Self Cleasing Velocity Calculator:
Enter the values of Manning’s coefficient of roughness n, radius, R, channel bed slope B, specific gravity, GS and sediment size ds to determine the value of self-cleansing velocity, Vs.
Self Cleasing Velocity Formula:
Self-cleansing velocity is the minimum velocity of fluid flow required to transport sediments effectively without allowing them to deposit and build up.
Self-cleansing velocity, Vs(m/s) in metres per seconds is calculated as the reciprocal of the Manning’s coefficient of roughness n, multiplied by the sixth root of the hydraulic radius, R(m) in metres and the square root of the product of the channel bed slope B, the difference in specific gravity between the sediment and the fluid (GS – 1), and the sediment size ds(mm) in millimetres.
Self-cleansing velocity, Vs(m/s) = (1/n) * R1/6 * √( B * (GS – 1) * ds)
Vs(m/s) = self-cleansing velocity in metres per seconds, m/s.
n = Manning’s coefficient of roughness.
R(m) = radiusin metres, m.
B = channel bed slope.
GS = specific gravity.
ds(mm) = sediment size in millimetres, mm.
Self Cleasing Velocity Calculation:
- Manning’s coefficient of roughness is 0.013, the hydraulic radius is 0.5, the channel bed slope is 0.002, the specific gravity of the sediment is 2.65, and the diameter of the sediment particles is 0.1mm, these parameters collectively determine the hydraulic characteristics necessary for maintaining a self-cleansing flow in a channel or pipe to prevent sediment buildup.
Given: n = 0.013, R(m) = 0.5m, B = 0.002, GS = 2.65, ds(mm) = 0.1mm.
Self-cleansing velocity, Vs(m/s) = (1/n) * R1/6 * √( B * (GS – 1) * ds)
Vs(m/s) = (1/0.013) * 0.51/6 * √(0.002 * (2.65 – 1) * 0.1)
Vs(m/s) = 76.92 * 0.895 * 0.01816
Vs(m/s) = 1.244m/s.
- Self-cleansing velocity required is 1.2m/s, the hydraulic radius of the channel is 0.4m, the channel bed slope is 0.0015, the specific gravity of the sediment is 2.6, and the diameter of the sediment particles is 0.15mm, these parameters are crucial for ensuring that the flow velocity is sufficient to transport sediment particles, thereby preventing their deposition and maintaining the channel’s efficiency.
Given: Vs(m/s) = 1.2m/s, R(m) = 0.4m, B = 0.0015, GS = 2.6, ds(mm) = 0.15mm.
R1/6 = 0.41/6 = 0.40.1667 = 0.831.
√( B * (GS – 1) * ds) = 0.0015 * (2.6 – 1) * 0.15 = 0.01899.
Self-cleansing velocity, Vs(m/s) = (1/n) * R1/6 * √( B * (GS – 1) * ds)
n = √( B * (GS – 1) * ds) * R1/6 / Vs(m/s)
n = 0.01899 * 0.831 / 1.2
n = 0.01315.