- #1
¡MR.AWESOME!
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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.
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.