Pressure in fluid mechanics of incompressible liquid

Question

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?

Answer 1

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!

Answer 2

How does this varying pressure conform with the constant pressure/density obtained from the equation of state?

It doesn't conform. It is a contradiction to reality: you cannot have an incompressible material. However, that doesn't mean the approximation isn't a useful one under many circumstances.

For example, consider water which as a density of roughly $\rho_0=1000$ $kg/m^3$ and a speed of sound of $c_{s0}=1,484$ $m/s$ at room temperature and standard atmospheric pressure. In order to see if compressibility is important to a given problem we might make a Taylor series approximation to the equation of state $$p(\rho, e) = c_{s0}^2 (\rho - \rho_0) + p_0$$ with $c_s$ being the speed of sound defined by $$c_{s0}^2 = \frac{\partial p}{\partial \rho}\Large{|}_{s,\rho_0}$$ Plugging in the speed of sound for water we see that $$\rho - \rho_0 = 4.54\times10^{-7}(p- p_0)$$ From this equation you can now estimate how much density change you can expect to see and evaluate how important that change really is to the problem at hand. For something like water flowing through a pipe at low Mach numbers you will find that changes in density contribute very little to the resulting flow features.

Answer 3

Note that your relation is for gases and it is a first order approximation to small pressure and density differences according to a Taylor series of the full definition $c^2=\partial p/\partial \rho$. For incompressible fluids c is infinite because $\partial p/\partial \rho$ goes to infinity.

Answer 4

If the velocity of the fluid is above 100 m/s, then we can regard the flow as incompressible, let's assume tat this is the case for section 2 (smaller surface area therefore greater velocity).

If the flow is steady, then, we can say that the mass flow rate (that is rhoUA) would remain constant.

Now, furthermore, if the liquid is isothermal, then we can take the following steps:

rho*U1*A1 = rho*U2*A2. rho= p/RT

p/RT*U1*A2 = p/RT*U2*A2. T & R cancel (both constant)

p1*V(fr)1 = p2*V(fr)2.$ where V(fr) is the volume flow rate i.e U x A

Now, comparing the pressures at the section 1 (bigger surface area) and at section 2 (smaller surface area), we can see that, the pressure at section 1 would need to be greater if the flow would continue moving.

This is because at section 1, the area is bigger and the velocity is slower.

Now, p1>p2 hence, why, if we rearrange the above equation $

(p1/p2)* V(fr)1 = V(fr)2. we can see that the volume flow rate at 2 would be greater than the volume flow rate at 1.

Now, this would make sense since the velocity would increase at section 2 due to the decrease in surface area.

The hardest part about this problem in uni for me was determining how the pressure would be greater at 1, and if you think in terms of the demand, which is to continue the flow of the fluid, then an increase in pressure would be the only plausible explanation for this example.

Answer 5

In liquids, pressure is not given by the above function of density (this is valid only for gases). The pressure will vary along the pipe according to the Bernoulli theorem and this variation is important to the description of the liquid flow. The density will vary slightly too, but this can be often neglected - the density can be assumed to have the same value everywhere.