Chapter 5, Problem 5.1P

To determine

The volume flow rate in m³ /h which causes the transition.

Answer to Problem 5.1P

The volume flow rate is $0.778{\text{m}}^{\text{3}}/\text{h}.$

Explanation of Solution

Given:

Diameter of the circular tube, $d=9cm=0.09m$

Reynolds number $R{e}_{}=2300$.

Fluid: Kerosene

State: 20°C

Mass density of kerosene $\rho =804kg/m^3$

Concept Used:

Reynolds number is given by,

$R{e}_{}=\frac{\rho \cdot v\cdot d}{\mu }$

Where, $\rho$ is the mass density of the fluid in kg/m³.

$v$ is the velocity of flow in m/s.

$d$ is the diameter of the circular tube in m.

$\mu$ is the dynamic viscosity of the fluid in kg/m-s.

Calculation:

From the given data, we know that we can use the Reynolds number to find the velocity of flow from which we can find the volume flow rate.

For kerosene at 20°C, $\rho =804kg/m^3$ and $\mu =0.00192\text{ kg/m-s}$

Reynolds number is given by,

$R{e}_{}=\frac{\rho \cdot v\cdot d}{\mu }$

Substituting the known values in the Reynolds number formula, we get

$2300=\frac{804\cdot v\cdot 0.05}{0.00192}$

$v=0.11m/s$

Now, we substitute the value of $v=0.11m/s$ in the volume flow rate equation.

$Q=Av=\frac{\pi d^2}{4}v$

$\frac{\pi {\left(0.05\right)}^2}{4}\text{x0}\text{.11}$

$=\text{2}{\text{.1598 x 10}}^{-4}{\text{ m}}^3/s$

$Q\approx 0.778{\text{ m}}^3/h$

Thus, the volume flow rate is $0.778{\text{ m}}^3/h$

Conclusion:

The volume flow rate is $0.778{\text{ m}}^3/h$.