"Roger Stafford" wrote in message <firstname.lastname@example.org>... > "prafull chauhan" <email@example.com> wrote in message <firstname.lastname@example.org>... > > [U(c)/[[(G(s)*g*d)]^0.5]]=c*h*[(d/R)^a]*(e^b) > > where U(c)= 0.23 > > G(s)=2.61 > > d=0.27 > > g=9.81 > > e=0.0066 > > R=0.185 > > h=1.36 > - - - - - - - - - - - > If you regard c, a, and b as all being unknowns, then you cannot solve for them using only one equation. In general you would need three equations to uniquely determine the three unknowns. The fact that U(c) is equal to 0.23 could be regarded as a second equation if U is a known function, but that still leaves you with two unknowns and only one equation. If either of these is regarded as a parameter, the other can be easily solved for in terms of the other as follows: > > Solve for a in terms of b and c as: > > (d/R)^a = U(c)/(G(s)*g*d)^0.5/c/h/e^b > a = log(U(c)/(G(s)*g*d)^0.5/c/h/e^b)/log(d/R) > > or solve for b in terms of a and c as: > > e^b = U(c)/(G(s)*g*d)^0.5/c/h/(d/R)^a > b = log(U(c)/(G(s)*g*d)^0.5/c/h/(d/R)^a)/log(e) > > I am afraid that is the best you can do in terms of solutions. Three unknowns and only one equation will in general constitute a two-dimensional surface in a three-dimensional space, or three unknowns with two equations would be a one-dimensional curve in three-dimensional space. Only when you furnish a third equation does that narrow down to a single point or perhaps a finite set of discrete points. > > Roger Stafford
Dear Roger, You are right. I am dealing with the equation is V_c/?((G_s-1)g*d_50 )=c*?_g *((d_50/R)^a)*((?)^b)
Thanks for your suggestion. I am working on it on Matlab. I wish to solve the problem with stepwise approximation. I tried the same equation with the sftool in matlab with custom equation. How can it be solved with c as function of a and b, where i may vary the values of a and b and c and get the right hand as unity. It is true that I need two more equations to solve it to get finite solutions. Well thanks your suggestion is really helpful.