# If the difference in pressure is zero between two points of a horizontal constant radius pipe, why isnt flow rate zero?

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?)

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

• If the fluid is inviscid, there is no flow resistance. Who told you that there is? – Chet Miller Mar 8 '18 at 0:21
• 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. – David White Mar 8 '18 at 1:02
• @DavidWhite You're assuming that the fluid cannot have an initial flow when entering this pipe. Remember it's an inviscid fluid. – JMac Mar 8 '18 at 1:26
• @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. – David White Mar 10 '18 at 15:34
• @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. – JMac Mar 10 '18 at 15:55

## 1 Answer

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.

• 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. – Chet Miller Mar 8 '18 at 2:59
• You're absolutely right. I'll add that to my answer. – arya_stark Mar 8 '18 at 3:12