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Definition of 0.8 bar?

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Old 02-14-2005 | 02:53 PM
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Tom 993tt
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Question Definition of 0.8 bar?

Is there a scientist (or meteorologist) in the house? I feel so inadequate...

What exactly are we measuring when we talk about 0.8 bar? Has the turbo compressed the gas/air mixture to 1.8 times the normal atmospheric pressure?

Inquiring minds want to know.

Thanks,
Old 02-14-2005 | 03:23 PM
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Hi. I think I can handle this one. The measurement "bar" is given to the pressure exerted by gas/gasseous mixture. Here in the USA, it's commonly expressed in pounds per square inch or lbs/in2. When you measure the pressure in your tires you think "pounds", but what you are really referring to is pounds per square inch.

Now, if we decided to speak in Metric or use some other foreign equivalent of "pounds per square inch" you could use the term "bar" instead. One bar is equivalent to 14.508 pounds per square inch. So, 0.5 bar would then be half of 14.508 lbs/inch2 (7.254), and so 0.8bar would be 11.606 lbs/inch2.

So, in your car when your turbo is making 0.8bar there is 11.606 lbs/in2 of pressure between the cold compressor all the way up to intercooler and eventually into the intake valves. If your intercooler had a tire valve, you could effectively take your tire pressure guage and measure this exactly as you would your tires.

hope this helps,

Jaime
Old 02-14-2005 | 04:23 PM
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Further you need to consider gauge pressure vs absolute pressure. The .8 bar you are referencing is gauge pressure, not absolute. Absolute pressure on a standard day at sea level is 14.7psi or 1bar or 101.3kpa, or 1013 millibars. As you increase altitude, you decrease pressure. So, your .8 bar of boost is really 180kPa absolute, 1.8bar absolute, or about 1800millibars at standard conditions.

If you are looking at the gauge, then you are measuring the additional pressure ABOVE atmospheric pressure. On a standard day at standard pressure, then the pressure ratio would be 1.8. If however, you are seeing .8bar on the gauge, and you are at 5000ft above sea level, then your pressure ratio is much higer than 1.8 because the atomospheric pressure would be lower than at sea level.

Last edited by Geoffrey; 02-14-2005 at 05:25 PM.
Old 02-14-2005 | 04:23 PM
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Roughly..1 Bar = 14.7 PSI = atmospheric pressure at sea level. And yes, when you say 0.8 Bar that would be above atmospheric pressure => 1.8 Bar.

I hope I am right !
Ok Geoffrey, we typed at the same time, only you are much more convincing
Old 02-14-2005 | 05:06 PM
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Thanks for the helpful responses.

So does all this mean that the quantity of gas and air molocules being crammed into the cylinder has increased by 80% when we are at 0.8 bar (as compared to the quantity of molocules when the gauge is at 0.0 bar)?
Old 02-14-2005 | 05:22 PM
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Put simply...No. The reading on the "boost" gauge only means that there is 0.8 BAR positive pressure in the intake plenum over the atmospheric pressure read by the ECU sensor outside the engine.

How many molecules of fuel/air actually make it into the cylinder for combustion is part of the whole art/science of engine dynamics. Many factors such as intake port flow, valve event timing, exhaust flow, piston velocities at what point in the valve timing, intake pulse effect, charge air mass moemtum, etc all factor in to how much actual combustable mixture gets compressed in the cylinder and ignited. Much of all this is highly debated and makes for the wonderful hobby we enjoy and argue over !!!

However you are interested in all this, do not stop asking questions ! The topic goes on and on. Just take all of the anwsers you get and know that at the end of the day, engines cannot defy the laws of physics, thermodynamics and if this was all quite simple, we would not have much of a hobby/interest to immerse ourselves in.

Enjoy the quest.
Old 02-14-2005 | 05:42 PM
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When you are talking about quanity of air, you really need to know the volume of air flow and the density to determine the mass of the air. The density of air on a standard day at sea level is 2.7 * pressure / Temp. So 2.7 * 14.7 / 520R = .076 lbs/cubic ft of air. If you have an engine that has 360cfm of air at a standard day, it would have 27.36lbs/min of mass flow. Now, double the pressure and you get 54.96lbs/min of mass flow which is about double. The volume stays the same, but because the density has increased, the mass does as well. This is one of the reasons an air flow meter does not provide consistent metering of fuel under all conditions. It only compensates for air temp, and not for pressure.
Old 02-14-2005 | 07:38 PM
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Tom,

I prefer your simplified approach. If you had a 250 HP engine running at 0 bar (normally aspirated), and then slapped on a blower or turbo with 15 PSI (1 Bar), you would expect that the HP would double to 500. You'd have to subtract the hp "cost" of driving the blower or turbo, and then consider some the inefficiencies mentioned above. I think that your simplified approach actually gets you pretty close.




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