what is the flow of air in cu. ft from a regulator with 3500 psi inlet and 150 psi outlet.?
359943
Would need to know the orifice diameter that is flowing air from this 150 psi source.........and if you are asking cu ft per minute.......\n\nThe rate of flow won't change whether there was 6 or 60 cylinders, but more overall delivered cu ft when the cascade was depleted.
805064
lol \nYou have homework that you don't know the answer to.  \n\nWhy would someone ask these here out of interest? Like you have these specific parameters and conditions that you are just raring to want to know just out of the blue.\n\n Quit wasting time writing and typing questions out waiting here for answers when you could be the next engineering graduate who actuallly UNDERSTANDS this stuff by doing it yourself.
865811
30.5 m of water.\n\nConvert units to desired units.\n\n1 atm is 14.?? psi is about 32 ft of water.\n\nLook in reference for more detailed unit conversions.\n\nOr use P= &#92;rho g H.
609875
first of all, 120/240 are standard voltages for AC power, not DC.  DC comes in all flavors.\n\nanyway: 2500 kilowatts DC?\n\nthat is 2.5 MEGAwatts.  \n\ni find this somehow hard to swallow.  when i worked at a particle accelerator there was ONE really big DC power supply that put out something like 1.5 megawatts.  it was the size of a small apartment, and required massive amounts of water cooling, was entirely custom built, and probably cost something on the order of a million dollars or more.\n\ni don't think there is any such thing as "usual" for a 2.5 megawatt power supply.  think about it - you're talking about nearly the power consumption of a small town.  anything liek that would have to be custom built.\n\n----\n\nok, having seen your additional details:\n\ni'm quite positive it's AC, not DC.  it also probably depends on what kind of generator it is.  to give you a rough estimate, here's an example of a 11.5MW turbodiesel generator on a cruise ship:\nhttp://www.ship-technology.com/projects/destiny/\nnote the output is at 6.6kV.  that would set the current at the output at around 1.8 amperes.  \n\nhere's some more information.  as you can see there is no real "usual".  \nhttp://www.ship-technology.com/contractors/electrical/abb2/\n\ncheers!
361244
first we must know turbine's inlet pressure and temperature (depend on design ) , mass flow of steam to turbine (for 2,5 MW)  and effiency of turbine (for calculate of exhaust conditions )\n\nexample : 100 MW double exhaust steam turbine need to 160 ton steam at 500 C and 50 bar for high pressure   and 40 ton steam at 200 C and 5 bar for low pressure. \n\nthere are many kind of boiler . if you dont know some require value then you can not calculate boiler pressure , temperarture and  stages (eco , eva , superheater) .\n\ni can say only about capacity of boiler  ; \n\ni f we accept turbine effiency = 70 % \nthen amount of approximately steam power = 2,5 . 100 / 70 = 3,571 MW \nand \nİf we accept boiler effiency = 85 % then real capacity of   \nboiler = 4,2 MW \nthis value near capacity of boiler. \n\nboiler inlet condition = 1 bar 50 C (i accept) \nenthalpy of inlet water = (from compressed water tables) = 200 kj/kg \nmass of flow = 5 ton/hour (i accept) = 5000 kg/hour = 5000/3600 kg/sn= 1,39 kg/sc.\n\n3,571 MW = 3571 kw = 1,39 . (hç-200) \n\nhç= outlet enthalpy of boiler \n\nhç = 2769 kj/kg (this value must stay at superheated region) \nwe must look to superheated region at thermodynamic tables. \nand we can find best outlet conditions for 2769 Kj/kg. and temperature can not be high than 500 C for 2769 kj/kg. \nwe must take care to pipe line length because some losses cause to change superheated steam to saturated steam or near conditions. \nif superheated steam early change conditions then blades of turbine can be damaged. at the end of turbine must be saturated steam or little bit big superheated steam for protection of blades. \nfinally \n4 bar , 160 C conditions enough for 2769 kj/kg.\n\n\nsummary : \nboiler inlet 1 bar 50 C \nboile r outlet 4 bar , 160 C \nmass of flow = 5 ton / hour \n\nalso next step , calculate of boiler's economizer , evaporatör , superheater , main steam line pipe dia. , pumps (circulating and feed water) , drums , recirculation lines , condenser , auxilary cooling water system (for cooling of condenser) , gland steam line , regulating system , control syst. , vacuum sys. (if necessary for condenser) , water treatment sys. , chemical dosing points ,  etc. etc. etc.\n\nturbine inlet conditions = 4 bar , 160 C 2769 kj/kg\n\nnet power = 2,5 mw = 2500 kw \n\n2500 = 1,39.(2769-hs) \nhs = 970 kj/kg (enthalpy of at the end of turbine ) \n\nthis entalphy shown turbine outlet conditions \n\n0,5 bar 85,4 C or 0,2 bar 85 C (vacuum)\n\nwater must be cooled up to 50 C (inlet of boiler)
1105216
This question has a simple answer observe\n V=IR now R=V/I\n  so i.e.120/0.6 =200 ohms \n  if you heating effects please see the joule,faradays heating effects of electricity
201698
The volume in cu.ft will be 70'x1'x1/3' = 23 1/3 cubic feet \n27cu.ft = 1 cubic yard\nYou'll need 0.86 cubic yard
452060
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.
644717
Rearranging the equation watts=volts*amps to solve for amps gives:\namps=watts/volts\nFor example, 1,200 watt hair dryer on the regular household voltage of 120V uses 10 amps.\nBe aware that your A/C may run on 220V or 240V, so check the data plate before calculating.\n\nAlso be aware that inductive devices like the electric motors in your A/C may have a power factor other than 1, but I would ignore that if I were you, because I find it confusing. If you don't want to ignore power factor, here are a couple of links.
437503
Q = m.s.ΔT\nHeat loss by the copper = heat gained by the metal.\nCopper (heat loss) = 200.0g x 0.38J/g.ºC x (195.5ºC-228.4ºC) = -2500.4J\ndivide the heat by the mass and ΔT of the metal to get the specific heat of the metal:\n2500.4J/(65.0g x (195.5ºC-18.3ºC)) = 0.2171 = 0.21 J/g.ºC (correct SF)
527581
Expansion coefficients for air can be determined using the attached link.\n\nBest wishes !
540563
weigh the graduated cylinder full with water.\ninvert the cylinder (still full) in a tub of water.\nblow bubbles into the cyliner to find the volume of the air you blew in.\nremove and reweigh the cyliner.\n\nit may be a little tricky, but it's not too bad.
1005734
That's a bit high...usually want to operate at about 40-45 C or less. Get yourself a better heat sink and cooling fan, and maybe a better cooled case.
799770
Sounds like a heat transfer question.  If so it depends on how the tube is being used.  \n\nFor a single pipe exposed to heat transfer to air, the outer diameter is used to determine the surface area.  For shell and tube exchangers they have charts for making the determination based on the tube configuration and baffling.
1124736
A person needs more info. What is ambient temp for design. What is the wall, floor and ceiling construction, insulation thickness and type. \n\nThese will determine the minimum design load, from that the condensor and evaporator coils can be sized. \n\nYou need an HVAC engineer or tech to do this.
474309
to reduce air pressure sudden surge
619720
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.