Secant method solving for pipe diameter - fortran

Question

I am trying to write a program to solve for pipe diameter for a pump system I've designed. I've done this on paper and understand the mechanics of the equations. I would appreciate any guidance.
EDIT: I have updated the code with some suggestions from users, still seeing quick divergence. The guesses in there are way too high. If I figure this out I will update it to working.
MODULE Sec
CONTAINS
SUBROUTINE Secant(fx,xold,xnew,xolder)
IMPLICIT NONE
INTEGER,PARAMETER::DP=selected_real_kind(15)
REAL(DP), PARAMETER:: gamma=62.4
REAL(DP)::z,phead,hf,L,Q,mu,rho,rough,eff,pump,nu,ppow,fric,pres,xnew,xold,xolder,D
INTEGER::I,maxit
INTERFACE
omitted
END INTERFACE
Q=0.0353196
Pres=-3600.0
z=-10.0
L=50.0
mu=0.0000273
rho=1.940
nu=0.5
rough=0.000005
ppow=412.50
xold=1.0
xolder=0.90
D=11.0
phead = (pres/gamma)
pump = (nu*ppow)/(gamma*Q)
hf = phead + z + pump
maxit=10
I = 1
DO
xnew=xold-((fx(xold,L,Q,hf,rho,mu,rough)*(xold-xolder))/ &
(fx(xold,L,Q,hf,rho,mu,rough)-fx(xolder,L,Q,hf,rho,mu,rough)))
xolder = xold
xold = xnew
I=I+1
WRITE(*,*) "Diameter = ", xnew
IF (ABS(fx(xnew,L,Q,hf,rho,mu,rough)) <= 1.0d-10) THEN
EXIT
END IF
IF (I >= maxit) THEN
EXIT
END IF
END DO
RETURN
END SUBROUTINE Secant
END MODULE Sec
PROGRAM Pipes
USE Sec
IMPLICIT NONE
INTEGER,PARAMETER::DP=selected_real_kind(15)
REAL(DP)::xold,xolder,xnew
INTERFACE
omitted
END INTERFACE
CALL Secant(f,xold,xnew,xolder)
END PROGRAM Pipes
FUNCTION f(D,L,Q,hf,rho,mu,rough)
IMPLICIT NONE
INTEGER,PARAMETER::DP=selected_real_kind(15)
REAL(DP), PARAMETER::pi=3.14159265d0, g=9.81d0
REAL(DP), INTENT(IN)::L,Q,rough,rho,mu,hf,D
REAL(DP)::f, fric, reynold, coef
fric=(hf/((L/D)*(((4.0*Q)/(pi*D**2))/2*g)))
reynold=((rho*(4.0*Q/pi*D**2)*D)/mu)
coef=(rough/(3.7d0*D))
f=(1/SQRT(fric))+2.0d0*log10(coef+(2.51d0/(reynold*SQRT(fric))))
END FUNCTION

Answer 1

You very clearly declare the function in the interface (and the implementation) as
FUNCTION f(L,D,Q,hf,rho,mu,rough)
IMPLICIT NONE
INTEGER,PARAMETER::DP=selected_real_kind(15)
REAL(DP), PARAMETER::pi=3.14159265, g=9.81
REAL(DP), INTENT(IN)::L,Q,rough,rho,mu,hf,D
REAL(DP)::fx
END FUNCTION
So you need to pass 7 arguments to it. And none of them are optional.
But when you call it, you call it as
xnew=xold-fx(xold)*((xolder-xold)/(fx(xolder)-fx(xold))
supplying a single argument to it. When you try to compile it with gfortran for example, the compiler will complain for not getting any argument for D (the second dummy argument), because it stops with the first error.

Answer 2

It seems that the initial values for xold and xolder are too far from the solution. If we change them as
xold = 3.0d-5
xolder = 9.0d-5
and changing the threshold for convergence more tightly as
IF (ABS(fx(xnew,L,Q,hf,rho,mu,rough)) <= 1.0d-10) THEN
then we get
...
Diameter = 7.8306011049894322E-005
Diameter = 7.4533171406818087E-005
Diameter = 7.2580746283970710E-005
Diameter = 7.2653611474296094E-005
Diameter = 7.2652684750264582E-005
Diameter = 7.2652684291155581E-005
Here, we note that the function f(x) is defined as
FUNCTION f(D,L,Q,hf,rho,mu,rough)
...
f = (1/(hf/((L/D)*((4*Q)/pi*D)))) !! (1)
+ 2.0 * log( (rough/(3.7*D)) + (2.51/(((rho*((4*Q)/pi*D))/mu) !! (2)
* (hf/((L/D)*((4*Q)/pi*D))))) !! (3)
)
END FUNCTION
where terms in Lines (1) and (3) are both constant, while terms in Line (2) are some constants over D. So, we see that f(D) = c1 - 2.0 * log( D / c2 ), so we can obtain the solution analytically as D = c2 * exp(c1/2.0) = 7.26526809959e-5, which agrees well with the numerical solution above. To get a rough idea of where the solution is, it is useful to plot f(D) as a function of D, e.g. using Gnuplot.
But I am afraid that the expression for f(D) itself (given in the Fortran code) might include some typo due to many parentheses. To avoid such issues, it is always useful to first arrange the expression for f(D) as simplest as possible before making a program.
(One TIP is to extract constant factors outside and pre-calculate them.)
Also, for debugging purposes it is sometimes useful to check the consistency of physical dimensions and physical units of various terms. Indeed, if the magnitude of the obtained solution is too large or too small, there might be some problem of conversion factors for physical units, for example.