Using the Swamee-Jain formula to estimate the friction factor f for Re = 2.0×10^5 and ε/D = 4.5×10^-4, which value is most appropriate?

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Multiple Choice

Using the Swamee-Jain formula to estimate the friction factor f for Re = 2.0×10^5 and ε/D = 4.5×10^-4, which value is most appropriate?

Explanation:
The Swamee-Jain relation gives you f directly from Re and ε/D without iteration: f = 0.25 / [log10( (ε/D)/3.7 + 5.74/Re^0.9 )]^2. Plugging in the values: (ε/D)/3.7 = 0.00045 / 3.7 ≈ 1.216×10^-4. For Re = 2×10^5, Re^0.9 ≈ 5.9×10^4, so 5.74/Re^0.9 ≈ 9.75×10^-5. The sum inside the log is 1.216×10^-4 + 9.75×10^-5 ≈ 2.19×10^-4. Taking the log: log10(2.19×10^-4) ≈ -3.66. Squaring gives ≈ 13.4. Therefore f ≈ 0.25 / 13.4 ≈ 0.0187. So the friction factor is about 0.0187, which is closest to 0.018.

The Swamee-Jain relation gives you f directly from Re and ε/D without iteration: f = 0.25 / [log10( (ε/D)/3.7 + 5.74/Re^0.9 )]^2.

Plugging in the values: (ε/D)/3.7 = 0.00045 / 3.7 ≈ 1.216×10^-4. For Re = 2×10^5, Re^0.9 ≈ 5.9×10^4, so 5.74/Re^0.9 ≈ 9.75×10^-5. The sum inside the log is 1.216×10^-4 + 9.75×10^-5 ≈ 2.19×10^-4.

Taking the log: log10(2.19×10^-4) ≈ -3.66. Squaring gives ≈ 13.4. Therefore f ≈ 0.25 / 13.4 ≈ 0.0187.

So the friction factor is about 0.0187, which is closest to 0.018.

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