# Why do fluids not accelerate?

A fluid flowing in a horizontal pipe must be flowing at a constant velocity because of the conservation of mass.

However, considering how there would be a pressure and hence force acting behind the fluid, for it to have a constant velocity, there must be an equal force slowing it down (depicted as $$F?$$).

I can't see a force that would be as big as the driving force. Can someone explain to me what this force is and how it's created?

• I feel there is context being omitted. For example, the weight of the water in the tank is decreasing so the force is decreasing. Where did you get the idea that it is constant velocity? Commented Aug 28, 2020 at 1:15
• @DKNguyen i think it would be reasonable to pretend maybe there's a tap that keeps the tank topped up such that the pressure is constant. It's just a thought experiment Commented Aug 28, 2020 at 1:49
• The force you are missing is viscous drag. Commented Aug 28, 2020 at 1:51
• Maybe I can see where you went wrong. The velocity is constant over space i.e. throughout the small pipe in your diagram due to the continuity equation but it is not constant over time. You can see this intuitively that when water in tank becomes less, the water velocity will become less as can be seen in leaking tank. When the tank is full and leaking, water forms an upside down parabola and gradually shrinks. It's not too much of a task if you want to prove it using the bernoulli equation. Commented Aug 28, 2020 at 4:01
• The title is missleading. The fluids can and do accelerate. The case discussed may not involve an accelerated fluid but this is no reason for the title which implies a (non existent) general property of fluids.
– nasu
Commented Aug 29, 2021 at 22:34

The fluid is accelerating. The continuity equation simply states that at any instant $$A_1v_1=A_2v_2$$ where $$A_1$$ and $$A_2$$ are the cross-sectional areas of the upper pipe and lower pipe respectively, with $$v_1$$ $$v_2$$ being the fluid velocities in the same. The potential energy of the fluid stored in the upper pipe is being converted to kinetic energy of the fluid flow in the bottom pipe. When the fluid in the upper pipe is at it's highest point, $$v_2$$ will be the greatest, and gradually this velocity decreases as the fluid height in the upper pipe decreases.

• Sorry, I think we might not be on the same page. I redrew my diagram to better reflect what I meant. Commented Aug 28, 2020 at 1:53
• The velocity (as a function of time) is not constant. There is an accelerating force. ie., Gravity. So when you say "for it to have a constant velocity" you are mistaken. Commented Aug 28, 2020 at 3:01
• @Drjh The OP is just talking about the fluid in the horizontal pipe, not comparing between the tank and the pipe. Commented Aug 28, 2020 at 12:23

The thing missing from your picture is the steep pressure gradients (and associated acceleration) at the inlet of the pipe. Far from the pipe inlet at the bottom of the tank, the pressure will be approximately hydrostatic; near the pipe inlet, it is anything but. The same thing can be observed at the drain of an emptying bathtub. Once in the pipe, if the flow in the pipe is inviscid and the pipe is horizontal, then it will be at atmospheric pressure everywhere in the pipe (ignoring some subtleties like the finite pipe height in the vertical direction) after the flow becomes parallel, and the flow will not accelerate further.

If there is wall friction in the pipe (which there will be in real pipes), then there will be a force resisting the flow, the result of which is a pressure gradient along the length of the pipe (a force is required to overcome the wall friction). The qualitative picture isn't changed by wall friction however.

Once into the pipe, there is no net acceleration (before it leaves the pipe), $$ma=0$$. But you have significant viscous sheering. This creates resistance to flow. Similar to drag. Viscous friction.

The only force driving flow from point A in the pipe to nearby point B in the pipe is the pressure gradient.

$$F= A ~\Delta P = A ~(P_A-P_B)$$

For lower viscosity that will be a lower pressure gradient. That pressure drop (force) increases with length and viscosity and velocity, and decreases with diameter (for a given volumetric flow $$Q$$ it does. I.e. yes it’s higher at the same $$V$$ and higher $$d$$, but lower at the same $$Q$$ and higher $$d$$ - point being a big pipe resists less for same volumetric flow than a little pipe).

But no, if it’s not going real fast, you don’t need a lot of pressure-force to drive flow. Depends on factors mentioned.

Yes you have basically constant velocity once you get into the pipe.

The pressure difference between bottom of tank (pipe inlet) and the atmosphere (pipe outlet) will drive flow fast enough that viscous-drag force equals pressure-difference force:

$$A~(P_i-P_f) = F_{\mu}$$

But $$F_{\mu}$$ is a function of flow rate. The flow rate will quickly settle at the point where that equation is true. The acceleration $$ma$$ is not a big part of it, and is only at the beginning.