There appear two versions of _Z1 and two versions of _Z2, and that is weird. Certainly a bug.

restart;
eq:=Diff(W(t), t) = -q*V*(sin(Phi)-sin(Psi[s]))/(2*h*Pi)
soln := solve([rhs(eq) = 0, Phi < 2*Pi], [Phi], allsolutions = true,explicit);
idts:=indets(soln,`local`);
nops(idts); #4
addressof~(idts);
nops(%); #4
getassumptions(idts[1]);#
getassumptions(idts[2]);#These are not the same assumptions
Lidts:=convert(idts,list);
S:=Lidts=~[z11,z12,2,3];
subs(S,soln);

Note: Another strange thing: If you replace Psi[s] with an unindexed name like psi, then you will notice that idts has five members: assumptions are made on psi. If you try replacing Psi[s] with Psi (which is an initially known function in Maple) you get an error message from solve saying
Error, (in assume) cannot assume on an assigned name

Please see a related discussion:
http://mapleprimes.com/questions/208328-Differential-Equation-And--Notation-Question

Get rid of the units (after having made them conform to each other) or introduce dimensionless variables. Then compute.

@tomleslie I'm not a big fan of units either. 
The expression exp(-4000*t) can be nothing but a pure number with no unit no matter what t is or what unit is used.
Just think of the series expansion for exp.
exp(x) = convert(exp(x),FormalPowerSeries,x);

If the unit of x is U what is the unit of exp(x)?

My thinking is: get your units straight from the start, perhaps with the help of Maple, then forget about them, and compute!

@tomleslie What would be the unit for t?

@shahid Sorry, I don't.

@shahid Since you have a Maple version earlier than Maple 18 eval[recurse] doesn't work.

This should work (I did this in Maple 15):

restart;
eq1:=u(x,y)*(diff(u(x,y), x))+v(x,y)*(diff(u(x,y), y)) = (diff(u(x,y), y, y))/(1+epsilon*theta(x,y))-epsilon*(diff(u(x,y), y))*(diff(theta(x,y), y))/(1+epsilon*theta(x,y))^2+theta(x,y);

eq2:=u(x,y)*(diff(theta(x,y), x))+v(x,y)*(diff(theta(x,y), y)) = (diff(theta(x,y), y, y))/Pr;

## Replace the original S with the following:
S1:={u(x,y) = diff(psi(x,y), y), v(x,y) = -(diff(psi(x,y), x))};
S2:={ psi(x,y) = x^(4/5)*f(x,eta)/(1+x)^(1/20), theta(x,y)=x^(1/5)*(1+x)^(-1/5)*h(x,y)};
S3:= eta = y/(x^(1/5)*(1+x)^(1/20));
S23:=eval(S2,S3);
eval([eq1,eq2],S1);
eval(%,S23);
subs(rhs(S3)=lhs(S3),%);
collect((lhs-rhs)~(%)=~0,[D,diff],factor);
res:=subs(y/(x^(1/5)*(1+x)^(1/20))=eta,%);

##To make it more readable you can continue with:
PDEtools:-declare(f(x,eta),h(x,y));
convert(res,diff);

About rewriting S1*S2-S3*S4.

You can rewrite the products of the inert sums S1*S2 and S3*S4 using the Cauchy product. Thus you get

S1*S2-S3*S4 = S12 - S34, where S12 is the Cauchy product of S1 and S2, and S34 is the Cauchy product of S3 and S4.

The difference S12-S34 can be combined to just one sum S:

S := 1/(2*Pi)*Sum(x^(3*k1+2)*3^(1/2+k1)*(Sum((3*GAMMA(k+2/3)*(-2/3-k1+k)*GAMMA(k1+1/3-k)+3*GAMMA(4/3+k)*GAMMA(2/3+k1-k))/(GAMMA(3*k+2)*GAMMA(3*k1-3*k+3)), k = 0 .. k1)), k1 = 0 .. infinity);

This can be written in short form as
1/(2*Pi)*Sum(x^(3*k1+2)*3^(1/2+k1)*m(k1), k1 = 0 .. infinity);

where m(k1) < 0 for all k1 >=0. It is given by:

m(k1) = Sum((3*GAMMA(k+2/3)*(-2/3-k1+k)*GAMMA(k1+1/3-k)+3*GAMMA(4/3+k)*GAMMA(2/3+k1-k))/(GAMMA(3*k+2)*GAMMA(3*k1-3*k+3)), k = 0 .. k1);

m(k1) is very rapidly decreasing in absolute value.

Now you don't have cancellation problems for x > 0 since all terms in S are negative.
But I wonder why x-values as high as x=100 are interesting to you. S will be huge (but negative) for x = 100.
For x=10 you get
evalf[50](eval(S,x=10));
    -4.7714630011989009659113681348806733279336141633480*10^8
and for x = 11:
evalf[50](eval(S,x=11));
    -1.1891747657696851151848674770800968156039786693327*10^10
### Notice that you will run into similar problems with the well known power series for exp(x):
restart;
S:=Sum(x^n/n!,n=0..infinity);
seq(evalf[500](eval(S,x=i*10)),i=0..10);

@shahid Here is what I got (last output shown only) when I just now copied my own code from above and executed the whole thing in Maple 2015.2. Actually, in my worksheet it is shown with the abbreviated notation used because of my use of PDEtools:-declare(f(x,eta),h(x,y)). But when I copy and paste it into MaplePrimes it shows up as you see below.

You could begin by telling us what Ni1 is. Maybe you also can tell us if a1 and a2 are assigned to anything.

@Bendesarts You wrote:

"Don't hesitate to let me know if you have new ideas for troubleshooting my issue of "drift" for the solution of this NL oscillator but in MapleSim."
(my emphasis)

I don't have MapleSim.

@shahid You forgot theta(x,y)=x^(1/5)*(1+x)^(-1/5)*h(x,y) it seems. I added it to S below.

restart;
eq1:=u(x,y)*(diff(u(x,y), x))+v(x,y)*(diff(u(x,y), y)) = (diff(u(x,y), y, y))/(1+epsilon*theta(x,y))-epsilon*(diff(u(x,y), y))*(diff(theta(x,y), y))/(1+epsilon*theta(x,y))^2+theta(x,y);

eq2:=u(x,y)*(diff(theta(x,y), x))+v(x,y)*(diff(theta(x,y), y)) = (diff(theta(x,y), y, y))/Pr;

S:={ psi(x,y) = x^(4/5)*f(x,eta)/(1+x)^(1/20), eta = y/(x^(1/5)*(1+x)^(1/20)), u(x,y) = diff(psi(x,y), y), v(x,y) = -(diff(psi(x,y), x)),theta(x,y)=x^(1/5)*(1+x)^(-1/5)*h(x,y)};

eval[recurse]([eq1,eq2],S);
collect((lhs-rhs)~(%)=~0,[D,diff],factor);
res:=subs(y/(x^(1/5)*(1+x)^(1/20))=eta,%);

##To make it more readable you can continue with:
PDEtools:-declare(f(x,eta),h(x,y));
convert(res,diff);

@Bendesarts Did you try to run your code in Maple itself, not only in MapleSim?

Clearly, we don't know (or rather I don't) whether there is actually a limit cycle. But at least it appears (or appeared) so.
You have an autonomous system of 8 odes of first order. If a given orbit stays bounded, the limit set may be quite complicated, just think of the Lorenz system of only 3 odes.
Even for a system of just 2 odes (where the possibilities for the limit set are well known) it may be difficult to determine numerically whether you have a limit cycle or just a very, very slow oscillating approach to an equilibrium point.

You may want to change the name of that attached file.

@shahid You should present your equations or expressions using Maple code as you did in your original question.

@Athar Shahabinejad You ought to tell us what it is you are running now, when you say

"Now when i run i get this answer ... "