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In a straight pipe of uniform cross-sectional area filled with a nonviscous, incompressible fluid, the pressure at one end is equal to the pressure at the other end. Which of the following statements is/are true?

I. The volume flow rate is zero because there is no net force on the fluid

II. The volume flow rate is constant throughout the pipe because the cross-sectional area is uniform

III. The pipe is not inclined relative to the horizontal since the pressure is constant.

My Logic:

Bernoulli's Equation tells us: $P_1 + \frac{1}{2} \rho v_{1}^2 + \rho gy_1 = P_2 + \frac{1}{2} \rho v_{2}^2 + \rho gy_2$.

Knowing that $Q = vA$, the velocity is the same on both sides since the radius is constant. I also know the volume flow rate is constant (Choice II) <- What are the conditions for using this equation?

Since the pressure is the same on both sides, $P_1 = P_2$ so the pipe has to be horizontal otherwise we wouldnt have conservation of energy. (Choice III).

I assumed that since $Q = \frac{dP}{R}$ that the flow rate would be zero but I'm told that isn't true. Why isn't it?


Part of my question is when these relationships apply:

Bernoulli's I can use when i have incompressible non viscous fluid.

$Q=vA$ (same requirements as Bernoulli?)

$Q= \frac{dP}{R}$ (?)

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    $\begingroup$ If the fluid is inviscid, there is no flow resistance. Who told you that there is? $\endgroup$ Mar 8, 2018 at 0:21
  • $\begingroup$ The correct answer is "I". A pressure differential is required to establish flow in a pipe. If dP is zero, there is no "driving force" for fluid flow. $\endgroup$ Mar 8, 2018 at 1:02
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    $\begingroup$ @DavidWhite You're assuming that the fluid cannot have an initial flow when entering this pipe. Remember it's an inviscid fluid. $\endgroup$
    – JMac
    Mar 8, 2018 at 1:26
  • $\begingroup$ @JMac, I am making no assumptions. If the fluid is flowing before it enters the pipe, there will be a pressure drop in the pipe due to fluid friction with the walls of the pipe. $\endgroup$ Mar 10, 2018 at 15:34
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    $\begingroup$ @DavidWhite If it's an inviscid fluid, it won't have any reason to experience pressure drop. A pressure drop can't occur from the "fluid friction" if there is no viscosity. $\endgroup$
    – JMac
    Mar 10, 2018 at 15:55

2 Answers 2

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First note that the lack of a net force does not imply zero flow velocity since the question points out that the flow is inviscid. Since the pipe has a uniform cross-section, and since there can be a flow ($v\gt 0$) then we can have $Q=Av\gt 0$ meaning answer (I) is not necessarily (always) true.

And if $Q$ is a constant and given $A$ is constant, then $v$ must also be constant, since $$Q=A_1v_1=A_2v_2=\text{constant} \\ \rightarrow A_1=A_2 \therefore v_1=v_2=\text{constant}$$ So we would chose (II).

For (III) to be true, then:

From Bernoulli's equation $$P_1 + \frac{1}{2} \rho v_{1}^2 + \rho gy_1 = P_2 + \frac{1}{2} \rho v_{2}^2 + \rho gy_2$$ we are told that the pressure is the same throughout the pipe, so the terms $P_1$ and $P_2$ are equal and so vanish, meaning $$\rho g \Delta y = \frac{1}{2}\rho (v_2^2-v_1^2)$$ For $v_1\ne v_2$ then $\Delta y\ne 0$ so that one end of the pipe is higher, but we established that $v_1=v_2$ so this means $\rho g\Delta y=0$ meaning the pipe is horizontal. This means that (III) can also be true.

To answer the other part of your question, we do not require that the continuity equation apply only for incompressible flows. The continuity equation applies to all fluids, whether they be compressible or incompressible flow because it expresses the law of conservation of mass which must be satisfied at every point in a flow. The Bernoulli's equation $$P+\frac{1}{2}\rho v^{2}+\rho gh=\text{constant}$$ does require incompressibility though.

Also, the equation$^1$ you quote $$Q= \frac{dP}{R}$$ does not appear to be correct assuming that $P$ is pressure and $R$ is a length. The quantity on the left $Q$ has units $\frac{m^3}{s}$ while the RHS appears to have units $\frac{N}{m^3}$ and so does this not appear to be a valid equation.

$^1$If you are using a variation of the Hagen-Poiseuille equation, $$P=\frac{8\mu LQ}{\pi R^4}$$ where $$R=\frac{8\mu L}{\pi r^4}$$ is a measure of flow resistance, then it is valid.

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  • $\begingroup$ I believe the final "incorrect" equation is a form of the Hagen-Pouseulle Equation, where $R$ is meant to be the de facto resistance (not radius). $\endgroup$ Oct 9, 2021 at 14:00
  • $\begingroup$ The Hagen-Poiseuillie equation is $$\Delta P=\frac{8\mu LQ}{\pi R^4}$$ where $R$ is the pipe radius and $\mu$ is the dynamic viscosity. The above equation $Q=dp/R$ is still dimensionally incorrect. $\endgroup$
    – joseph h
    Oct 9, 2021 at 20:00
  • $\begingroup$ The Hagen-Poiseuille Equation can be viewed as analogous to Ohm's Law (∆V=IR), which are both special cases of the more general relationship (driving force) = (flow rate) × (resistance). In this context (common in physiology/biology), the HP Equation takes the form ∆P=QR, where R is the resistance to flow, as I previously stated (of course depending on viscosity, pipe length and pipe radius). I assume this abbreviated version of the HP Equation is what OP was presenting when the mentioned Q=∆P/R. It is dimensionally consistent with appropriate definition of R. $\endgroup$ Oct 10, 2021 at 21:19
  • $\begingroup$ So what would be the appropriate definition and units of R in this case where Q=P/R given Q is the flow rate and P is the pressure? The equation is still incorrect unless there is a dimensionally correct definition of R. $\endgroup$
    – joseph h
    Oct 10, 2021 at 22:49
  • $\begingroup$ I'm sure you can see the resistance would be defined as $R=\frac{8\mu L}{\pi r^4}$. I feel like this conversation is digressing. My goal was to clarify for anyone who reads that the equation the OP presented ($Q=\Delta P/R$) is indeed a valid relationship that I have seen published in textbooks, and that in this equation $R$ represents not the pipe radius but resistance to flow, whose definition can be readily observed by understanding that $Q=\Delta P/R$ is a simplified manifestation of the Hagen-Poiseuille Equation. $\endgroup$ Oct 10, 2021 at 23:33
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We can't really say much about the volume flow rate, it may or may not be zero. If the pipe is horizontal, however, and the flow is streamline (that's when you can use Bernoulli's equation) - the volume flow rate will be zero beyond question.

If the fluid is flowing, then velocity ≠ 0, and consequently the difference in height for a particular streamline should change - as a result, the pipe must be inclined at an angle, not purely horizontal.

Does that clear it up?

Edit: In the comments section, Chester Miller correctly pointed out the following -

Choices II and III can be true, although they don't necessarily have to be true. Choice II can also be true if the cross section varies, as long as the entrance and exit cross sections are equal. And choice III can also be true if the pipe is curved in the vertical as long as the entrant and exit cross sections are at the same elevation.

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    $\begingroup$ I disagree with this. See @JMac's comment. It seems to me, choices II and III can be true, although they don't necessarily have to be true. Choice II can also be true if the cross section varies, as long as the entrance and exit cross sections are equal. And choice III can also be true if the pipe is curved in the vertical as long as the entrant and exit cross sections are at the same elevation. $\endgroup$ Mar 8, 2018 at 2:59
  • $\begingroup$ You're absolutely right. I'll add that to my answer. $\endgroup$ Mar 8, 2018 at 3:12

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