# Different Bernoulli equation from $F=dp/dt$ [closed]

If any part of the question needs clarifying, there might be explanation in the post-script and of course ask if needed.

I understand how the Bernoulli equation is derived for incompressible fluid in a pipe. I will write the equation for a system where elevation is effectively constant. $$P_1+\frac{\rho}{2}v_1^2=P_2+\frac{\rho}{2}v_2^2$$ Now, I confused myself when considering the case where $$P_2=0$$ and starting from $$F=\frac{dp}{dt}$$ . I am thinking the infinitesimal change in momentum of the system as the piston moves by $$dx$$ (see post-script) is: $$dp=\rho A_1 dx (v_2-v_1)$$ And so: $$\frac{dp}{dt}=F=P_1 A_1=\rho A_1 \frac{dx}{dt}(v_2-v_1) = \rho A_1 v_1(v_2-v_1)$$ And so $$P_1=\rho v_1(v_2-v_1) \neq \frac{\rho}{2}(v_2^2-v_1^2)$$

My question is: where is my mistake?

Post-script:

The physical picture I was thinking of was a cylindrical pipe whose cross-section abruptly goes from [larger] $$A_1$$ to [smaller] $$A_2$$. The pressure in the volume with larger cross section is achieved by a piston, and there is no piston pushing on the volume of lower cross section.

• The Bernoulli equation is not transient. It is for steady state. You need to research derivations of the transient version of the Bernoulli equation. Oct 2 at 11:12

It is better to work with small elements and later take the limit.

$$\Delta p=\rho A_1\Delta x(v_2−v_1) \implies \frac{\Delta p}{\Delta t} = \rho A_1\frac{\Delta x}{\Delta t}(v_2−v_1)$$

The average velocity of the small element is:$$v_e = \frac{\Delta x}{\Delta t} = \frac{v_1 + v_2}{2}$$

So, when the $$\Delta$$'s go to zero: $$F = PA_1 = \rho A_1\frac{v_1 + v_2}{2}(v_2−v_1) = \frac{1}{2}\rho A_1(v_2^2−v_1^2)$$

• I like this answer a lot and upvoted it, but am still hanging myself up. $\Delta$ x is associated with the piston and not the fluid at the junction. Also, the idea of the piston’s force being partially balanced makes sense when picturing an elastic, annular membrane joining the two pipes. I sure could be wrong, might be interesting to experiment with a membrane having well known elastic properties to see if $F_{annulus}=-\frac{\rho}{2}(v_2-v_1)^2$ Oct 2 at 0:48

Thank you to whoever telepathically answered my question. The answer can be found by asking yourself what happens if there is a piston pushing in the opposite direction on the volume of smaller cross section.

The force applied on the larger cross section must increase by $$P_2 A_2$$ in order for the velocity to be kept at $$v_1$$. Indeed, $$v_1=0$$ if $$P_1 A_1=P_2 A_2$$ since the forces balance … and then we must consider possible forces at the annular junction between region 1 and 2.

I have no idea how to derive said force from geometry and velocities, but using the real Bernoulli equation we get: $$F_{annulus}=-\frac{\rho}{2}(v_2-v_1)^2$$