Texx
Super Moderator
Original Article Credit: pig666eon
pressure ratio of compressor
pr/compressor = (boost+ ambient pressure)
__________________
ambient pressure
eg:
comp creating 10 psig and and ambient pressure is that of sea level
(10+14.5)
_______
14.5
= 1.68
it is important to take into account pressure drops such as air filter and intercooler
pressure drop of air filter + intercooler = -1
eg:
(10-1 +14.5)
_______
14.5
= 1.61 with pressure drops
outlet air temperature
all equations here are in deg f (ˆ is the power of)
T out = [(temp in+460) × compressor pr ˆ 0.283] - 460
eg:
= [(85f+460) × 1.68ˆ 0.283] - 460 = 171f
171f is the air temp leaving the compressor assuming the compressor is at 100% efficiency
density ratios
dr = pr × (temp in + 460)
____________
(temp out + 460)
eg: pr is 1.61 and air out after cooling is 120f
1.61 × (85+460)
______
(120+460)
density ratio = 1.51
engine displacement
disp= (#cylinders)×(stroke)×(bore)squared
______________________________
1.27
eg:
= 4 × 3.571 × ( 3.425×3.425)
_______________________
1.27
disp = 131.9 cid
quick cc conversion × 16.39
eg:
131.9 × 16.39 = 2155 cubic centimeters
volumetric flow rates
(ve = volumetric efficiency of a engine at a cretin rpm)
vfr @ 100% = (displacement)×rpm)
________________
3456
vfr actual = (vfr@100%) × (ve)
eg:
= 131.9 cid × 6000rpm
___________________
3456
= 228.3 @ 100%
=(228.3 × 0.9) = 205.5 cfm
air density and mass flow rates
air dens = (2.703 × pressure)
_______________
(temp + 460)
eg:
=(2.702×14.7)
__________
(85×460)
=0.073lb/ft cubed
mass flow rate
mfr= (2.703)×(pressure)×(vfr)
__________________
(temp)+(460)
eg:
= (2.703×14.7×205.5cmf)
___________________
(85+460)
= 14.98 lb/minute
for a turbo mfr simply multiply the density ratio by the mfr
eg:
= 14.98 × 1.51 = 22.6 lb/minute
horse power
hp crank = (mfr×60)
___________
(air fuel ratio)×(bsfc)
bsfc = brake specific fuel consumption
asuming we have a air fuel mix of 12:1 and a bsfc of 0.5
eg:
= 22.6 × 60
______________
12 × 0.5
=226 crank horse power
once the mass flow rate has been found a general rule is to multiply by 10 to get a estimate of crank horse power
all equations in this will just give you good estimates and never take the figures you get as actual
any figures in any equations which are there without expiation they are a given and are standard
any faults in the maths or in figures let me know and i will change them
all figures above were not of a toyota
any questions on any of the above i will do my best to answer them and help out
it looks complicated but trust me its not if you follow the equations and fill in the numbers were there ment to be you cant go too wrong!!!
pressure ratio of compressor
pr/compressor = (boost+ ambient pressure)
__________________
ambient pressure
eg:
comp creating 10 psig and and ambient pressure is that of sea level
(10+14.5)
_______
14.5
= 1.68
it is important to take into account pressure drops such as air filter and intercooler
pressure drop of air filter + intercooler = -1
eg:
(10-1 +14.5)
_______
14.5
= 1.61 with pressure drops
outlet air temperature
all equations here are in deg f (ˆ is the power of)
T out = [(temp in+460) × compressor pr ˆ 0.283] - 460
eg:
= [(85f+460) × 1.68ˆ 0.283] - 460 = 171f
171f is the air temp leaving the compressor assuming the compressor is at 100% efficiency
density ratios
dr = pr × (temp in + 460)
____________
(temp out + 460)
eg: pr is 1.61 and air out after cooling is 120f
1.61 × (85+460)
______
(120+460)
density ratio = 1.51
engine displacement
disp= (#cylinders)×(stroke)×(bore)squared
______________________________
1.27
eg:
= 4 × 3.571 × ( 3.425×3.425)
_______________________
1.27
disp = 131.9 cid
quick cc conversion × 16.39
eg:
131.9 × 16.39 = 2155 cubic centimeters
volumetric flow rates
(ve = volumetric efficiency of a engine at a cretin rpm)
vfr @ 100% = (displacement)×rpm)
________________
3456
vfr actual = (vfr@100%) × (ve)
eg:
= 131.9 cid × 6000rpm
___________________
3456
= 228.3 @ 100%
=(228.3 × 0.9) = 205.5 cfm
air density and mass flow rates
air dens = (2.703 × pressure)
_______________
(temp + 460)
eg:
=(2.702×14.7)
__________
(85×460)
=0.073lb/ft cubed
mass flow rate
mfr= (2.703)×(pressure)×(vfr)
__________________
(temp)+(460)
eg:
= (2.703×14.7×205.5cmf)
___________________
(85+460)
= 14.98 lb/minute
for a turbo mfr simply multiply the density ratio by the mfr
eg:
= 14.98 × 1.51 = 22.6 lb/minute
horse power
hp crank = (mfr×60)
___________
(air fuel ratio)×(bsfc)
bsfc = brake specific fuel consumption
asuming we have a air fuel mix of 12:1 and a bsfc of 0.5
eg:
= 22.6 × 60
______________
12 × 0.5
=226 crank horse power
once the mass flow rate has been found a general rule is to multiply by 10 to get a estimate of crank horse power
all equations in this will just give you good estimates and never take the figures you get as actual
any figures in any equations which are there without expiation they are a given and are standard
any faults in the maths or in figures let me know and i will change them
all figures above were not of a toyota
any questions on any of the above i will do my best to answer them and help out
it looks complicated but trust me its not if you follow the equations and fill in the numbers were there ment to be you cant go too wrong!!!