Colebrook equation for friction factor in turbulent flow is given by, f-0.5 = -4 loge [(ε/D) + (1.26/NRe √F). It reduces to Nikuradse equation for a value of (ϵ/D) equal to __________________?
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Answer:
Colebrook equation for friction factor in turbulent flow is given by, f-0.5 = -4 loge [(ε/D) + (1.26/NRe √F). It reduces to Nikuradse equation for a value of (ϵ/D) equal to 1.
