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This might be a simple question for the most of you, but I am very new to mathematica so please bear with me. I'll refer to the code below:

H = 10
T0 = 3600
Cd = 0.0573*Sqrt[1 + 0.148*U0]
Iv = 0.06*(1 + 0.043*U0)*(z/10)^-0.22
Solve[Utz == U0*(1 + Cd*Log[z/H])*(1 - 0.41*Iv*Log[T/T0]) && U0 > 0, U0]

I would like to Solve the above formula for U0, so I get U0=... with Utz, z and T as the variables. I only need solutions >0. However, the notebook keeps 'running' when I use this. I have also used input values for Utz, z and T, this works fine and the correct value for U0 is provided.

Am I doing this correctly and does the Solve simply take a lot of time, or is there another explanation?

Thanks in advance,

Hans

*Edit1:

Thanks Sjoerd for your comment. The problem lies in the fact that Cd and Iv are both defined by variables U0 and z. That is why I want to incorporate those in the Solve for U0, something like?:

Solve[Utz == U0*(1 + (0.0573*Sqrt[1 + 0.148*U0])*Log[z/H])*(1 - 0.41*(0.06*(1 + 0.043*U0)*(z/10)^-0.22)*Log[T/T0]) && U0 > 0 && T > 0 && z/H > 0, U0]

However, again this results in 'running' process.

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In this case, it helps if you leave your parameters undefined (Solve doesn't like the inexact quantities you use) and add some assumptions.

So, let's first clear your variables

H =.; T0 =.; Cd =.; Iv =.

Then

Solve[Utz == U0*(1 + Cd*Log[z/H])*(1 - 41/100*Iv*Log[T/T0]) && 
      U0 > 0 && T > 0 && z/H > 0, U0, Reals]

Note that I also changed the 0.41 to 41/100

Almost instantaneously, this gives you a long conditional expression which I show shortened below:

{{U0 -> ConditionalExpression[-((100*Utz)/(-100 + 41*Iv*Log[T/T0] - 100*Cd*Log[z/H] + 41*Cd*Iv*Log[T/T0]*Log[z/H])), << 1 >> ]}}

You can now fill in the values for the parameters.

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