I am reading a book of Fundamental Energy Systems. The author describes the rate of change in head for a turbomachine as:

$$ \frac{1}{2}[(V_1^2-V_2^2)+(U_1^2-U_2^2)+(V_{R2}^2-V_{R1}^2)] = H =U_1V_{u1} - U_2V_{u2} $$ and the static effect as: $$SE =(U_1^2-U_2^2)+(V_{R2}^2-V_{R1}^2) $$

However he later says: enter image description here

His wording states $ R = SE/H $. But he actually wrote it as we see above: $$ R = \frac{SE}{2H} = \frac{(1/2)SE}{(1/2)2H} = \frac{(1/2)SE}{H} $$

I tried deriving his result from $ R = SE/H $ but it was not possible, I just want to make sure my conclusion is correct (R = (0.5*SE)/H ; instead of R = SE/H ) as the author does not seem to mind his own wording.

Thanks for illustrating me.

  • $\begingroup$ The highlighted part says it is a ratio... and if you divide the other halves of the expressions it comes out. $\endgroup$
    – Solar Mike
    Jan 15, 2021 at 19:03
  • $\begingroup$ Yes, the ratio would be SE/H. Which is the ratio of the total SE, to the total head H. But that is not SE/2H = (1/2)SE/H which is what the author wrote. $\endgroup$
    – RMS
    Jan 15, 2021 at 19:23
  • $\begingroup$ No, multiply first eqn on both sides by 2 gives 2H and the U and Vr terms then cancel with the lhs terms for SE leaving the velocity difference which is head. $\endgroup$
    – Solar Mike
    Jan 15, 2021 at 19:27
  • $\begingroup$ So you are saying R = SE/2H ? SE/2H = [ (U1^2- U2^2) + (Vr1^2 - Vr1^2) ] /[ (U1^2- U2^2) + (Vr1^2 - Vr1^2) ] [1 + (V1^2-V2^2)/((U1^2- U2^2) + (Vr1^2 - Vr1^2)) ] = 1/(1 + (V1^2-V2^2)/((U1^2- U2^2) + (Vr1^2 - Vr1^2)) ). That is not what he got, I am not sure I am following you. His words say SE/H. With SE/H and the definitions provided please show how to reach that form. I appreciate the help. $\endgroup$
    – RMS
    Jan 15, 2021 at 19:40
  • 1
    $\begingroup$ I think this is just a typo, and the definition of SE should be $\frac 1 2[(U^2_1−U^2_2)+(V^2_{R2}−V^2_{R1})]$. Compare with the previous formula. The total head should be the SE + some extra terms. $\endgroup$
    – alephzero
    Jan 15, 2021 at 20:21

1 Answer 1


As @alephzero commented, the Static Effects = 1/2 (Terms). Which I had interpreted to be SE = (Terms). The author does is not clear, as he only highlighted the (Terms) when talking about the SE.


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