To double an engine output.
#1
To double an engine output.
On a forced induction motor, 14.7 psi of boost is considered x2 the engine NA output.
So if a given 350ci engine made 400 hp NA at 6000 rpms. Would the same engine size make 800 hp at 12000 ( with a better flowing induction and exhaust system of course). Would this be right???
So if a given 350ci engine made 400 hp NA at 6000 rpms. Would the same engine size make 800 hp at 12000 ( with a better flowing induction and exhaust system of course). Would this be right???
#3
Originally Posted by 2002_Z28_Six_Speed
Why do you come up with 14.7 psi as double the power? Sure that 1atm plus 14.7 psi is two bar but I would think that there is more to it than that.
#4
Originally Posted by Wnts2Go10O
at 14.7psi it is at double the atmospheres and basically makes your motor act like twice the size. so instead of having a 346 you would in all technicallity have a 692ci motor.
theres way more to it than that. you must have better flowing heads, a cam designed around FI, and the intake and exhaust to go with it. the efficiency of the turbo or super would also play a role. plus, there is less than double the air, because it is heated and takes up more space.
#5
Also it depends om what type of FI you chose.
A turbocharger does it´s work almost for free, using the othervise wasted exhaust energy. A mechanicaly driven SC on the other hand, make the pistons and crank produce twice as much Hp that what is measured on the fly/rear wheel. It (the SC) takes as much power to drive as it is adds.
This has the effect that in theory you will have to allocate half the pressure/boost increase to drive the supercharger.
Take my 50% statement with a grain of salt, it used to be accurate for the old Roots SC´s, the new ones are propably more efficient, but the principle holds true.
Br//
A turbocharger does it´s work almost for free, using the othervise wasted exhaust energy. A mechanicaly driven SC on the other hand, make the pistons and crank produce twice as much Hp that what is measured on the fly/rear wheel. It (the SC) takes as much power to drive as it is adds.
This has the effect that in theory you will have to allocate half the pressure/boost increase to drive the supercharger.
Take my 50% statement with a grain of salt, it used to be accurate for the old Roots SC´s, the new ones are propably more efficient, but the principle holds true.
Br//
#6
Turbo's do use up energy
To say a blower needs 100 hp to create 100 hp is pretty far off even for old Roots blowers.
We are talking a LOT of power to make a 8 rib belt slip.
A compact car´s AC compressor can easyly consume 8-10 Hp, the much bigger Ac compressors in large SUV´s can consume a lot moore. There are never any issues on belt slip with Ac compressors.....
http://en.wikipedia.org/wiki/Supercharger
Positive displacement superchargers may absorb as much as a third of the total crankshaft power of the engine, and in many applications are less efficient than turbochargers
Br//
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#8
Twice the boost will have nothing at all to do with doubling the rpms. You may double the output, but at the same general range of rpms. It does not always take double the boost to double the hp. Especially on turbo engines. Sometimes it takes more; sometimes less. Depending on how the motor is set up.
#9
Originally Posted by topend
On a forced induction motor, 14.7 psi of boost is considered x2 the engine NA output.
So if a given 350ci engine made 400 hp NA at 6000 rpms. Would the same engine size make 800 hp at 12000 ( with a better flowing induction and exhaust system of course). Would this be right???
So if a given 350ci engine made 400 hp NA at 6000 rpms. Would the same engine size make 800 hp at 12000 ( with a better flowing induction and exhaust system of course). Would this be right???
If it were that simple, 1 more lb of boost would always give us X 1.069% more HP. It's always less then that for the reasons already stated and whatever % it is decreases as the boost increases.
#10
your looking at it all wrong, boost is just a measurement of restriction, you cant judge power by it. you need to be looking at airflow if you wanna put some kind of power statement to it
#12
Don't forget a power band flattens out at some point. The HP needs air to continue increasing. When it doesn't it starts to fall off.
I have the perfect example. I had a 557" BBC with a set of heads. It ran 9.32.
We put the exact same heads on a 477" BBC. It ran a 9.31.
I ran out of air flow, per the inch's available.
I have the perfect example. I had a 557" BBC with a set of heads. It ran 9.32.
We put the exact same heads on a 477" BBC. It ran a 9.31.
I ran out of air flow, per the inch's available.
#13
To double power to the wheels, you have to burn twice the air and fuel... basically. Depending on atmospheric, you may have double the mainfold pressure than a naturally aspirated engine does at 14.7 pounds of boost... It is very unlikely that you've doubled the airflow (by mass NOT volume... hence the temperature effects), or have no additional parasitic losses, either directly from the mechanical drive of a supercharger, or the increased backpressure (pumping losses) involved with a turbo when a forced induction power adder is added... The whole original premise is wrong. The goal with forced induction is NOT to make "boost"... (unless you're a ricer).
#14
Originally Posted by mrdragster1970
Don't forget a power band flattens out at some point. The HP needs air to continue increasing. When it doesn't it starts to fall off.
I have the perfect example. I had a 557" BBC with a set of heads. It ran 9.32.
We put the exact same heads on a 477" BBC. It ran a 9.31.
I ran out of air flow, per the inch's available.
I have the perfect example. I had a 557" BBC with a set of heads. It ran 9.32.
We put the exact same heads on a 477" BBC. It ran a 9.31.
I ran out of air flow, per the inch's available.
#15
Originally Posted by EdmontonSS
To double power to the wheels, you have to burn twice the air and fuel... basically. Depending on atmospheric, you may have double the mainfold pressure than a naturally aspirated engine does at 14.7 pounds of boost... It is very unlikely that you've doubled the airflow (by mass NOT volume... hence the temperature effects), or have no additional parasitic losses, either directly from the mechanical drive of a supercharger, or the increased backpressure (pumping losses) involved with a turbo when a forced induction power adder is added... The whole original premise is wrong. The goal with forced induction is NOT to make "boost"... (unless you're a ricer).
My question is ,if u spin a motor 2x the peak power rpm would it make 2x the power??
#16
Actually I still say it was less stroke, less TQ that helped the most.
I had a stock wheel well and could only fit a 29 X 10 tire.
The small motor had a better 60'.
The small motor did spin a little higher, not much, but I think that was because the converter was slipping more. I have several converters.
I had a stock wheel well and could only fit a 29 X 10 tire.
The small motor had a better 60'.
The small motor did spin a little higher, not much, but I think that was because the converter was slipping more. I have several converters.
#17
Originally Posted by topend
i agree with u to double the power u need to burn twice the fuel.
My question is ,if u spin a motor 2x the peak power rpm would it make 2x the power??
My question is ,if u spin a motor 2x the peak power rpm would it make 2x the power??
"Peak power" is peak power, power drops from "peak power" because of added frictional and pumping losses past "peak power"...
Power is torque x RPM... Most LS1 make "peak torque" at 4000-5000 RPMS, this is where volumetric efficiency, pumping losses and frictional losses all combine to make the engine most efficient. Torque generally falls past this point, but the increased RPM allows the engine to continue to produce more power. Past the "peak power" the engine become less efficient at filling the cylinders with air/fuel charge (volumetric efficiency), friction and pumping losses increase, past "Peak Power" the RPM increase no longer makes up for the loss of Torque.
If you could double the engine RPM AND keep the engine Torque constant would double power, but that's the only way...
#18
Originally Posted by 3.4camaro
theres way more to it than that. you must have better flowing heads, a cam designed around FI, and the intake and exhaust to go with it. the efficiency of the turbo or super would also play a role. plus, there is less than double the air, because it is heated and takes up more space.
pretty sure this is a theoretical question...the guy is talking about getting his sbc to rev to 12,000...no easy feat.
#19
Originally Posted by topend
i agree with u to double the power u need to burn twice the fuel.
My question is ,if u spin a motor 2x the peak power rpm would it make 2x the power??
My question is ,if u spin a motor 2x the peak power rpm would it make 2x the power??
i dont think the power/rpm is quite as linear as a power/boost (ceteris paribus)
...but then again F1 cars make 1,000hp when they spin 20,000rpms
cansidering a moderately efficient 3.0l makes about 350hp@7,000rpms...i dont know where i was going with that
#20
Power/boost is not linear either...
If you said Power/Air Density.. i'd be more inclined to belive it
Compressing Air heats the air.. due to "inefficiencies" in the compression process. So first off 14.7lbs of boost is likely only about a 1.6 density increase.. (if that, depending on intercooler of course).
Next, Friction in the engine does not go up the same as doubing engine size. You in effect have less friction due to the smaller engine (i.e. not doubled)... although.. the friction does increase do to increased peak cylinder pressures.. but valvetrain and such remains largely unchanged.
Boost system power.. yes.. it takes power to make that boost.. effciency in the boost system will have a large effect as the power used directly comes out of the crank. oh.. and turbos do take power.. contrary to general publics belief.. it comes out as pumping losses
so in the end.. even doubleing density ratio doesnt always equate to doubled brake power.
However, it does roughly equate to doubled indicated power (twice oxygen content mixed with twice fuel content.. assuming perfect combustion and MBT for both conditions) But then again.. there are even factors that make that statement not 100% true... (i.e. flow through cylinder head, exhaust manifold pressures, etc..)
If you said Power/Air Density.. i'd be more inclined to belive it
Compressing Air heats the air.. due to "inefficiencies" in the compression process. So first off 14.7lbs of boost is likely only about a 1.6 density increase.. (if that, depending on intercooler of course).
Next, Friction in the engine does not go up the same as doubing engine size. You in effect have less friction due to the smaller engine (i.e. not doubled)... although.. the friction does increase do to increased peak cylinder pressures.. but valvetrain and such remains largely unchanged.
Boost system power.. yes.. it takes power to make that boost.. effciency in the boost system will have a large effect as the power used directly comes out of the crank. oh.. and turbos do take power.. contrary to general publics belief.. it comes out as pumping losses
so in the end.. even doubleing density ratio doesnt always equate to doubled brake power.
However, it does roughly equate to doubled indicated power (twice oxygen content mixed with twice fuel content.. assuming perfect combustion and MBT for both conditions) But then again.. there are even factors that make that statement not 100% true... (i.e. flow through cylinder head, exhaust manifold pressures, etc..)
Last edited by XBR24; 12-11-2006 at 08:38 PM.