Consider a constant-diameter L-shaped pipe with an entrance and an exit. If water is being pumped into the pipe at a constant rate, the flow rate of the water at the exit must equal that at the entrance, correct? In addition, there must also be a loss of kinetic energy of the water due to friction loss and the bend of the pipe, also correct? If both of those statements are true, then the only way I can think of that the water could be losing kinetic energy is through a decrease in turbulence so I am wondering if it can be said that the turbulence of a fluid flowing through a pipe decreases over the distance of the pipe. If the flow through the pipe is completely laminar, is it possible for the fluid to lose kinetic energy?
No, your reasoning is incorrect. There are two forms of energy in this flow, with kinetic energy of the fluid particles being just one of them. The other one is pressure. In your example loss of energy occurs mostly through the pressure drop along the pipe. In the laminar flow case (kinetic) energy can be lost through viscous effects (causing friction), or kinetic energy can be converted into pressure: For inviscid flow, if the cross section of the pipe increases, the fluid velocity will decrease, accompanied by a corresponding increase in pressure, as described by the Bernoulli equation.
If the turbulent flow is statistically steady, then you must be putting in energy (using pump, say) at the same rate as the flow is losing it. Therefore turbulence does not necessarily decay along the pipe's length; it can exist indefinitely so far as there is sufficient input of energy (at the large scales). What the turbulence then does is that it transfers energy input at large scales to small scale motion where it is dissipated away by viscous effects.
Laminar flow also loses energy due to viscous dissipation, but then again for flow to occur at all, there must be input of energy from an external agency such as a pump to make up for that loss.