I am guessing it would take about double that amount of cigars to make up one cubic foot. so Approx 125 BTUs at a temp increase of 1 degree per second (another guess). So based on the detection temp and the current temp= 60 seconds?
Congrats on passing your test, btw.![]()
Mama said a lot of things and be thankful was the one she never minded saying twice
--Drive-By Truckers
Fark? I guess that is a re-deployment of the old DOS attack lingo...
*sigh*
Yet another forum (I hang at BOTL and Puff) where I hafta establish some sorta bona fides before being allowed to play.
Good luck guys!
Here's a "simpler" equation:
t = (X*H^.1.33)/release rate^(.33)
Unfortunately, X is something fairly complex...
That said (and I'll likely do the math for someone's benefit), it looks like the response time is under 1 second...
The average size of the cigars is 5x50 cause you really dig those Robustos.
The Heat Release Rate of 1 cubic foot of tequila-soaked cigars can be assumed to be 250 BTU/s.
The inside temperature is 75 degrees F because your wife is always cold.
The detector alarm temperature is 135 degrees F.
The ceiling height in the living room is 10 ft and the ceiling is flat.
The heat detector is located on the living room ceiling, approximately 2 ft. from the centerline of the fire.
The Response Time Index (RTI) of the heat detector is 100.
The fire can be assumed to be Steady State.
Hmmm
okay, we need something like /ceiling jet lag time (assume 1.2) = .833
we need radial distance to detector (stated as 2 feet)
we need height of ceiling (10 feet)
and heat release rate (given = 250 btus)
raise radius to 11/6 = 2^11/6 = 3.56
Now calc release rate^1.3 times height^1/2
BUT: convert BTU to KW (what, like... BTU/minute equals 0.01757 kilowatt so 250*60/0.01757 =~ 854,000kw/min)
I'll prolly hafta get back to heat release conversions...
T=1/(jet lag) * 3.56/10^(0.5)*Heat Release^(.333) = ~3/(3.2*Heat Release)
I kin already see Imma WRONG cuz this is gonna make a rilly smaaaallll #....
Back ta the drawing board
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