Question

g A smooth pipeline with a diameter of 5 cm carries glycerin at 20 degrees Celsius. The flow rate in the pipe is 0.01 m3/s. What is the friction factor

Answer 1

The friction factor is 0.303.

Step-by-step explanation:

The flow velocity (
`v`), measured in meters per second, is determined by the following expression:

`v = (4\cdot \dot V)/(\pi \cdot D^(2))` (1)

Where:

`\dot V` - Flow rate, measured in cubic meters per second.

`D` - Diameter, measured in meters.

If we know that
`\dot V = 0.01\,(m^(3))/(s)` and
`D = 0.05\,m`, then the flow velocity is:

`v = (4\cdot \left(0.01\,(m^(3))/(s) \right))/(\pi\cdot (0.05\,m)^(2))`

`v \approx 5.093\,(m)/(s)`

The density and dinamic viscosity of the glycerin at 20 ºC are
`\rho = 1260\,(kg)/(m^(3))` and
`\mu = 1.5\,(kg)/(m\cdot s)`, then the Reynolds number (
`Re`), dimensionless, which is used to define the flow regime of the fluid, is used:

`Re = (\rho\cdot v \cdot D)/(\mu)` (2)

If we know that
`\rho = 1260\,(kg)/(m^(3))`,
`\mu = 1.519\,(kg)/(m\cdot s)`,
`v \approx 5.093\,(m)/(s)` and
`D = 0.05\,m`, then the Reynolds number is:

`Re = (\left(1260\,(kg)/(m^(3)) \right)\cdot \left(5.093\,(m)/(s) \right)\cdot (0.05\,m))/(1.519 (kg)/(m\cdot s) )`

`Re = 211.230`

A pipeline is in turbulent flow when
`Re > 4000`, otherwise it is in laminar flow. Given that flow has a laminar regime, the friction factor (
`f`), dimensionless, is determined by the following expression:

`f = (64)/(Re)`

If we get that
`Re = 211.230`, then the friction factor is:

`f = (64)/(211.230)`

`f = 0.303`

The friction factor is 0.303.