To find the minimum diameter of the pipeline, we can use the Bernoulli's equation, which relates the pressure, velocity, and elevation of a fluid at two points along a streamline:
P1 + (1/2)ρv1^2 + ρgh1 = P2 + (1/2)ρv2^2 + ρgh2
where P is the pressure, ρ is the density, v is the velocity, g is the acceleration due to gravity, and h is the height.
In this case, we can assume that the velocity of the water at the outlet of the pump is negligible, so v2 = 0. We also know the density of the water (1000 kg/m^3), the height difference (7.6 m), the length of the pipeline (183 m), and the pressures at the outlet of the pump (550 kPa) and at the rotating jets (410 kPa). We want to find the minimum diameter of the pipeline, which will affect the velocity of the water and therefore the pressure drop along the pipeline.
To simplify the equation, we can assume that the pipeline is horizontal (h1 = h2) and that the pressure drop due to friction along the length of the pipeline is negligible compared to the pressure drop at the rotating jets. This allows us to solve for the velocity of the water at the outlet of the pump:
P1 + (1/2)ρv1^2 = P2
v1 = sqrt[(P2 - P1)/(1/2ρ)]
v1 = sqrt[(410000 Pa - 550000 Pa)/(1/2 * 1000 kg/m^3)]
v1 = 14.14 m/s
Now we can use the equation for the volumetric flow rate Q, which relates the velocity of the water to the cross-sectional area of the pipeline A and the volumetric flow rate Q:
Q = Av1
We want to find the minimum diameter of the pipeline, so we can use the equation for the cross-sectional area of a circular pipe:
A = (πd^2)/4
Q = (πd^2)/4 * v1
d = sqrt[(4Q)/(πv1)]
d = sqrt[(4 * 51 m^3/hr)/(π * 14.14 m/s)]
d = 0.163 m or 163 mm
Therefore, the minimum diameter of the pipeline to meet the pumping requirements is 163 mm.