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Assume $50\text{ gpm}$ at inlet of a pipe with pressure drop as $10\text{ psig}$. Will the flow at outlet still be same $(50 \text{ gpm})$?

Will a larger pressure drop will decrease the flow rate at outlet?

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    $\begingroup$ If the flow rate at the entrance is different than the flow rate at the exit, what is happening to the missing water? $\endgroup$ Commented Jul 10 at 15:27
  • $\begingroup$ @RC_23 It could be compressed or rarified, which would change the volumetric flow rate (though not the mass flow rate.) And while water can be treated as incompressible in many circumstances, the OP never said anything about the fluid being water. ;-) $\endgroup$ Commented Jul 10 at 18:52

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In steady state, mass must be flowing into the pipe at the same rate that mass is flowing out of the pipe; which means that $$ \rho_\text{in} Q_\text{in} = \rho_\text{out} Q_\text{out} \quad \Rightarrow \quad \frac{\rho_\text{in}}{\rho_\text{out}} = \frac{Q_\text{out}}{Q_\text{in}} $$ where $\rho_i$ stands for the density at each end of the pipe and $Q_i$ stands for the volumetric flow rate.

So the only way that $Q_\text{out}$ can differ significantly from $Q_\text{in}$ in the steady state is if the density of the fluid changes significantly. Density is generally an increasing function of pressure, so a lower outlet pressure will generally correspond to a higher outlet flow rate.

How much of a difference this makes, however, depends on the fluid you're dealing with. For liquid water, the density does not change significantly unless the pressure difference is many times greater than atmospheric pressure, and so we have $Q_\text{in} \approx Q_\text{out}$ to an excellent approximation. However, for an ideal gas at a fixed temperature, the density is proportional to the (absolute) pressure. If we are working at absolute pressures near atmospheric pressure, then the pressure difference would be a significant fraction of the absolute pressures involved, and we would expect the flow rate to increase noticeably.

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