was driving down the freeway and had a small leak that left a trail of gasoline on the road behind him and it accidentally caught fire 100 ft. behind him going at a speed of 45 mph. How fast would the driver have to drive for the flame not to catch up. (variables: sunny, 72 degrees F, 45% humitiy, pressure: stable, flat road, and the truck leaked an average of 1 gallon per minute.) I dont know if this is enough info to solve a problem like this. Let me know if you need any more info.
2006-06-23 08:54:59 · 6 answers · asked by billgutsky 3 in Science & Mathematics ➔ Physics
87 octane. no wind.
2006-06-23 09:06:27 · update #1
Stream is 2" on average
2006-06-23 09:06:55 · update #2
The concept is "flame propagation speed." See the reports below. Unfortunately my reference books are about 15 miles from here, but the data in the reports sounds about the same with some other work I've done
Using the rough estimate of 1 meter/per second flame speed, that would work out to a bit over 2 miles per hour (mph), or a medium walking pace.
2006-06-23 09:13:36 · answer #1 · answered by techyphilosopher2 4 · 0⤊ 1⤋
As techy... pointed out, flame front numbers for gasoline are available for different situations. But, one more variable needs to be thrown in. What time of day is it? I mean, the temperature of the road surface will determine how quickly the gasoline evaporates. This could impact the speed as it tries to overtake the truck. Anyhow, this motorcycle link claims flame fronts of max .5 m/s. That is 1.1 mph. The other link is scientific, but for methane. Basically the same numbers.
Added:
I just realized, if the road surface if hot enough, all the gas could evaporate in one spot before the flames get there.
2006-06-23 09:51:26 · answer #2 · answered by Karman V 3 · 0⤊ 0⤋
"...a gasoline-air mixture creates a flame front speed that ranges from around 70 feet/second up to around 170 feet/second..."
Assuming the lower number (for lower octane), that means the flame will travel 48 miles per hour, slightly faster than the truck. The flame is 3 mph faster, or 15840 ft/hr faster, or 264 ft/min faster, or 4.4 ft/sec faster, than the truck.
Since the flame starts out only 100 feet behind the truck, and it travels 4.4 ft/sec faster, it will catch up to the truck in about 22.7 seconds.
However, something would actually have to ignite the gasoline first, because it wouldn't burn by itself in 72F temperatures.
NOTE: the burn speed is for "normal engines," not a normal daytime outdoor setting. I was looking all around, and this was the best I could come up with.
2006-06-23 09:02:11 · answer #3 · answered by iwastypingthat 4 · 0⤊ 0⤋
well this wuld depend on the speed at which the gasoline burned which depends on the width and thicknes of the gasoline film (the wider and less thick the faster). by the way in any spill you need to know the wind speed and direction...
anyway this is a movie like problem so ask hollywod for a really blowing scene ^_^
2006-06-23 09:04:23 · answer #4 · answered by michael_gdl 4 · 0⤊ 0⤋
In diehard-2, the burning fuel caught up with a jet liner during takeoff, so that truck is doomed.
2006-06-23 09:07:02 · answer #5 · answered by Anonymous · 0⤊ 0⤋
The thing is, gasoline does not explode, but in gas form it does, so unless you are driving a truck full of gas vapor it probably wont explode unless it crashes.
2006-06-23 09:02:25 · answer #6 · answered by Anonymous · 0⤊ 0⤋