Disclaimer – I’m in the fields making hay and didn’t devote a lot of time to this…. I think all the numbers are correct but if you find a math error – well that happens – this is more about the concept than anything.
Ok guys – it seams as if there has been some confusion about injector oil consumption and I hope that the following series of posts will help shed some light on this. Starting out I’m going to go slow and keep it simple and build upon the concepts. All of the data starting out here will be best case scenario and in the real world is subjected to inefficiencies inherent in each of the subsystems
Lets start out by looking at a simple set of 238cc hybrid injectors with 80% tips. Here is a graph showing the fuel injection for these injectors based upon
ICP and
PW. As a standard disclaimer these injectors were flowed using a “Gen3” style gear pump coupled to an electric motor spinning at 1750RPMs???? – I think. This will become important later and will fill in more details.
Ok now comes the simple part – we know that it is a hybrid injector and that for every cc of fuel expelled the injector HAS to use 5 times that volume in oil. Picture the above graph with the volumes multiplied by 5.
Now using this data we have the actual oil consumption for the injectors – now lets take a closer look at pump used to produce these actual flow rates – remember I said these numbers were produced on an injector flow bench running the same basic gear pump that is used on the Gen3 system – that is 0.61 c.i. per pump revolution or for those metric gurus 9.99cc per revolution – since the pump is spinning at a known 1750 RPMs (I need to verify that…) we know its flowing 17482cc of oil per minute (or 4.71 gallons per minute). For arguments sake I want to convert this oil flow rate to something that we care about, say a standard 17* pump that displaces 7.2cc per revolution. Straight up we get to 2428 pump RPMs. Considering the pump on a PSD (single stocker lets not complicate things further….yet) turns at 85% of the crank speed we get an engine RPM of 2856 RPMs. If you take into account some inefficiency in the system, we will just round it to a nice clean 3100 engine RPMs to produce the same result as found on the bench. (RPMs get rounded up due to inefficiencies because of the reduced the cc/rev that the pump puts out).
Now here comes the rub – suppose you want to run that 3.5ms of
PW at 2000 RPMs and 3000psi  you might not be able to do so…. How about at 3500 RPMs? – now that is still possible….. Let me elaborate a little….
We know that the 17* pump WILL work at 3100 RPMs because that is basically how these injectors were tested. At 3100 RPMs the 17* pump will produce 4.71 gallons per minute. At 3500 RPMs that 17* pump will be putting out 5.15 GPM (considering inefficiencies). That is more than enough to supply the injectors based on the flow sheet – in fact the duty cycle of the
IPR will be lower to be able to achieve the required pressure than it was at the lower RPMs.
Lets look at 2000 RPMs – the 17* pump can only supply 2.94 GPM at that RPM! Now you ask – “But is that enough for these injectors?” Lets find out that answer. We know we want 3.5ms of
PW and we want 3000psi – a quick lookup at the graphs and that determines that we want to inject 238cc of fuel (per 1000 shots for those that do not know). This means that we need 1190cc of oil per 1000 injection events –or 1.19cc of oil for one event. This event takes place in 3.5ms. This gives us 1.19cc of oil in 3.5ms… that right there is a flow rate. There are 12 crankshaft degrees per millisecond at 2000 RPMs – or 42 degrees of crankshaft rotation – or 35.7 degrees of rotation of your 17* oil pump. So 1.19cc per 35.7 degrees of crankshaft rotation is 12cc per pump revolution! Guess what – BAM P1211! That pump will NEVER keep up at that pressure with tuning that calls for fuel like this. In fact at that RPM I would be surprised if that 17* pump keeps up calling for 1000psi with 3.5ms. (coincidentally that is 142cc of fuel at 2000 RPMs – which should be good for about 280350 HP at 2000 RPMs…..)
Lets talk about the
IPR for a moment. Everyone talk about how important it is, and talks about taking it out and cleaning it, etc… but few talk about how it actually works. Think about it like this – it is basically an electric solenoid that is being fed a signal from the computer. This signal is a series of high and low voltages that dictate the strength the magnetic field contained within the solenoid. By varying the strength of this field the
IPR is regulating the amount of oil that can bleed off from the system. A strong field is trying to hold as much oil in the system as possible; a weaker field is trying to bleed off more oil through the
IPR to help maintain a pressure. So when comparing injectors, HPOPs, tuning, etc… that is why everyone talks about the duty cycle of the
IPR – that duty cycle is basically how strong the
IPR is. So in essence if you swap nothing else but pumps – and your
IPR duty cycle drops under similar conditions – then your new pump has a higher volume output than your old pump –OR something about the pump/setup has less of a pressure drop.
OK so far we have looked at a WOT scenario and looked how many different variables effect the pressure (volume) of oil available to the injectors at any given time, engine RPM,
IPR Duty Cycle, Commanded
PW, Commanded
ICP, etc… but unless you’re a competition rig – how does this effect you? Lets take a look at what goes on when you are hooked to a trailer pulling a grade. To save a little time I am going to borrow some work from Charles done in Smack Talk – some things have been changed to protect the innocent and it has been cleaned up a bit, but none the less it is still his scenario:
Quote:
For an example of work, take a scenario of an PSD pulling a trailer climbing a 6% mountain grade for 6 miles with a loaded trailer behind it as would be the case pulling Mont Eagle grade in Tn.
There is a specific amount of work involved with that task.
Lets say the GCVW is 20,000lbs. Now if we forgo the complications of friction, wind resistance and all that for the moment, we can figure the primary force acting on this truck and trailer which happens to be gravity.
A 6% grade is a 2.7* angle. From this we can figure the component of gravity that is acting on the vehicle attempting to pull it down the slope.
We would have 20,000lbs straight into the earth and a component of 942lbs headed down the slope.
So the force involved here is the 942lbs, or 4190 N (Newtons).
The distance is 6 miles, or 9504 M (Meters).
Knowing this we can calculate the work required to move this vehicle to the top of Mont Eagle.
Since Work is Force x Distance, it's as simple as 4190 N x 9504 M = 39,821,760 NM (Newton Meters, units of work).
So that's your work. That's literally what must be done. Any engine can do it, the only question is how long will it take. That's where power comes in.
For instance....
Lets say you managed 45mph up this pass in this truck. Traveling 6 miles at 45mph would take you 8 minutes from bottom to top. Now we have our Time value. The important one...
8 minutes is 480 seconds.
We know that Power = Work / Time, so it's as simple as 39,821,760 NM / 480 s = 82,962 Watts of power required to do that task in that time.
What is Watt you might ask? It's a unit of power, just like kilowatts (1000 watts) or Horsepower, or Joules per second, so on and so forth.
So how many Horsepower is 82,962 Watts? Well it's 112 Horsepower.
Now lets say – I don’t want to go 45mph up that hill…. I want to do 70mph up that hill!
The only thing that changes is the amount of time it takes to climb the hill – only 5.1 minutes to reach the summit or 308 seconds. That boils it down to 129291 watts or 174hp.

Now you might want to know where I am headed with all of this and how it applies to what injectors, HPOPs, tuning, etc….
Lets examine the situations just discussed – 112hp at 45mph and 174hp at 70mph. Looking at some gear ratios – lets just say it’s a 6 speed truck to eliminate any of the argument for the TC slippage, etc… it can also show that things like the P1211 code can be as much of a driving style and tuning as it is an actual problem. The truck also has 33” tires and 3.73 gears.
First the 45mph scenario – in OD this can be done at about 1300 RPM, In 5th gear you will see it about 1725rpm, in 4th gear you can see it 2225rpm, and not practically done in 3rd gear by a “stockish” truck running a set of hybrids and a turbo towing a trailer. For arguments sake I’m going to assume 2.2hp per cc of fuel injected. So to make 112hp we will need to inject 51cc of fuel. At 500psi
ICP that’s about 2.75ms, at 750psi that is about 1.75ms, at 1000psi that is about 1.5ms, at 1250psi that is 1.4ms, at 1500psi that is 1.3ms at 2000psi its 1.2ms, at 2500psi its 1.1ms, at 3000psi its about 1ms.
For the most part the goal is to inject fuel at the highest pressure possible to get the best atomization possible – but this is also a balance of being able to produce the pressure, as well as inject the fuel in an appropriate time window for a controlled burn. That is an entire thread all by itself – maybe some other day…..
So now I have all these data points of pressures and PWs that can inject the fuel I need to do the task – and a handful of RPMs where I can meet my speed goal. So now what…. Well those RPMs dictate the flowrate of the pump in a volume per time – those
PW numbers dictate the rate that oil is required to use the associated
ICP – Basically what that means is if the oil flow requirement for the injection event of a certain
PW is not met by the pump – then you can’t maintain the pressure that is called for and BAM – the dreaded P1211…
Lets look at these data points – given some inefficiencies in the pump, max possible
IPR duty cycle, etc…. at 1300 RPMs we can not inject fuel any faster than about 1.5ms and maintain 1250psi, at 1725rpms we should be able to manage in the 1.2ms with about 2000psi, and can do pretty much anything we would like to do in this scenario at the 2225rpm mark. Asking for any more than about 2000psi at 1750rpms is just going to result in popping a code – at 1300rpms don’t think about any more than 1250psi or expect a nice yellow light on your dash.
The 70mph scenario yields about 2025rpms in OD, 2650rpms in 5th gear, and is not practical in 4th gear. To make the 174hp we will need to inject 79cc of fuel. At 750psi that is roughly 2.5ms, at 1000psi its about 2.1ms, at 1250psi its about 1.9, at 1500psi its about 1.6ms, at 2000psi its about 1.5ms, at 2500psi that is 1.4ms, and at 3000psi that is about 1.25ms. Looking at these data points at 2025 we should be able to run about a 1.5ms
PW at 1000psi, at 2650 rpms we can hit any of our targets.
Its important to note that for any of the above data we can extend the
PW longer, lower the
ICP demanded and inject the same amount of fuel required for the task with no risk of not being able to meet the commanded
ICP. None of the above takes into account anything with your turbocharger, or any other variables of driving up and down the road – but it does go to show some of the requirements that the injectors can exert on a high pressure oil system. There are lots of variables and this does not hold perfectly true in every case due to some of these variables – I hinted here and there about some things that can make these numbers seem out of place.
That’s the quick and loose version of how the injector setup can place demands on the high pressure oil system. It will be a few days before I can take this to the next level and show the effects of different injectors and different pumps on this system that must work together – For arguments sake I will stick to only 238cc hybrids for now – I will just change tips around – or pumps around – not total injector capacity. The moral of the story will be the faster injectors demand more oil (because they can expel more fuel per time) and the larger the oil pump the lower in the rpm band you can hold pressure on a set of injectors. Additional food for thought – just because you can dump all your fuel at full
ICP pressure just off idle, doesn’t make it a good idea….