enter image description here

According to bernoulli equation

$$ P_1+1/2 \rho v_1^2+ \gamma h_1= P_2+1/2 \rho v_2^2+ \gamma h_2. $$

Because $$ v_1=0\quad h_1=0 $$ So $$ 0= 1/2 \rho V^2+ \gamma h. $$ It is wrong I know, but why? Please help, thanks.


After a short time to review my book, I have an idea. The $\gamma h$ in bernoulli equation cannot be seen as pressure. It must be seen as potential energy. Am I right? Then, in the water surface $\gamma h \neq 0$ as the pressure but $\gamma h > 0$ as potential energy.

  • $\begingroup$ If h1=0 at the top, then h2<0 at the bottom. As you can see, the absolute heights don't matter. What matters is only the height difference. $\endgroup$ – CuriousOne Jan 6 '15 at 3:02
  • $\begingroup$ @CuriousOne Thanks for replying. But $\gamma h$ isn't a hydrostatic pressure? It increases with water depth. $\endgroup$ – ytyyutianyun Jan 6 '15 at 3:23
  • $\begingroup$ hydrostatic pressure - The pressure exerted by gravity at a given point within a fluid that is at equilibrium, increasing in proportion to depth from the surface. $\endgroup$ – LDC3 Jan 6 '15 at 3:31

Like stated in the comment of CuriousOne, the positive direction of the heights is upwards.

Another way of looking at this is when all velocities are zero and you say that $h_1=0$, so by looking at position 1 and 3. Now you can use hydrostatic pressure to express $P_3$ in terms of $P_1$, $\gamma$ and absolute height difference $h'=h-l$.

$$ P_3 = P_1 + \gamma h' $$

If you now fill in the Bernoulli equation you get:

$$ P_1 + \frac{1}{2}\rho v_1^2 + \gamma h_1 = P_3 + \frac{1}{2}\rho v_3^2 + \gamma h_3, $$

but we set all velocities and $h_1$ to zero, thus

$$ P_1 = P_3 + \gamma h_3. $$

If we combine this with my first equation it can be derived that:

$$ h_3 = -h'. $$

You are correct that this height terms is potential energy, however the zero of this potential can be positioned anywhere you like, since in the Bernoulli equation only the height difference matters.

| cite | improve this answer | |
  • $\begingroup$ Very thanks, it works with clearly explanation. I must confuse the two issues between fluid statics and fluid dynamics. $\endgroup$ – ytyyutianyun Jan 8 '15 at 10:46

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.