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