Water is flowing through a cylindrical pipe of diameter 1.5 m. The coefficient of viscosity of water is 80 `Ns//m^2` and the Reynold's number is 1500. What is the maximum velocity of water, to avoid a turbulent flow?

Options

A. 60 m/s
B. 80 m/s
C. 100 m/s
D. 40 m/s

Explanation

To find the maximum velocity of water in a cylindrical pipe to avoid turbulent flow, we can use the formula for Reynolds number (Re):

\[ Re = \frac{\rho V D}{\mu} \]

Where:
- \( \rho \) = density of the fluid (for water, approximately \( 1000 \, kg/m^3 \))
- \( V \) = velocity of the fluid (what we need to find)
- \( D \) = diameter of the pipe (given as \( 1.5 \, m \))
- \( \mu \) = dynamic viscosity of the fluid (given as \( 80 \, Ns/m^2 \))

Given that the Reynolds number is \( 1500 \), we can rearrange the formula to solve for \( V \):

1. Substitute the known values into the Reynolds number formula:
\[ 1500 = \frac{(1000 \, kg/m^3) \cdot V \cdot (1.5 \, m)}{80 \, Ns/m^2} \]

2. Rearranging the equation to solve for \( V \):
\[ V = \frac{1500 \cdot 80}{1000 \cdot 1.5} \]

3. Calculate the numerator:
\[ 1500 \cdot 80 = 120000 \]

4. Calculate the denominator:
\[ 1000 \cdot 1.5 = 1500 \]

5. Now, divide the numerator by the denominator:
\[ V = \frac{120000}{1500} = 80 \, m/s \]

Thus, the maximum velocity of water to avoid turbulent flow is \( 80 \, m/s \).