turbos cost hp like superchargers? IMHO no.
01-09-2003, 11:08 PM
#1
TECH Regular
Thread Starter
Join Date: Apr 2002
Location: ...
Posts: 462
Likes: 0
Received 0 Likes
on
0 Posts
turbos cost hp like superchargers? IMHO no. I think the hp used to turn a turbo is minimal compared to a supercharger, becuase:
as we know the back pressure in the manifold is just a little higher at all times then the positive pressure in the intake for a turbo car, so one might think that's the hp cost of the turbo. The engine has to work to push the exhaust out, that's why you gain hp with low exhaust backpressure as well as less exhaust remaining in the piston for the next powerstroke. thus you assume that the turbo car loses the hp from the high back pressure, but you forget that the same principles apply to front pressure. An NA motor has to work in order to get air/fuel into the pistons. Under boost the air is being pushed harder then NA so the vacumn in the piston as it moves to top dead center disapears faster and may even be pushed by the intake mixture at some point with some even being pushout out the exhaust valve (which means there is very little exhaust left in the piston). The point is the turbo car loses less hp when pulling air into the piston then the NA car but loses more hp when pushing exhaust out. Kind of evens out a little. What do you guys think?
as we know the back pressure in the manifold is just a little higher at all times then the positive pressure in the intake for a turbo car, so one might think that's the hp cost of the turbo. The engine has to work to push the exhaust out, that's why you gain hp with low exhaust backpressure as well as less exhaust remaining in the piston for the next powerstroke. thus you assume that the turbo car loses the hp from the high back pressure, but you forget that the same principles apply to front pressure. An NA motor has to work in order to get air/fuel into the pistons. Under boost the air is being pushed harder then NA so the vacumn in the piston as it moves to top dead center disapears faster and may even be pushed by the intake mixture at some point with some even being pushout out the exhaust valve (which means there is very little exhaust left in the piston). The point is the turbo car loses less hp when pulling air into the piston then the NA car but loses more hp when pushing exhaust out. Kind of evens out a little. What do you guys think?
01-10-2003, 01:05 AM
#3
9 Second Club
iTrader: (1)
Join Date: Nov 2001
Location: Chicago
Posts: 782
Likes: 0
Received 0 Likes
on
0 Posts
Re: turbos cost hp like superchargers? IMHO no. It's a pretty well known accepted fact that Turbos will lose around 5% from the back pressure present due to the turbocharger while a SC will lose around 30% due to the parasitic losses of driving a compressor off the crankshaft.
01-10-2003, 07:55 PM
#4
Staging Lane
Join Date: Jul 2002
Location: Katy/Houston/San Marcos/Austin, TX
Posts: 94
Likes: 0
Received 0 Likes
on
0 Posts
Re: turbos cost hp like superchargers? IMHO no. all depends on belt size for SC...
roots blowers have more loss than centrifugal due to a much larger belt.
roots blowers have more loss than centrifugal due to a much larger belt.
01-10-2003, 08:53 PM
#5
TECH Regular
Thread Starter
Join Date: Apr 2002
Location: ...
Posts: 462
Likes: 0
Received 0 Likes
on
0 Posts
Re: turbos cost hp like superchargers? IMHO no. I just got back from being out when I wrote this thread, man I was buzzing <img border="0" alt="[chug]" title="" src="graemlins/gr_chug.gif" /> <img border="0" alt="[chug]" title="" src="graemlins/gr_chug.gif" />
You can think of an engine as a pump also. It has to pump the exhaust out and pump in the intake mixture then it has to compress it down, this all pulls from the net hp.
The air passing through the filter and through the intake system builds up momentum. The air doesn't move-stop-move-stop as the intake valves open, it actually builds constant momentum often in pulses because it's moving through a tube-like space. In an NA engine the motor itself has to motivate the air to pass through the filter, by creating an area of lower pressure within then the outside air pressure (about 14.7psi). When the intake valve opens the cylinder space within is nearly a vacumn, thus the gas mixture outside pushes its way in, however if you can imagine if this space wants to be filled it will pull from both directions, the gas and the piston will be pulled together, thus the energy to move that air into the engine comes from the piston and so the engine acts as an air pump. The movement of the air originates from the engine. With a turbo or supercharger it's much different, the intake air is actually already moving in at high velocity from an outside source. The same principles govern this process, when the intake valve opens the piston is pulled toward the valve for a split second as the lower pressure area atempts to fill in however because the intake air is being pumped in from an outside source (turbo) it ends up ramming into the piston and pushing it down. So there is a very small time in which the piston is being pulled away from the direction it's moving (takeing away hp) and a longer period where the momentum of the mixture is pushing the piston down(adding power). It takes energy to turn the turbo as the motor has to pump it, but in turn the turbo also pumps the engine with the intake air. So my point is, with a turbo or supercharger, the energy used to pump in the air is not all wasted, some is transformed back into positive hp. It takes a huge amount of hp to turn the belt of a supercharger and some of that is gained back as it pumps the engine, however the amount of hp lost by pumping the turbo (high backpressure in the exhaust manifold) is gained back through the same process.
This is part of the reason why turbos, although they create larger back pressure then NA or supercharger cars, actually rob very little power from the engine.
What do you guys think about this rambleing? ouch my head hurts <img border="0" alt="[bang head]" title="" src="graemlins/gr_banghead.gif" />
You can think of an engine as a pump also. It has to pump the exhaust out and pump in the intake mixture then it has to compress it down, this all pulls from the net hp.
The air passing through the filter and through the intake system builds up momentum. The air doesn't move-stop-move-stop as the intake valves open, it actually builds constant momentum often in pulses because it's moving through a tube-like space. In an NA engine the motor itself has to motivate the air to pass through the filter, by creating an area of lower pressure within then the outside air pressure (about 14.7psi). When the intake valve opens the cylinder space within is nearly a vacumn, thus the gas mixture outside pushes its way in, however if you can imagine if this space wants to be filled it will pull from both directions, the gas and the piston will be pulled together, thus the energy to move that air into the engine comes from the piston and so the engine acts as an air pump. The movement of the air originates from the engine. With a turbo or supercharger it's much different, the intake air is actually already moving in at high velocity from an outside source. The same principles govern this process, when the intake valve opens the piston is pulled toward the valve for a split second as the lower pressure area atempts to fill in however because the intake air is being pumped in from an outside source (turbo) it ends up ramming into the piston and pushing it down. So there is a very small time in which the piston is being pulled away from the direction it's moving (takeing away hp) and a longer period where the momentum of the mixture is pushing the piston down(adding power). It takes energy to turn the turbo as the motor has to pump it, but in turn the turbo also pumps the engine with the intake air. So my point is, with a turbo or supercharger, the energy used to pump in the air is not all wasted, some is transformed back into positive hp. It takes a huge amount of hp to turn the belt of a supercharger and some of that is gained back as it pumps the engine, however the amount of hp lost by pumping the turbo (high backpressure in the exhaust manifold) is gained back through the same process.
This is part of the reason why turbos, although they create larger back pressure then NA or supercharger cars, actually rob very little power from the engine.
What do you guys think about this rambleing? ouch my head hurts <img border="0" alt="[bang head]" title="" src="graemlins/gr_banghead.gif" />
01-11-2003, 03:49 PM
#6
9 Second Club
iTrader: (1)
Join Date: Nov 2001
Location: Chicago
Posts: 782
Likes: 0
Received 0 Likes
on
0 Posts
Re: turbos cost hp like superchargers? IMHO no. Could you tell us all what your point was again?
I'm have no idea where you're going with this.
around 5% losses for Turbo
around 30% losses for Supercharger
If your trying to understand why then it's simple.
Turbos do cause some backpressure.
Superchargers cause NO backpressure but do take away HP from the motor due to driving the compressor off the crank.
All other forces are so small there's not much to talk about. <img border="0" alt="[boring]" title="" src="graemlins/gr_zzz.gif" />
I'm have no idea where you're going with this.
around 5% losses for Turbo
around 30% losses for Supercharger
If your trying to understand why then it's simple.
Turbos do cause some backpressure.
Superchargers cause NO backpressure but do take away HP from the motor due to driving the compressor off the crank.
All other forces are so small there's not much to talk about. <img border="0" alt="[boring]" title="" src="graemlins/gr_zzz.gif" />
01-11-2003, 05:35 PM
#7
TECH Regular
Join Date: Feb 2002
Location: the moon
Posts: 414
Likes: 0
Received 0 Likes
on
0 Posts
Re: turbos cost hp like superchargers? IMHO no. modern centrifugal superchargers operate at MUCH less than a 30% loss.
Roots superchargers do probably take that much to drive.
But I won't argue the Turbo is more efficient and I think we all know that.
Roots superchargers do probably take that much to drive.
But I won't argue the Turbo is more efficient and I think we all know that.
Trending Topics
01-11-2003, 09:18 PM
#8
9 Second Club
iTrader: (1)
Join Date: Nov 2001
Location: Chicago
Posts: 782
Likes: 0
Received 0 Likes
on
0 Posts
Re: turbos cost hp like superchargers? IMHO no. I'd be very interested to get the actual loss numbers from supercharger manufacturers.
Here's an excerpt from Banks which may be a but biased...not sure.
Taken from: http://www.bankspower.com/im_articles/HotRod_May01.cfm
----------------------------
Supercharging is the process of force-feeding air into a combustion engine; in fact, early turbochargers were referred to as exhaust-driven superchargers. However, modern use of the term supercharger usually involves a compressor that is driven off of the crankshaft, either using pulleys and a belt or a direct shaft. These designs draw power from the crankshaft, often in the range of 30 percent of the engine's output at maximum boost. A turbocharger has no direct connection to the engine, but rather uses the force contained in the exiting exhaust to drive the charge air compressor, usually drawing only 5 to 10 percent of engine output.
-----------------------------------
I've been searching for emperical data to support my numbers but no one seems to have published them.
Here's an excerpt from Banks which may be a but biased...not sure.
Taken from: http://www.bankspower.com/im_articles/HotRod_May01.cfm
----------------------------
Supercharging is the process of force-feeding air into a combustion engine; in fact, early turbochargers were referred to as exhaust-driven superchargers. However, modern use of the term supercharger usually involves a compressor that is driven off of the crankshaft, either using pulleys and a belt or a direct shaft. These designs draw power from the crankshaft, often in the range of 30 percent of the engine's output at maximum boost. A turbocharger has no direct connection to the engine, but rather uses the force contained in the exiting exhaust to drive the charge air compressor, usually drawing only 5 to 10 percent of engine output.
-----------------------------------
I've been searching for emperical data to support my numbers but no one seems to have published them.
01-11-2003, 10:00 PM
#9
9 Second Club
iTrader: (1)
Join Date: Nov 2001
Location: Chicago
Posts: 782
Likes: 0
Received 0 Likes
on
0 Posts
Re: turbos cost hp like superchargers? IMHO no. Found an article from Mercedes Benz who used one of the most expensive well engineered superchargers available today. They did a test and found the losses are a hair over 15%.
This was with teflon coated parts and other cutting edge technologies. Not sure where the SC's are at that we use on our LS1's.
This was with teflon coated parts and other cutting edge technologies. Not sure where the SC's are at that we use on our LS1's.
01-11-2003, 10:36 PM
#10
TECH Regular
Thread Starter
Join Date: Apr 2002
Location: ...
Posts: 462
Likes: 0
Received 0 Likes
on
0 Posts
Re: turbos cost hp like superchargers? IMHO no. </font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by chasgiv3:
<strong> Could you tell us all what your point was again?
I'm have no idea where you're going with this.
around 5% losses for Turbo
around 30% losses for Supercharger
If your trying to understand why then it's simple.
Turbos do cause some backpressure.
Superchargers cause NO backpressure but do take away HP from the motor due to driving the compressor off the crank.
All other forces are so small there's not much to talk about. <img border="0" alt="[boring]" title="" src="graemlins/gr_zzz.gif" /> </strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">No. I don't think it's simple at all, if you take into account every force placed on the crank by a particular rod throught one rev (2 strokes) of the piston. Not simple at all.
I'm not really trying to understand as much as I'm trying to start up some tech talk here on ls1tech.com.
Might be changing the subject a little but...
Take a sheet of paper and draw a graph on it, the x axis goes from zero to 360 drgrees, that represents one rev of a piston. at zero is TDC(top dead center) and at 90 is BDC (bottom dead center) the forces placed on the rod and crank by ONLY the inertial loads (force) of the rotating assembly is a -cos(x)*(peakinertialloadvalue) function down the graph. So while reving the engine without any applied power you are putting these forces on the crank. At TDC it's a pull on the crank (hp taken away) so it's negative value which is followed by a positive value (+ power applied) in which the rod+piston is pushing on the crank. These values are independent of other forces and cancel each other out if you add up the area above and below the curve. Now add in the power stroke, intake stroke, exhaust stroke, and compression stroke and all the respective forces and add that crap all up, some negative some positive, and one big positive and you get the total force on the rod. Take the average force throughout 0-90 degrees, 90-180, 180-270,270-360. Add all 4 together for a value in lbs (lbs is a unit of force) then multiply it by the stroke but convert it to feet. You get WORK (not torque) which is in lbft. The reason we add the avg forces in all 4 phases is so that we can account for the forces we often ignore, like work needed to pump the exhaust out of the piston, or the work needed to compress the mixture. IMHO these forces are not too small to ignore becuase even 500 hp drag cars care about an extra 20 hp.
Who thinks the rwtq values are real. They arn't. The "torque" we see on dynographs is actually work.
So I don't think anything about engines, turbos, superchargers,and hp is simple.
Max velocity of the piston is directly between TDC and BDC. Max force on the rod is right on TDC or BDC. The inertial load at TDC is in fact negative and the force of the explosion phase starting at 16 degrees before TDC (if 16 degrees advance) is positive. So if the force of the power stroke were to be exactly matched with the inertial load throughout that phase, there would be in fact zero load on the crank in the power stroke, yet the BDC positive inertial load would still be there and have nothing to cancel it out, thus an overall positive hp would be achieved without any more force on the engine at WOT at 5000 rpm then if just sitting in neutral and reving at 5000rpm. A peak force in the PS of twice the inertial load force at that point is safe, because it would simply mean the same amount of force is applied to the rod, but in the other direction (positive direction) then the inertial force. Any force above 2 times the inertial load force in the powerstroke is dangerous if you are reving near the highest safe rpm at that time. Think about it.
This is why turbos or boost is most often not as hard on the engine as high compression NA and or nitrous. Because the inertial load is the slope of the cos function, and a boosted powerstroke produces less peak force then a high compression PS(powerstroke) and follows a less steep decline in force until BDC so it follows the cos curve of the inertial load better. But because the deline in force on a turbocharged PS is less steep, the total work can be greater then the PS of a high compression ratio NA engine because it's the area under the curve that counts.
Oh yeah and superchargers put a constant load on the crank a negative load. crank load from supercharger + TDC after compression stroke(that's a negative inertial load without powerstroke to cancel it out) at WOT near highest safe rpm of engine all motor = BOOOOOOM... rods goes through your block as crank bends and takes out the whole motor(well most of the time it's not this dramatic but you get my point, something ends up breaking). A supercharger lowers max safe WOT rpm of your engine.
<strong> Could you tell us all what your point was again?
I'm have no idea where you're going with this.
around 5% losses for Turbo
around 30% losses for Supercharger
If your trying to understand why then it's simple.
Turbos do cause some backpressure.
Superchargers cause NO backpressure but do take away HP from the motor due to driving the compressor off the crank.
All other forces are so small there's not much to talk about. <img border="0" alt="[boring]" title="" src="graemlins/gr_zzz.gif" /> </strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">No. I don't think it's simple at all, if you take into account every force placed on the crank by a particular rod throught one rev (2 strokes) of the piston. Not simple at all.
I'm not really trying to understand as much as I'm trying to start up some tech talk here on ls1tech.com.
Might be changing the subject a little but...
Take a sheet of paper and draw a graph on it, the x axis goes from zero to 360 drgrees, that represents one rev of a piston. at zero is TDC(top dead center) and at 90 is BDC (bottom dead center) the forces placed on the rod and crank by ONLY the inertial loads (force) of the rotating assembly is a -cos(x)*(peakinertialloadvalue) function down the graph. So while reving the engine without any applied power you are putting these forces on the crank. At TDC it's a pull on the crank (hp taken away) so it's negative value which is followed by a positive value (+ power applied) in which the rod+piston is pushing on the crank. These values are independent of other forces and cancel each other out if you add up the area above and below the curve. Now add in the power stroke, intake stroke, exhaust stroke, and compression stroke and all the respective forces and add that crap all up, some negative some positive, and one big positive and you get the total force on the rod. Take the average force throughout 0-90 degrees, 90-180, 180-270,270-360. Add all 4 together for a value in lbs (lbs is a unit of force) then multiply it by the stroke but convert it to feet. You get WORK (not torque) which is in lbft. The reason we add the avg forces in all 4 phases is so that we can account for the forces we often ignore, like work needed to pump the exhaust out of the piston, or the work needed to compress the mixture. IMHO these forces are not too small to ignore becuase even 500 hp drag cars care about an extra 20 hp.
Who thinks the rwtq values are real. They arn't. The "torque" we see on dynographs is actually work.
So I don't think anything about engines, turbos, superchargers,and hp is simple.
Max velocity of the piston is directly between TDC and BDC. Max force on the rod is right on TDC or BDC. The inertial load at TDC is in fact negative and the force of the explosion phase starting at 16 degrees before TDC (if 16 degrees advance) is positive. So if the force of the power stroke were to be exactly matched with the inertial load throughout that phase, there would be in fact zero load on the crank in the power stroke, yet the BDC positive inertial load would still be there and have nothing to cancel it out, thus an overall positive hp would be achieved without any more force on the engine at WOT at 5000 rpm then if just sitting in neutral and reving at 5000rpm. A peak force in the PS of twice the inertial load force at that point is safe, because it would simply mean the same amount of force is applied to the rod, but in the other direction (positive direction) then the inertial force. Any force above 2 times the inertial load force in the powerstroke is dangerous if you are reving near the highest safe rpm at that time. Think about it.
This is why turbos or boost is most often not as hard on the engine as high compression NA and or nitrous. Because the inertial load is the slope of the cos function, and a boosted powerstroke produces less peak force then a high compression PS(powerstroke) and follows a less steep decline in force until BDC so it follows the cos curve of the inertial load better. But because the deline in force on a turbocharged PS is less steep, the total work can be greater then the PS of a high compression ratio NA engine because it's the area under the curve that counts.
Oh yeah and superchargers put a constant load on the crank a negative load. crank load from supercharger + TDC after compression stroke(that's a negative inertial load without powerstroke to cancel it out) at WOT near highest safe rpm of engine all motor = BOOOOOOM... rods goes through your block as crank bends and takes out the whole motor(well most of the time it's not this dramatic but you get my point, something ends up breaking). A supercharger lowers max safe WOT rpm of your engine.
01-11-2003, 10:44 PM
#11
TECH Regular
Thread Starter
Join Date: Apr 2002
Location: ...
Posts: 462
Likes: 0
Received 0 Likes
on
0 Posts
Re: turbos cost hp like superchargers? IMHO no. what I'm trying to say is turbos are much better then superchargers, and both of which are better then nitrous. So go turbo!! <img border="0" title="" alt="[Big Grin]" src="gr_grin.gif" />
01-12-2003, 01:45 AM
#12
9 Second Club
iTrader: (1)
Join Date: Nov 2001
Location: Chicago
Posts: 782
Likes: 0
Received 0 Likes
on
0 Posts
Re: turbos cost hp like superchargers? IMHO no. ROTFLOL...you may be over analyzing your ME books. But so be it.
You should try to summarize your points into up front statements to get readers interested. <img border="0" title="" alt="[Big Grin]" src="gr_grin.gif" />
You should try to summarize your points into up front statements to get readers interested. <img border="0" title="" alt="[Big Grin]" src="gr_grin.gif" />
01-12-2003, 02:22 AM
#13
Re: turbos cost hp like superchargers? IMHO no. dude.. your thinking way too hard.... your gonna bust an O-ring.... turbos rule.. it's as simple as that... CHEERS!! <img border="0" alt="[chug]" title="" src="graemlins/gr_chug.gif" /> <img border="0" alt="[chug]" title="" src="graemlins/gr_chug.gif" /> <img border="0" alt="[chug]" title="" src="graemlins/gr_chug.gif" />
01-12-2003, 06:24 PM
#15
Staging Lane
Join Date: Jun 2002
Location: Layton, UT
Posts: 62
Likes: 0
Received 0 Likes
on
0 Posts
Re: turbos cost hp like superchargers? IMHO no. 4 degrees of initial timing, before any type of vacuum advance. Remembver , it s a stang so the computer is a bit low-tek compared to ours. At WOT he should pick up the proper amount of advance.
01-12-2003, 08:16 PM
#16
TECH Fanatic
Join Date: Nov 2001
Location: Chi-Town
Posts: 1,055
Likes: 0
Received 0 Likes
on
0 Posts
Re: turbos cost hp like superchargers? IMHO no. not only is the size of the belt an issue, but also the size of the blower. Its all rotating parts....the harder it is to turn them the more hp it will cost
01-12-2003, 09:48 PM
#17
Re: turbos cost hp like superchargers? IMHO no. </font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by chasgiv3:
<strong> I'd be very interested to get the actual loss numbers from supercharger manufacturers.
I've been searching for emperical data to support my numbers but no one seems to have published them. </strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">I did some calcs the other night when we had a SC car on the dyno. It looked as though it took about 80 HP to drive the compressor at full boost (~8psi) FWIW.
He put down 518.
ProCharger P1SC
<strong> I'd be very interested to get the actual loss numbers from supercharger manufacturers.
I've been searching for emperical data to support my numbers but no one seems to have published them. </strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">I did some calcs the other night when we had a SC car on the dyno. It looked as though it took about 80 HP to drive the compressor at full boost (~8psi) FWIW.
He put down 518.
ProCharger P1SC
01-13-2003, 05:44 AM
#19
TECH Addict
Join Date: Nov 2001
Location: College Station, Tx
Posts: 2,207
Likes: 0
Received 0 Likes
on
0 Posts
Re: turbos cost hp like superchargers? IMHO no. </font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Who thinks the rwtq values are real. They arn't. The "torque" we see on dynographs is actually work. (...) Add all 4 together for a value in lbs (lbs is a unit of force) then multiply it by the stroke but convert it to feet. You get WORK (not torque) which is in lbft.</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">
I disagree with that - you are measuring torque, even in your example. Torque is the rotational analogue of force, and is force * length of the moment arm (1/2 the stroke actually), which is exactly what you describe above. Work is the distance over which force is applied - e.g. I push a box with 5lbs of force over 10 feet.
Think about it - I can apply 10lbs of force to a 1' radius crankshaft and not have it move. I have still applied a torque of 10 ft-lbs, but no work has been done as it hasn't mooved. Or even if it does move then the work would be the force applied * the arc length traveled by the object at the end of the crankshaft (assuming everything else has 0 mass, to make it simple).
If your method is correct then the work done rotating an object 25 times is no different than rotation it 1 time - as long as the mass and radius are the same. (and obviously that isn't correct).
</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Max velocity of the piston is directly between TDC and BDC. Max force on the rod is right on TDC or BDC.</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">
Yep, though exactly where you get max velocity depends on your r/s ratio - here is a spreadsheet I had done awhile ago that graphs the piston position and it's derivatives through a crank revolution (from TDC to TDC) - you can change the first value in the data table heading to try different combinations.
http://www.slowcar.net/shared/pistonmotion.xls
</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">This is why turbos or boost is most often not as hard on the engine as high compression NA and or nitrous. Because the inertial load is the slope of the cos function, and a boosted powerstroke produces less peak force then a high compression PS(powerstroke) and follows a less steep decline in force until BDC so it follows the cos curve of the inertial load better. But because the deline in force on a turbocharged PS is less steep, the total work can be greater then the PS of a high compression ratio NA engine because it's the area under the curve that counts.</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">I don't really accept this statement either, especially the bolded part (emphasis mine). When you talk about the peak forces on the power stroke, and we assume the same piston area, you are basically talking about cylinder pressure, which is directly analgous to torque.
You can make FAR more cylinder pressure/torque with a turbo (or blower setup) for that matter than you can with a NA setup - so how can the FI setup help but have more cylinder pressure and thus force on the power stroke. It seems to me you claim that it is because a smaller force is averaged out over a longer period, but the first part of that is not true - it is a not a smaller force. It will not be twice as much at twice the torque levels, because of the area under the curve effect you mention, but it will be greater. Here is a diagram from corky bell's book
<img src="http://www.slowcar.net/shared/cylprs.gif" alt=" - " />
This clearly shows cylinder pressure (thus force) going up across the board - peak and all.
</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Oh yeah and superchargers put a constant load on the crank a negative load. </font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">
How is the supercharger load a negative load?
Everything else aside, turbo's cost horsepower also - if only from increased pumping losses (having to work against a pressurized exhaust side). Yes they are definitely more efficient, as the derive a good bit of their energy from a source that is otherwise wasted, so you improve the overall effeciency of the motor. But they still take horsepower to turn, and still "cost" some in a theoretical sense (and in that same sense a blower costs more) - but both obviously make net power.
Chris
I disagree with that - you are measuring torque, even in your example. Torque is the rotational analogue of force, and is force * length of the moment arm (1/2 the stroke actually), which is exactly what you describe above. Work is the distance over which force is applied - e.g. I push a box with 5lbs of force over 10 feet.
Think about it - I can apply 10lbs of force to a 1' radius crankshaft and not have it move. I have still applied a torque of 10 ft-lbs, but no work has been done as it hasn't mooved. Or even if it does move then the work would be the force applied * the arc length traveled by the object at the end of the crankshaft (assuming everything else has 0 mass, to make it simple).
If your method is correct then the work done rotating an object 25 times is no different than rotation it 1 time - as long as the mass and radius are the same. (and obviously that isn't correct).
</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Max velocity of the piston is directly between TDC and BDC. Max force on the rod is right on TDC or BDC.</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">
Yep, though exactly where you get max velocity depends on your r/s ratio - here is a spreadsheet I had done awhile ago that graphs the piston position and it's derivatives through a crank revolution (from TDC to TDC) - you can change the first value in the data table heading to try different combinations.
http://www.slowcar.net/shared/pistonmotion.xls
</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">This is why turbos or boost is most often not as hard on the engine as high compression NA and or nitrous. Because the inertial load is the slope of the cos function, and a boosted powerstroke produces less peak force then a high compression PS(powerstroke) and follows a less steep decline in force until BDC so it follows the cos curve of the inertial load better. But because the deline in force on a turbocharged PS is less steep, the total work can be greater then the PS of a high compression ratio NA engine because it's the area under the curve that counts.</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">I don't really accept this statement either, especially the bolded part (emphasis mine). When you talk about the peak forces on the power stroke, and we assume the same piston area, you are basically talking about cylinder pressure, which is directly analgous to torque.
You can make FAR more cylinder pressure/torque with a turbo (or blower setup) for that matter than you can with a NA setup - so how can the FI setup help but have more cylinder pressure and thus force on the power stroke. It seems to me you claim that it is because a smaller force is averaged out over a longer period, but the first part of that is not true - it is a not a smaller force. It will not be twice as much at twice the torque levels, because of the area under the curve effect you mention, but it will be greater. Here is a diagram from corky bell's book
<img src="http://www.slowcar.net/shared/cylprs.gif" alt=" - " />
This clearly shows cylinder pressure (thus force) going up across the board - peak and all.
</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Oh yeah and superchargers put a constant load on the crank a negative load. </font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">
How is the supercharger load a negative load?
Everything else aside, turbo's cost horsepower also - if only from increased pumping losses (having to work against a pressurized exhaust side). Yes they are definitely more efficient, as the derive a good bit of their energy from a source that is otherwise wasted, so you improve the overall effeciency of the motor. But they still take horsepower to turn, and still "cost" some in a theoretical sense (and in that same sense a blower costs more) - but both obviously make net power.
Chris
01-13-2003, 01:13 PM
#20
TECH Fanatic
Re: turbos cost hp like superchargers? IMHO no. JimmyKash:
Interesting. Anyone know the S-Trim gear ratio? I'd like to see a P1-SC vs. D1-SC vs. S-Trim vs. T-Trim. I think it's good to see how much load each one puts on the crank. Normally the Vortechs don't have to turn as hard with their aftercooler vs. blowing thru all the intercooler tubing.
Interesting. Anyone know the S-Trim gear ratio? I'd like to see a P1-SC vs. D1-SC vs. S-Trim vs. T-Trim. I think it's good to see how much load each one puts on the crank. Normally the Vortechs don't have to turn as hard with their aftercooler vs. blowing thru all the intercooler tubing.