# Outflow from a tank

Consider a tank with an outflow a the bottom: with the unsteady Bernoulli equation:$$\int_S \frac{\partial v}{\partial t} \ dS+\frac{v^2}{2}+\frac{p}{\rho}+gz=0$$where $s$ is the path between $S_1$ and $S_3$. Suppose a inviscid fluid with constant density. Why does the solution of the Bernoulli equation change depending on whether $l$ very small or not? In other words, why is there a term $\frac{\partial v}{\partial t}$ introduced when $l$ is not negligibly small?

• -1. Unclear. Please cite a source for the Bernoulli equation which includes such a term. – sammy gerbil Mar 13 '17 at 0:03
• I'm not 100% sure, but I think this term should come from the friction. – Alpha001 Mar 13 '17 at 0:04
• @Alpha001 OP says the fluid is inviscid and incompressible. – sammy gerbil Mar 13 '17 at 1:25
• Are you referring to the transient version of the Bernoulli equation? – Chet Miller Mar 13 '17 at 1:30
• @miller edited, yes i am – Ben L Mar 13 '17 at 7:12

$S_1 \to S_2 = S_2 \to S_3$. The only change here is $v$ and $p$ at $S_1$ and $S_2$. $p_1>p_2$ at $S_1$ and $p_2<p_1$ at $S_2$, $v_1<v_2$ at $S_1$ and $v_2>v_1$. In the Bernoulli E substitute the proper values and that satisfy the same result with L as in question being used in equation on the right. $\rho+\rho g h+\frac{1}{2}\rho v^2 = \rho +\rho g \ell +\frac{1}{2}\rho v^2$. As for the derivative $dv/dt$ it will indicate only the time for the fluid to move in H and L with {H=l} in time; the values of Bernoulli equation do not change because, if the values change then Bernoulli will not be in the books.