# Reconciling $Q=PR$ and $Q=vA$

Say we have a closed loop system, where the flow $Q$ is constant through a pipe.

Then, at one segment of the pipe, we make the radius smaller. This causes the resistance $R$ to be greater, so the pressure $P$ must be higher as well. So, at this stretch of pipe, the fluid is under greater pressure.

Meanwhile, we've made the radius smaller, so that cross-sectional area $A$ is lower. Then, to compensate, $v$ has to be greater, to keep $Q$ equal.

So, in the same stretch of pipe, the velocity $v$ is greater, but so is the pressure.

Is this at odds with the Bernoulli principle, which says that when pressure and velocity of a fluid are inversely proportional?

I have a very limited understanding of the Bernoulli equation, so it's very possible that I'm misinterpreting it. But my thought was that if a fluid flows faster through a pipe, the collapsing pressure on that pipe would be greater. However, this seems at odds with the fact that the fluid appears to be under more pressure when it flows faster.

• what is $Q$? what is $R$? what is $P$? what is $v$? "This causes $R$ to be lower, so $P$ must be higher" why? Jan 8, 2016 at 17:07
• pressure and velocity are not inversely proportional in Bernoulli equation Jan 8, 2016 at 17:08
• You want to look at the Darcy Weisbach equation: en.wikipedia.org/wiki/Darcy%E2%80%93Weisbach_equation
– Gert
Jan 8, 2016 at 17:36
• Where did you get $Q=pR$?any reference? Jan 9, 2016 at 14:48

I have a very limited understanding of the Bernoulli equation, so it's very possible that I'm misinterpreting it. But my thought was that if a fluid flows faster through a pipe, the collapsing pressure on that pipe would be greater.

The following derivation will show that the pressure gradient along the flow line of the system depends also on the lengths of the pipes, not only on the diameters of the pipes.

Consider the following diagram for a horizontal pipe (no changes in potential energy): As the fluid flows the total pressure loss $\Delta p$ is:

$$\Delta p = (p_3-p_2)+(p_2-p_1)$$

Using Darcy-Weisbach, the pressure drop $\Delta p_i$ over each segment of pipe, assuming laminar flow (see edit for turbulent flow) is given by:

$$\Delta p_i=\frac{128\eta L_i Q}{\pi D_i^4}$$

$\eta$ is the viscosity and $Q$ the volumetric through-put. So that:

$$(p_3-p_2)+(p_2-p_1)=\frac{128\eta L_1 Q}{\pi D_1^4}+\frac{128\eta L_2 Q}{\pi D_2^4}$$

So that the overall pressure drop can be calculated:

$$p_3-p_1=\frac{128\eta L_1 Q}{\pi D_1^4}+\frac{128\eta L_2 Q}{\pi D_2^4}$$

Assuming the flow is into atmospheric pressure, then $p_3$ is known and by calculation also $p_1$.

$p_2$ can then be calculated by:

$$p_2=p_1+\frac{128\eta L_2 Q}{\pi D_2^4}$$

The pressure gradient will be something like (schematic, not to scale):

Edit: on turbulent flow.

For turbulent flow the expression for $\Delta p_i$ is replaced with:

$$\Delta p_i=\lambda_i\frac{L_i}{D_i}\frac{v_i^2}{2g}$$

Where $v_i$ is the mean flow velocity and $\lambda_i$ the friction coefficient for turbulent flow, which can be calculated from the equations on this page.

The derivation then proceeds analogously to the case of laminar flow.

• This analysis only addresses laminar viscous pressure variations, while the OP's question seems to be focusing on inertial effects (by his allusion to Bernoulli's equation). Of course, both can be important, and he does not seem to know how to include both in an analysis. Incidentally, that 64 in your equations should be 128. Jan 9, 2016 at 4:06
• @ChesterMiller: please see edit. Please also note that the derivations of Darcy-Weisbach (and it's turbulent equivalent) are obtained from Bernoulli's equation.
– Gert
Jan 9, 2016 at 14:47
• I respectfully disagree with regard to the derivation being obtained from the Bernoulli equation. Yes, it is possible to include a drag term in the Bernoulli equation, but the turbulent equivalent of the Darcy-Weisbach is based exclusively on experimental determination of the relationship between the friction factor and the Reynolds number. Jan 9, 2016 at 15:11
• @ChesterMiller: there is in my derivation only one minor thing missing: head loss at the sudden expansion (point $L_2$). I've not included it because the correction is usually small and because it would 'muddy the waters' for an OP who seems slightly confused already.
– Gert
Jan 9, 2016 at 15:23
• I agree.... And, "slightly confused" is quite an understatement. Jan 9, 2016 at 15:38

Then, at one segment of the pipe, we make the radius smaller. This causes the resistance R to be greater, so the pressure P must be higher as well

No. It is just the opposite. The total energy of a given mass of fluid is constant, being the sum of a potential term $$PV$$ and a kinetic term $$\frac{1}{2}mv^2$$.

Exactly because the velocity increases in the narrow region, the pressure decreases.

Considering the friction of the fluid with the pipe walls, we can say that the pressure drop for the same length is greater in that region, what is another issue.