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How does the author get that formula(d alpha formula) using the momentum equation . The book here is Mechanics and thermodynamics of propulsion by Philip Hill and Carl Peterson.

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It involves simple differentiation operation to get there.

We have $\frac{dP}{\rho} = -UdU$ and $\frac{dP}{\rho} = -udu$. These equations can be re-organised as $$\frac{-dP}{\rho U} = dU, ~\frac{-dP}{\rho u} = du$$ Given, $\alpha = \frac{u}{U}$. Differentiate this relation on both sides, we get $$d\alpha = \frac{du}{U} - \frac{u}{U^2}dU$$ Now, substitute the expressions for $du$ and $dU$ in the above, we have \begin{eqnarray*} d\alpha &= &\frac{1}{U}\frac{-dP}{\rho u} - \frac{u}{U^2}\frac{-dP}{\rho U}\\ &= &-\frac{1}{U^2}\frac{U}{u}\frac{dP}{\rho} + \frac{1}{U^2} \frac{u}{U}\frac{dP}{\rho} \\ &= &-\frac{1}{U^2}\frac{1}{\alpha}\frac{dP}{\rho} + \frac{1}{U^2} \alpha\frac{dP}{\rho} \\ &= & \frac{dP}{\rho U^2}\left(\alpha - \frac{1}{\alpha}\right)\\ &= & \frac{dP}{\rho U^2}\left(\frac{\alpha^2 - 1}{\alpha}\right) \end{eqnarray*}

Hope this answers!

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    $\begingroup$ Thank you Krishna .. :) $\endgroup$
    – Toshith
    Jun 7 '20 at 5:57

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