How can you caluclate how fast a A/C unit can cool a room? For instance if you had a a/c that can give you 10000Btu/Hr and a room of 1000Cubic Feet, and the tempature is 100F in the room and you want to bring it to 70F. How long would that take? Also how can you find the instintanious temp of the room at any point during the test?
I would say: i) if the room volume is constant and there is no door opened, the density is constant. ii) Conservation of Energy: [tex] \rho c_v \frac{dT}{dt}=-Q_{AC}=-xxxxx Watts[/tex] C_v represents the constant volume heat capacity of the air. Integrating the equation you will have a linear evolution of T. It's a simple model.
Assuming no condensation, Clausius2's equation will work in theory. In reality, it depends on so many unquantifiable variables that its not something that's ever calculated.
Yea, to be more exact, your probably going to have to use a psychometric chart to account for humidity and things like that. If you were to assume it was dry air, then just find how many BTUs need to be removed, use enthalpy since it's a closed system. Then divide by the your BTU/hr of your A/C. Also, you may want to see if that's actual performance of the A/C or just what it uses. If it is just what it uses, then you will have to divide by some efficiency to get actual BTU "removal." I would say since it's only a 10'x10'x10', you could maybe say there is a fan and that the temperature in the room is equal throughout. Oh, also, you will need to find out the mass of the air in the room (so Btu = Btu instead of Btu/lbm, which is what you'll look up). To find that, you can use the ideal gas law, simple stuff there.
You can do a very rough calculation if you know how many CFMs are recirculated, and how much power is generated within the room. <edit : just noticed russ' presence...backs away gingerly>
Do you mean condensation of the inner air?. I don't think that will be the main difference. Although I know you work in this stuff in your professional life, I am going to jump into the pool and say that the main "error" of the balance I posted is to assume that the heat exchanged is constant. As the refrigeration cycle states, the efficiency of the AC system will depend on both inner and outer temperatures, and so the total heat exchanged. So that, the heat exchanged is coupled with time in some way I don't feel like to calculate .
Plus you have to take into account the items in the room, southerly facing walls, how many windows, if the room is carpeted....This is definitely one case that I am glad there are many tables involved.
Well here is the information I have to deal with. 1. The room is 10x10x10 and it is empty 2. The air being circulated taken from the room put into the AC and then put back into the room at the rate of 200 cubic feet/min 3. BTU/hr changes with time over a 3min cycle from about 2000BTU/hr to 5000BTU/hr 4. the room starts at 95F and it needs to get to 70 5. Humidity is around 20% at 95F I dont know any of the calculations to get anywhere with this... Any help is apreciated. George
Get out a psychometric chart and look up 20% Relative Humidity and 95°F. Follow it over to find your humidity ratio (i think), its omega either way. That will give you a mass ratio of water to air in the room. Now, cool the air down to 70° by moving straight left until you hit the saturation line, then follow that line down to 70°F. The point that you hit the saturation line is called the Dew Point, any cooling beyond that point will result in condensation. Anyways, the initial and at the final point, if you trace those points "northwest" you will be able to find enthalpy. From there I would think you could just get your difference in enthalpy and will then proceed to use your energy balance. OK OK, I just remembered that I have my Thermo Book here at work. Here's what you do: First, write a Dry Air and Water mass balance m(a1) = m(a2) = m(a) m(a1)w(1) = m(a2)w(2) + m(w) where terms in () are subscripts and a is air, w is water, and w is omega equals the humidity ratio in kg water/kg dry air gotten from the psychometric chart (sorry, if anyone wants to Latex this, they are more than welcome) from the water balance you get m(w) = m(a)[w(1) - w(2)] also note that all m's and Q's are mass and heat flow respectively, not just mass Plugging into energy balance, we get: Qout = m(a)[h1 - h2] - m(w)h(w) *where h1 and h2 are from the psychometric chart, and h(w) js h(f) at your final temperature So, get your chart out and get h's, omegas and specific volumes for both points. m(a) mass flow of air = V1/v1 (V is volume flow) m(w) = m(a)*(w1-w2) Qout(flow) = m(a)[h1-h2] - m(w)h(w) and this gives you units of btu/min Then I would guess you take the Btu of your AC, divided by that to give you units of minutes. I apologize for this being unclear and probably making no sense, but maybe you'll get a couple of tidbits outta it that will help you solve the problem.
Grab a psychrometric chart. Mark the point that intersects the graph at 20% humidity and 95 deg F. All you have to do in this case is sensible cooling. your air will end up at the 70 deg F point directly to the left of the 95 deg F point about at the 45% RH mark. Get your Enthalpy starting and ending (approx 31.2 btu/lb at 95 Deg F down to approx 24.6 btu/lb at 70 deg F). Here is a site that will show you what the line will look like Psychrometric chart All you are really dealing with is cooling a certain weight of air. Look at the Psych chart again at the 95 deg F point. You will see it is about 14.15 cubic ft/lb so divide your room Cubic feet by this and you have (1000/14.15)=70.7 lbs of air. Multiply this by the difference in enthalpy 31.2-24.6 and you get 6.6 btu/lb * 70.7 = 466.62 btu. What does this line refer to? Heat gain? Without any other form of heat gain into the space, the system only requires 467 btu to cool the room air. If there is heat gain, then the question of time does not apply. The system will have to cycle to maintain the temperature. You will reach 70 degrees in a few minutes. Since your AC unit at 10,000 btuh gives about 166.7 btu/min it should only take about 467btu/167btum = 2.8 minutes to cool the air to 70 deg F.
I don't know what you mean by "inner air" - I mean in the cooling coil of the A/C unit. Taking air from 90 degrees down to 70 degrees at the cooling coil takes exactly the same amount of energy as taking 70 degree air down to 50 degrees if the air is dry. If the wet-bulb temperature is 60 degrees, then it takes more than twice as much energy to take that 70 degree air down to 50 degrees because you have to condense a lot of water out of it. Yes, those are the types of other "unquantifiable variables" I mentioned. To do anything here, we probably have to assume an empty box, perfectly insulated. Ok, that's a start - I don't understand number 3, though. Are you just averaging the capacity of the a/c unit? With starting and ending conditions and a heat-flow rate, you just plot the points on a psych chart (I can walk you through that....) and subtract to find the difference in enthalpy - then convert that to total BTU (the psych chart has the density and you have the volume of the room). Then just divide by the heat-flow rate. edit: ehh, since humidity isn't an issue, you can ignore the psych chart and just use the heat capacity of air: about .17 btu/lb-F and specific volume of about 14.1 ft^3/lb. One simplification: condensation (discussed above) is not an issue: 20% RH at 95F means you have to cool that air to 48F before you get much condensation. Most A/C units bottom-out at 52-55F, though since the coils are much colder, you will get a little bit of condensation even though the average air temperature doesn't really get cold enough.
I don't get that one either. My findings are about the same as yours. All we are really considering here is the sensible at that low starting humidity.
Since you've heard from the engineers, here's another side of the story, from a physicist !! All of the above calculations make one implicit assumption : that the time constant for changes in the room is large compared to the required cooling time. Since the cycle time for your room is 5 minutes (1000 cu.ft/200 cfm), and Artman's calculation gives 2.8 minutes, this assumption is just barely valid (and so, I don't expect that Artman's result would be off by very much). However, if the capacity of your AC unit were indeed less than 5000 btuh, the same calculation would possibly be off by 50% or more (still, it's easily good enough for an order of magnitude estimate). The reason for this error is that the above calculation does not account for the air circulation rate. If you do account for this you will get an exponentially decaying temperature that levels off at the output temperature of the air conditioner. Another effect is the rate at which the temperature homogenizes in the room. In your case, the room is small and 200 cfms will provide enough convection that thermal gradients of say 5F won't last more than seconds, so that's not much of a correction really. Of course, all of this is neglecting other effects. Thermal conduction through walls can mostly be neglected since the time constant there will be of the order of many minutes (likewise with the cooling of large objects in the room). The dominant effect from the objects in the room will be the first order correction to the volume of air in the room. Finally, a big assumption made here is that there are no sources of heat in the room. Since 10,000 btuh = 3 kW, heat sources within the room (lights, computers/equipment, people, etc.) may be neglected only if they put out only a tiny fraction of a kW - this is unlikely unless the room is empty and there's not much equipment running.
Having invited upon myself the wrath of the engineers, I'd better walk the walk : I'm going to use SI units for intermediate calculations so I can calculate by hand. [tex]10,000 ~btu/hr = 3~ kW [/tex] [tex]V = 1000~cu.ft = 30 ~m^3 [/tex] [tex]\dot{V} = 200~ cfm \approx 0.1~ m^3/s [/tex] [tex]C_{air} \approx 1.0~ kJ/kg-K [/tex] [tex] \rho _{air} = 1.2 ~kg/m^3 [/tex] [tex]P_{AC} = 10,000~ btuh \approx 3~kW [/tex] From the cooling power of the AC, the incoming air properties and the flow rate, the output temperature from the AC can be calculated. [tex] P = \dot{H} = \dot{m}C \Delta T=\rho*\dot{V}C \Delta T [/tex] gives [tex]3000 = 1.2 * 0.1 * 1000*\Delta T [/tex] With [itex]T_{in} = 95^{\circ}F [/tex] this gives [itex]T_{out} = 50^{\circ}F [/itex]. Since the dew point of 20% air at 95F is 47.7F, we can neglect dehumidification. (In reality, there will be some small amount of water removal, making the output temperature slightly greater than the calculated 50F.) Next we set up the differential equation for the rate of change of temperature. Over a small time [itex]\delta t [/itex], let the mass of air entering and leaving the room be [itex]\delta m[/itex]. If the total mass of air in the room is M, the temperature is T, and calling the temperature of the air from the air conditioner T' (=50F) we have : [tex] T + \delta T = \frac {MT - \delta m T + \delta m T'}{M} [/tex] Cross-multiplying, canceling terms, dividing by [itex]\delta t [/itex] and taking the limit of small [itex]\delta t[/itex] gives : [tex]M \dot{T} = -\dot{m}(T-T')[/tex] , where [itex]\dot{m}[/itex] is the mass flow rate. Writing this as : [tex]\frac{d}{dt}(T-T') = -\frac{\dot{m}}{M}(T-T') [/tex] gives the integral form : [tex]\int_{T_1}^{T_2} d(T-T') = -\frac {\dot{m}}{M} \int_0^\tau dt [/tex] The solution is simply : [tex]\tau = \frac{M}{\dot{m}} ln \left( \frac{T_1 - T'}{T_2-T'} \right) [/tex] Plugging in numbers, (and neglecting the change in density of air with temperature - I'm too lazy, but you don't have to be) we get : [tex]\tau = \frac{1000}{200} ln \left( \frac {95-50}{95-70} \right) = 5*ln(1.8) = 2.94 ~min[/tex] Ouch ! What did Artman get ? To answer the second part of your question (T as a function of time t) just take exponents in the solution above to get : [tex]\frac {T-T'}{T_1-T'} = e^{-t \dot{m}/M} [/tex] which gives : [tex]T(t) = 50 + 45e^{-t/5} [/tex] <ducking into bomb shelter and waiting out the cold, long winter>
Ooooh ! Mine only took a few milliseconds to work out in my head - actually microseconds, if you don't count the unit conversions and such nonsense AND I answered the second question too... <neener neener neener :tongue:>
Yeah, but you're good at math. I suck at it. I even overestimate how long it takes to put a couple of dots on a psych chart. You know, I forgot he asked a second question. I would have suggested he look at a thermostat. :tongue: