Amount of Energy Required to Cool the Intake Air

In summary, the Camaro's A/C only uses a heat source to do the cooling, and that got the author thinking about applications for the car. First, the car only uses a cabin air A/C, not what the author cares about. Secondly, as a way to cool the intake air, the author wants to be able to cool the intake air a significant amount. The car will only suck 15,000L/min of ambient air at most, but we'll say it's always sucking this much. Let's say we're at sea level. The author calculates that it will take 175,833 BTU/min to cool the intake air a 9.7 degrees Celsius. The "certain time" is a
  • #1
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I recently learned about absorption refrigerators, which only use a heat source to do the cooling, and that got me thinking about applications for the Camaro. First, just for the cabin A/C (not what I care about). Secondly, as a way to cool the intake air (what I do care about). I want to be able to cool the intake air a significant amount. So I think I need to know how much energy it will take to cool the intake air a certain amount in a certain time. The car will suck 15,000L/min of ambient air at most but we'll say it's always sucking this much. Let's say we're at sea level. The ammonia which is used as the coolant has a boiling temp of -33[tex]\circ[/tex]C. So i guess that's as cold as the intake air could possibly get. I guess variables would be intake air temp and the change in the temp of the intake air. The "certain time" is a minute because of the measure of the air flow.

So this is what I did.
Q=mc[tex]\Delta[/tex]T
I used 10,000 BTU's as an energy amount because that was what I saw some A/C's rated as. And it was rated in BTU/hr so I divided it by 60 to get BTU/min and then multiplied by 1,055 to get Joules.
Q=175,833
m=15,000L*1.2g/L=18,000g
c=1.007J/g

175,833/(18,000*1.007)=[tex]\Delta[/tex]T=9.7[tex]\circ[/tex]C

So that means that the air will be 9.7 degrees cooler after passing by the A/C on it's way to the intake manifold, right? This doesn't feel right to me and I know that the specific heat of the air changes with the temp of the air but I don't know how to use that. I also think that the rate of heat exchange between two things will be greater the greater the difference between the two things' temperatures are. I'm sure there are also many other things that I'm leaving out. I spent all day trying to figure this out. I just don't know enough and can't find the info. Basically, I want to be able to calculate how much energy it will take to cool the air a to-be-determined amount in a minute when the air will be a to-be-determined temperature and the coolant will be a static temp (-33C). I'm not afraid to learn, btw. So don't hold back.

I understand that it probably isn't really practical to use an A/C to cool intake air, btw, so don't worry that I'm disillusioned with some fantasy miracle solution.
 
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  • #2
the short answer is ..it takes more energy to do the cooling , than the engine would produce. not worth the effort..what is desired is a way to improve the volumetric efficiency of the Internal Combustion engine. All an IC does is act as a heat pump. converting chemical energy, gas, into heat and useable power. the more fuel you can pack into the combustion chamber, the faster you go. now a days, the trick is to make the same amount of power with less $ 4 gallon fuel. thia leads off to another subject but returning to the cool intake question..we did similar work 40 years ago and last year. for years racers used a cool can to cool the. fuel. is simply a can with a coil of aluminum fule line mounted in it. the racer would put ice in the can before a race.theoretically, the fuel is cooled more than ambient temperature and is a denser mixture of fuel / air thus more power. this is why older cars had chokes on the carburators, to "choke off" the air supply in real cold weather. the choke would kick off once the IC reached operating temp. Cool can- i can testify, after much testing that this is not a good solution. not enuff time or cold enuff media to cool the fuel.Back when drag racing in the 1960s, we went so far as to run acetone and dry ice in the can..froze the fuel line up to the fire wall..ran like a bat out of he--... but..our driver was overcome by fumes and started our in high gear at the starting line..have you ever heard a 426 cubic inch motor lug down when the hammer is dropped at 7000 RPM? Expen$$$$$$ive...

back to the question..to cool your intake to generate enough power ( fuel economy) to justify the modification, the technology and energy required to do the job,cost effectively, is not currently available. ifin it were, Detroit would have done it already.
 
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  • #3
I know it's not practical. I stated that at the end of my first post. If it were, it would be done. I also know about liquid to air intercoolers, for a FI setup. And nitrous and CO2 spraying and all that. I was just giving some background. I want to know how to do the math, and it would be greatly appreciated if some one could help me out here.
 
  • #4
I checked your math - your delta-T calculations are correct.

One note: over the temperature range you are dealing with, I don't think the specific heat would change much. Not certain, though - I don't know what the air temp would be, I just can't imagine it would be enough to matter.
 
  • #5
Thanks Russ, that gives me confidence. I've been doing a lot of other math to find this and that as well. If I have any more questions, I will post them in a new, appropriately titled thread. Thanks for your help.
 

1. What is the purpose of cooling the intake air in a scientific experiment?

The purpose of cooling the intake air is to lower the temperature of the air entering the system, which can have a significant impact on the overall energy efficiency and performance of the experiment. This is because cooler air is more dense and contains more oxygen, allowing for better combustion and more power output.

2. How is the amount of energy required to cool the intake air measured?

The amount of energy required to cool the intake air is typically measured in joules or watts. This can be calculated by measuring the change in temperature of the air as it passes through the cooling system and applying the specific heat capacity of air to determine the amount of energy needed to cool it to the desired temperature.

3. What factors can affect the amount of energy required to cool the intake air?

The amount of energy required to cool the intake air can be affected by a variety of factors, including the initial temperature of the air, the specific heat capacity of the cooling system, the flow rate of the air, and the desired temperature of the cooled air. Other factors such as the efficiency of the cooling system and any external influences like ambient temperature or humidity can also play a role.

4. How does the amount of energy required to cool the intake air impact the overall efficiency of the experiment?

The amount of energy required to cool the intake air can have a significant impact on the overall efficiency of the experiment. This is because the energy used to cool the intake air is subtracted from the total energy produced by the experiment, which can lower the net energy output. Therefore, it is important to minimize the amount of energy required to cool the intake air to maximize the overall efficiency of the experiment.

5. Can alternative methods be used to cool the intake air and reduce the amount of energy required?

Yes, there are alternative methods that can be used to cool the intake air and reduce the amount of energy required. For example, using a heat exchanger or a more efficient cooling system can help reduce the energy needed to cool the air. Additionally, implementing insulation or reflective materials in the intake system can also help reduce the amount of heat absorbed by the air, resulting in less energy needed for cooling.

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