Deriving Bernoulli's equation with the kinetic gas theory: Missing 1/2 factor

Inspired by Eugene Khutoryansky video about Bernoullis law (https://www.youtube.com/watch?v=TcMgkU3pFBY), I decided to put the thoughts and ideas from her video into mathematical terms. My derivation of Bernoullis law is as follows:

Lets define the x-axis parallel to the fluid’s velocity and the z-axis perpendicular to it.

Since Force is defined as the time-derivative of the impulse (here $tilde{p}$), we can rewrite the pressure $p$ as
$$p=frac{dtilde p}{dtA}$$
We know that in a collision of $N_z$ particles with a stationary object an impulse of $dp=2mv_zN_z$ is being transferred onto it. Furthermore The amount $N_z$ of particles colliding with the wall can be rewritten as
$$N_z=n_zdV=n_zAv_zdt$$
Where $n_z$ is the density of particles moving towards the z-direction and $v_z$ is the vertical speed component of the particle. Both formulas yield
$$p=2mnv_z^2rightarrow v_z^2=frac{p}{2mn}$$
We also know that $n_z=frac12n$, since in a gas or fluid moving along the x-axis there is no preferred direction along the z-axis, so that equal amounts are flowing in the opposite directions. Also, $mn=rho$. This gives us
$$v_z^2=frac{p}{rho}$$
Energy conservation gives us
$$E_{kin}=E_{kin}’rightarrow v^2=v’^2rightarrow v_x^2+v_z^2=v_x’^2+v_z’^2$$
This ultimately gives us
$$p+rho v_x^2=p’+rho v_x’^2$$
or
$$p+rho v_x^2=p_0$$
Which is almost equal to the bernoulli-equation
$$p+frac12rho v_x^2=p_0$$
I was already pretty satisfied with my result and thought that the missing factor of 1/2 is a mathematical error of some sort. However, I couldn’t find any. So I thought that perhaps I have made some wrong assumptions. I couldn’t find any neither, since both of the assumptions I made where a factor of 1/2 is present, namely the equations
$$dp=2mv_zN_ztext{ and }n_z=frac12n$$
are not wrong (at least not by my understanding). I would really appreciate it if someone could point out any mistakes I could’ve made.