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Consider a (p)rototype consisting in an incompressible and newtonian fluid flowing in a pipe of diameter $D_P$, studied by similarity in a (m)odel of the same fluid in another pipe of diameter $D_m<D_p$.

Equating the Reynolds numbers: $D_m v_m = D_p v_p$. That is

$$v_m = \frac{D_p}{D_m}v_p > v_p$$

so the fluid moves faster in the model which is smaller in order to archive 'some kind' of similarity.

Q1 Which kind of similarity?

The Reynolds number arises after adimiensionalize the viscosity using the density ($\rho$), velocity ($v$) and a lenght ($D$). Using the same variables to adimensionalize the forces $F$ and the powers $P$,

$$ F^* = \frac{F}{\rho v^2 D^2} \qquad P^* = \frac{P}{\rho v^3 D^2}$$

If the adimensional forces and powers are the same for the model and the prototype ($F^*_m = F^*_p$) and ($P^*_m = P^*_p$), it is ready found that

$$ \begin{align} F_p &= F_m\\ P_p &= \frac{D_m}{D_p} P_m \end{align}$$

which means that the forces are equal in the model and prototype, and the power of the prototype (relative to the model) is smaller than the model.

The conclusion above must to be false.

Q2 What is wrong here? I think that it is related to lack of time correspondence between the model and the prototype, but I am not sure about how to treat this.

Doing the same for energy, I found that $E_p = \frac{D_p}{D_m}E_m$, but I would expect that the proportionality is with volume or mass and not with diameter.

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  • $\begingroup$ +1 VERY good question. I might add a bounty since it has no answers. I noticed you followed my Materials Modeling proposal. We are now in the commitment stage and it would be very very helpful if you could click commit! area51.stackexchange.com/proposals/122958/… Only 2.7% of the people fulfill commitment on new proposals anyway so it doesn't matter if you can't participate much, but it would be very very helpful if you can commit! $\endgroup$ Commented Mar 13, 2020 at 9:14
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    $\begingroup$ @user1271772, Thank you. I am going to commit later today (I am preparing a 4 hours class since 5am to be given this morning, very out of time right now) $\endgroup$ Commented Mar 13, 2020 at 9:35
  • $\begingroup$ Thank you very much! $\endgroup$ Commented Mar 13, 2020 at 9:36

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