there is car parking in basement of a building , so fresh air is required by fans and ducting . how much is fresh air required ?
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.
144461
I only had time to do the first one, btu the answer is \ne-2.  Check your email for the work.
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]
153822
That could be a temperature issue. Make sure the air vents are clean and clear and that your fans are running.
409234
To get the most out of your room air conditioner, you'll need to consider the size of the unit that best suits your needs.\nThe cooling capacity of room air conditioners is measured in British thermal units per hour, or Btu/h. To figure out the maximum cooling capacity for a room, you will first need to calculate the size of the room where the air conditioner will be placed.\nTo do this, simply multiply the width of the room by the length of the room.  Use the following information and formulars to help you with calculating your air conditioning requirements:\n\n\nHeat Load\n\nThe amount of heat generated is known as the heat gain or heat load. Heat is measured in either British Thermal Units (BTU) or Kilowatts (KW). 1KW is equivalent to 3412BTUs.  The heat load depends on a number of factors, by taking into account those that apply in your circumstances and adding them together a reasonably accurate measure of the total heat can be calculated. Factors include: \n\nThe floor area of the room\n The size and position of windows, and whether they have blinds or shades \nThe number of room occupants (if any) \nThe heat generated by equipment \nThe heat generated by lighting \nArea of Room\n\nThe amount of cooling required depends on the area of the room. To calculate the area in square metres:\n\n\nRoom Area BTU = Length (m) x Width (m) x 337\n\n\nWindow Size and Position\n\nIf, your room has no windows, you can ignore this part of the calculation. If, however there are windows you need to take the size and orientation into account.\n\n\nSouth Window BTU = South Facing window Length (m) x Width (m) x 870\nNorth Window BTU = North Facing windows Length (m) x Width (m) x 165\nIf there are no blinds on the windows multiply the result(s) by 1.5.\nAdd together all the BTUs for the windows.\n\nWindows BTU = South Window(s) BTU + North Window(s) BTU\n\n\nOccupants\n\nYou will have to take that into account people who normally working in the space. The heat output is around 400 BTU per person.  \n\nTotal Occupant BTU = Number of occupants x 400\n\n\nEquipment\n\nClearly most heat in a room is generated by the equipment. This is trickier to calculate that you might think. The wattage on equipment is the maximum power consumption rating, the actual power consumed may be less. However it is probably safer to overestimate the wattage than underestimate it.\n\n\nEquipment BTU = Total wattage for all equipment x 3.5\n\n\nLighting\n\nTake the total wattage of the lighting and multiply by 4.25.\n\n\nLighting BTU = Total wattage for all lighting x 4.25\n\n\nTotal Cooling Required\n\nAdd all the BTUs together.\n\nTotal Heat Load = Room Area BTU + Windows BTU + Total Occupant BTU + Equipment BTU + Lighting BTU\n\nThis is the amount of cooling required so you need one or more air conditioning units to handle that amount of heat.\n\n--------------------\n\nGeneral thumb rule \n\n100 sqr ft area with roof height of 9 - 10 ft will need 1 Ton AC unit.
428509
dA = ½ (b.dh + h.db)\nSo δA ≈ ½ (b.δh + h.δb)\n= ½ (100 x ±1 + 10 x ±2)\n= ½  * ±120\n= ±60ft²\n\nCheck:\nMin area = ½ * 98 * 9 = 441ft²\nMax area = ½ * 102 * 11 = 561ft²\n\nNow 500±60 ft² = (440 to 560)ft²\n\nSo estimate is pretty close!!!
648202
Is he stupid?  Please show him this question and all the answers saying he an idiot.  Air conditioners run based on how hot it is inside the building and if you are constantly letting more heat into your building, the air conditioner will keep running.  Not only is that a huge waste of money, but it also doesn't effectively cool the building.
201925
The top concerns that come to my mind are Radon gas and mold/mildew if the basement is damp.  Both can be tested for if you are concerned.
392382
enough vapour to run a car
477399
How about a small fan with a bowl of water in front of it to increase air humidity.  It may be worth trying...;
140880
Cooling, Heating, and Dehumidifying\n\n3in1Fig1.jpg (23385 bytes)\n\n    As you turn on the unit,  the refrigerant in the system (in liquid form and with low temperature and low pressure), pressurized by the compressor, turns into gas. The gas then enters into the condenser and perform a process called heat exchange. In this process, refrigerant turns back into liquid by releasing heat energy to the air (Airflow 2) and becomes cooler than before. The liquid refrigerant with cooler temperature enters into the evaporator and perform another heat exchange with the air (Airflow 1). In this process, the heat energy in the air is absorbed by the refrigerant in the evaporator and the resulting cooler air is exhaust through the front vent. Note: Airflow 1 and Airflow 2 are independent of each other.\n\n     \n\n    Because the evaporator is lower in temperature than the air, water vapor will condense if the evaporator's temperature is lower than the ambient dew point. This cycle of condensation will lower the Relative Humidity of the room and thereby achieve the goal of dehumidification.\n\n    3in1Fig3.jpg (16322 bytes)  3in1Fig4.jpg (19482 bytes)\n\n    However, if the unit is used as a dehumidifier, remember to remove the air exhaust duct from the back in order to effectively dehumidify the room. If not, constant air flow from the outside (in order to compensate the air flow to the outside through the air exhaust duct) will make the dehumidifying process ineffective.\n\n     \n\n    Heating function (optional in most models) is included with the intention to make this unit an all-season air-conditioning unit. Operationally, this function is just the same as other features. You can operate the fan, louver, and timer, just as you can for cooling.\n\n________________________________________________________________
603864
http://en.wikipedia.org/wiki/Air_conditioner#Air_Conditioning_System_Equipment\nAs for the water in the air, you have to remember, that this water is taken out of all the air that is passing through the air conditioner for all the hours it is running.  That's a LOT of air.  Air can hold more water than you think.
45664
Hey what is your climate? You know what? It doesn't matter. 150 sq ft is like a walk in closet. Get the smallest window unit you can find. Or get a great big one from a thrift store or flea market. A hand held fan blowing over an ice tray could cool that room. That must be a school question or something so you probley want pounds. 15lbs
150161
You need to know the velocity of the air entering the pipe.  In your case that will depend on the speed of your vehicle.  \n\ncfm= cubic feet per min  (strange units given that your pipe is metric)\nradius of pipe = 36mm / 2 = 7.0866 in\nArea of the crosssection of the pipe is: 3.1415 x (7.0866)^2 = 157.8 in^2 = 1.0956 ft^2\n\nFlow of air into pipe, Q = 1.0956 ft^2 * velocity of air entering pipe in ft/min = Q in cfm\n\nvelocity of air entering pipe is probably the same as the velocity of your car.
226303
You have a 1.5 HP AC that is always running and a 2.5 HP AC that trips every 5 min. All things being equal which AC will consume the least power. \n\nAC’s have the compressor and a fan. The compressor consumes most of the power when it is on. I would guess that the difference in power consumption between the two AC fans is fairly close (a fraction of a HP each), which effectively factors the fans out. That leaves the compressors. A 2.5 HP AC will consume 1.7 times more power than the 1.5 HP AC. Therefore, if the 2.5 runs 5 min before trip than the 1.5 should go 8 or 9 min before tripping. If the 1.5 stays on all day then the 2.5 is more economical by far. \n\nPerhaps there is a malfunction either AC.
311509
depends on the transit time to thermal transfer rate ratio.\n1) Adiabatic if the transit time is much less than the thermal equalization time. \n2) PV = nRT if the transit time is much greater than the thermal equalization time. \n\nSo for the sake of simplicity, for option 2\nV = nRT/P\nV1/V2 = T1P2/T2P1\n\nT1/T2 = (273 + 7)/(273 + 20) = 0.9556\nP2/P1 = 33/49 =0.6735 [33 ft depth is +1 atmospheres]\n\nV1/V2 = 0.9556 * 0.6735 = 0.6436\n\nV2 = V1/0.6436 = 3.12 cm3\n\nAs the transit time becomes less than the thermal equilibrium time, the air bubble is cooler so the volume is less than the calculated value.
484651
dry air has a density of ρSTP = 1.293 g/L.