Answer:
The friction factor is 0.303.
Explanation:
The flow velocity (
), measured in meters per second, is determined by the following expression:
(1)
Where:
- Flow rate, measured in cubic meters per second.
- Diameter, measured in meters.
If we know that
and
, then the flow velocity is:
![v = \frac{4\cdot \left(0.01\,\frac{m^{3}}{s} \right)}{\pi\cdot (0.05\,m)^{2}}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B4%5Ccdot%20%5Cleft%280.01%5C%2C%5Cfrac%7Bm%5E%7B3%7D%7D%7Bs%7D%20%5Cright%29%7D%7B%5Cpi%5Ccdot%20%280.05%5C%2Cm%29%5E%7B2%7D%7D)
![v \approx 5.093\,\frac{m}{s}](https://tex.z-dn.net/?f=v%20%5Capprox%205.093%5C%2C%5Cfrac%7Bm%7D%7Bs%7D)
The density and dinamic viscosity of the glycerin at 20 ºC are
and
, then the Reynolds number (
), dimensionless, which is used to define the flow regime of the fluid, is used:
(2)
If we know that
,
,
and
, then the Reynolds number is:
![Re = \frac{\left(1260\,\frac{kg}{m^{3}} \right)\cdot \left(5.093\,\frac{m}{s} \right)\cdot (0.05\,m)}{1.519 \frac{kg}{m\cdot s} }](https://tex.z-dn.net/?f=Re%20%3D%20%5Cfrac%7B%5Cleft%281260%5C%2C%5Cfrac%7Bkg%7D%7Bm%5E%7B3%7D%7D%20%5Cright%29%5Ccdot%20%5Cleft%285.093%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%5Cright%29%5Ccdot%20%280.05%5C%2Cm%29%7D%7B1.519%20%5Cfrac%7Bkg%7D%7Bm%5Ccdot%20s%7D%20%7D)
![Re = 211.230](https://tex.z-dn.net/?f=Re%20%3D%20211.230)
A pipeline is in turbulent flow when
, otherwise it is in laminar flow. Given that flow has a laminar regime, the friction factor (
), dimensionless, is determined by the following expression:
![f = \frac{64}{Re}](https://tex.z-dn.net/?f=f%20%3D%20%5Cfrac%7B64%7D%7BRe%7D)
If we get that
, then the friction factor is:
![f = \frac{64}{211.230}](https://tex.z-dn.net/?f=f%20%3D%20%5Cfrac%7B64%7D%7B211.230%7D)
![f = 0.303](https://tex.z-dn.net/?f=f%20%3D%200.303)
The friction factor is 0.303.