enter image description here

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.


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!

  • 1
    $\begingroup$ Thank you Krishna .. :) $\endgroup$
    – Toshith
    Jun 7 '20 at 5:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.