An Archimedean screw water lift carries water uphill in pockets formed by rotating helical flights and a surrounding trough or casing. The animation shows the water moving from a lower source to an elevated outlet. Enter flow and lift to calculate ideal hydraulic power, or change the inclination to compare the straight-axis length needed to span that height.
Archimedean Screw Water Lift Interactive Calculator
Vary flow, lift height, and screw angle to see hydraulic lifting power, daily volume, and screw-axis geometry update live.
Equation Used
The calculator uses the standard hydraulic lifting-power equation for an Archimedean screw: flow rate times lift height gives the water power required. The screw-axis length is estimated from the lift height and installation angle.
- Fresh water density is 1000 kg/m3.
- Hydraulic power is ideal water-lifting power before mechanical losses.
- Flow is steady and the screw carries discrete trapped pockets upward.
- Screw angle is measured from horizontal.
Water carried up an incline
The screw’s lower end receives water from the source. As the shaft and flights rotate inside the trough, water collects in pockets and travels toward the higher outlet. The drive supplies the energy needed to raise the water’s elevation.
The spaces are not perfectly sealed. Clearance, water levels and geometry affect actual filling and leakage. The illustrated shaft diameter, helix spacing and three starts explain the motion but do not determine the entered flow rate. The blue markers show transport at an illustrative rotation rate.
Parts to identify
- Helical flights: form the moving sides of the water pockets.
- Central shaft: supports the flights and transmits torque.
- Trough: surrounds the lower portion of the rotating assembly.
- Inlet and outlet: mark the lower and upper water levels.
- Bearings and drive: support and rotate the screw.
In the installation geometry, height H is the vertical separation and θ is the angle from horizontal. A straight axis rising through H has length H/sin θ. A complete machine also needs its particular inlet, outlet, bearing and drive geometry, so the displayed length is not a fabrication dimension.
Comparing water-lifting duties
Screw pumps are used in drainage and water-handling systems. A useful first energy question is how much water must be raised through what elevation difference. The same ideal power relation applies regardless of whether the eventual pump is a screw or another type.
Actual screw-pump selection requires geometry and performance data beyond those energy inputs. Spaans Babcock’s design guidance discusses the interaction of dimensions, inclination, filling and speed. This calculator does not choose those parameters or promise a flow capacity.
A screw turbine extracts energy from descending water, while a pump supplies energy to lift it. The similar appearance does not establish that an existing machine can be converted between duties. Suitability for debris or fish passage also needs evidence for the particular installation.
Hydraulic power, mass flow and installation geometry
For freshwater density ρ = 1000 kg/m³, gravity g = 9.81 m/s², flow Q in m³/s and vertical lift H in meters, Hydraulic Power = ρgQH/1000 kilowatts. Mass Flow = ρQ kg/s and Daily Volume = 86400Q m³/day for uninterrupted flow over 24 hours.
Axis Length = H/sin θ, with θ measured from horizontal. Across the available 22–38° control range, this is ordinary right-triangle geometry. A shallower angle needs a longer axis for the same rise.
Angle changes length while leaving ideal power, mass flow and daily volume unchanged at fixed Q and H. This is expected: the ideal elevation-energy requirement depends on height, not the route taken. In a real screw, angle can affect achievable flow and efficiency; here Q is an independent input.
Hydraulic Power is power delivered to the raised water, not electrical motor input. With a specified overall efficiency η, input power would be hydraulic power divided by η, but efficiency is not entered or estimated here. The model also omits leakage, drive losses, additional head losses and any change in flow through a day.
Worked example: flow and lift
Use the initial settings Q = 5 m³/s, H = 4 m and θ = 26°. Hydraulic Power = 1000 × 9.81 × 5 × 4/1000 = 196.2 kW. Mass Flow is 5000 kg/s, and continuous 24-hour volume is 432000 m³.
Axis Length = 4/sin 26° = 9.124688 m, displayed as 9.12 m. Changing angle to 22° gives 10.677869 m, displayed as 10.68 m; changing it to 38° gives 6.497077 m, displayed as 6.50 m. The other outputs stay fixed.
Reduce Q to 2.5 m³/s while keeping H = 4 m. Ideal power becomes 98.1 kW, mass flow 2500 kg/s and daily volume 216000 m³. Axis length is unchanged. These numbers describe a requested duty; they do not show that the illustrated screw has sufficient capacity.
If overall efficiency were independently established as 0.75, the original 196.2 kW hydraulic duty would require 261.6 kW input. That efficiency is a hypothetical example, not a value supplied by this calculator or a motor selection.
What the three inputs control
| Change with other inputs fixed | Effect in the calculator |
|---|---|
| Increase flow | Power, mass flow and daily volume increase in proportion. |
| Increase lift | Power and required straight-axis length increase in proportion. |
| Increase inclination | The axis is shorter for the same vertical rise. |
These relationships do not rank real pumps by efficiency, price, bearing life or solids handling. That comparison needs the operating duty and documented equipment performance.
Water-lift calculation questions
Flow and lift are independent inputs. Ideal elevation power is ρgQH, while angle determines the geometric length. Actual screw efficiency and achievable flow may depend on angle, but those are not calculated.
No. It is the straight-line length corresponding to H/sin θ. Inlet submergence, discharge arrangement, bearings, drive and other dimensions need separate definition.
No. Flow is a specified input. Diameter, pitch, starts, rpm, water levels and hydraulic performance are not solved.
No. It excludes mechanical and electrical losses and does not select a motor or gearbox.
No. It assumes the entered flow continues for all 86400 seconds in a day. A variable-flow or intermittent duty requires integration over its actual operating times.
No. It is the initial setting, not an optimization result. This tool compares length but has no bucket-volume or efficiency model.
No. It has no bearing, shaft dynamics, temperature or aeration model. It cannot assign a fault or a repair tolerance.
No. This calculation contains no biological or passage assessment. A survival claim needs evidence for the species, size range and specific installation.
Further reading
Spaans Babcock: Screw Pump Design Parameters. The manufacturer’s selection discussion concerns actual performance; the power and triangle calculations here describe the entered lifting duty and geometry.
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