A6-inch diameter smooth plastic pipe 3100 ft in length delivers 2.2 cfs of water between two reservoirs. water viscosity is 11.11 x 10-6 ft2/s. what is the elevation difference (ft) in water surface between the reservoirs

h = 6.39 m

Explanation:

radius = 6/2 =3in= 3/12 in/ft = 0.25 ft

apply continuity equation to evaluate the velocity

flow rate = area * velocity

        2.2 = π(0.25^2) * V  

⇒        V = 11.2 ft/s

*assuming there is no applied pressure difference along the pipe

applying Energy conservation law

initial potential energy = final kinetic energy

                              pgh = (1/2)*pv^2

where p  = density

            v = velocity  

plugging in the values:

p cancels out on both the sides

⇒ 9.81 * h = (11.2^2)/2

              h = 6.39 m

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r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.