Question

The Reynolds number for the flow of a fluid in a circular tube is specified at 2500.What will be the Reynolds number when the tube diameter is increased by 20% and the fluid velocity is decreased by 20% keeping fluid the same?​

Answer 1

Answer:

Nothing will occur because the size tube has increased the it doesn't matter

Explanation:

When tube increase it diameter fluid also speed of flow

Answer 2

Ques. The Reynolds number for the flow of a fluid in a circular tube is specified at 2500.What will be the Reynolds number when the tube diameter is increased by 20% and the fluid velocity is decreased by 40% keeping fluid the same?​

a. 1200

b. 1800

c. 3600

d. 200

Answer:

The correct answer is option b. $1800$.

Explanation:

Concept:

The Reynolds number is a dimensionless number used to classify fluid systems where viscosity plays a significant role in regulating fluid velocities or flow patterns.

The mathematical term for fluid motion is flow velocity, which is a vector field. The term "flow speed" refers to the entire length of the flow velocity. The vector field that gives fluids' velocities at specific times and locations is known as flow velocity.

Given:

The Reynolds count for the flow of a fluid  $= 2500$

The tube diameter is increased by $= 20$%

The fluid velocity is decreased by $=20$%

To find:

We have to find the required Reynolds number.

Solution:

$Re$ $=$ Reynolds number

$\rho$ $=$ Density of the fluid

$V$ $=$ Flow speed

$D$ $=$ Characteristic linear dimension

$\mu$ $=$ Dynamic viscosity of the fluid

Now, put the values in the formula,

$\begin{aligned}&R_{e}=\frac{\rho V D}{\mu} \text { (For pipe flow) } \\&R_{e}^{\prime}=\frac{\rho}{\mu} \times 1.2 D \times 0.6 \times V \\&R_{e}^{\prime}=\frac{\rho D V \times 1.2 \times 0.6}{\mu} \\&R_{e}^{\prime}=2500 \times 1.2 \times 0.6 \\&R_{e}^{\prime}=1800\end{aligned}$

#SPJ3