The question goes like this (picture included): Two large, open reservoirs A and F contain the same liquid. The liquid in A escapes the reservoir through a tube BCD, with a narrowing in C ($D_b = \sqrt{2}D_c)$. A and F are connected trough a tube from C to E, this tube is filled with air ($\rho_{liquid}>>\rho_{air})$ and $H_1$ is the height of A, $H_2$ the height of F. These are the questions:

A) Find the escape speed of the liquid in D

B) Find $H_2$ in function of $H_1$ enter image description here


A) I solved this using Bernoulli's formula: $P_0 + \rho\cdot g \cdot H_1 = P_0 + \frac{1}{2}\cdot \rho \cdot v^2$ where the left side is measured from A and the right side from D. This equation becomes $v_d = \sqrt{2\cdot g \cdot H_1}$. I personally think this solution is correct, but verification is welcome.

B) I haven't found a solution for this one, because I'm not sure what formulas I can apply and what formulas I can't. For example, I found two possible solutions for the speed of the liquid in C. Using Bernoulli, I'v found that $v_c = \sqrt{2\cdot g \cdot H_1}$ (as in A), but using the continuity equation ($A_c\cdot v_c = A_d\cdot v_d$ and we know that $A_c = \frac{2\cdot \pi \cdot D_c^2}{8}$ and $A_d= \frac{2\cdot \pi \cdot D_d^2}{4}$) I have that $v_c = 2\cdot \sqrt{2\cdot g \cdot H_1}$. I personally think that the second one is correct, but I can use some explanation on that. With the $v_c$, I feel like you could somehow use Bernoulli again, but I'm kinda stuck here. I could use any hint, solution or correction. Thanks

  • $\begingroup$ This is not a homework help site. $\endgroup$ – sammy gerbil Jan 28 '17 at 20:31

Your second solution for B is correct. The first one isn't, because for that solution you have assumed that $p_C=p_D$, which is obviously incorrect. Notice that there would have been nothing wrong with using the Bernoulli equation correctly for part B; this would have required you to take into account the reduced pressure at point $C$. Note that there's a factor of $g$ missing underneath those square roots for the velocity.

So, yes, after you have determined the velocity at $C$ from continuity, you can now use the Bernoulli equation again, most conveniently between $C$ and $D$. You know all the velocities, and elevations are identical, so you can determine the pressure difference between $C$ and $D$.

  • $\begingroup$ Thanks, would it be correct that the pressure in C is $P_c = P_0 - \frac{3}{2} \cdot \rho \cdot g \cdot H_1\$? Edit: Just realised this isn't correct $\endgroup$ – BMike Jan 28 '17 at 18:23
  • $\begingroup$ Indeed, I just noticed it $\endgroup$ – BMike Jan 28 '17 at 18:28
  • $\begingroup$ I corrected it to $P_c = P_0 - 3 \cdot \rho \cdot g \cdot H_1\$ $\endgroup$ – BMike Jan 28 '17 at 18:30
  • $\begingroup$ Ok, I think I got it now. We know the pressure at C. The pressure at E should be $P_e = P_0 - \rho \cdot g \cdot H_2$, sine the pressure at F is $P_0$ and E is $H_2$ than F. $P_c = P_e$, and this leads to $3 \cdot H_1 = H_2$. What do you think? $\endgroup$ – BMike Jan 28 '17 at 18:38

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