just out of curiosity, iv been trying to find any info or a math equation for figuring out the amount of gallons per second of blast time or time it would take for a k3 to get from 200psi to 90psi with 10 gallons of air. also let say i want to add more air, id like to know who many gallons i would have to add to equal 1 second of blast time.
There is a formula for how many cubic feet of air it takes to pressurize a given size tank to a given pressure.
Then here is a rough chart for air consuption:
The problem is there are so many variables. That chart is pressure at the inlet base while blowing the horn, but it’s continually changing & losing pressure…unless you are regulating the pressure down and have a large tank.
By the chart, if you could maintain 90psi measured at the inlet base while blowing, a K3 would consume about 70CFM. So you can do the math with how many cubic feet are stuffed in your tank.
I have never seen anybody test the pressure @ the horn while blowing, let alone list out their valve size, airline size & length, fitting count & I.D., etc…all the things that affect pressure before it gets to the horn.
The K3LA consumes something like 90 CFM of air and using my own experiences will eat through 5 gallons of air in roughly 5 seconds from 175psi to 60psi. If you regulate it, you’d end up with more time.
So for 10 gallons of air at 200psi, regulated down to 150psi for the horns, you could possibly see 15 seconds of honk time, which in reality is a crazy amount of time to blast someone.
-Kris
hmmm…
ok so… here’s the math.
V=ᴨr²h
V=(3.142)(4.75²)(20.5)
1453.08 Cubic inches
divide by 12 and
121.09 cubic feet for one 5 gallon tank
so 242.18 cubic feet for 10 gallons
so to calculate blast time
242.18 ÷ 70 = 3.45 minuets… of course from full to empty.
not sure if any of this right but…
plus i cant seem to find an equation to calculate how long it would take to get from 200psi to 90psi with 10gallons at 70cfm
1 cu ft = 1728 cu inches = 7.48 gallons
10 gallons / 7.48 = 1.337 cu ft
Using your tank measurements… 1453 cu inches / 1728 = .84 cu feet or 6.28 gallons
(The tank is less because of the rounded ends & actual inside dimensions.)
But anyway, a 10 gallon tank has 1.34 cu ft of air at atmospheric pressure. You stuff a lot more volume into the tank when pumping to 200psi. Boyle’s Law states that P1V1=P2V2.
200psi * 1.34 cu ft = 14.7psi * V2
V2 = 200*1.34/14.7
V2 = 18.23 cu ft when released back to atmosphere.
18 cu ft / 70 cu ft per min = .26 minutes or 15 seconds ----- THEORETICALLY
I’m such a nerd…theoretically
Using the above: 18.23 cu ft stuffed into 10 gallons = 200psi
do the same calculation for 90psi the subtract from 18.23.
90psi * 1.34 cu ft = 14.7psi * V2
V2 = 90*1.34/14.7
V2 = 8.2 cu ft still in the tank
(18.23 cu ft @ 200psi) - (8.2 cu ft @ 90psi) = 10.03 cu ft consumed
10 cu ft / 70 cu ft per minute = .143 minutes or 8.6 seconds.
There are a couple reason why this seems low.
I would definitely recommend regulating the air down to 150 psi. It will grant you more honk time for what you have in your tank and will be better for the horns themselves.
I have a 12 gallon tank, pumps to 200psi, regulated to 140psi, all 1/2" hose and I can blow about. 17 seconds without it weakening
^^^ That’s good to know. I’m so anxious to test my 14 gallons regulated.
So after 17 seconds what is your tank pressure? I can calculate how many cu ft were used.
That I don’t know. My gauge is in my tool box. I am about to do pillar gauges and add air to my pillar
i knew i did something wrong… but makes more sense now.
i just ordered a regulator and should come next week:)
According to that chart the K3 is louder than a K5. I’ve read that the K3 is louder, that chart seems to verify it.
Yeah, that chart gets posted when that argument breaks out…
EDIT:
…and I’ve never gotten into that argument. BUT if that chart is God, then notice that any given pressure at the base always correlates to a certain db level no matter which horn. For example 75psi at the base is always 113db no matter what horn. It’s all about flow and keeping that pressure up at the inlet while blowing.
The weird thing about the chart is that the K3 is graphed (tested?) higher.
I see your point. I’m not so sure that chart is the last word.
If the chart is correct on air flow it makes it appear as though most K5s are limited by the size of the air valve used (assuming large air lines are used).
Even if the numbers are right, the K3LA is supposed to be 156db while the K5LA is supposed to be 154db. A 2db difference to our ears at any volume will not be noticeable.
-Kris
I don’t understand how the K3LA would be louder than the K5LA since the K5LA uses the same bells, plus 2 more. I figure the difference is the K5LA uses so much air that in most cases it’s starving a bit.
I wonder if a Graham White valve even flows enough air to fully feed a K5LA at 140psi. I’ve read that a GW valve flows 70cfm wide open at 140psi. That chart shows the K5LA using over 90cfm at around 80psi.
Thoughts?
I could see where the K3 is louder when the K5 is being starved…and that might be the case for the average hobby setup. But if you could fully feed a K5, there’s no way it’s not louder.
I agree with your point. It’s just like big car stereos where woofer cone area and power contribute to total output. 5 woofers with 100watts each is louder than 3 woofers with 100watts each. More bells operating with the same pressure = more output.
MarineOne I am not discrediting those numbers as I have seen them quoted a lot. But where did they come from and what were the air setups and mic placement? Even knowing that the chart above is a reading @ 100 feet doesn’t tell us what direction the horns and mic are facing. After reading Donovan572’s point, it kinda makes me wonder too if they were tested with a Graham White.
I dunno, but like you I see those same numbers thrown out all the time hence the “supposed to” … now if we could get a K3LA and a K5LA tuned and in a lab for testing, those would be results I’d love to see.
-Kris
^^me too.
And I’ve thought about my stereo/woofer/power example and it is not a fair comparison. All woofers would be “in phase” playing the same frequency which the horns are not.