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How to Calculate Head Loss for a Rapid Sand Filter

A rapid sand filter typically employs three or more layers containing different types of filter materials, which helps filter out multiple types of contaminants. This type of filter is "rapid" because water can flow through it up to 40 times faster than through a slow sand filter. The larger velocity results in larger head loss as well -- head loss measures the decrease in pressure as water flows through a system. You must take head loss into account when designing the filter bed for a rapid sand filter.

Instructions

- 1
  Multiply the water's velocity times the density of water and divide the result by the viscosity of the water. For room temperature water -- about 20 degrees Celsius or 70 degrees Fahrenheit -- use 998.2 kg/m^3 for the density and 0.001 N-s/m^2 for the viscosity. Look the density and viscosity up in density and viscosity charts for other temperatures. With a velocity of 0.005 m/s at room temperature, the calculation is (0.005 m/s)*(998.2 kg/m^3)/(0.001 N-s/m^2) = 4,991/m.
- 2
  Multiply the particle shape factor times the grain diameter in meters for each layer of the rapid sand filter. Use 1 for the particle shape factor for spherical sand, 0.73 for angular sand or crushed coal, 0.89 for worn sand, 0.65 for crushed glass and 0.28 for mica flakes. If the filter has a layer of crushed coal with a particle diameter of 0.0005m, a layer of worn sand with a particle diameter of 0.0008 m and layer of spherical sand with a particle diameter of 0.005 m, for example, the respective results are 0.73*(0.0005m) = 0.000365m, 0.89*(0.0008m) = 0.000712m, and 1*(0.005m) = 0.005 m.
- 3
  Multiply the particle shape and grain diameter values for each layer of the filter by the result of step 1 to find the Reynold's numbers for each layer. For example, the Reynold's number for the crushed coal layer would be (4,991/m)*(0.000365m) = 1.82. The Reynold's numbers for the worn and spherical sand layers would be 3.55 and 24.96, respectively.
- 4
  Subtract the porosity of each filter layer from one, divide by the Reynold's number for that layer, multiply by 150 and add 1.75 to get the friction factor for each layer. If the crushed coal had a porosity of 0.45, for example, the friction factor for that layer would be [150*(1 - 0.45)]/1.82 + 1.75 = 47.1. If the porosities of the worn and spherical sand layers were 0.55 and 0.65, the respective friction factors would be 20.77 and 3.85.
- 5
  Cube the porosity for each filter layer, then multiply the result by the particle shape factor and the grain diameter for that layer. For the crushed coal layer, the result would be [(0.45)^3]*0.73*0.0005m = 0.00003m. The results for the worn and spherical sand layers would be 0.00012m and 0.00137m.
- 6
  Subtract the porosity of each layer from one and multiply the result by the friction factor and depth for that layer, then divide by the result of Step 5. If the crushed coal layer has a depth of 1.5m, the calculation would be [(1 - 0.45)*47.1*1.5 m]/0.00003m = 1,295,250. If the worn and spherical sand layers each had a depth of 1m, the results would be 77,887 and 984, respectively.
- 7
  Add together the results of Step 6. For example, 1,295,250 + 77,887 + 984 = 1,374,121.
- 8
  Square the velocity of the water flowing thorough the filter and divide the result by the acceleration due to gravity -- 9.81m/s^2. For a velocity of 0.005m/s, [(0.005 m/s)^2]/9.81m/s^2 = 0.000003m.
- 9
  Multiply the result of Step 7 by the result of Step 8 to find the total head loss in the rapid sand filter. For example, 1,374,121*0.000003m equals a head loss of 4.1m.

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