Pipeflow Example Type 3 Application (continued)

f = π²gD⁵ / 8Q² Here there are two unknowns, f and D.

In general f depends on the Reynolds number, R_ey and the relative roughness, ε/D.

Now R_ey = VD/ν = (Q/A)(D/ν) = 4Q/πDν

Now Q = 0.5 ft³/sec which gives f = π²gD⁵ / 2 (1)

and R_ey = 2/πDν

From tables for water, ν = 1.21 x 10^-5 ft²/sec So R_ey = 52613/D (2)

Here you need an iterative scheme using the Moody Chart to obtain a solution for D.

Strategy: Start by guessing a value for D, calculate R_ey from (2) then determine value of

friction factor, f, from the Moody Chart. Compare this value of f with the value obtained

from (1). Once these values are sufficiently close, the solution is achieved.

First guess pick D = 0.1 ft. then R_ey = 526130 and for a “smooth pipe” f = 0.009.

From (1) f = 0.0015 Not close enough to f = 0.009.

Next pick D = 0.15, then R_ey = 350750 then for a “smooth pipe” f = 0.014

And from (1) f = 0.012 which is closer. Next pick D = 0.155

R_ey = 339400 then for a “smooth pipe” f = 0.0142

and from (1) f = 0.0142 with the final result that D = 0.155 ft.

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