I am starting this thread to make sure I am doing this right (as there are ppl on here that know a heck of a lot more than I do), and for those that don't want to take someone's word for how much fuel they are gonna need to produce a certain amount of power.
I am trying to nail down the cc/1000 shots that I need to turn my pump to for my desired horsepower. On the way to this, we can also find how many lb/min of fuel we will need though out the RPM range, which, with your A/F ratio, will give your lb/min of air for turbo sizing (not factoring VE).
I am going to use the 4BTA CPL tag 1963 which is pretty much the same as my 1839 tag as far as the p pump goes. These tags max out at 2500 RPM, so I will interpolate the data at higher RPM.
The BSFC (Brake Specific Fuel Consumption) is the rate of fuel consumption divided by the power produced. On the EPA tag it is given for us as:
lb/(HP*hr)
and
g/(kW*hr)
I will be using imperial for this example.
I want 300 HP out of my 4BTA @ 3200 RPM
On the EPA tag, we can follow the BSFC curve up and approximate it at .42 lb/(HP*hr), which is the fuel consumption at full load at that RPM. So now let's find the lb/hr of fuel:
300 HP * .42 lb/(HP*hr) = 126 lb/hr of fuel (divide that by 60 min and multiply your A/F ratio and you got your tentative airflow)
126/4 cylinders = 31.5 lb/hr per cylinder
31.5 lb/hr * (1 liter/1.834 lb) * (1000 cc/ 1 liter) * (1 hr/ 60 min) = 286 cc/min assuming 1 liter of diesel weighs 1.834 lbs
286 cc/min * (1 min/ 1600 shots) * (1000) = 179 cc/ 1000 shots 4 stroke engines inject fuel every other stroke 1600 = 3200/2
That is how much fuel needed to get 300 HP with those conditions. Now let's look at max torque. I am aiming for 600+ ft*lbs @ around 1800 RPM.
Formula for HP from RPM: HP = Torque*RPM/5252 = 600*1800/5252 = 205 HP
205 HP * .334 lb/(HP*hr) / 4 cylinders = 17.1 lb/hr per cyl
17.1 lb/hr * (1 liter/1.834 lb) * (1000cc/1 liter) * (1hr/ 60 min) = 155 cc/min
155cc/min * (1 min/ 900 shots) * (1000) = 172 cc/ 1000 shots
So I will need 172 cc/ 1000 shots to get 600 ft*lbs of torque at 1800 RPM.
This is pretty simple, a couple weeks back I thought there was some secret to get the cc/1000 shots, but someone smarter than I am pointed me in the right direction (i.e. the EPA tag) and this is what I got. Let me know if I am missing anything here.
-Ryan
I am trying to nail down the cc/1000 shots that I need to turn my pump to for my desired horsepower. On the way to this, we can also find how many lb/min of fuel we will need though out the RPM range, which, with your A/F ratio, will give your lb/min of air for turbo sizing (not factoring VE).
I am going to use the 4BTA CPL tag 1963 which is pretty much the same as my 1839 tag as far as the p pump goes. These tags max out at 2500 RPM, so I will interpolate the data at higher RPM.
The BSFC (Brake Specific Fuel Consumption) is the rate of fuel consumption divided by the power produced. On the EPA tag it is given for us as:
lb/(HP*hr)
and
g/(kW*hr)
I will be using imperial for this example.
I want 300 HP out of my 4BTA @ 3200 RPM
On the EPA tag, we can follow the BSFC curve up and approximate it at .42 lb/(HP*hr), which is the fuel consumption at full load at that RPM. So now let's find the lb/hr of fuel:
300 HP * .42 lb/(HP*hr) = 126 lb/hr of fuel (divide that by 60 min and multiply your A/F ratio and you got your tentative airflow)
126/4 cylinders = 31.5 lb/hr per cylinder
31.5 lb/hr * (1 liter/1.834 lb) * (1000 cc/ 1 liter) * (1 hr/ 60 min) = 286 cc/min assuming 1 liter of diesel weighs 1.834 lbs
286 cc/min * (1 min/ 1600 shots) * (1000) = 179 cc/ 1000 shots 4 stroke engines inject fuel every other stroke 1600 = 3200/2
That is how much fuel needed to get 300 HP with those conditions. Now let's look at max torque. I am aiming for 600+ ft*lbs @ around 1800 RPM.
Formula for HP from RPM: HP = Torque*RPM/5252 = 600*1800/5252 = 205 HP
205 HP * .334 lb/(HP*hr) / 4 cylinders = 17.1 lb/hr per cyl
17.1 lb/hr * (1 liter/1.834 lb) * (1000cc/1 liter) * (1hr/ 60 min) = 155 cc/min
155cc/min * (1 min/ 900 shots) * (1000) = 172 cc/ 1000 shots
So I will need 172 cc/ 1000 shots to get 600 ft*lbs of torque at 1800 RPM.
This is pretty simple, a couple weeks back I thought there was some secret to get the cc/1000 shots, but someone smarter than I am pointed me in the right direction (i.e. the EPA tag) and this is what I got. Let me know if I am missing anything here.
-Ryan