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Imagine a pipe which is bent to form a circular arc and a liquid is force through this pipe. How will you find the net force exerted by the fluid on pipe? Is the only force acting on fluid centripetal force or is there a tangential force too?

No drag forces or viscous forces

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  • $\begingroup$ It depends; is any assumption being made about drag forces between the fluid and the pipe? $\endgroup$ – joshphysics Dec 4 '13 at 18:21
  • $\begingroup$ Your question is rather vague, but it sounds like a simple control volume analysis problem. The final result should give you that the force acting on the pipe is equal and opposite to the change in momentum of the fluid and pressure force acting over the entrance and exits. Assuming incompressible, steady state flow: $\vec{F} = A(P_2 \vec{n}_{out}-P_1 \vec{n}_{in})+\rho \bar{U}^2(\vec{n}_{out}-\vec{n}_{in})$. To figure out $P_2$ you will need to know something about the flow in the pipe. If you assume you can ignore viscous forces then $P_2 = P_1$. $\endgroup$ – SimpleLikeAnEgg Dec 4 '13 at 18:47
  • $\begingroup$ In the absence of drag and viscosity, I would guess there is no tangential force, see D'Alembert's paradox. $\endgroup$ – Doru Constantin Dec 5 '13 at 21:40
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Fluid dynamics problems such as this are generally best approached by control volume analysis. Consideration of conservation of mass, momentum, energy, and sometimes angular momentum for an isolated control volume system generally provide an engineering answer. To figure out the force exerted on the pipe by the fluid it would seem appealing to isolate the fluid with a control volume. However, since the pipe is stationary, it is entirely equivalent to find the force acting on the pipe externally to keep it in place. Assuming steady flow within the pipe, conservation of mass for an isolated pipe-fluid control volume gives $$\dot{m}_{in}A_{in}=\dot{m}_{out}A_{out}$$ with $\dot{m}_{in}=\overline{\rho_{in} \vec{u}_{in}\cdot\vec{n}_{in}}$ and $\dot{m}_{out} = \overline{\rho_{out} \vec{u}_{out}\cdot\vec{n}_{out}}$ where $\vec{n}$ are surface unit normal vectors and $\vec{u}$ is the surface velocity vector and $A$ is the pipe cross section area. Conservation of momentum in vector form yields $$\Sigma F = \vec{f}_{pipe}+\vec{f}_{gravity}+\vec{f}_{other}+\overline{P}_{in}A_{in}\vec{n}_{in}+\overline{P}_{out}A_{out}\vec{n}_{out} = \dot{m}\overline{\vec{u}_{in}}A_{in}-\dot{m}\overline{\vec{u}_{out}}A_{out}$$ If you further assume that the only force acting on the pipe is to resist the change in momentum of the fluid and that the pressure drop from entrance to exit is negligible, then you will arrive at the approximation $$\vec{f}_{pipe}= \dot{m}\overline{\vec{u}_{in}}A_{in}-\dot{m}\overline{\vec{u}_{out}}A_{out}$$

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Find the net momentum of a cross-section of fluid entering the pipe. Write this as a vector, like $\vec{p_{in}} = \dot{m_1}(u_1\hat{i} + u_2\hat{j} + u_3\hat{k})$. Next, find the net momentum of the fluid leaving the pipe. Write this as $\vec{p_{out}} = \dot{m_2}(v_1\hat{i} + v_2\hat{j} + v_3\hat{k})$. Suppose the fluid element you're considering takes a time of $\Delta t$ to travel through the pipe

The net force being exerted on the liquid is the change in momentum per unit time.

$$ \vec{F} = \frac{\vec{p_{out}} - \vec{p_{in}}}{\Delta t} $$

By Newton's third law, the same force is exerted on the pipe. Remember, $\vec{F}$ and $\vec{p_{in}}$ and $\vec{p_{out}}$ are vectors

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  • $\begingroup$ "Suppose the fluid element you're considering takes a time of Δt to travel through the pipe." The time it takes to travel through the pipe is irrelevant. $\endgroup$ – SimpleLikeAnEgg Dec 5 '13 at 7:19

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