Most liquids can be approximated to be incompressible, since the Mach-number is much smaller than 1. That means that the density variations are negligible and from the relation between pressure p and density ρ,
$$
p=c_s^2 \rho
$$
we see that the pressure in constant as well. Now, say that I look at a pipe with the following geometry:
From Bernoulli's equation we get that the pressure and velocity will be different between the large-radius part of the pipe and the small-radius part. How does this varying pressure conform with the constant pressure/density obtained from the equation of state?
Best Answer
Using Bernoulli you get:
$$ \frac{P_1}{\rho} + \frac{1}{2}v_1^2 = \frac{P_2}{\rho} + \frac{1}{2}v_2^2 $$
Using your formula:
$\displaystyle{c_s^2 + \frac{1}{2}v_1^2 = c_s^2 + \frac{1}{2}v_2^2}$ and this implies: $v_1 = v_2$
From mass continuity: $v_1 \times A_1 = v1 \times A_2$, so $A_1=A_2$, which is a false. There is clearly a misinterpretation.
Your equation is a state function that relates pressure and density, but the constant is extremely small for liquids, and Bernoulli's equation assumes that the fluid is not compressible, so your state function can't be used together with Bernoulli's.
I hope this helps!