Mr Robertson considers as variables parameters that Mr Wang does not want to include as variables. I asked Mr Wang to define ranges because if the symbolic analysis does not help, then having a look at the equations may be useful, let me explain:
Since Mr Wang does not say anything else about the parameters, let me list them
eq1: Iclo assumed constant
eq2: facl and Tc assumed constant
eq3: Tr, Ta and hc assumed constant
eq4: Tsk(i) assumed constant
eq5: as in eq3, hc assumed constant
So, we have 5 variables to solve: Fcl hr To Tcl h
that I rename to: x y z w v
The simplified version of the 5 initial equations is:
eq1: x=1/(1+v)
eq2: y=((1+z)*.5+2)^3
eq3: z=(y+1)/v
eq4: w=z+x*(1-z)
eq5: v=1+y
I have define 5 ranges, that some of them may or may not help solve the system, but at least it's a start point:
x_min=-10;x_max=10;x_step=.1;x_range=[x_min:x_step:x_max]
y_min=-10;y_max=10; y_step=.1; y_range=[y_min:y_step:y_max]
z_min=-10;z_max=10; z_step=.1; z_range=[z_min:z_step:z_max]
w_min=-10;w_max=10; w_step=.1; w_range=[w_min:w_step:w_max]
v_min=-10;v_max=10; v_step=.1; v_range=[v_min:v_step:v_max]
now
v=v_range
x=1./(1+v)
figure(1);plot(v,x);grid on;hold all
z=z_range
y=((1+z)*.5+2).^3
figure(2);plot(z,y);grid on;w=z+x*(1-z)
simplifying some equations is possible:
eq3 and eq5 yield z=(y+1)/(1+y) nearly constant
eq4 with z nearly constant means w shape is w=1+x
I said 'nearly constant' because the simplified equations do not include the the parameters
Including parameters: Iclo facl Tc Tr Ta Tsk(i) hc
renamed: k1 k2 k3 k4 k5 k6 k7
eq1: x=1/(1+k1*v)
eq2: y=k2*((k3+z)*.5+2)^3
eq3: z=(k4*y+k5*k7)/v
eq4: w=z+x*(k6-z)
eq5: v=k7+y
give values to parameters, and find out if the 5 equations have a solution, or if perhaps the only way to solve this equations system is for values of parameters and variables within certain range windows.
Let me know if you progress with anything above explained
John
