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 ([tex]Re_{D}[/tex]) is a dimensionless criterion use for flow regime of fluids, which is defined as:
[tex]Re_{D} = \frac{\rho \cdot v\cdot D}{\mu}[/tex] (Eq. 1)
Where:
[tex]\rho[/tex] - Density, measured in kilograms per cubic meter.
[tex]\mu[/tex] - Dynamic viscosity, measured in kilograms per meter-second.
[tex]v[/tex] - Average flow velocity, measured in meters per second.
[tex]D[/tex] - 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 [tex]Re_{D,gas} = Re_{D,w}[/tex]. That is:
[tex]\frac{\rho_{w}\cdot v_{w}\cdot D_{w}}{\mu_{w}} = \frac{\rho_{gas}\cdot v_{gas}\cdot D_{gas}}{\mu_{gas}}[/tex] (Eq. 2)
Where subindex [tex]w[/tex] is used for water and [tex]gas[/tex] for gasoline.
If we know that [tex]\rho_{gas} = 690\,\frac{kg}{m^{2}}[/tex], [tex]\mu_{gas} = 0.006\,\frac{kg}{m\cdot s}[/tex], [tex]v_{gas} = 0.5\,\frac{m}{s}[/tex], [tex]D_{gas} = 1\,m[/tex], [tex]\rho_{w} = 1000\,\frac{kg}{m^{3}}[/tex], [tex]\mu_{w} = 0.0018\,\frac{kg}{m\cdot s}[/tex] and [tex]D_{w} = 0.05\,m[/tex], then we get the following formula:
[tex]57500 = 27777.778\cdot v_{w}[/tex]
The fluid velocity for the prototype system is:
[tex]v_{w} = 2.07\,\frac{m}{y}[/tex]
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.
