Question

If the Darcey's friction factor is 0.008 and the acceleration due to gravity is 10 m/s$^2$, then chezey's constant is

• A. 120
• B. 100
• C. 80
• D. 60

Hint:

Remember the relationship $C = \sqrt{8g/f}$.
Also, be aware that there are two friction factors in use:
- Darcy friction factor ($f$): Used in the Darcy-Weisbach equation. The value is typically 0.01-0.05.
- Fanning friction factor ($f_F$): Used in some chemical engineering contexts. $f_F = f/4$.
Make sure you are using the correct factor ($f$) in the formula for Chezy's constant.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
f=0.008, g=10 m/s². Find Chezy's C.

Step 2: Key Relationship (Alternate):
C = √(8g/f). Links Darcy-Weisbach and Chezy equations.

Step 3: Detailed Explanation:
C = √(8×10/0.008) = √(80/0.008) = √10000 = 100. Manning's n relates: C = R^(1/6)/n. Higher C means smoother channel.

Step 4: Final Answer:
100.