Question

The critical velocity of water flowing through 4 cm internal diameter pipe is _____ m/s​

Answer 1

Answer:

water is flowing in a pipe of diameter 4cm with a velocity3m//s. the water then enters into a tube of diameter 2 cm. the velocity of water in a another pipe.

Answer 2

To find:

The critical velocity of water flowing through 4 cm internal diameter pipe.

Calculation:

Critical Velocity a flowing water is the maximum velocity up to which the fluid dynamics remains streamlined. After that Velocity , the fluid flow becomes turbulent.

For that , Reynolds Number is 2000.

$\eta$ for water = 8.9 × 10^(-4) Pa s.

$\boxed{ \rm{R_{e} = \dfrac{ \rho \times v \times d}{ \eta} }}$

Putting available values in SI unit:

$= > \rm{R_{e} = \dfrac{ 1000 \times v \times (\frac{4}{100}) }{ 8.9 \times {10}^{ - 4} } }$

$= > \rm{2000= \dfrac{ 1000 \times v \times (\frac{4}{100}) }{ 8.9 \times {10}^{ - 4} } }$

$= > \rm{2= \dfrac{ v \times (\frac{4}{100}) }{ 8.9 \times {10}^{ - 4} } }$

$= > \rm{2= \dfrac{ v \times 4 }{ 8.9 \times {10}^{ - 2} } }$

$= > \rm{1= \dfrac{ v \times 2 }{ 8.9 \times {10}^{ - 2} } }$

$= > \rm{v = \dfrac{8.9}{2} \times {10}^{ - 2} }$

$= > \rm{v = 4.45 \times {10}^{ - 2} \: m {s}^{ - 1} }$

$= > \rm{v = 4.45 \: cm \: {s}^{ - 1} }$

So, final answer is:

$\boxed{\bf{v = 4.45 \: cm \: {s}^{ - 1} }}$