Recently I was doing a pipe network problem without Hardy Cross Method (approximate method )

I have obtained 10 desired equations with 10 unknowns as shown in this image below highlighted by red pen. I tried solving manually but stumped .. then i thought to use mathematical software... but except wolframalpha i dont know any other softwares

Can anyone know how to solve them... as wolframalpha feels lazy to show answer :D

How to solve using software also ?

  • $\begingroup$ without hardy cross it would be difficult definitely $\endgroup$ – Aman Rajput Jun 2 at 7:12
  • 2
    $\begingroup$ The general method is using Gröbner bases. $\endgroup$ – Wuestenfux Jun 2 at 7:44
  • $\begingroup$ Okay let me check that.. $\endgroup$ – learningstudent Jun 2 at 7:58
  • $\begingroup$ I don't understand any of this head loss business, but your system of pipes has $0.2+Q_1+Q_6$ flowing in, and $0.2$ flowing out. It would seem that $Q_1=Q_6=0$, and hence $Q_2=Q_3=0$ and $Q_4=Q_5=0.2$. This might not be consistent with the last six equations, as Quasi's calculations seem to show. $\endgroup$ – vadim123 Jun 2 at 17:52

Elementary algebra is all that's needed -- tedious, but routine.

Eliminating the variables one at a time, we get \begin{cases} Q_2=Q_1 - {\large{\frac{K+125}{10K}}} \qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\; \\[4pt] Q_3={\large{\frac{K+125}{10K}}}\\[4pt] Q_4=Q_1 + {\large{\frac{K-125}{10K}}}\\[4pt] Q_5={\large{\frac{K-125}{10K}}}\\[4pt] Q_6=-Q_1\\[4pt] \end{cases} \begin{cases} \,h_1=25-KQ_1 \\[4pt] \,h_2= -2KQ_1^2 + \left({\large{\frac{K+125}{5}}}\right)Q_1 - {\large{\frac{K^2-2250K+15625}{100K}}} \\[4pt] \,h_3=20-KQ_1^2\\[4pt] \,h_4= -KQ_1^2 - {\large{\frac{K^2-2250K+15625}{100K}}} \\[4pt] \end{cases} where $Q_1$ satisfies the quadratic equation $$(200K^2)Q_1^2-(5000K)Q_1+(K^2-250K+15625)=0$$ But for the given value of $K$, one can verify that the discriminant of the above quadratic is negative, hence the given system of equations has no real solutions.


Sorry i am not able to add comment , but i think that we can make these equations into a 10×10 matrix and then we can reduce the matrix to reduced row echelon(Gauss Jordon method) form to get the solution. Then we will get values of H(i) and all Q^2 .From Q^2 we will get values of all Q's


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.