BTU readings for Air Conditioners

[Los11]

I'm a summer intern and I'm working on electrical energy conervation. One area I would like to improve electrical usage in would be the air conditioners. Here at the plant, we use chiller systems and package units for some buildings. With the little bit of thermo that i have had so far, I managed to calculate the amount of power needed to heat a certain volume of air from 45 degrees to X degree in the chiller unit. I used Air density = 1.29 kg/m3; Cp = 1000W/kg-K ; and substituted K with C b/c it would be the change in T anyways. The only part im missing is the time it will take for it to actually heat this volume to the temperature. How is that calculated? Also, does the BTU listed on AC units represent the amount of energy needed to raise the room size range 1 degree?

With the package units that operate just like your conventional home unit, does the BTU represent the same as the above question? Also, how would i go above calculating how long it will take for a volume to cool to specified temeprature?


This would be very help for my project. Thanks Alot

 

[ChasBean1]

You usually don't worry about the time it takes. The analysis is for a steady-flow, open system and we assume equilibrium conditions. If you're really concerned with the time you can relate it to air exchange rate: Rate of air exchanged is the volume flow rate of air in divided by room volume. If you have a 90°F air mass in a room it would take 55°F entering air one air exchange (e.g., one minute for a 1,000 ft3 room with 1,000 cfm of air entering) to reduce the room to 55°F. Of course this is over-simplified because there is a heat generation rate within the space.

The BTU rating on the A/C units you mention is BTU per hour. It's a rate of heat transfer. About 15 BTU/hr per square foot is a pretty good size for residential.

Also, if you're calculating power with respect to cooling, use enthalpy differences versus temperature differences to account for latent heat.

 

[lilliput1]

Read up more on energy conservation retrofits like using variable frequency drives, using central chilled water fan coil units instead of package air cooled DX units, relamping with energy efficient light fixtures and or using motion detectors for light switching, space temperature setback during unoccupied period, variable air volume air conditioning, variable volume hoods, using high efficient motors, cutting pump impellers, etc. Avoid start/stop, duty cycling since they cause wear & tear on equipment.

AC controls automatically start/stop the compressor in respond to the load. They are sized for the maximum load. So I am not sure where you are leading with regards to time to heat the space.

 

[MintJulep]

The BTU (really BTH/hr) rating on any piece of direct expansion HVAC equipment is not a constant. It is the rating at some set of conditions that may be(but usually are not) stated on the name plate. The cooling produced by, and power drawn by dx equipment is a function of the temperature of the air entering the condenser and evaporator coils, as well as other factors.

Calculating the time required to produce a given change in temperature (pull-down time) is a VERY complicated thing to do, especially as you almost never have all of the information that you need.

As ChasBean notes, most cooling calculations are done assuming steady-state operation. (Probably because the transient calculations are such a bitch.)

However, if you really want to take a stab at calculating pull down time here is a VERY VERY crude approximation:

Calculate the heat load (rate at which heat is entering) the space of interest.

Calculate the total volume of air in the space of interest.

Assume that the cooling capacity of your equipment is a constant, as noted on the nameplate.

The excess cooling capacity is avaiable to reduce the temperature of the space.

For sensible heat only (in American units) capacity = 1.1 x CFM x delta T. BTU/hr = (ft^3/minute) x degrees (The 1.1 contains lots of unit conversions to change cubic feet to pounds at standard conditions. I don't remember all of them, and am too lazy to look it up at the moment. Its not important at the moment.) So, now you have BTU/hr and cubic feet Shuffle terms around a bit and you arrive at the number of minutes it takes to cool that many cubic feet of air down 1 degree.

Of course latent heat removal will lengthen this time. Also, in real life you are not cooling only the air, you are cooling all of the equipment in the room. If you know the specific heat and mass of all the things in the room, you can throw those into the equation as well. (This is where I inevitably give up on this calculation, as it is redicuously time consuming to collate all of this data.)

 

[ChasBean1]

Q = m cp dT (excluding latent heat). You have mass flow (m) which is density * volume flow, so Q = rho v cp dT. Numbers you find recurring for standard conditions are rho =.075 lbm/ft3; cp = .24 BTU/lbm°R; and 60 minutes per hour (to change BTU/min to BTU/hr). They multiply together to get 1.08 cfm dT...

We usually take 1.08 cfm dT for granted but not using the root equation can give bad results at different temperatures or high altitudes, etc. Figured this wouldn't hurt to post.

 

[MintJulep]

ChasBean,

Thanks for the dimensional analysis. I was just too lazy to work through it or look it up last night.

You are absolutly correct that 1.08 cfm dT can get you into trouble if you are significantly away for standard temperatue and pressure. When it matters use the right values for density and specific heat.

 

[ChasBean1]

Mint, this was meant as an add-on, not to sharpshoot you. I do that all the time - give standard eqns. without derivation. This particular one is pretty common though so it's good to be reminded of its source as it might be different sometimes. Best regards, -CB