# Bernoulli with friction

I'm currently stuck trying to solve an exercise for fluid mechanics and I hope you can help me.

Here' s the exercise: They want me to determine the decrease in pressure caused by the turbine ("Kraftwerk"). According to the text waters is only flowing from "Stausee" to "Kraftwerk". "Wasserschloss" is a measurement to keep the pressure at a distinct level. All the variables are already known.

I tried to solve this problem with the Bernoulli-Equation from the top ("Stausee") down to the turbine ("Kraftwerk") giving me the following equation:

$$p_{\infty}+\rho g H_{\mathrm{WS}}=\frac{\rho}{2} w_2^2\left(1+\zeta_{\mathrm{U2}}+\lambda_2\frac{L_2}{D_2}\right) +p_{\mathrm{KW}}+\rho g H_{\mathrm{KW}}+\frac{\rho}{2}w_1^2 \left(\zeta_{\mathrm{Ein}} +\zeta_{\mathrm{U1}} +\lambda_1 \frac{L1}{D1}\right) + p_{\mathrm{turbine}}$$

According to the official solution I should unse the Bernoulli-Equation from $U$ to the turbine ("Kraftwerk) instead. Giving me the following equation:

$$\frac{\rho}{2} w_1^2 + p_{\mathrm{U}}+ \rho g H_{\mathrm{U}} =\frac{\rho}{2} w_2^2 \left(1+\zeta_{\mathrm{U2}} + \lambda_2 \frac{L2}{D2}\right) + p_{\mathrm{KW}} + \rho g H_{\mathrm{KW}} + p_{\mathrm{turbine}}$$

I don't really understand why i get two different results.

I hope you can help me.

-
Please, for subscripts, use the latex notation $\$p\_\{kw\}\$$to render p_{kw}. What are \xi u1,\lambda 1,L1,D1,\xi u2,\lambda 2,L2,D2 ? What is the difference between pturbine and pkw ? What are exactly w1 and w2 ? – Trimok Sep 26 at 18:41 \xi_{ein} is an coefficient giving the loss in pressure due to the entrance in pipe 1. – kornnflake Sep 27 at 10:02 \xi_{ein} is an coefficient giving the loss in pressure due to the entrance in pipe 1.\xi_{U1} and \xi_{U2} are used to calculate the loss in pressure while exiting pipe 1 and entering pipe 2. \lambda is the loss of pressure because of friction. The loss in pressure caused by friction is depending on the length of the pipe (L) and the diameter (D). p_{kw} is the static pressure of the fluid exiting the turbine while is describing the loss in pressure by the turbine p_{turbine}. – kornnflake Sep 27 at 10:10 In your first equation, in the left member, I think it is p_S and H_S, and not H_{WS}, because your are beginning with the top tank ("Stausee") . Right ? – Trimok Sep 27 at 10:28 add comment ## 1 Answer In supposing my last comment correct, there is no contradiction between your first equation and the second equation. Applying the Bernoulli equation from S to U, you have :$$p_{S}+\rho g H_{\mathrm{S}}=\frac{\rho}{2}w_1^2 \left(1+\zeta_{\mathrm{Ein}} +\zeta_{\mathrm{U1}} +\lambda_1 \frac{L1}{D1}\right) + p_{U}+ \rho g H_{\mathrm{U}}\tag{1}$$Your second equation is applying the Bernoulli equation from U to KW :$$\frac{\rho}{2} w_1^2 + p_{\mathrm{U}}+ \rho g H_{\mathrm{U}} =\frac{\rho}{2} w_2^2 \left(1+\zeta_{\mathrm{U2}} + \lambda_2 \frac{L2}{D2}\right) + p_{\mathrm{KW}} + \rho g H_{\mathrm{KW}} + p_{\mathrm{turbine}}\tag{2}$$Summing equations (1) and (2), you get :$$p_{S}+\rho g H_{\mathrm{S}}=\frac{\rho}{2}w_1^2 \left(\zeta_{\mathrm{Ein}} +\zeta_{\mathrm{U1}} +\lambda_1 \frac{L1}{D1}\right)\tag{3} + \frac{\rho}{2} w_2^2 \left(1+\zeta_{\mathrm{U2}} + \lambda_2 \frac{L2}{D2}\right) + p_{\mathrm{KW}} + \rho g H_{\mathrm{KW}} + p_{\mathrm{turbine}}

But this is precisely your first equation.

-