Question

Your company is planning to build a pipeline to transport gasoline from the refinery to a field of storage tanks. The parameters for the prototype system are a pipe diameter of 1 m, with a flow velocity of 0.5 m/s at 25°C. The model system will use water at STP with a geometric scaling factor of 1 : 20. What fluid velocity is required in the model system to guarantee kinematic similarity in the form of equal Reynolds numbers?

Answer 1

Answer:

The model system will need water flowing at a velocity of 2.07 meters per second to guarantee kinematic similarity in the form of equal Reynolds numbers.

Explanation:

The Reynolds number ($Re_{D}$) is a dimensionless criterion use for flow regime of fluids, which is defined as:

$Re_{D} = \frac{\rho \cdot v\cdot D}{\mu}$ (Eq. 1)

Where:

$\rho$ - Density, measured in kilograms per cubic meter.

$\mu$ - Dynamic viscosity, measured in kilograms per meter-second.

$v$ - Average flow velocity, measured in meters per second.

$D$ - Pipe diameter, measured in meters.

We need to find the equivalent velocity of water used in the prototype system. In this case, we assume that $Re_{D,gas} = Re_{D,w}$. That is:

$\frac{\rho_{w}\cdot v_{w}\cdot D_{w}}{\mu_{w}} = \frac{\rho_{gas}\cdot v_{gas}\cdot D_{gas}}{\mu_{gas}}$ (Eq. 2)

Where subindex $w$ is used for water and $gas$ for gasoline.

If we know that $\rho_{gas} = 690\,\frac{kg}{m^{2}}$, $\mu_{gas} = 0.006\,\frac{kg}{m\cdot s}$, $v_{gas} = 0.5\,\frac{m}{s}$, $D_{gas} = 1\,m$, $\rho_{w} = 1000\,\frac{kg}{m^{3}}$, $\mu_{w} = 0.0018\,\frac{kg}{m\cdot s}$ and $D_{w} = 0.05\,m$, then we get the following formula:

$57500 = 27777.778\cdot v_{w}$

The fluid velocity for the prototype system is:

$v_{w} = 2.07\,\frac{m}{y}$

The model system will need water flowing at a velocity of 2.07 meters per second to guarantee kinematic similarity in the form of equal Reynolds numbers.