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:
The question asks to calculate Chezy's constant ($C$) given the Darcy-Weisbach friction factor ($f$).

Step 2: Key Formula or Approach:
The Darcy-Weisbach friction factor ($f$) and Chezy's constant ($C$) are two different coefficients used to quantify friction in pipe flow and open channel flow. They are related by the following equation:
\[ C = \sqrt{\frac{8g}{f}} \] where:
$C$ = Chezy's constant
$g$ = acceleration due to gravity
$f$ = Darcy-Weisbach friction factor

Step 3: Detailed Explanation:
We are given:
- Friction factor ($f$) = 0.008
- Acceleration due to gravity ($g$) = 10 m/s$^2$
Substitute these values into the formula:
\[ C = \sqrt{\frac{8 \times 10}{0.008}} \] \[ C = \sqrt{\frac{80}{0.008}} \] \[ C = \sqrt{\frac{80}{8/1000}} = \sqrt{10 \times 1000} = \sqrt{10000} \] \[ C = 100 \] The units of Chezy's constant are m$^{1/2}$/s.

Step 4: Final Answer:
Chezy's constant is 100.