AC Delete?

[wilfongtsi]

So since I never ever use my AC I was wondering if its possible to replace my engine belt with a non AC belt and not remove any of the AC components, just in case I ever change my mind and want to put it back on, So I've found a non AC climate control and bought a non AC engine belt, but cant really see what needs to be done to take the belt off? Is there anything else I'm missing to put the belt on? Thanks in advance.

[joecvt]

Why bother doing this if you are not going to remove the components like the compressor and such that weigh the most. An A/C left off does not use really any power from the engine at all. It is freely rotating until the A/C clutch is engaged. The compressor never rotates if the A/C and defrost are off. If you don't remove the additional weight of the A/C components, you will not see a real difference in MPG.

JoeCVT = Just your average CVT owner

[wilfongtsi]

Its more something I was going to try, which is why I wanted to be ble to put it back if needed, I know there is extra weight from everything, but how much extra weight is it actually? I also realize it doesn't really cause any huge draw if its not running but at the same time if it doesn't need to be spinning it, it should help a lil? Thanks for your reply tho, but I'm still at a loss if this could be done by leaving the compressor on there?

[Hitech]

Looking at the manual would apear to show the the Non AC uses the same idler and a shorter belt. So try the non-AC belt.

[JimIsbell]

While you may never use the AC, removing it or disabling it will also kill your defrost or severely limit it. When you get fog on the inside of the windshield, you will not be able to remove it quickly without the AC. Personally I think it would be a safety hazard to disable the AC.

[Hitech]

I wish there was a quick way of removing the fog from outside of the windshield without running the wipers and smearing the glass (washers make it worse). Running full hot defrost takes about 5-6 min (South Florida summer problem)

[Fl_Fyr_Fytr]

(South Florida summer problem)

Wow, there are a few of us here, eh?

[baboracus]

wilfongtsi, you have 2 choices:
1. you can remove the outer pulley on the AC compressor which is a hassle because it has one of those circular clips on it.
2. you can bring the car to a shop and drain the refrigerant for around $15 and then remove the compressor (6 bolts). First, you need to remove the underbody splash guards however.

If you leave the compressor on the vehicle the non-AC belt WILL DRAG on the compressor pulley.

I'm a big cheapskate but I still would HIGHLY recommend option 2, as it is a lot easier than option #1, and you probably won't break anything.

[Hybrid-Battery-Repair]

Originally Posted by wilfongtsi

I know there is extra weight from everything, but how much extra weight is it actually?

If you keep your tank below 3/4 at all times, you'll probably save more weight.

Seriously, between the compressor, evaporator, condenser and lines, I don't think you break 25 lbs (but that's just a guess).

[baboracus]

the AC compressor + outer spinning component = 13 lbs.

outer spinning component = 3.56 lbs = 1.616 kg
outer spinning component radius = 2.5 inches = .0635 meters

at 2000 RPM (average engine revolution) the spinning component is spinning at 209 rad/s

energy = 1/2 * mass * radius^2 * (rad/s)^2

= .5 * 1.616 kg * .004 meters^2 * 43681 rad^2/seconds^2

= 142 kg * m^2/s^s = 142 J

1 J/s = 1 W
746W = 1 HP


So, if HP on engines is calculated by the second, and for each second you are using 142 J.

Then you are using 142W of power to spin the AC compressor, which is roughly 142/746 = 0.2 HP.

If there are any mistakes in those calculations please correct them! But otherwise, 13 lbs and +0.2HP is the gain I can conclude from removing the AC compressor.

0.2 HP may not sound like much but if we compare it at the max rating of 73 HP using 5700 RPM then the HP gain comes out to be:

.5*1.616*.004*596.9^2 = 1152W = 1.5 HP

To get even more fancy 1.5 HP is approximately 2.1% of the engine's power.