How to derive $$v=\sqrt{\left(\frac{2\left(P_{1}-P_{2}\right)}{\rho}\right)}\left[\left(\frac{A}{a}\right)^{2}-1\right]^{-\frac{1}{2}}~?$$ I came across it while studying fluids mechanics in medicine and I am not sure how to derive it. It is related to Venturi effect. I tried using equation of continuity $Av=av'$ and Bernoulli's theorem $$P_1+ \rho gh_1+ ½\rho v^2=P_2+ \rho gh_2+ ½\rho v'^2$$ but to no avail. How is it related to Venturi effect?
Venturimeter Flow Rate – Flow Rate of Venturimeter Calculation
bernoulli-equationflowfluid dynamicsmedical-physics
Related Solutions
I used a numerical model to estimate flow parameters. Here are a few options for pipes of different diameters at the outlet. The problem is unsteady, the Bernoulli integral is obviously not preserved. The output velocity first increases, then reaches a maximum and decreases. It is not clear which speed should be limited. In the above figures, the flow velocity is normalized to the speed of sound at the input, so even the smallest output velocity is about 10 m / s. The animation shows how the modulus of the flow velocity changes with time.
Fig. 2 shows the modulus of the current velocity (left), the maximum output velocity depending on time and the distribution $p+\frac {1}{2}\rho \vec {v}^2$ at the last moment. It can be seen that the Bernoulli integral is not conserved due to the large density and pressure gradient in the thin pipe.
And a couple of pictures that show the same thing, but the Reynolds number is increased 10 times. It can be seen that a vortex was formed in the thick pipe, which led to the blocking of part of the pipe.
Summary of discussion on this question between AugieJavax98, philip_0008 and sammy_gerbil :
$P_1$ is not equal to $P_2$. The question says that there is a "pressure difference" between $A_1$ and $A_2$. The height $h$ is assumed to be the same at both ends (the pipe is lying horizontally). The difference in area causes a difference in flow speed (continuity equation) and a difference in pressure (Bernoulli equation).
The illustration suggests that $A_2 > A_1$. Therefore $P_2 > P_1$. However, this leads to an imaginary value for $v_2$, the speed through $A_2$.
However, the illustration could be misleading. The text does not state that $A_2 > A_1$. Assuming instead that $A_1 > A_2$ and therefore $P_1 > P_2$ leads to a real (and realistic) value for $v_2$ :
$$P_1 + \frac{1}{2} \rho v^2_1 = P_2 + \frac{1}{2} \rho v^2_2$$
$$P_1 - P_2 = \frac{1}{2} \rho (v^2_2 - v_1^2)$$
$$7500Pa = \frac{1}{2} (1000kg/m^3)(v^2_2 - (3.25m/s)^2)$$
which gives $$v=5.06m/s$$.
Best Answer
If the tube narrows (i.e $a<A$) we have Bernoulli $$ \frac 12 v_1^2 +\frac{P_1}{\rho}= \frac 12 v_2^2+ \frac {P_2}{\rho}. $$ Using mass conservation $Av_1=av_2$ this becomes $$ \frac 12 v_1^2 +\frac{P_1}{\rho}= \frac 12 v_1^2\left(\frac A a\right)+ \frac {P_2}{\rho}, $$ which is easily arranged into your formula for $v\equiv v_1$.