Have a opening for a paint booth.  OSHA requires a flow into the booth of 100 lineal feet per minute.  Need an equation to estimate the cfm's fan size in cubic feet per minute using square feet booth opening and 100 lineal feet per minute.
136796
LFM * Area = CFM\n\nIf your booth has a 32" x 7 ft opening, then it's"\n\n2.67 * 7 * 100 = 1869 CFM
353674
The question asks for the difference in pressure so it is not necessary to know how high the air column extends above the building or assume the bottom is at sea level.  and you have the correct equation.  \n\nPair top floor- P air bottom floor= rho x g x h
403478
You need to use kinematics and Newton's 2nd law to solve the problem.  In the horizontal direction, Newton's 2nd law states:\n\nForce - friction = m*a.  Since you are given the values, you can solve for the accleration:\n\n1149 - 909 = (2700)*a, solving for a, you get:\n\na = 0.089 m/s^2\n\n----------------\n\nNow use kinematics to solve for the distance needed to reach the desired final velocity.  Use the following kinematic equation:\n\n2ax = v_f^2 - v_i^2.  You know that:\n\nv_f = 3.2 m/s\nv_i = 0 m/s (starting from rest)\na = 0.089 m/s^2\n\nNiw solve for x:\n\nx = [v_f^2 - v_i^2]/(2a) = 3.2^2/(2*0.089) = 57.6 meters\n\n---------------\n\nHope this helps
647428
Quick and dirty: Calculate cost of employee wages ($ wage x time). Now come up with workers comp cost based on this, typically its 37% of wages for janitorial work. Take these two numbers and add the cost of supplies. Calculate some overhead, maybe 2% of wages and add it. This will give you a rough cost estimate. Take the total cost and add 35% to 40% of it as profit margin (includes contingency) and this will give you the price to quote.
491185
The axial velocity of the fan is:\nV= w*r ; where w= revolutions per sec. and r= the radius of the impeller. This velocity is different from the absolute inlet and outlet velocities. With the geometry of the blade one can design the fan to give a certain flow rate which is :\nQ= v*A where v is the resultant absolute velocity and A is the cross sectional area of the duct. The total pressure is equal to the static pressure plus the dynamic pressure. From the dynamic pressure one can obtain the velocity ,v.\np=1/2 d*v^2 ; where d=density of the fluid.
203603
One must ask what is the air volume of the structure in Cu Ft (or Cu M) then determine what the volume of exhaust gasses are to be expected that remain within safe concentrations versus the total volume of exhaust gas versus total structure volume, then determine how often the total air mass is to be exchanged to maintain said safe consentrations.\n\nJust a guess.
674292
4 miles (give or take)\n\nyou basically work out how many miles the firetruck goes in 1 minute (which is .83 50/60) then multiply by the 5 mins travelling time. 5 x .83 = 4.16 miles...\n\nformula would be: (a/60) x b = c\n\nwhere:\n\na is the constant speed\nb is the time taken to arrive\nand\nc is the answer
137299
f1=20h\nf2=20000h\nv.sound=300m/s\nwavelength wil be represented by "LL"\n\nLL1=v.sound/f1\nLL2=v.sound/f2\n\nLL1/2=length of pipe 1=7.5m\nLL2/2=length of pipe 2= 7.5x10-3m
99390
The link discusses the operating curves for typical fans including centrifugal fans.  The article also discusses the fan laws that are necessary to understand fan operation and selection.  Good luck.
296630
Well, the volume of water (in cubic inches) within a foot-long piece of pipe will be:\n\nV = 3.14159*r^2*(12in)\n\n(where r is the inner radius of the pipe in inches, 1" and 2" respectively)\n\nThen convert that volume from cubic inches into liquid gallons using the site listed below.\n\nedit: That sounds like an interesting idea.  You are heating the pipes via a solar panel?  In any case, good luck on saving energy bill money.
400674
The 5 gallon tank measures 0.6685 cubic feet.  The compression ration is 15 (150 psi / 10 psi) so the tank holds 15 times as much air supply as its volume, or 0.6685 x 15 = 10.03 cubic feet. So, at 8000 cfm, you run out in 10.03 CF/8000cfm = .00125 minutes or 0.0752 seconds.
227785
Hostile work environment, I'd go with that, or consider calling OSHA (Occupational Health and Safety.) Noxious gases and high levels of bacteria are cause for concern for the workplace, right?\n\nDid they pass out eye protection in your office?\n\nHow about Scott Air Packs or other respirators? Are they in easy reach of anyone who might require them?\n\nIs there an adequate alarm system? I suggest that rotating flashing lights above the bathroom door (warning the unwary not to enter) plus sirens and a loudspeaker announcing when it is safe to re-enter the workspace might be something to consider.\n\nWhat about running "fire" drills. If you see the Comptroller heading in the general direction of the loo, your health and safety officer is completely within his rights to pull the alarm and announce a general evacuation of the premises. Then he should conduct a role call at the designated meeting areas.\n\nIt might be wise to practice administering CPR (preferably on the prettiest women or guys, if that's your preference) \n\nIt might not be too extreme to demand that the Comptroller's favorite stall be marked off with biohazard tape and flooded with strong UV. Ask about installing a huge fan to whisk toxic fumes out of the building. In fact, an executive "washroom" with a special lock should be built just for this guy....\n\nIn the next county.
70304
use the fmla.... Q=msT\n\nwhere Q is the quantity of heat required, m is mass, s is specific heat capacity of water, T is the temperature difference. i dont hav a calculator with me now, u do the computations plz...
427719
I tried to import a spreadheet for you but I coouldn't make it come through, if you email me I'll send it to you, like1m@yahoo.com  \n\nYou can fill in the two variables, just write it in excel and you will be able to calculate your values for voltage and current dependent on the resistance for any given values.  All answers here are correct as it is the basic premise of \nOhms law, but you asked for a chart.  You do need to know the resistance though.  The interrelation is such that for at a given voltage with infinate resistance there will be no current (flow), for that same voltage with minimum resistance there will be maximum current (flow) this would be referred to as a dead short and is comparable basically to welding. Either extreme is not normally practical unless you are turning the power "off" in one case or welding in the other (dead short).  Dependent on the requirement usually the interrelation is desired to be somewhere in the middle.  There is also the element of reactance which is slightly more dynamic that static resistance. but for now this is what your auto calc chart should look like....  \n\nInput ohms & volts to autocalc Amps\n\nInput OhmsInput VoltsI=V/R\n      200                     24                 0.12\n\nInput ohms & Amps to autocalc Volts\n\nInput OhmsInput AmpsV=I*R\n    200                     0.12                   24
150737
Use the formula PV=rnT\n\np= pressure in atm\nV= volume in L or dm^3\nr=number of moles\nn= constant: 0.0821 atm x L/ moles x K (This constant is used to balance out units)\nt= temperature in K\n\nSo for your problem:\n2.3 atm x 120 L = r x (0.0821 atm x L/ moles x K) x 340 K\n\nthe constant should allow you to be left with moles as your unit\nusing algebra, r is left, which also stands for the number of moles (meaning the problem coordinates with your question)\n\nusing this setup, i got 9.89 moles
206600
HI been a compulsive Modder/Performance Freak me and two Friends was sat talking (Computers of course) and we came up with an solution to fan filters a box (internal) fro a 80mm fan think of a rubber pipe collar round or a square box section that will fit over the fan, then you need (don't laugh) woman's tights cut a round or square peace out to fit over the walls of the tube/box and secure with wire, this works my mate has this now on his input fans, no restriction of air flow and a very clean case, SO some thing 20mm deep with no ends that will span the fan and a section of Lady's tights stretched over the end furthest from the fan. Good luck, Cybercat.
364369
1. volume = h*2*pi*r^2 =1000 so r^2= 1000/2/pi/h\n\nsurface area = 2*pi*(h+2*r^2)\nsurface area / volume = (h+2*r^2)/h/r^2 \n=1/r^2+2/h  = 2*pi*h/1000+2/h  which is a minimum when h is about 17.84  [(1000/pi)^0.5]