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Snow Gun Output Rate

Also known as snow gun output · snowmaking production rate · snow gun gpm · how much snow per hour · snowmaking capacity · snow production rate · acre feet per day snowmaking · cubic metres of snow per hour

V˙s=V˙w ρwρs\dot{V}_s = \dot{V}_w \, \frac{\rho_w}{\rho_s}

Enter your known values, leave one input blank, and solves for the missing one. Try different units for next level excitement!

Constant used — built into this formula, no need to enter
ρice=917 kg/m3\rho_{\mathrm{ice}} = 917\ \text{kg/m}^{3}Density of Ice at 0 °C
See also
Snow Volume from Water Volume — The same conversion on a total volume rather than a rate.Snow Made in a Wet-Bulb Window — What this rate becomes over a night, with the losses and the wet-bulb condition included.Wet-Bulb Temperature (Stull 2011) — Whether the gun may run at all, which the flow rate has nothing to say about.Hydronic Heat Transfer (Water) — The other place a water flow rate is the whole calculation — same units, entirely different question.

Learning zone

Snow guns are specified in gallons or litres per minute of water. Hills are planned in cubic metres or acre-feet of snow. This is the conversion, and it is the same mass balance as the volume page with a clock attached: the water flow multiplied by the density ratio gives a snow production rate.

A single fan gun taking 500 gpm at 400 kg/m³ makes about 284 m³ of snow an hour, or roughly 3,400 m³ over a twelve-hour night — enough to cover 11,000 m² of trail to 30 cm, which is about a third of a kilometre of a 30 m run. That is one gun. A hill covering its terrain in a season runs dozens at once, which is why snowmaking plant is one of the largest capital items a ski area owns.

The water flow is not the constraint people assume it is. Three other things gate a gun before the pump does.

Wet bulb decides whether it may run at all, and no amount of water compensates for a wet bulb above about −2.5 °C. Below about −5 °C production is reliable; below −10 °C the guns can be opened up and the snow comes out dry and light. In between, the operator is trading quantity against quality in real time. That trade is the entire skill of the job and it has nothing to do with flow rate.

Compressed air is the other utility, and on many hills it runs out first. An air-water gun uses compressed air twice: to shatter the water into droplets small enough to freeze in the time they spend in the air, and to expand through the nozzle, which cools the plume by a few kelvin and seeds it with tiny ice nuclei. A gun of this size wants 100 to 300 cfm of it. Fan guns cut that dependence hard by using a big electric fan for the throw and only a small compressor for the nucleators — which is why they took over, and why the plant's electrical service became the binding constraint instead.

And wind decides where the output lands. A fan gun throws its plume 20 or 30 m to buy hang time; a crosswind puts a good fraction of a perfectly made night's production into the trees. That loss shows up nowhere in this equation, which is content to tell you what left the nozzle.

Read the rate as an ideal, then take the losses off. Ten to thirty percent of the water leaves as vapour or drifts, so a real gun delivers less to the ground than this figure — and about 12% of that is a hard physical floor, the fraction of each droplet that must evaporate to freeze the rest. Run backwards, the equation sizes a system: name the snow the hill must have on a date, get a flow, and then find out whether the pond, the pump head and the pipe will carry it. The pipe usually decides, and peak demand is every gun at once on the coldest night, not the seasonal average.

Snow Gun Output Rate
V˙s=V˙w ρwρs\dot{V}_s = \dot{V}_w \, \frac{\rho_w}{\rho_s}
V̇wV̇sρs
Where
  • V˙s\dot{V}_s= Snow production rate (m³/h)
  • V˙w\dot{V}_w= Water flow to the gun (L/s)
  • ρs\rho_s= Density of the snow made (kg/m³)
Missing one of these? Work it out first, then come back