1000 HP 20B Street Car Project
Administrative Me
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From: Kansas
Originally Posted by jimlab
I think you might be mixing up air flow (cfm) and fuel flow (lbs./hr.).
The rule of thumb for a naturally aspirated piston engine is 2 x intake runner air flow (cfm) at maximum valve lift = maximum horsepower. For an engine with intake ports flowing 330 cfm at maximum lift, maximum horsepower would be about 660. In other words, 1 cfm = 2.06 flywheel horsepower. For a rotary, I'm not sure what the equivalent would be, except that a rotary consumes about twice what its rated displacement would imply.
As far as fuel consumption, a naturally aspirated piston engine has a BSFC (brake specific fuel consumption) of about 0.50, or 0.5 lbs. per hour per horsepower produced. The same 660 horsepower engine would therefore require 330 lbs. of fuel per hour (at WOT). Divide by 8 and divide again by 0.8 (80% duty cycle) and you end up with 51.56 lbs./hr. Therefore, 10 horsepower = 0.78 lbs./hr.
With forced induction or a naturally aspirated rotary, raise BSFC to 0.6-0.7. For a turbocharged rotary, use 0.7-0.8. Convert to cc/min. by multiplying lbs./hr. by 10.2.
The rule of thumb for a naturally aspirated piston engine is 2 x intake runner air flow (cfm) at maximum valve lift = maximum horsepower. For an engine with intake ports flowing 330 cfm at maximum lift, maximum horsepower would be about 660. In other words, 1 cfm = 2.06 flywheel horsepower. For a rotary, I'm not sure what the equivalent would be, except that a rotary consumes about twice what its rated displacement would imply.
As far as fuel consumption, a naturally aspirated piston engine has a BSFC (brake specific fuel consumption) of about 0.50, or 0.5 lbs. per hour per horsepower produced. The same 660 horsepower engine would therefore require 330 lbs. of fuel per hour (at WOT). Divide by 8 and divide again by 0.8 (80% duty cycle) and you end up with 51.56 lbs./hr. Therefore, 10 horsepower = 0.78 lbs./hr.
With forced induction or a naturally aspirated rotary, raise BSFC to 0.6-0.7. For a turbocharged rotary, use 0.7-0.8. Convert to cc/min. by multiplying lbs./hr. by 10.2.
They say that 1.92 cfm is equal to 1 rwhp. Therefore, if you take your rwhp (630) and multiply that by 1.92; this gives you the CFM (1209.6). Then, to convert the CFM to lbs/min; you divide 144.17 from it and then multiply that number by 10 (for there is 144.17 cfm in 10 lbs/min). Thus, when I do that, I get 83.9 lbs / min airflow.
Here are some of the links that talk about it, which you might find interesting:
https://www.rx7club.com/3rd-generation-specific-1993-2002-16/how-much-cfms-does-fd-intake-exhaust-flow-300929/
https://www.rx7club.com/single-turbo-rx-7s-23/compressor-map-analysis-292937/
https://www.rx7club.com/single-turbo-rx-7s-23/3-rotor-t72-640rwhp-%4013psi-chance-change-342677/
https://www.rx7club.com/single-turbo-rx-7s-23/cfm-formula-use-turbo-sizing-208448/
https://www.rx7club.com/single-turbo-rx-7s-23/lbs-min-bsfc-13bt-13bre-262514/
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From: Moon Twp. Pennsylvania
Originally Posted by RX-Heven
Wow, that looks nice.
What tb are you using?
Do the vertical runners on the intake need to be so tall?
They look almost as tall as the stock lim and there is tons of room between the engine and uim because of this.
Is it possible to take a section out of the lim to lower the position of the uim to allow you to locate the injector rails on top?
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From: l.a.
Originally Posted by rx7tt95
FD Racer, question. VE peak usually coincides with torque peak or where torque and hp cross on a graph. If we pick a VE peak at 6K rpm (theoretical), the engine is still capable of flowing more air with higher rpm. Peak VE is never at rpm max as a general rule in most engines.... So one would have to take into account the increase in CFM with RPM in sizing a turbo. So if Mike's turbine is pushing off the right side of the graph, he's not running into surge. That's the line on the left, sometimes referred to as stall. There are also plenty of larger compressor wheels which are in their range at "paltry" boost levels, it just depends on how they're shaped. Now if the turbo he chose ran shaft speeds beyond what the CHRA was designed for to satisfy boost requirements, he'd still be off the right side of the graph and not the left. If the engine's CFM requirement is to the left of the surge line, the turbo wouldn't work. If you look at the compressor map the surge line clearly has the shaft speed labeled. Shaft speed has more to do with boost levels than supplying CFM which is why it's indicated on the pressure axis, no? The engine would dictate CFM, not the turbo correct? And the number would be static for any given rpm range. I'm sort of working through things in my head as I type so I apologize for the jumbled thoughts. I've always understood that the BB CHRA's were able to tolerate higher shaft speeds than standard bearing center housings. If the bearings burned up, it wasn't due to underuse. I don't see how that's possible? If it's having to spin at a greater rpm, one beyond what it was designed for in order to supply 12psi for his engine, then failure could occur. But I do concur it's hard to believe that 12psi's worth of work did this turbo in unless the compressor just wasn't capable of flowing the CFM's needed by the engine. I know it's a big compressor but again, if you follow the plot out, (and if Mike's CFM numbers are correct), it's off the deep end.
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From: Moon Twp. Pennsylvania
Originally Posted by BLACK_RX
Man, just read through the posts, that thing is sick. Keep up the good work.
I will be starting on the project again in the next couple of weeks and posted a lot more pics as it goes along!
Photo Diety
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From: Florida
Ah, ok FD. Thought you were saying that the turbo was being overspun :-) I still have a hard time believing it's being underrun however, given the flow rate of his 20B. Oiling issues would be far more likely.
Banned
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From: Rochester, MI
Originally Posted by Auto Illusions
i am actually not trying to save on weight to much.
Mine main goals are a show car with a ton of power and a functional chasis.
The car has brembo 14" front and 12.9" rear custom brake kits.
JIC FLTA-2 coil overs
mazda comp hollow front and heavier rear sway bars
power steering, heat and A/C are to be kept since it is a street car!
A full stereo system with sub and amps since i own and operate a performance/audio shop and the car is to be used as a demo car for all we can do for a customer.
full interior with custom fiberglass rear section.
streetable and driveable!
Any way, here is a pic of the last time the motor was in the car. It is actually at Headway Performance right now being built with all new internals, pinned and a killer race port!
Headway Performance is one of the best places for rotary engine porting. He uses a flow bench to test all his port work, not just the grind and hope theory!
Mine main goals are a show car with a ton of power and a functional chasis.
The car has brembo 14" front and 12.9" rear custom brake kits.
JIC FLTA-2 coil overs
mazda comp hollow front and heavier rear sway bars
power steering, heat and A/C are to be kept since it is a street car!
A full stereo system with sub and amps since i own and operate a performance/audio shop and the car is to be used as a demo car for all we can do for a customer.
full interior with custom fiberglass rear section.
streetable and driveable!
Any way, here is a pic of the last time the motor was in the car. It is actually at Headway Performance right now being built with all new internals, pinned and a killer race port!
Headway Performance is one of the best places for rotary engine porting. He uses a flow bench to test all his port work, not just the grind and hope theory!
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Joined: Oct 2003
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From: Toronto, ON
Originally Posted by rx7tt95
FD Racer, question. VE peak usually coincides with torque peak or where torque and hp cross on a graph.
HP = (TQ * RPM) / 5252
So, when RPM = 5252, HP = TQ.
But the other part is right - VE peak and max torque should coincide.
Administrative Me
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From: Kansas
Originally Posted by rx7tt95
Ah, ok FD. Thought you were saying that the turbo was being overspun :-) I still have a hard time believing it's being underrun however, given the flow rate of his 20B. Oiling issues would be far more likely.
Rotary Enthusiast
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From: Union, NJ
>he calculated the RPM shaft speed to be a little over 100k at the airflow/pressure ratio I was running<
OK, what pressure ratio were you running.....better yet....how much boost?
Anything under 15 psi or pressure ratio of ~2.00 you're nowhere near a 100k shaft speed !!!!!!!!!!
my .02
JD
OK, what pressure ratio were you running.....better yet....how much boost?
Anything under 15 psi or pressure ratio of ~2.00 you're nowhere near a 100k shaft speed !!!!!!!!!!
my .02
JD
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From: Kansas
Originally Posted by Boostn7
>he calculated the RPM shaft speed to be a little over 100k at the airflow/pressure ratio I was running<
OK, what pressure ratio were you running.....better yet....how much boost?
Anything under 15 psi or pressure ratio of ~2.00 you're nowhere near a 100k shaft speed !!!!!!!!!!
my .02
JD
OK, what pressure ratio were you running.....better yet....how much boost?
Anything under 15 psi or pressure ratio of ~2.00 you're nowhere near a 100k shaft speed !!!!!!!!!!
my .02
JD
The pressure ratio was between 1.7 and 1.90
Photo Diety
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From: Florida
Copied and pasted off the internet...
1 horsepower is defined as 550 foot-pounds per second (read How Horsepower Works to find out how they got that number). The units of torque are pound-feet. So to get from torque to horsepower, you need the "per second" term. You get that by multiplying the torque by the engine speed.
But engine speed is normally referred to in revolutions per minute (RPM). Since we want a "per second," we need to convert RPMs to "something per second." The seconds are easy -- we just divide by 60 to get from minutes to seconds. Now what we need is a dimensionless unit for revolutions: a radian. A radian is actually a ratio of the length of an arc divided by the length of a radius, so the units of length cancel out and you're left with a dimensionless measure.
You can think of a revolution as a measurement of an angle. One revolution is 360 degrees of a circle. Since the circumference of a circle is (2 x pi x radius), there are 2-pi radians in a revolution. To convert revolutions per minute to radians per second, you multiply RPM by (2-pi/60), which equals 0.10472 radians per second. This gives us the "per second" we need to calculate horsepower.
Let's put this all together. We need to get to horsepower, which is 550 foot-pounds per second, using torque (pound-feet) and engine speed (RPM). If we divide the 550 foot-pounds by the 0.10472 radians per second (engine speed), we get 550/0.10472, which equals 5,252.
So if you multiply torque (in pound-feet) by engine speed (in RPM) and divide the product by 5,252, RPM is converted to "radians per second" and you can get from torque to horsepower -- from "pound-feet" to "foot-pounds per second."
1 horsepower is defined as 550 foot-pounds per second (read How Horsepower Works to find out how they got that number). The units of torque are pound-feet. So to get from torque to horsepower, you need the "per second" term. You get that by multiplying the torque by the engine speed.
But engine speed is normally referred to in revolutions per minute (RPM). Since we want a "per second," we need to convert RPMs to "something per second." The seconds are easy -- we just divide by 60 to get from minutes to seconds. Now what we need is a dimensionless unit for revolutions: a radian. A radian is actually a ratio of the length of an arc divided by the length of a radius, so the units of length cancel out and you're left with a dimensionless measure.
You can think of a revolution as a measurement of an angle. One revolution is 360 degrees of a circle. Since the circumference of a circle is (2 x pi x radius), there are 2-pi radians in a revolution. To convert revolutions per minute to radians per second, you multiply RPM by (2-pi/60), which equals 0.10472 radians per second. This gives us the "per second" we need to calculate horsepower.
Let's put this all together. We need to get to horsepower, which is 550 foot-pounds per second, using torque (pound-feet) and engine speed (RPM). If we divide the 550 foot-pounds by the 0.10472 radians per second (engine speed), we get 550/0.10472, which equals 5,252.
So if you multiply torque (in pound-feet) by engine speed (in RPM) and divide the product by 5,252, RPM is converted to "radians per second" and you can get from torque to horsepower -- from "pound-feet" to "foot-pounds per second."
Rotary Enthusiast
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From: Union, NJ
>Do you realize that the shaft speed lines extend downwards? The range you see on most maps are just where the ideal areas of turbo flow. They don't map them down where my airflow is, for they don't forsee people running in those areas.<
Very much aware of that.....downwards but not backwards....
Either way...your new choice of turbo is a much better match for the 20B.
I'm sure you'll let us know how it works out......
Auto Illusions: WOW !!! nice work....
Really enjoy pics of such a project from begining to end....can't wait to see the final product. :-)
JD
Very much aware of that.....downwards but not backwards....
Either way...your new choice of turbo is a much better match for the 20B.
I'm sure you'll let us know how it works out......
Auto Illusions: WOW !!! nice work....
Really enjoy pics of such a project from begining to end....can't wait to see the final product. :-)
JD
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From: Kansas
Originally Posted by Boostn7
>Do you realize that the shaft speed lines extend downwards? The range you see on most maps are just where the ideal areas of turbo flow. They don't map them down where my airflow is, for they don't forsee people running in those areas.<
Very much aware of that.....downwards but not backwards....
Either way...your new choice of turbo is a much better match for the 20B.
I'm sure you'll let us know how it works out......
JD
Very much aware of that.....downwards but not backwards....
Either way...your new choice of turbo is a much better match for the 20B.
I'm sure you'll let us know how it works out......
JD
What do you mean? To help show the picture, let me show you the lines as they extend downwards:
Look at the peek CFM, and tell me where it falls? According to my general understanding of all this, along with the people who manufacture these turbos, the RPM shaft was at 100k RPMs. This isn't going backwards by any means, from what I see. Can you elaborate?
Rotary Freak
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From: l.a.
i know its are the turbo experts, but conceptually i don't see how that's the case. the exhaust flowing through the ports goes into the turbine and spools the turbo. the more exaust you flow the faster it spins and the more boost you build. but when you open the wg, flow will be diverted from the turbine, which would spin the turbo only at a set rpm to maintain whatever boost you determined. so i don't know how your turbo could be spinning that fast and yet maintain only 10 lbs. of boost. even if your exhaust flowed 10,000cfm, as long as it's going out the wg you're not spinning the turbo any faster. and if it didn't go out the wg then you'd see an accompanying rise in boost. there has to be a direct relationship between turbo rpm and boost pressure, that's how a turbo works. remember, the wg doesn't relieve boost pressure, it just diverts exhaust flow which stops the turbo from spinning any faster, which in turn stops boost pressure from increasing.
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From: Kansas
Originally Posted by fdracer
i know its are the turbo experts, but conceptually i don't see how that's the case. the exhaust flowing through the ports goes into the turbine and spools the turbo. the more exaust you flow the faster it spins and the more boost you build. but when you open the wg, flow will be diverted from the turbine, which would spin the turbo only at a set rpm to maintain whatever boost you determined. so i don't know how your turbo could be spinning that fast and yet maintain only 10 lbs. of boost. even if your exhaust flowed 10,000cfm, as long as it's going out the wg you're not spinning the turbo any faster. and if it didn't go out the wg then you'd see an accompanying rise in boost. there has to be a direct relationship between turbo rpm and boost pressure, that's how a turbo works. remember, the wg doesn't relieve boost pressure, it just diverts exhaust flow which stops the turbo from spinning any faster, which in turn stops boost pressure from increasing.
I don't know how to respond. I guess I can call ITS and ask them, but by the logic you presented, that means at any point the wg opens; the flow numbers can no longer be mapped. Unless, the wastegate holds the flow in the turbo at 84 lbs/min, and no more. I can't see this being the case, for then it would be impossible to map any of run, for the wastegate opened at 2900 rpms on the dyno. This doesn't make since.
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From: Orange, California
Originally Posted by fdracer
i know its are the turbo experts, but conceptually i don't see how that's the case. the exhaust flowing through the ports goes into the turbine and spools the turbo. the more exaust you flow the faster it spins and the more boost you build. but when you open the wg, flow will be diverted from the turbine, which would spin the turbo only at a set rpm to maintain whatever boost you determined. so i don't know how your turbo could be spinning that fast and yet maintain only 10 lbs. of boost. even if your exhaust flowed 10,000cfm, as long as it's going out the wg you're not spinning the turbo any faster. and if it didn't go out the wg then you'd see an accompanying rise in boost. there has to be a direct relationship between turbo rpm and boost pressure, that's how a turbo works. remember, the wg doesn't relieve boost pressure, it just diverts exhaust flow which stops the turbo from spinning any faster, which in turn stops boost pressure from increasing.
If you look at the plot on the graph you can see his motor wants 80 lbs/min of air at 10psi, that turbo can't do that, yet thats what he was trying to get out of it.
Think of riding a bike, going down hill if you try and pedal, if the hill is real steep, the bike will move faster than u can spin the wheels, it's the same thing with the compressor trying to maintain (to low of a) pressure at to high of a flow rate.
-Dustin
Last edited by thedguy; Sep 21, 2004 at 01:54 PM.
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From: Kansas
Originally Posted by Boostn7
OK....looking at your line you're telling me you're making over 800 flywheel horsepower @ ~13.5psi?????
JD
JD
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From: Orange, California
Originally Posted by Boostn7
OK....looking at your line you're telling me you're making over 800 flywheel horsepower @ ~13.5psi?????
JD
JD
Red maybe someones math is a bit off? making (or tryingto make) 800hp at 13.5psi at 6100rpms seems a bit high.
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From: l.a.
Originally Posted by thedguy
Reds trouble doesn't have to do with the exhaust side, it's the compressor, he's running too low of a pressure for the amount of air the engine wants to suck up. Basically the compressor is spinning extremely fast to try and create 10psi, BUT the engine taking the air faster than the turbo can compress, so it spins faster to try and keep up.
If you look at the plot on the graph you can see his motor wants 80 lbs/min of air at 10psi, that turbo can't do that, yet thats what he was trying to get out of it.
Think of riding a bike, going down hill if you try and pedal, if the hill is real steep, the bike will move faster than u can spin the wheels, it's the same thing with the compressor trying to maintain (to low of a) pressure at to high of a flow rate.
-Dustin
If you look at the plot on the graph you can see his motor wants 80 lbs/min of air at 10psi, that turbo can't do that, yet thats what he was trying to get out of it.
Think of riding a bike, going down hill if you try and pedal, if the hill is real steep, the bike will move faster than u can spin the wheels, it's the same thing with the compressor trying to maintain (to low of a) pressure at to high of a flow rate.
-Dustin