10-14-2012, 02:03 PM
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#11 (permalink)
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Master EcoModder
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Quote:
Originally Posted by arcosine
Humm, intake, compression, power, exhaust, that's two revolutions, one intake stroke. Piston is working agaist the vacuum 1/4 of the time, doesn't matter how many cylinders.
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Yup, and that would be part of your pumping losses, not power. For your power equation to be correct you need to have the correct number of intake strokes, which is dependent on the number of cylinders. And you're using pressure, not vacuum to calculate power.
Or, you can just skip the intake stroke calc and divide the engine displacement by 2 since that is the volume the intake manifold pressure is seeing each revolution.
Remember, the displacement of the engine is calculated by determining the volume of one cylinder (piston top dead center to piston bottom dead center), ignoring whether the design is 4 stroke or 2 stroke, then multiplying the volume of that one cylinder times the number of cylinders. So if you have a 2 liter 4 stroke engine it only pulls in 1 liter of air per revolution (a 2 stroke would pull in 2 liters, which is why they usually have a higher power density).
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10-14-2012, 02:30 PM
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#12 (permalink)
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Master EcoModder
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for you to be making 36HP..... assuming a BSFC of .5, then you would be consuming fuel at a rate of 18 lb of gasoline per hour. that's almost 3GPH. at 60MPH, that's 20MPG.
you CANNOT calculate horsepower from that kind of equation. there are way too many variables that are missing, some of which you can't really measure outside of a dyno either.
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10-14-2012, 04:06 PM
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#13 (permalink)
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You need to take into account the energy regained on the compression stroke, even if we assume constant pressure and all to make things simpler.
2042rpm is 2042/60=34.03 revolutions per second. The engine draws in half its rated displacement per revolution, so this is 1.905L * 17.01 (let's just drop the thousandths) = 32.3L/second.
The intake stroke consumes 32.3L*42.8kPa=1382 J per second, or 1.85hp. I don't know where your 36hp is coming from.
To sweeten the deal, the compression stroke now regains some of that power, probably around 40% of it. So the total loss is about 1hp. Seems about right for a 1.9L engine that runs at those speeds.
BMW only saw 10% improvement in fuel economy on naturally aspirated engines with Valvetronic, the throttle is not as evil as some people make it out to be.
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10-14-2012, 04:14 PM
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#14 (permalink)
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Master EcoModder
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Quote:
Originally Posted by arcosine
saturn sc1:
14.2-8 = 6.2 psi differential pressure
1905 liter displacement
4 stroke engine
2042 RPM at 65 mph
Pv = pressure * volume/ time
Pv = (6.2 * 4.448/.0254^3) * (1905/1000/4) * (2042/60) * 1.34/1000 = 36 hp
units:
Pv = (psi* N/lbf * m^3/in^3) * (L/m^3/intake strokes per revolution) * rpm/(min/s) = N *m/s
N*m/s /1000*1.34 = hp
6.2 psid
1.905 liter displacment
2042.14153122327 rpm
power, KW 27.2
HP vacuum 36.5
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I'm going to do this in English units since it's what I'm most familiar and comfortable with. I'll carry the units through so we can see what we're dealing with. Remember that power is work per unit time, work is force times distance, so power is force times distance per unit time. This is commonly expressed as 1 HP = 33,000 ft-lbs/minute or 550 ft-lbs/sec.
1905 cc = 114.3 cu. in.
Start with your formula:
Pv = pressure * volume/ time
6.2 psi * 114.3 cu. in./2 (this 2 is based on my previous assertion that you only have one intake stroke per 2 revolutions so you have to divide the displacement by 2) * 2042/60 = 12059 in-lbs/sec.
12059/12 (inches in one foot) = 1004.9 ft-lbs/sec.
1004.9/550 (ft-lbs/sec in 1 HP) = 1.83 HP.
So the "vacuum horsepower" is only 1.83. I think this is measuring (if it's really measuring anything) the pumping loss of actually getting the air into the engine, not the engine output. Remember the pressure in the engine on the power stroke is much higher than 6 psi.
Please point out if I made any mistakes in the equations - I've had a lot of interruptions while I wrote this.
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10-14-2012, 05:06 PM
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#15 (permalink)
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Master Ecomadman
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1.8 hp makes more sense, since when coasting in gear it doesn't make hardly any difference of the throttle is open or closed. I see the error, pressure should be divided by m^2 not m^3... that would make it 1.8 hp, if I divide the displacement by 2. Anyway the engine is operating between 1/2 and 2/3 maximum BMEP at 2042 RPM at 65 mph and a 15 mph headwind, probably a good point on the BSFC map.
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10-14-2012, 05:12 PM
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#16 (permalink)
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herp derp Apprentice
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i dont know if this is correct but i also came up with 1.83 hp using a lazier method
an online calculator at On-Line Fan Calculation
under FAN APPLICATION FORMULAS
to figure Fan Brake Horsepower
Flowrate (CFM): i entered 70 -roughly 2000liters per minute
Static Pressure at Discharge (in. Water): i entered 166 -roughly 6psi
Static Fan Efficiency (%): i entered 100
and it spit out 1.83hp
but that 1.83 hp on the table is why people can see gains with hot air intakes
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10-14-2012, 09:51 PM
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#17 (permalink)
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Corporate imperialist
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So 1.8 horsepower wasted you are never going to be able to harnes or delete.
Thats close to what it takes to run power steering or a thermal clutch radaitor fan when its "free wheeling".
That isobaric part of the otto cycle really sucks.
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10-15-2012, 01:13 PM
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#18 (permalink)
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...beats walking...
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...it's WORK sucking air into the engine (through carburetter past throttle plate), which takes ENERGY and thus POWER.
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10-15-2012, 03:07 PM
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#19 (permalink)
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EcoModding Lurcher
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Quote:
Originally Posted by Patrick
1004.9/550 (ft-lbs/sec in 1 HP) = 1.83 HP.
So the "vacuum horsepower" is only 1.83.
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Wow, I was in the shower and missed this thread. What you left out is that air is compressible and that changes the work done pumping air through a throttle. The actual formula for throttling power is:
HP = D * RPM * Pm * (k / k - 1) / (12 * 2) * ((Pa / Pm) ^ (k - 1 / k) - 1) * (1 / 33000)
Where D is displacement in cu in.
Pa is 1 atm, 14.7 psi.
Pm is the manifold pressure
k is 1.41 for air (adiabatic index or ratio of specific heats)
divide by 12 to change inches to feet, divide by 2 because 4 stroke, and 1 HP = 33,000 ft-lbs/minute
Using this example (114.3 cid, 2042 rpm, 6.2 psi map) the result is 1.8 hp, which shows your (Patrick) formula was good enough, and this post was superfluous.
-mort
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10-15-2012, 06:36 PM
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#20 (permalink)
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Master EcoModder
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Quote:
Originally Posted by oil pan 4
So 1.8 horsepower wasted you are never going to be able to harnes or delete.
Thats close to what it takes to run power steering or a thermal clutch radaitor fan when its "free wheeling".
That isobaric part of the otto cycle really sucks.
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Well the 1.8 horsepower lost to the vacuum is partially recovered on compression, and probably helps fuel atomization slightly at low engine speed, but yea even if it's 1 horsepower lost that's more than all the accessories minus AC. Good thing cooled EGR is making it to more engines now.'
On the bright side though, engine and drivetrain friction is probably eating up something like 3-4hp on the highway, so 1hp isn't all that bad in a way.
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