Water is pumped between two large open reservoirs through 1.5 km of smooth pipe. The surfaces in the two reservoirs are at the same elevation. When the pump adds 20kW to the water, the flowrate is 1 m3/s. If minor losses are negligible. Determine the pipe diameter.

0.92 m

Explanation:

Assuming the head gain is equal to the head loss, we have

flV²/2gD = W'(s) /γΔV

We can get the velocity using the continuity equation

V = ΔV/A,

V = 4 * 1 / πD²

V = 1.273/D², this value for B would be inserted into the primary equation I stated above

[f * 1500 * (1.273/D²)²]/ [2 * 9.8 * D] = (20000/9810)

[f * 123.9325 / D^5] = 2.039, making D the subject of formula, we have

(f * 123.9325) / 2.039 = D^5

Now we find the 5th root of both sides.

D = 2.274.f^(1/5)

R(E) = ρVD/μ

R(E) = (1.136 * 10^6)/D

Next we use the Colebrook Formula,.

See the attachment for the calculation

oil flow is a highly convective flow compared to water flow at normal temperature and pressure.

answer:

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