Injector Calculators: https://sssquid.com/v3/injector-calculator/

Fuel Pump Sizing: Fuel Pump Sizing (SSSQUID WIKI)

### THE EASY WAY

There are a lot of injector calculators on the internet, but none seem to calculate very well and the few we’ve used suggested a stock M50B25 should be running no less than 24 lb/hr injectors (stock is 17.5, and those handle an extra 30 Hp easily) and a stock M20B25 should be running no less than 20 lb/hr (stock is about 14.5 lb/hr). So let’s learn a quick and easy way to choose injectors with just a little math.

This is a very simple method and does not account for many factors, so just use it as a way to estimate what you’ll need. That being said, it does tend to be quite accurate!

Simply take your target horsepower at the wheels*, divide by number of cylinders, and then divide by two.

i = ( h / c ) / 2

###### * to find WHp: multiply BHp (brake horsepower; “crank horsepower”) by 0.92-0.88 for FWD, 0.9 to 0.85 for RWD, or 0.88 to 0.82 for AWD. What is WHP/BHP? (external link)

As proof of concept, we will look at a few examples.

First the RWD M50B25TU (with VANOS):

Horsepower (h): 172 WHp (192 BHp)
Cylinders (c): 6
Stock Injector Size: 17.5 lb/hr @ 50.75 PSI

i = ( 172 / 6 ) / 2
i = 28.66 / 2
i = 14.33

Since 14.33 is less than 17.5, we know it can handle the specified power level. However, it’s a good idea to add an operating error margin of 15% to this number for safety. To do this: multiply 14.33 by 1.15. This brings us to 16.5, quite close to the stock number but still leaving us with room to grow.

Speaking of growing: the completely stock S50B30, which BMW claimed output 243 BHp, actually tends to output around 195-205 WHp (210-220 BHp). It also used the same 17.5 lb/hr injectors as the M50. Generally the stock injectors don’t allow much room for growth over stock, but can take the vehicle to about 210 WHp before upgrading is necessary. (There are a few dyno examples you can see for yourself (external link))

Let’s use the equation to see if it matches our observed maximum supported power.

i = ( h / c ) / 2
i = ( 210 / 6 ) / 2
i = 35 / 2
i = 17.5

Excellent, it does! However this is absolutely maxing out the stock injectors and there is no safety margin.

One more you say? Okay! Let’s look at the M54B30, which comes with 23.5 lb/hr injectors stock and is rated at 231 BHp (or 207 WHp with our brake-to-wheel conversion of 0.9, which it makes exactly according to a dynojet).

i = ( 207 / 6 ) / 2
i = ( 34.5 ) / 2
i = 17.25 ; 19.84 with safety margin

Fantastic! So we know we have room to grow here before upgrading injectors. Sure enough, we’re usually able to see up to a maximum of about 250 WHp with this injector size.

Let’s work backwards with this next example. We’ll be looking at how much horsepower our 42 lb/hr injectors support on a 6-cylinder engine (too many people running 200 WHp engines seem to think they need 42 lb/hr injectors for some reason).

We take flow in lb/hr, multiply by two, and then multiply again by number of cylinders.

h = i * 2 * c

h = wheel horsepower
i = injector flow in lb/hr
c = cylinder count

h = 42 * 2 * 6
h = 42 * 12
h = 504

504! That’s quite a lot! But let’s be safe and apply our 15% safety margin by multiplying this number by 0.85. This gives us 428 WHp safely supported.

That’s right, your 42 lb/hr injectors support a whopping 428 WHp with a safety margin, or 504 completely maxed under ideal conditions.

Now, we can get more complicated with all of this based on airflow values and AFR, but this method works quite well for the basics. If you’d like to be more specific, keep reading!

But wait! Why can’t I just get massive injectors even if they’re more than I need? Who cares?!

As your injector’s flow rate increases, the time at which the injector opens becomes increasingly small in order to produce the same air-to-fuel ratio. Generally (but not always) the larger the injector flow rate, the more time the it takes for the injector to open and close fully (known colloquially as “dead time”). At idle and low engine speeds this can lead to issues, specifically poor air/fuel mixing. This increases likelihood of knock and poor overall operation. It also decreases the total length of piston stroke over which fuel is added to the cylinder. Increasing the spread of fuel over the length of the stroke allows for better mixing of fuel and air, which leads to a decreased possibility of knock and pre-detonation, as well as better fuel and engine efficiency (this means the possibility of more power and increased fuel economy).

### THE LONG-WAY AROUND

Okay, the methods we’ve used previously assume an average engine efficiency, but not all engines are equal. Some engines have greater efficiency of material used to output power, and we aim for different AFR (air-fuel ratio) depending on our needs. Let’s take a look at how these things affect our fueling needs.

The stock 565 cc/min (53.8 lb/hr) injectors on the 2005-2014 STi tend to max out at 360-380 WHp.

Horsepower (h): 380
Cylinders (c): 4
Stock Injector Size: 53.8 lb/hr

i = ( 380 / 4 ) / 1.8
i = 95 / 1.8
i = 52.77

Well darn if that isn’t close. But maybe you want to get a little more in-depth with it by taking into consideration AFR? Let’s change the final denominator to a function….

i = ( h / c ) / d
d = 1 + sqrt( a / 14.7 ) – ( r / 3600 / ( c * 10) )

Where:
i = injector flow
h = WHp
c = cylinder count
a = target peak power AFR
r = peak power RPM (or redline if you are unsure)
sqrt means “square root”

Let’s use this new equation to check the EJ255/257 with 565 CC/min (53.8 lb/hr) injectors again.

i = ( 380 / 4 ) / d
i = 95 / d
d = 1 + sqrt( 10.5 / 14.7 ) – ( 6000 / 3600 / ( 4 * 10 ) )
d = 1 + sqrt( 0.714 ) – ( 1.944 / ( 40 ) )
d = 1.845 – 0.0486
d = 1.7964
i = 95 / 1.7964
i = 52.67

Okay, that’s just about spot on, right? But what about engine efficiency (not to be confused with volumetric efficiency)? Well, this adds an extra level of complication, because there is no such thing as a singular, static “engine efficiency”. Sure, you’ve read joe-website that says an engine has a very specific engine efficiency coefficient… well, I hate to break this to you, but that’s complete bunk. Your engine will have different efficiency coefficients at different rotational speeds and airflow rates. It will change from gear to gear, from RPM to RPM, from air temperature to air temperature. You can take a generalized estimate to get a general idea, but it will not be exact without having direct access to engineering white papers on your engine, which are nearly impossible to obtain if you aren’t a combustion engine engineer that has designed or tested the engine, and then that only applies in its stock factory form. Engine modifications can change this calculation, sometimes significantly.

So what to do? Since we have to estimate all of these things anyway, the above equations will usually work quite well. Otherwise we’ll have to guess and estimate the efficiency of our engine at its peak power output. We’re going to be working with the “d” variable once again, to which we’ll add the following items in bold text:

d = 1 + sqrt( a / 14.7 * ( T * fr ) ) ( r / 3600 / ( c * 10 ) )

Where:
fr = full-throttle engine efficiency coefficient f @ RPM r
= Theodorus’ constant — this is the square root of 3, which can be effectively simplified as 1.860025.

A perfect engine that is able to convert 100% of the combustion phase into rotational energy would have an engine efficiency coefficient of 1.00. This is virtually impossible to achieve. Most vehicles will have a coefficient in the range of 0.40 to 0.60 at peak engine output. Remember, we are calculating this at peak power, where peak injector flow will almost always happen (though not always).

Okay, let’s take the same EJ257 engine we’ve been working with above and convert this to a fuel requirement. We’ll be using an fr of 0.55 (@ 6,000 RPM), which is a pretty standard estimate for this engine.

d = 1 + sqrt( 10.5 / 14.7 * ( 1.860025 * 0.55 ) ) – ( 6000 / 3600 / ( 4 * 10 ) )
d = 1.813
i = ( 380 / 4 ) / 1.813
i = 95 / 1.813
i = 52.39

As you can see, even with all of these complicated variables, the equation produces a number that is very, very close to the number we get from using the more simplified versions discussed above.

You can also adjust the variables to make sure the injectors are still sufficient at redline (you’ll use WHp at redline for this).

Another proof of concept, we can size up the stock N/A M20B25TU again with an engine efficiency coefficient of about 0.51. Plugging in 172 WHp, 13 AFR, and 6000 RPM we get:

d = 1.9126
fr
= 0.51
i = 28.667 / 1.9126
i = 15.18 ; 17.45 with 15% safety margin

There we go, with safety margin our calculated flow rate is almost exactly what the stock injectors flow! Bada-bing, bada-boom, right?

Then why did BMW use the same injectors with the S50 engine? For one, it has a greater peak power engine efficiency coefficient than the M50B25 has, about 0.58 compared to 0.51. How does this affect our fueling necessities? We’ll change a few variables in this case: fr of 0.58, r of 6,600, and h of 205). We then get:

i = 17.55

Again, we know that these injectors are pretty much near maximum at stock power levels, which correlates with the results of this equation very neatly.

### FORCED INDUCTION

In the most simplistic form of our equation, there’s just one simple, small change we make to the equation for forced induction — we change the final denominator from 2.0 to 1.8. So the new equation looks like this:

i = ( h / c ) / 1.8

Too simple for you? Try the more complicated equations above. But below you will be shown that this simple equation is usually good enough!

Let’s quickly size injectors for a turbocharged M50B25 @ 16 PSI outputting an estimated generous 330 WHp.

i = (h / c ) / 1.8
i = ( 330 / 6 ) / 1.8
i = 55 / 1.8
i = 30.55 ; 35.14 with safety margin applied

Taking it a bit further with our “probably more than you need to worry about” equation detailed in the section above, we plug in 11.2 AFR and 6600 RPM.

i = ( 330 / 6 ) / d
i = 55 / d
d = 1 + sqrt( 11.2 / 14.7 ) – ( 6600 / 3600 / ( 6 * 10 ) )
d = 1.8174
i = 55 / 1.8174
i = 30.26 ; 34.80 with 15% safety margin

And using our complete equation, we plug in an estimate engine efficiency coefficient of 0.51 @ 6,600 RPM:

d = 1 + sqrt( 11.2 / 14.7 * ( 1.860025 * 0.51 ) ) – ( 6600 / 3600 / ( 6 * 10 ) )
d = 1.819
i = ( 330 / 6 ) / 1.822
i = 55 / 1.819
i = 30.22

There ya go! We can see that all of these equations output very similar results, and any should be good enough for most peoples’ needs.

If you have any questions or concerns, please feel free to send us a message and we’ll reply as quickly as possible!

#### Auxiliary: How much HP will I make by going forced induction?

One of the most common questions we’re asked is about what size injectors to choose for a turbocharger or supercharger build. In order to understand this, there is the need to explain how much power you’ll actually be gaining from going forced induction.

The general rule-of-thumb is that forced induction will add about 3-4 WHp per pound of dynamic pressure (pressure above atmosphere; also known as positive pressure) for ever litre of engine displacement. For example, a 2.5 litre vehicle with a turbocharger or supercharger producing 10 PSI of dynamic pressure will add approximately 75 to 100 WHp.

An M50B25 can run up to 16 PSI on TD03/04 with simply ARP head studs and an MLS gasket. At 170 WHp stock, this will add 120-160 WHp and put you near 270-330 WHp. (A completely stock M50B25 with stock head bolts and gasket should not be running more than 8-10 PSI for safety.)

Please use our DynoGen webapp to estimate your build’s output!

## Injector Spray Patterns (Nozzles)

There is another thing to consider when purchasing new injectors: spray pattern. This difference is most noticeable when comparing older injectors to 1991+ injectors. Older injectors tend to have a single exposed pintle (Fig. 1), whereas more modern injectors tend to have an internal pintle with a patterned nozzle (Fig. 2) that helps better disperse fuel into the intake air.