Say that we have water flowing horizontally in a pipe of constant diameter. The continuity equation for incompressible fluids, which is a statement of mass conservation, guarantees that the flow rate will be the same throughout the pipe. Given that the diameter is constant it also guarantees that the velocity will be the same throughout.

Now say the pipe reaches a point where it is tilted. The tilted segment leads downhill and reaches another horizontal segment.

Water is now going downhill and so it is accelerating. Let us assume an ideal situation with negligible friction and hydraulic resistance. If the water is accelerating under gravity then velocity must change so I assume the diameter of the water flow itself must shrink and the water will "stretch" in order to maintain continuity. Is this actually the case?

What happens at the bottom when the water flow becomes horizontal again? Does it gradually "bulge up" at the bottom in order to reestablish the horizontal flow of constant diameter in the lower horizontal pipe? Assuming that the conservation of mass holds and continuity is preserved, the horizontal flow and speeds must be the same in both horizontal pipes.

One additional remark on this situation. What if the water were being pumped uphill in a pipe in the same configuration? Is it possible to pump water uphill and maintain a constant velocity in a constant diameter pipe? How can you demonstrate whether it is or isn't from first principles? It seems like water in the pipe would have to slow down when it encounters an uphill segment, but it can't pile up anywhere in the pipe either, and certainly doesn't expand, so it must also preserve continuity somehow I would assume? I feel like I am really missing something.

I apologize if this question seems trivial or silly. My backgrounds are in chemistry and in geology and I could not find any explanations on the internet regarding this. My fluid mechanics, physical hydrology, and hydrogeology textbooks also were of no help.

  • $\begingroup$ Are you neglecting pressure on purpose? $\endgroup$
    – Kyle Kanos
    Aug 11, 2022 at 19:32
  • $\begingroup$ No, I just don't know enough about this. I considered pressure but don't know what to do with it in this situation. pressure differences drive a flow but I am too confused to apply it here. $\endgroup$
    – MattGeo
    Aug 11, 2022 at 19:37
  • $\begingroup$ This is literally part of the derivation of Bernoulli's equation. I find it hard to believe your fluid mechanics book doesn't cover this. Which book is it? $\endgroup$
    – Kyle Kanos
    Aug 11, 2022 at 20:17
  • $\begingroup$ Fundamentals of Fluid Mechanics, 5th Edition by Munson, Young, and Okiishi $\endgroup$
    – MattGeo
    Aug 11, 2022 at 20:19
  • $\begingroup$ it covers bernoulli of course. I just don't understand the situation involving constant diameter and downhill flow $\endgroup$
    – MattGeo
    Aug 11, 2022 at 20:20

2 Answers 2


The water does not necessarily accelerate going through a downwards stretch of pipe. It is prevented from doing so by the other water ahead of and behind it, and by the walls of the pipe. Just as you do not accelerate when descending in an elevator at constant speed. But what does happen on a downhill slope is the water loses potential energy, but gains an equal amount of mechanical energy (pressure).

In a steady pipe flow situation, the flow rate is constant throughout the full length of the pipe, and the rate of flow is determined by the total pressure (energy) change from the beginning to the end. You have to look at the entire system to analyze it in this way, and determine the flow rate that will occur.

  • $\begingroup$ Is it correct to say then that it is not accelerating in the downslope part of the pipe because as it loses potential energy the static pressure of the fluid is increasing and this offsets the velocity gain it would have otherwise experienced if this were a free-fall type scenario? $\endgroup$
    – MattGeo
    Aug 12, 2022 at 1:55
  • $\begingroup$ I suppose that is correct. Perhaps a more precise way to say it is: in a free fall scenario (e.g. a fountain), the gravitational potential energy (GPE) is converted to kinetic energy (KE). In a downhill pipe scenario, the GPE is converted to a combination of KE and static pressure. $\endgroup$
    – RC_23
    Aug 12, 2022 at 2:49
  • $\begingroup$ There is a force balance on the fluid in the tilted section involving 3 forces: the gravitational body force, the pressure forces at the ends, and the viscous drag force at the walls. The viscous drag for prevents the fluid from accelerating, even for a horizontal pipe. For a horizontal pipe, the gravitational body force does not contribute to the horizontal force balance. $\endgroup$ Aug 12, 2022 at 11:20
  • $\begingroup$ I actually get it now. It seems so simple at this point I almost feel embarrassed for not just seeing it to begin with. Not sure what my mind was doing. $\endgroup$
    – MattGeo
    Aug 12, 2022 at 13:47
  • $\begingroup$ Never be embarrassed for asking even seemingly obvious questions. Einstein did that by asking "why does a man falling feel weightless," and ended up revolutionizing gravity and astronomy. $\endgroup$
    – RC_23
    Aug 12, 2022 at 23:17

Very good question!

I myself would love to have a good explanation. For now, though, I can give you an explanation which is not good but may satisfy you.

To describe fluid motion, we use the field theory. Probably, US educated people will call it field calculus. In any case scenario, I mean just the mathematics of fields.

I believe that peculiarity of the field theory gives rise to your (and - admittedly - my) confusion.

First, let's go through the field theory. In particular, what its mathematics means physically.

Imagine a flow in a pipe. Assume we can measure, say, velocity at some points in the flow. To make that assumption visual, imagine we have a fish net. Now, imagine you spread your fish net on the floor. It is flat and even and is lying on the floor. Now put another fish net on top (making sure their nods match). Connect matching nods of the nets with threads (the length of a thread equals to the side of a cell in the net). Repeat it several times. Now, pull the layers apart. You've got a 3-D fish net. You can skip all that if it seems messy to you - the purpose is for you to imagine a finite 3-D net. Now assume we have some sort of tiny velocity measuring devices at each nod of our 3-D net. Those devices can, wirelessly, transmit their velocity readings to a computer. Now, put the 3-D net into the flow somehow. Do not look at the flow now. Turn around and look at the computer. On the computer's screen, you'll see velocity readings from every nod of the net. Remember that the net is staying still in the flow (assume you anchored it to the pipe). The flow flows through the net. Assume you put an incredibly tiny particle into the flow. The particle is neutrally buoyant. Its mass is tiny. It follows the flow exactly. Its velocity is the velocity of the flow at the point where it is currently situated. This particle passes by a nod number $n$ in the net. The nod number #n# detects its velocity $v_n$ and transmits it to the computer where you read it. The particle keeps moving and reaches the next nod: the nod number $n+1$. The nod number $n+1$ detects its velocity $v_{n+1}$ and transmits it to the computer where you read it. Velocities $v_n$ and $v_{n+1}$ can be different (take my word for it at this point). Now assume you put another particle which is the exact copy of the first particle. The flow is absolutely the same (don't think too much what that means just get it intuitively). Therefore, when the second particle passes by the nod number $n$, the nod measures the same velocity $v_n$ as for the first particle that had passed by it before. The second particle keeps moving, reaches the nod number $n+1$ and this nod measures exactly the same velocity $v_{n+1}$ as for the first particle that had passed by it before. You can add more particles, and - if the flow stays the same - you get the velocity magnitude $v_n$ for all of them at the nod number $n$ and the same velocity magnitude $v_{n+1}$ at the nod number $n+1$. And remember that I asked you to assume $v_n \neq v_{n+1}$. In other words, velocity of the flow can change from point to point and, simultaneously, stay the same at a given point. That is where the mathematics of the field theory meets physics of the field theory. In school physics course we talk about physical bodies. For instance, a body can be thrown at some angle to the horizon and we want to know its trajectory. That means that we are looking at the body as the whole thing which is moving in space. It is useful when our body is a rocket for example. Bodies can accelerate and decelerate as a whole (i.e., the entire body changes velocity - not only a portion of it). We had gotten used to working with whole physical bodies by the time we start learning fluid mechanics at university. That concept is so natural to us that we don't even think of it. And suddenly the professor starts talking about flow fields. We feel something's wrong but we don't understand what exactly. I can tell you what's wrong: we can't understand where a physical body is here that we can look at as it is moving through the space and define a trajectory for it. Unwittingly, we transfer our skills of working with whole bodies to fluids. And a mess occurs. And the reason for the mess is because we are not looking at flows as at whole bodies moving through space. Can you define a body for a fluid flow? If it is a piece of a very soft jelly that you playfully throw at your friend at a party, then you may define a body moving in space. But if you have a pump, a pipe, a tank and a constant flow of water in this system, then what is the body that is moving is space here? And even if we could define the body, what is the use of finding its trajectory? Its trajectory is the pipe. Here's the fundamental thing about fluid mechanics: we don't care about fluid trajectories. We are not interested in how fluid moves in space. We are not looking at it at all. Instead, we are looking at the readings from the nods of the net. That is why I told you before, that you put the net in the flow, turn around, stop looking at it and look only at the computer. That's an allusion on telling you that you are not interested in the flow movement. If we, a priory, know the trajectory of a fluid flow - which is just the pipe - then what we are interested in? We are interested in how our flow affects the pipe. That's very useful when you pick a pipe for the fuel in a rocket: the pipe should better not crack. That is why we are not looking at the flow. We are looking at the readings from the nods of the net: if the nods close to the pipe's wall detect high velocities, then the pipe can erode (I made a deliberate mistake here for the ease of explanation: in fact we are interested in velocity gradients not just velocities, which are zero at the wall by the way). And now we are ready to partially address your concern. When you say that flow is accelerating or decelerating in the tilted part of the pipe you, unwittingly, keep trying to implement the nice and visually-clear concept from the school physics course: you are trying to say that the fluid is a body which is moving in space. You have to stop looking at the fluid flow and look at the flow field given to you by the net. You are not following the particle in the flow as if you were considering the flow as a body. You are measuring velocities of different particles which keep passing by a point in space where the nod of the net is situated. Even if the fluid body is accelerating, its flow field is not, meaning that velocities at the points do not change, they only change between the points. I.e., if the flow stays the same while you are taking velocity measurements, then even in the tilted part of your pipe the nod number $n$ will always be giving you the same velocity $v_n$ and the nod number $n+1$ will always be giving you the same velocity $v_{n+1}$ for all the seeding particles that keep passing with the flow by the nods. The rule that your refer to in your question about diameter change when velocity changes was derived from the equations of fluid mechanics where were written using the field theory. And in the field theory, "velocity change" means that velocity at a point changes with time. If it doesn't change with time, then we say that the flow is steady, the velocity doesn't change and, therefore, diameter doesn't change. How can $v_n$ and $v_{n+1}$ change with time? That's difficult to explain, here's my best try. I asked you before to assume that "the flow is the same" and I asked you to get that intuitively. Now - intuitively again - assume that when you have pumping fluid constantly it means that the flow is the same. And likewise, if you were to pump fluid with a manual pump you get "not the same flow": on the power stroke you pump the fluid, then the fluid moves inertially while you're moving the piston back. Then you pump the fluid again on the power stroke and then, again, it moves inertially while you're moving the piston back. Inertial and power movement of the fluid are intuitively different. During the time of inertial movement you can have one velocity reading from the particles passing by the nod number $n$, during the time of power fluid movement you can get a different velocity reading from the nod number $n$. Thus, you have time dependent $v_n$.

Second, let's briefly go through the mathematics of continuity.

Recall your net. But now imagine the cells are infinitely small. You have two such nets: one for density measurements and one for velocity measurements. Put the both nets in the flow. Points (particles) of the fluid will be passing by the nods of the nets. In general case, both density and velocity at each point will be changing with time as well as they will be different between the nods of the nets at any given time moment. Remember: they do not belong to the same particles, instead they are just measurement for a magnitude of different particles. In other words, both velocity and density are nothing more than mathematical functions that change with time and space (space being the nods of the nets): $$\rho = \rho(x, y, z, t)$$ $$\vec v = \vec v(x, y, z, t)$$ The continuity equation is written for these fields (a field is nothing more than a function, like the functions that I wrote immediately above). Here it goes: $$\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \vec v) = 0$$ Here goes the crucial assumption: the density field is steady. Densities at all the points do not change with time. Densities only change from point to point (i.e., in space). That means that the density time derivative is zero and we have the following: $$\nabla \cdot (\rho \vec v) = 0$$ Integrate over all the coordinates: $$\int \int \int \nabla \cdot (\rho \vec v) dx dy dz = \int \int \int 0 dx dy dz$$ Use divergence theorem to convert the left hand side to a surface integral. Also remember that integral of zero is constant. $$\int \int (\rho \vec v) d \vec S = constant$$ Consider a very simple case, when $S$ is a cross section in your pipe, density doesn't change across this cross section, velocity always changes across the cross section but can be safely changed to some equivalent (average) velocity which is the same for all the points in the cross section. Then the previous integral reduces to the following: $$\rho v_{equivalent} \int \int dS = constant$$ $$\rho v_{equivalent} S = constant$$ The previous expression implies that the constant is the same for all the cross sections in your pipe at a given time. Therefore, you can pick any two cross sections at a given time and write down the following (but pay attention that velocity is time invariant now): $$\rho_1 v_{equivalent_1} S_1 = \rho_2 v_{equivalent_2} S_2$$ Let's say that fluid is water and is very reluctant to change its density. Therefore, density doesn't change at all. Which means $\rho_1 = \rho_2$. Let's also drop the $equivalent$ subscript. And let's use $A$ for cross section area. $$v_1 A_1 = v_2 A_2$$ That is a very simple form of continuity which you reference in your question. That is how it can be derived. Remember not to look at velocities in this formula as at velocities of a solid body which is accelerating or decelerating. Instead, velocity in that formula is the average magnitude of the field function $\vec v_{{normal}_{component}} = function(x, y, z=invar, t=invar)$.

Now consider your inclined part of the pipe. Its cross section is constant. Therefore for two cross sections at the top and at the bottom, velocities must be the same. That doesn't seem very intuitive. Let's see how it can happen.

First, let's pick a system of coordinates. Let it be a Cartisian (i.e., Descartes) system of coordinates. Let's say $x$ axis is directed along the central line of the inclined part of the pipe.

Let's write down the Navier-Stokes equation. Projection on the $x$ axis is enough (no need for the $y$ and $z$ projections). I'm going to use a bit simplified version.

$$\rho (\frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y} + w \frac{\partial u}{\partial z}) = - \frac{\partial P}{\partial x} + \rho \frac{g}{sin \alpha} + \mu (\frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2} + \frac{\partial^2u}{\partial z^2})$$

Now - just for the ease of explanation - assume a 2-D flow: make all $z$-derivatives equal to zero and $w=0$. Apart from that, since we want velocities at the top and bottom cross sections to be equal to $0$, all the $x$-derivatives are zero. We don't have velocity in the $y$ direction, therefore, $v=0$. We assume steady flow: the time derivative is zero. Therefore, we are left with the following equation: $$- \frac{\partial P}{\partial x} + \rho \frac{g}{sin \alpha} + \mu \frac{\partial^2u}{\partial y^2} = 0$$ You see: the pressure, the gravity and the friction terms must balance each other if we want to have the same velocities at the top and at the bottom cross sections.

By the way, this equation is used to derive the velocity profile in a circular tube. You need to solve it for $u$. In order to do that, you first have to express $P$ through $u$. You can do that using continuity.

If you want to, I can show you how it can be done.

  • $\begingroup$ How does this wall of text address the pressure OP isn't accounting for in their calculations? $\endgroup$
    – Kyle Kanos
    Aug 11, 2022 at 23:36
  • $\begingroup$ so does it have anything to do with the simple fact that if steady state flow is allowed to fully develop, since the pressure at the bottom of the downslope segment is larger than at the top, which comes at the expense of elevation head / potential energy, it will require the velocity to remain constant throughout? I had thought that before I even asked this question on this site but it too feels wrong to me. Is the higher pressure at the bottom keeping it from being accelerated by gravity? You have to pardon my ignorance. Most of my work with water involves Darcian flows. $\endgroup$
    – MattGeo
    Aug 11, 2022 at 23:48
  • $\begingroup$ the static pressure P is not the same as hydrostatic pressure though. Isn't the pressure P simply the pressure due to the fluid itself in the absence of gravity? When you lose potential energy going downhill in the pipe you gain static pressure, so this interchange allows velocity to remain constant. In the beginning I assumed this all was explained by Bernoulli's equation and then I started doubting it because of my acceleration confusion. $\endgroup$
    – MattGeo
    Aug 12, 2022 at 1:53
  • $\begingroup$ @MattGeo, yes, I'm sorry, my bad. Static pressure is not the same as the hydro static head. You're right. I'll delete my comment. Apart from that, I edited my response to show what terms are balanced to get the same velocities at the top and bottom cross sections. See at the bottom of my response. Let me know if you need more explanations about how I did it. $\endgroup$ Aug 12, 2022 at 2:21

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