# Convert water measurement into speed measurement?October 5, 2014 8:58 PM   Subscribe

I have a hose that can put out water at 60 litres per second (16 gallons per second) and I want to know if this hose can roughly replicate a wind speed of 275 kilometres per hour (170mph). We will put an object right in front of the hose - so there will be little pressure fall-off from the hose. Is this conversion measurement at all possible? I don't mind if the answer is a little approxiamate.
posted by meech to Science & Nature (22 answers total)

I would imagine you would need the length and diameter of the hose to make this calculation.
posted by Joh at 9:26 PM on October 5, 2014

I'm not the math person, no doubt one will be along soon, but so long as you know the diameter of your hose this should be a simple calculation assuming "air" without humidiity at Sea level and water of a certain temperature. Are you looking to replicate the "wind speed of 275 kilometres per hour" or are you attempting to do a simulation using water in the stead of air and looking to replicate the pressure? The latter is somewhat more complicated. It's a doable calculation though definitely not by me.

Some details of what you are trying to accomplish will make it easier to help I think.
posted by vapidave at 9:33 PM on October 5, 2014

Velocity = Flow rate / cross-sectional area of hose

You need to know the hose diameter. Assuming normal 5/8" hose:

60 liters / second * 10 ^-3 meters / liter
--------------------------------------- =
pi * (7.9375 10 ^-3 m ) ^2

= .06
-------
.0001979326

= 303.1 m/s, or 670 miles an hour.

...if I did my math right. That number seems far too high unless you are talking about a pressure washer or something.
posted by zug at 9:34 PM on October 5, 2014 [2 favorites]

Do you mean 60 liters per minute? Because envelope math says 60 liters/sec would fill a typical in-ground swimming pool in less than an hour.
posted by ROU_Xenophobe at 9:37 PM on October 5, 2014

Going back a few years to my college days...
If you want to match dynamic pressure the object sees, then the ratio of the densities of the two fluids must be inversely proportional to the ratio of the squares of the velocities: density1*V1^2 = density2*V2^2.
Water is roughly 1000/1.2 x the density of air, so V(water)= V(air)*sqrt[(1.2/1000)] = ~9.5 km/hr, which is attainable from a garden hose, though you would need to narrow down the opening some.

Too accurately match the forces for this level of approximation, the water flow would need to be spread out evenly over the same area that was seeing the wind flow. Whether you can do this with your hose would depend on the size of the thing you're blasting.

A purist would also want to match Reynolds numbers, which would involve scale models. 275 km/hr winds can be safely described as fully turbulent though, so if you can ensure the water flow is turbulent, the Reynolds number is not critical.

But it's late and it's been 15 years, so I might have forgotten something.
posted by cardboard at 9:43 PM on October 5, 2014 [3 favorites]

ROU_X beat me to it. 16 gps is around 2cfs, which is about right for firehose or irrigation line on a farm, but easily an order of magnitude larger than anything possible in a household.

Regardless, knowing the flow doesn't give you velocity without knowing the size of the pipe -- there is missing information needed.
posted by Dip Flash at 9:44 PM on October 5, 2014

I'll be curious to see if any engineers or physicists can give you an answer on your question. I can’t imagine how it would be possible to draw a comparison of water at 8 lbs/gal, or 64 lbs/cu.ft. to air with almost no mass, even at the highest velocity, to water at an undetermined velocity. A standard fire hose and nozzle combination for structure firefighting would be an inch and a half hose with around 125 psi pump pressure feeding to a 100 to 200 gpm nozzle. You’re indicating a discharge of 960 gallons/minute. What pressure would you have at your pump or hose discharge? A deck gun or monitor mounted on a fire truck would be capable of 500 to 1000 gpm or higher. It would be solidly attached to the truck because of recoil and typically would be supplied by at a pipe of at least one and a half inches.
posted by X4ster at 9:48 PM on October 5, 2014

Here's a link to some firefighter hydraulics calculations if it might help with your calculations.
posted by X4ster at 9:54 PM on October 5, 2014

Air is compressible; water is incompressible. Their fluid dynamics are radically different. Huge density difference, too. Viscosity as well, but that's not as important at high flow velocities.

Water is not a good model for wind.
posted by mr_roboto at 9:59 PM on October 5, 2014 [3 favorites]

Response by poster: Thanks all. It is indeed a fire hose - roughly 10cm in diameter.

mr_roboto I get what you're saying. I'm looking for a visual equivalence - rather than hard science. But isnt getting hit by air at 275kph roughly (and i mean really roughly) the same as getting hit by air at 275kph? Isn't it like - which is heavier a ton of bricks or a ton of feathers?
posted by meech at 11:31 PM on October 5, 2014

> But isnt getting hit by air at 275kph roughly (and i mean really roughly) the same as getting hit by air at 275kph

No - not at all. Because water is so much denser than air, you will feel a lot more force from the water if the speed is the same.

You'll need to quantify what you mean by "replicate a wind speed of 275 kilometres per hour" before you'll be able to answer your question.

For example, do you mean that the flow speed should be the same? Or do you mean that an object of a certain shape and size should experience the same forces due to the flow? Or something else?
posted by richb at 12:02 AM on October 6, 2014

But isnt getting hit by air at 275kph roughly (and i mean really roughly) the same as getting hit by air at 275kph? Isn't it like - which is heavier a ton of bricks or a ton of feathers?

No.

First of all, it all depends on how much water. 1 nL of water at 275 kph will be imperceptible. 100 L of water will knock you over.

And it's not a ton of bricks vs. a ton of feathers. Air is a lot less dense than water, so a similar volume of air at the same velocity as a volume of water carries much less momentum.

And the fluids behave completely differently, especially at high velocity. Air will compress when you push it against a barrier. Its density will change, and some of the energy in the flowing air will go into increasing the density. Water doesn't do this. It's incompressible; its density stays constant. It's totally different.

Even "roughly" speaking, it's not a good model. It's a bad model. It's not "a little approximate". It's wrong.
posted by mr_roboto at 12:09 AM on October 6, 2014 [1 favorite]

It would be a tonne of feathers/tonne of bricks situation if you had the same mass of each - if you weighed the amount of air hitting the object, and had the same weight of water hit the object, spread out over the same area, it would be closer. As it is, the air that would be hitting the object at 275 kph would be very light - so it's a few kilos of feathers/tonne of bricks situation. A jet of water at 275kph will do a lot of damage.
posted by twirlypen at 2:27 AM on October 6, 2014

I'm not a physicist or engineer or anything, but I can tell you from direct personal experience that the pressure of air at 100 mph is dramatically different to the pressure of water at 100 mph. Dramatically different. The difference would be even more dramatic at 170 mph. If you are trying to simulate the effect of wind on an object, a compressor and air hose would likely be much better than a fire hose.
posted by dg at 3:25 AM on October 6, 2014

The compressibility of air is negligible and traditionally ignored until you reach 1/2 the speed of sound. That is ballpark 170 m/s or 600 km/h, so at 275 km/h you're fine.
posted by cardboard at 3:35 AM on October 6, 2014

= 303.1 m/s, or 670 miles an hour.

If you just dump "60 liters/sec /(pi*(5/8in)^2)" into google and let it do the unit conversions, you get 75.78 m/s or 169 mph.

If you just apply cardboard's formula, being hit with 60liters/sec of water moving at ~8 meters/sec would be like being hit by air moving at about 780 km/hr?
posted by ROU_Xenophobe at 7:05 AM on October 6, 2014

I think the ton of bricks / ton of feathers thing is pretty apt here. Yeah they weigh the same but you'd definitely want to get hit with those feathers rather than bricks.
posted by stinkfoot at 9:51 AM on October 6, 2014

If you just dump "60 liters/sec /(pi*(5/8in)^2)" into google and let it do the unit conversions, you get 75.78 m/s or 169 mph.

That's AWESOME because I hate unit conversions. But 5/8 is the hose diameter, so you need to divide by 2 before inputting to google:

60 liters/sec /(pi*((5/8in/2))^2) which gives 303.133477 m/s (so my unit conversions were correct). Since the fire hose is actually 10cm, the google-approved rate becomes 7.6 m/s or 17mph.

This still doesn't get at what the poster wants, since the force of fluid isn't the same as the force of air, but it at least gets the speed.
posted by zug at 9:52 AM on October 6, 2014

"But isnt getting hit by air at 275kph roughly (and i mean really roughly) the same as getting hit by air at 275kph?"

As has been pointed out, there is no comparison. Here's an easy way to test that, if you have a boat. Get in the boat on a calm day, bring it up to a few knots, and maintain a steady speed (or if you're in a canoe or kayak, just stop paddling and coast. Hold your hand just above the surface of the water and note how much resistance the wind causes. Now stick your hand in the water and note how much more resistance it offers.
posted by brianogilvie at 10:01 AM on October 6, 2014

What you really want to know is the force that the water or air imparts, but I think this is impossible to know because the water that comes out of the hose is aerated, so there's no way to know the mass.
posted by zug at 10:43 AM on October 6, 2014

Cardboard's responses looks the best to me.
posted by SemiSalt at 1:58 PM on October 6, 2014

Couldn't one calculate the PSI force come out of the hose? Is there a way to convert GPM through a 10" opening into PSI (and then to PSF)?

170 MPH wind is approximately 74 PSF (170*170*.00256).
posted by yeti at 2:40 PM on October 6, 2014

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