Why do we need to do work to maintain motion of a fluid?

Question

On page 355 of principles of physics by Resnick walker and Halliday, the author writes:

Work must also be done on the system (at the input end) to push the entering fluid into the tube and by the system (at the output end) to push forward the fluid that is located ahead of the emerging fluid. In general , the work done by a force of magnitude $textit{F}$, acting on a fluid sample contained in a tube of area $textit{A}$ to move the fluid through a distance $Delta x$, is
$$ F Delta x = pA Delta x = p(A Delta x) = p Delta V $$

According to newton's first law, an object in motion remains in motion unless acted by an unbalanced external force .So, if we take the bulk of fluid as an object, then it would continue motion unless forces act on it. Then why is that we always have a pressure difference at ends of the tube that we take? As in, the paragraph I quoted suggests that we need to do some work to push fluid whether stationary or moving over a displacement.
As a side note, I recall in thermodynamics that we do $ int P dV$ to attain work, however this is an expansion work. So, would a correct interpretation of the formula be that work is done for expanding the fluids volume? Precisely speaking, as the fluid flows, it is expanding it's own volume (in a way)

Gert · Accepted Answer

Work must also be done on the system (at the input end) to push the
entering fluid into the tube and by the system (at the output end) to
push forward the fluid that is located ahead of the emerging fluid.

This pressure can be estimated, or at least understood, by means of the Bernoulli Principle (so for an ideal, inviscid fluid).

$$p_1+frac12 rho v_1^2+rho gz_1=p_2+frac12 rho v_2^2+rho gz_2$$
Assume $v_1approx 0$ (or at least $v_2 gg v_1$) and the potential energy changes negligible, then the above equation reduces to:
$$Delta papprox -frac12 rho v_2^2$$
where $v_2$ is calculated from the volumetric flow rate. $Delta p$ is negative because it is a pressure drop.
So $Delta p$ is required to impart kinetic energy to the fluid by the pump to keep fluid entering the pipe.

As a side note, I recall in thermodynamics that we do ∫PdV to attain
work, however this is an expansion work. So, would a correct
interpretation of the formula be that work is done for expanding the
fluids volume?

The fluid is considered incompressible ($frac{partial V}{partial p}=0$) so there's no compression or expansion going on ($Delta V=0$)

talrefae · Answer

The answer is ultimately due to viscosity. Viscosity serves as an energy "sink" – it constantly dissipates energy out of the system. Viscosity is due to the friction between two layers of fluids moving at different speeds and hence "dragging" each other. With ideal fluids (ones with no viscosity), kinetic energy is conserved, assuming no external forces.
The mathematical justification for this is a little involved, but I'll leave it here for anyone who's interested. Consider the Navier-Stokes equation:
begin{equation}
frac{partial mathbf{u}}{partial t}+(mathbf{u} cdot nabla) mathbf{u}=-nabla(p / rho)+v nabla^{2} mathbf{u}.
end{equation}
Dot both sides of this equation with the field velocity $mathbf{u}$. With some manipulation, this is
begin{equation}
frac{partial}{partial t}left[frac{1}{2} mathbf{u}^{2}right]=-nabla cdotleft[left(frac{1}{2} mathbf{u}^{2}+p / rhoright) mathbf{u}right]+frac{partial}{partial x_{j}}left[tau_{i j} u_{i} / rhoright]-2 v S_{i j} S_{i j}
end{equation}
where
begin{equation}
tau_{i j}=rho nuleft(frac{partial u_{i}}{partial x_{j}}+frac{partial u_{j}}{partial x_{j}}right) = 2 rho nu S_{ij}.
end{equation}
Take the integral of both sides so that we get the time rate of change of the total kinetic energy in the system on the LHS. Using the divergence theorem, and the assumption that $mathbf{u}$ dies off either at some boundary or at infinity, we are left with:
begin{equation}
frac{mathrm{d}}{mathrm{d} t} int frac{1}{2} mathbf{u}^{2} mathrm{d} V=-int 2 nu S_{i j} S_{i j} mathrm{d} Vequiv-int varepsilon mathrm{d} V
end{equation}
where $varepsilon = 2 nu S_{ij} S_{ij} geq 0$. Thus, kinetic energy is continuously being lost out due to viscosity, and work is needed to keep a fluid moving.

niels nielsen · Answer

The question is a bit unclear, but here is my response. I will delete if it's not on-topic.
Assume no losses due to friction. Then note that every small volume of fluid that we wish to push into the pipe must be given a certain flow velocity, after which that fluid element will coast its way down the length of the pipe with that velocity, and leave the end of the pipe with that velocity.
Therefore, the pump at the head end of the pipe is continuously performing work on the fluid it is pushing into the pipe, to accelerate it up from zero to the flow velocity of the rest of the fluid in the pipe. That acceleration work is present even if there is no viscuous dissipation occurring inside the pipe.

Why do we need to do work to maintain motion of a fluid?

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