-1,J)))+(CONST2*DT*(TS(I,J)-T3(I,J)))+T3(I,J)
	END DO
	
		
! Shell pass equations
	DO I=10,1,-1
	TS(I,J+1)=(CONSS1*(DT/DX)*(TS(I+1,J)-TS(I,J)))-(CONSS2*DT*((3*TS(I,J))-T1(I,J)-T2(I,J)-T3(I,J)))+TS(I,J)
	END DO
    
	TIMEJ=DT*(J+1)
	WRITE(9,40) TIMEJ, T1(11,J+1), T2(1,J+1), T3(11,J+1), TS(1,J+1)  
    K=J  
    DO I=11,1,-1
		ERRO=ABS(TS(I,J+1)-TS(I,J))
        IF(ERRO.GT.EPS) GOTO 100
	END DO
	GO TO 150
100	END DO !End of Grand DO LOOP

150	WRITE(8,10) 'T1','T2', 'T3', 'TSHELL'
    DO I=1,11
      	DISTANCE=(I-1)*DX
		WRITE(8,20) DISTANCE, T1(I,K), T2(I,K), T3(I,K), TS(I,K)
    END DO
10	FORMAT(4(A8))
20	FORMAT(F4.1,4(F8.3))
	TIME=DT*K
    WRITE(8,30) 'TIME NEEDED TO REACH STEADY STATE:', TIME
30	FORMAT(A35,F8.3)
40	FORMAT(5(F8.3))
50	FORMAT(5(A8)) 
    
	END PROGRAM SHELLTUBE

A couple of comments on your program, that are intended to help you for next time.

You do not appear to need past itteration solutions. As this is a forward single step solution, you only need the last solution. You could have the dimension statement: DOUBLE PRECISION, DIMENSION(0:10) :: T1,T2,T3,TS, T1L,T2L,T3L,TSL where T1L is the solution estimate at the last step.

Your convergence test does not require storage of the error, only calculation and accumulation of a failure test, such as:

!   Define errors here and let the loop stop if all satisfied 
!    Why do you store the errors ?
!
      DO I=0,10 
         IF (ABS (T1(I)-T1L(I)) >= EPS) EXIT 
         IF (ABS (T2(I)-T2L(I)) >= EPS) EXIT 
         IF (ABS (T3(I)-T3L(I)) >= EPS) EXIT 
         IF (ABS (TS(I)-TSL(I)) >= EPS) EXIT 
      END DO 
      if (I > 10) EXIT   ! Exit loop if convergence was reached for all tests
!
!   Store this itteration as last estimate
      DO I=0,10 
         T1L(I) = T1(I)
         T2L(I) = T2(I)
         T3L(I) = T3(I)
         TSL(I) = TS(I)
      END DO 
!
   END DO  ! End the grand DO (deltime) 

Your code makes extensive use of unnecessary ( ). I'd suggest that you could remove these and change the layout of the equations to make them easier to check. Also, from your coding, as T1 is only a fn of T1N (the previous estimate), the order of the DO loop is not important. Below is an example of how I changed your equation to improve clarity.

! Set initial temperatures of tubes and shell side along the system 
!
  DO I = 0,10 
    T1L(I) = 295.0 
    T2L(I) = 295.0 
    T3L(I) = 295.0 
    TSL(I) = 295.0 
  END DO 
!
  TSL(10) = 420.0
  TS(10)  = 420.0 
!
  DO J = 0,50000
!
!   Calculate the temperature of pass 1 
      DO I = 0,8 
!       Original
!        T1(I,J+1)=(-1.0)*CONST1*(DT/(2*DX))*((-1.0)*3.0*(T1(I,J))+4.0*(T1(I+1,J))-T1(I+2,J))+& 
!        (DT*CONST2)*(TS(I,J)-T1(I,J))+T1(I,J) 
!
!       becomes by removing ',J'
!        T1(I) = (-1.0)*CONST1*(DT/(2*DX))*((-1.0)*3.0*(T1L(I))+4.0*(T1L(I+1))-T1L(I+2))  & 
!              + (DT*CONST2)*(TSL(I)-T1L(I))+T1L(I) 
!
!       becomes by removing unnesecary '()'
!        T1(I) = -1.0 * CONST1 * DT/(2*DX) * ( -3.0 * T1L(I) + 4.0 * T1L(I+1) - T1L(I+2) )  & 
!              + DT*CONST2 * (TSL(I)-T1L(I)) + T1L(I) 
!
!       becomes by rearranging
         T1(I) = T1L(I)                                                                  &
               + CONST1 * DT / (2*DX) * ( 3.0 * T1L(I) - 4.0 * T1L(I+1) + T1L(I+2) )     & 
               + CONST2 * DT * (TSL(I)-T1L(I))  
      END DO 

I hope this helps.

John

Thanks John for that great support. I will be using your suggestion for newly assigned term project 😃 I will share the new code with you