Calculating Injector Size
Mar 17, 2004, 04:19 PM
#1
Calculating Injector Size - Revisited!
**See My Post below**
Calculating injector sizing.
This is taken from various places on the web.
Horsepower * B.S.C.F
# of Cyl. * duty cycle
B.S.C.F is Brake Specific Fuel Comsumption
On Turbo or Supercharged engines it is around .55-.60
Specifically it is the number of pounds of fuel it takes to make 1 HP for 1 hour.
Low to medium performance = .50
Performance engine with head work = .45
Performance engine with expert head work = .40 - .45
Supercharged and turbocharged engines = .55 - .60
The more efficient the engine design the lower the number, forced induction requires more fuel for cooling the cylinders.
Duty cycle is the maximum amount of on time for the injectors. 80% is a commonly used number for this calculation. This leaves some duty cycle and is not running the injectors at their max duty cycle.
Let’s look at the stock Evo. If the flywheel HP is 271 then at 80% duty cycle the injectors would be
271*.60=162.6
4 * 80% = 3.2
162.6 / 3.2 = 50.8125 lb/hr.
50.8125 lb/hr * 10.5 = 533 cc
Stock injectors are around 525cc So it’s damn close with just the rough numbers. And the stock injectors are probably more taxed then 80% under load so the BSCF number can come down some.
Note: lb/hr conversion is 10.5 either divide or multiply in either direction.
Therefore if you wanted to make 400 flywheel hp which is roughly 330 bhp then:
400*.6=240
4*80% = 3.2
240/3.2 = 75 lb/hr
75 * 10.5 = 788 or 780 cc injectors
These calculations are done at 43.5 psi of fuel pressure which is commonly used to flow rate injectors.
Calculating injector sizing.
This is taken from various places on the web.
Horsepower * B.S.C.F
# of Cyl. * duty cycle
B.S.C.F is Brake Specific Fuel Comsumption
On Turbo or Supercharged engines it is around .55-.60
Specifically it is the number of pounds of fuel it takes to make 1 HP for 1 hour.
Low to medium performance = .50
Performance engine with head work = .45
Performance engine with expert head work = .40 - .45
Supercharged and turbocharged engines = .55 - .60
The more efficient the engine design the lower the number, forced induction requires more fuel for cooling the cylinders.
Duty cycle is the maximum amount of on time for the injectors. 80% is a commonly used number for this calculation. This leaves some duty cycle and is not running the injectors at their max duty cycle.
Let’s look at the stock Evo. If the flywheel HP is 271 then at 80% duty cycle the injectors would be
271*.60=162.6
4 * 80% = 3.2
162.6 / 3.2 = 50.8125 lb/hr.
50.8125 lb/hr * 10.5 = 533 cc
Stock injectors are around 525cc So it’s damn close with just the rough numbers. And the stock injectors are probably more taxed then 80% under load so the BSCF number can come down some.
Note: lb/hr conversion is 10.5 either divide or multiply in either direction.
Therefore if you wanted to make 400 flywheel hp which is roughly 330 bhp then:
400*.6=240
4*80% = 3.2
240/3.2 = 75 lb/hr
75 * 10.5 = 788 or 780 cc injectors
These calculations are done at 43.5 psi of fuel pressure which is commonly used to flow rate injectors.
Last edited by timzcat; Jun 13, 2004 at 02:05 AM.
Mar 17, 2004, 04:42 PM
#2
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Re: Calculating Injector Size
Originally posted by timzcat
Calculating injector sizing.
271*.60=162.6
4 * 80% = 3.2
149.05 / 3.2 = 50.8125 lb/hr.
50.8125 lb/hr * 10.5 = 533 cc
Calculating injector sizing.
271*.60=162.6
4 * 80% = 3.2
149.05 / 3.2 = 50.8125 lb/hr.
50.8125 lb/hr * 10.5 = 533 cc
162.6/3.2= 50.8125
Rest looks good.
This means that we should upgrade injectors sooner than we think. If all your assumptions are right.
Edited it, typo.
Last edited by timzcat; Mar 17, 2004 at 05:07 PM.
Mar 17, 2004, 05:29 PM
#3
Disclaimer: I don't know too much about this particular subject.
Why are you using 43.5 psi for your calculations? Doesn't the fact that we have a rising rate fuel pressure regulator factor into injector sizing? So shouldn't you be using 63.5 psi for your calculations since at WOT, boost is reading 20psi (the rising rate is 1:1 on the stock FPR)?
I always thought that the injector calculators were designed for N/A applications in mind.
Why are you using 43.5 psi for your calculations? Doesn't the fact that we have a rising rate fuel pressure regulator factor into injector sizing? So shouldn't you be using 63.5 psi for your calculations since at WOT, boost is reading 20psi (the rising rate is 1:1 on the stock FPR)?
I always thought that the injector calculators were designed for N/A applications in mind.
Mar 17, 2004, 05:57 PM
#4
You are correct. Injector flow rates are published at 43.5 psi.
To calculate the flow at 63.5 psi, do the following:
Find the square root of 43.5 which is 6.595
Find the square root of 63.5 which is 7.9675
Divide the 7.9675 by 6.595 = 1.208
Multiply this by the injector size 50.8125 * 1.208 = 61.3815 (lb/hr)
So the same 533cc injector will flow 645 cc at 63.5 psi.
Therefore a 680 cc injector is good for 821 cc at 63.5 psi which is more then the required 780 cc above to make 400 HP.
To calculate the flow at 63.5 psi, do the following:
Find the square root of 43.5 which is 6.595
Find the square root of 63.5 which is 7.9675
Divide the 7.9675 by 6.595 = 1.208
Multiply this by the injector size 50.8125 * 1.208 = 61.3815 (lb/hr)
So the same 533cc injector will flow 645 cc at 63.5 psi.
Therefore a 680 cc injector is good for 821 cc at 63.5 psi which is more then the required 780 cc above to make 400 HP.
Jun 13, 2004, 01:54 AM
#5
Okay I finally got around to plugging everything into a spreadsheet.
This is all based on commonly used calculations to determine injector size.
For the actual Excel spreadsheet you need to download the Injector Chart.txt file and change the file extension to .xls
This is all based on commonly used calculations to determine injector size.
For the actual Excel spreadsheet you need to download the Injector Chart.txt file and change the file extension to .xls
Jun 13, 2004, 05:46 PM
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Timzcat,
The reason there is a rising rate regulator on a turbo charged engine is to keep a stable fuel pressure differential over the injector. ie 43.5 psi When your boost pressure rises in the inlet manifold the fuel pressure must also rise by the same amount to keep this effective injector pressure constant. You can raise your base fuel pressure higher to squeeze more out of your injectors but you do this at the expense of fuel pump reliability, and the possibility of adversely affecting the injector spray pattern.
The reason there is a rising rate regulator on a turbo charged engine is to keep a stable fuel pressure differential over the injector. ie 43.5 psi When your boost pressure rises in the inlet manifold the fuel pressure must also rise by the same amount to keep this effective injector pressure constant. You can raise your base fuel pressure higher to squeeze more out of your injectors but you do this at the expense of fuel pump reliability, and the possibility of adversely affecting the injector spray pattern.
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Jun 14, 2004, 05:31 PM
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Yes, what AussieRSX points out is vital.
One of the really important details that seems to get omitted from fuel injector discussions, RC Engineering's web site included, is that when a "reference" fuel pressure is usually stated, it's given as a differential pressure measurement, or in the case of an injector tested by itself outside an engine, the gauge pressure. Paying attention to the difference between absolute and differential pressure is of paramount importance.
This is in fact, the only way it is possible for one to use direct ratios of the square root of "fuel pressure" to get injector flowrate. To understand why, we need to understand the following:
For a spray nozzle discharging an incompressible fluid (e.g. a liquid such as water or fuel), the mass flowrate is a function of internal nozzle geometric parameters, fluid density, and the pressure difference across it. So, mathematically for a given fluid spray nozzle,
flowrate = K * Sqrt(delta_P) where K is a constant for the nozzle that includes the vital fluid dynamic geometry parameters as well as the fluid parameter of density.
For a bigger injector flowing the same fuel, the K will be correspondingly larger but it will adhere to the same equation.
So, it is critical to note that the pressure term is delta_P and not P_absolute. There would really be a lot less confusion if people would say "This injector flows X cc/min at Y psid" (d = differential).
This of course means that if you want to use your injectors at 60 psid for example, the FPR will have to pressurize the fuel rail to an absolute pressure of ambient_pressure + boost_pressure + 60 psi.
Also, one more point -- if the FPR follows a rate of 1:1 it is not a "rising rate" regulator. It is a manifold-referenced regulator that will effectively provide the same delta_P across in the injector despite above-atmospheric (i.e. boost) or below-atmospheric (i.e. vacuum) manifold pressure. A rising rate FPR increases the delta_P the injector sees as the intake manifold sees boost pressure. In other words, as the engine sees boost the RRFPR increases the fuel flow rate over what you get with the base regulated fuel pressure.
One of the really important details that seems to get omitted from fuel injector discussions, RC Engineering's web site included, is that when a "reference" fuel pressure is usually stated, it's given as a differential pressure measurement, or in the case of an injector tested by itself outside an engine, the gauge pressure. Paying attention to the difference between absolute and differential pressure is of paramount importance.
This is in fact, the only way it is possible for one to use direct ratios of the square root of "fuel pressure" to get injector flowrate. To understand why, we need to understand the following:
For a spray nozzle discharging an incompressible fluid (e.g. a liquid such as water or fuel), the mass flowrate is a function of internal nozzle geometric parameters, fluid density, and the pressure difference across it. So, mathematically for a given fluid spray nozzle,
flowrate = K * Sqrt(delta_P) where K is a constant for the nozzle that includes the vital fluid dynamic geometry parameters as well as the fluid parameter of density.
For a bigger injector flowing the same fuel, the K will be correspondingly larger but it will adhere to the same equation.
So, it is critical to note that the pressure term is delta_P and not P_absolute. There would really be a lot less confusion if people would say "This injector flows X cc/min at Y psid" (d = differential).
This of course means that if you want to use your injectors at 60 psid for example, the FPR will have to pressurize the fuel rail to an absolute pressure of ambient_pressure + boost_pressure + 60 psi.
Also, one more point -- if the FPR follows a rate of 1:1 it is not a "rising rate" regulator. It is a manifold-referenced regulator that will effectively provide the same delta_P across in the injector despite above-atmospheric (i.e. boost) or below-atmospheric (i.e. vacuum) manifold pressure. A rising rate FPR increases the delta_P the injector sees as the intake manifold sees boost pressure. In other words, as the engine sees boost the RRFPR increases the fuel flow rate over what you get with the base regulated fuel pressure.
Jun 14, 2004, 09:50 PM
#9
Stock injectors are around 525cc So it’s damn close with just the rough numbers. And the stock injectors are probably more taxed then 80% under load so the BSCF number can come down some.
http://www.evomoto.com/product_info....roducts_id=111
etc, etc.
Last edited by Zeus; Jun 14, 2004 at 09:54 PM.
Jun 14, 2004, 10:40 PM
#10
When I originally creatd the post I did not have solid information on the injector size, OE.
Okay so at 560cc using a .55 BSFC and 85% Duty Cycle you'd be maxed at 330 hp.
With 100% duty cycle, which obviously can't be, they are maxed out at 388hp.
My point is 85% is a good reference for the duty cycle vs. max hp.
Given that it shows the limits of the stock injectors. Of course there is some extending their capability with more pump.
My whole point was to provide the calculations and since I have done this I am starting to question 660 cc injectors with 272s, etc. Especially since it is highly likely I will go with a larger turbo at some point in the future. I am really starting to look at the 720cc Denso injectors instead. An injector can open less not more.
I more then welcome any information that pertains to this type of sizing with cams and support for larger turbos.
Okay so at 560cc using a .55 BSFC and 85% Duty Cycle you'd be maxed at 330 hp.
With 100% duty cycle, which obviously can't be, they are maxed out at 388hp.
My point is 85% is a good reference for the duty cycle vs. max hp.
Given that it shows the limits of the stock injectors. Of course there is some extending their capability with more pump.
My whole point was to provide the calculations and since I have done this I am starting to question 660 cc injectors with 272s, etc. Especially since it is highly likely I will go with a larger turbo at some point in the future. I am really starting to look at the 720cc Denso injectors instead. An injector can open less not more.
I more then welcome any information that pertains to this type of sizing with cams and support for larger turbos.
Last edited by timzcat; Jun 14, 2004 at 10:43 PM.
Jun 29, 2004, 08:25 AM
#11
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Something does not seem to add up to me in negating the rising pressure of the fuel. From emperical data it seems that it must at least have some effect on fuel output.
My 1993 MKIV Supra with a single turbo has twin walbro fuel pumps, 1/2" fuel line feeding an HKS fuel rail with Greddy 720 cc injectors. My base fuel pressure is 41 psi. My car made 664 rwhp on the dyno at 27 psi of boost. I have a techtom MDM-100 that allows me to read engine parameters of the OBD-1 computer. During these dyno runs, I checked duty cycle, it was in the upper 70's to low 80's . The air fuel ratio was in the 11.3 range.
Assuming a 15% drivetrain loss, the car should be making ~780 bhp (flywheel). Putting these numbers into your equations above and negating the fuel pressure increase, the numbers just dont add up.
calculations:
780 bhp * .55 BSFC = 429 lbs/hr ( I even assumed a lower BSFC to make things a little better)
6 injectors * .82 = 4.92
429 lbs/hr / 4.92 = 87.2 lbs/hr
Converting to cc/min 87.2 * 10.5 = 915 cc/min
I only have 720cc injectors (rated a 43.5 psi)
Compensating for fuel pressure increase (if you agree that it works)
sqrt (68 psi/43.5 psi) = 1.250 * 720 cc injectors = 900 cc/min (effective size)
I do understand the delta P that you are talking about, it just doesn't seem to stand up to emperical data.
Brian
My 1993 MKIV Supra with a single turbo has twin walbro fuel pumps, 1/2" fuel line feeding an HKS fuel rail with Greddy 720 cc injectors. My base fuel pressure is 41 psi. My car made 664 rwhp on the dyno at 27 psi of boost. I have a techtom MDM-100 that allows me to read engine parameters of the OBD-1 computer. During these dyno runs, I checked duty cycle, it was in the upper 70's to low 80's . The air fuel ratio was in the 11.3 range.
Assuming a 15% drivetrain loss, the car should be making ~780 bhp (flywheel). Putting these numbers into your equations above and negating the fuel pressure increase, the numbers just dont add up.
calculations:
780 bhp * .55 BSFC = 429 lbs/hr ( I even assumed a lower BSFC to make things a little better)
6 injectors * .82 = 4.92
429 lbs/hr / 4.92 = 87.2 lbs/hr
Converting to cc/min 87.2 * 10.5 = 915 cc/min
I only have 720cc injectors (rated a 43.5 psi)
Compensating for fuel pressure increase (if you agree that it works)
sqrt (68 psi/43.5 psi) = 1.250 * 720 cc injectors = 900 cc/min (effective size)
I do understand the delta P that you are talking about, it just doesn't seem to stand up to emperical data.
Brian
Jun 29, 2004, 04:33 PM
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Originally Posted by TURBODAWG
calculations:
780 bhp * .55 BSFC = 429 lbs/hr ( I even assumed a lower BSFC to make things a little better)
6 injectors * .82 = 4.92
429 lbs/hr / 4.92 = 87.2 lbs/hr
Converting to cc/min 87.2 * 10.5 = 915 cc/min
I only have 720cc injectors (rated a 43.5 psi)
780 bhp * .55 BSFC = 429 lbs/hr ( I even assumed a lower BSFC to make things a little better)
6 injectors * .82 = 4.92
429 lbs/hr / 4.92 = 87.2 lbs/hr
Converting to cc/min 87.2 * 10.5 = 915 cc/min
I only have 720cc injectors (rated a 43.5 psi)
Are you sure that your FPR is 1:1 and is not the rising rate type? My point is that if the FPR is 1:1 then by definition its purpose to maintain the same pressure differential across the injectors across varying intake manifold pressures. If you do have a rising rate type, then you would certainly expect to see increased delta_p as the FPR saw boost.
If the differential pressure across the injector is not used as the "fuel pressure", the square root relationship cannot be valid.
Also, not that I would expect it to be hugely different, but how sure are you of the accuracy of your WHP measurement? Any corrections applied to it?
Jun 29, 2004, 06:09 PM
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The FPR is a TRD fuel pressure regulator which is 1:1 ratio increase in fuel pressure. I also have an electronic fuel pressure gauge and have watched it and it goes from 41 psi at 0 psi boost pressure to 68 psi at 27 psi of boost pressure.
The dyno was done on a Dynojet dyno. We have tested two dynos in this area and they read almost exactly the same. That is about right whp wise for a car with those mods. It did have corrections applied, but they went against me. The uncorrected whp was 674 whp.
Brian
The dyno was done on a Dynojet dyno. We have tested two dynos in this area and they read almost exactly the same. That is about right whp wise for a car with those mods. It did have corrections applied, but they went against me. The uncorrected whp was 674 whp.
Brian
Jun 29, 2004, 08:59 PM
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Originally Posted by TURBODAWG
The FPR is a TRD fuel pressure regulator which is 1:1 ratio increase in fuel pressure. I also have an electronic fuel pressure gauge and have watched it and it goes from 41 psi at 0 psi boost pressure to 68 psi at 27 psi of boost pressure.
Fluid dynamics is pretty clear that the pressure differential is the relevant pressure. Consistent with what I have been saying are: http://www.2gnt.com/www/corbin/foolafc.htm , http://www.clarks-garage.com/shop-manual/fuel-01.htm , and http://www.bigblocksix.com/Eddie/injectorswap/ which all point out that fuel pressure is measured in psig (gauge, or atmospheric referenced), not psia (absolute). So if we were to use an absolute pressure transducer on the fuel rail, at precisely atmospheric pressure in the intake manifold (or with the pressure reference hose disconnected from the FPR), we would see fuel_absolute_pressure=FPR_base_pressure + atmospheric_pressure .
If all you have to do is take the gauge pressure at the fuel rail and use the square root of the ratio of it against the reference pressure, then basically you are saying that the pressure that the injector tip (outlet) sees is immaterial to flow rate.
If I somehow ran a boost pressure of 60 psi (so ~75 psia at the intake manifold), but my gauge fuel pressure was 55 psi (so ~70 psia at the injector feed), all I would have to do is take Sqrt(55/43.5)*Original_Flow_Rate to get the new flow rate? What if I limited boost to 40 psi but adjusted the FPR to give me the same 55 psig? Would that also be the same flow value? I think intuition alone will tell you that it doesn't make sense to only consider the upstream pressure. Not to mention that in the first case you have greater pressure at the injector tip than at the feed -- you will not flow fuel from that injector.
I agree with you that running the calculations with the numbers you have provided doesn't agree with the constant differential pressure=constant flowrate principle. The only thing I can come up with is that there must be some sort of error in the numbers or some additional source of fuel that is being missed. Or else I'm overlooking something that would change or impact the numbers you gave.
The constant differential pressure "idea" is neither my own original nor new. Some independent discussion of the same thing in the context of fuel pressure regulators: http://www.mirafiori.com/~thad/fi/fpress.html , http://yarchive.net/car/fuel_pressure.html
I would say that you have a happy coincidence on your hands with respect to the numbers you have working out if you just consider gauge fuel pressure in absence of involving intake manifold pressure. However, it is a dangerous coincidence.
If I haven't convinced you, you could always place a call to a recognized authority, like RC Engineering. The question to ask them is: if a certain size injector is used on a forced-induction engine with a return-based fuel system and a 1:1 FPR, then as boost builds, does the flowrate across the injector change?
Last edited by DystopiaMagnate; Jun 30, 2004 at 04:30 AM.
Jun 30, 2004, 08:42 AM
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All Fuel pressure gauge that I know off all read gauge pressure. So psia = patm + pgauge. I am sure my Defi D series Fuel pressure gauge reads gauge pressure also.
Another MKIV supra that I know off has a completely stock fuel system on with an SP63 turbo kit. The stock supra has (6) 550 cc side feed injectors and a big enough fuel pump to out flow the injectors. He made 540 rwhp on the stock fuel system. His injector duty cycle was ~97%. Virtually maxed out injectors. Plugging his number into the equations looks like this:
540 rwhp (15% drivetrain loss) = 635 bhp
635 bhp x .55 lb/hr. bhp = 349.3 lb/hr
6 injectors x 100% dc = 6 ( assuming 100% duty cycle to help out)
349.3 lb/hr / 6 = 58.2 lb/hr
Converting to cc/min 58.2 lb/hr x 10.5 = 611 cc/min
This also throws a monkey wrench into the equation as well.
However,
By reducing the BSFC down to .5 lb/hr.bhp then the equation works out to 556 cc/min which is within striking distance. Also in using the conversion rate to cc/min it assumes an average density for gasoline. If this density is off slightly then it can make slight error as well. If I recall correctly race gas has a different density than regular gas also. Both my dyno pulls and his were both done with leaded ras gas. The BSFC also depends on the air/fuel ratio that is used when the runs are done. There is one guy who made in the 560's rwhp on the stock fuel system, he must have been running extremely lean with very good gas.
Brian
Another MKIV supra that I know off has a completely stock fuel system on with an SP63 turbo kit. The stock supra has (6) 550 cc side feed injectors and a big enough fuel pump to out flow the injectors. He made 540 rwhp on the stock fuel system. His injector duty cycle was ~97%. Virtually maxed out injectors. Plugging his number into the equations looks like this:
540 rwhp (15% drivetrain loss) = 635 bhp
635 bhp x .55 lb/hr. bhp = 349.3 lb/hr
6 injectors x 100% dc = 6 ( assuming 100% duty cycle to help out)
349.3 lb/hr / 6 = 58.2 lb/hr
Converting to cc/min 58.2 lb/hr x 10.5 = 611 cc/min
This also throws a monkey wrench into the equation as well.
However,
By reducing the BSFC down to .5 lb/hr.bhp then the equation works out to 556 cc/min which is within striking distance. Also in using the conversion rate to cc/min it assumes an average density for gasoline. If this density is off slightly then it can make slight error as well. If I recall correctly race gas has a different density than regular gas also. Both my dyno pulls and his were both done with leaded ras gas. The BSFC also depends on the air/fuel ratio that is used when the runs are done. There is one guy who made in the 560's rwhp on the stock fuel system, he must have been running extremely lean with very good gas.
Brian