how to Read compressor maps and matching turbos to your engine.
Reading compressor maps and matching turbos to your engine.
A lot of people going for bigger turbos recently so I thought I will take to explain how to read a compressor map for the said turbo. This will show you how to find surge lines, max boost, max power achieved by a turbo. And when you can get (X) boost at ( Y) rpm
First we have to consider the most common engine specs that we have 2 l and 2.3l I will do a rough calucations for both
volume of air (cu ft/min)
= engine rpm x engine cid
------------------------------
(1728 x 2)
This equation is for finding the volume of air going into the engine. The displacement for a 2 l is 122 cu.in and 2.3 l is 140. cu.in . We have a four stroke engine; the intake valve on a cylinder opens once every 2 revolutions of the engine. So, for every 2 revs the engine takes in 122 cu.in for a 2 l and 140cu.in for 2.3l of air. What boost are we running to achieve that will depend depends on the pressure and temperature of the air in the intake manifold. But the volume is always the same every 2 rpm.
So for a 2 l
Vol of air at 3000revs = 3000x 122/(1728x2)= 106 CFM
Vol at 4000revs = 141 cfm
Vol oat 5000revs = 176.5 CFM
For a 2.3 same equation will give you
121 @ 3000revs, 162cfm@ 4000revs, 202cfm @ 5000revs
The Ideal Gas Law what an equation, pretty much rules everything . It relates the air pressure, temperature, volume, and mass (ie, boost in pounds) of air. If you know any three of these, you can calculate the fourth. The equation is written:
PV=nRT
where P is the absolute pressure (not the gauge pressure), V is the volume, n is related to the number of air molecules, which is an indication of the mass (or pounds) of air, R is a constant number, and T is the absolute temperature.
The Ideal Gas Law can be rearranged to calculate any of the variables. For example, if you know the pressure, temperature, and volume of air you can calculate the pounds of air:
n=PV/(RT)
That is useful, since we know the pressure (boost pressure), the volume (which we calculate as shown in the first section "Engine Volumetric Flow"), and we can make a good guess on the temperature. So we can figure out how many pounds of air the engine is moving. And the more pounds of air you move the more power you will make.
Here is the Ideal Gas Law
To get pounds of air:
n(lbs/min)
= P(psia) x V(cu.ft./min) x 29
-----------------------------------
(10.73 x T(deg R))
the reason this is handy is because most compressor maps are actually displayed in lbs/min so this will make easy reference.
Now lets work this out on a car running 1.5bar of boost, Using the figures that we have worked out earlier for volume, for both the 2 and 2.3l configurations. But we will need work out the absolute boost pressure which is the pressure you get on your gauge in psi plus 14.7. so if you are running 1.5bar which is 22psi then add 14.7 = 37.7psi absolute.
Temperature is also need to be absolute which is temperature in the intake manifold in F plus 460. so if you are running a respectable intercooler you shouldn't get more than 45c in the intake that is 113f + 460f = 573F absolute. Now we can apply the equation to get lb/min at various boost levels.
2l first
3000revs@ 1.5bar 18.84lb.min
4000revs@ 1.5bar 25lb.min
5000revs @1.5bar 31.38lbs.min
2.3l
3000revs@ 1.5bar 21.5lb.min
4000revs@ 1.5bar 28.8lb.min
5000revs @1.5bar 35.9lbs.min
I am sure I have lost most people by now but there is light at the end of the tunnel so bear with me.
Now if you look at a compressor map
There are two different sets of curves in the graph; efficiency curves and rpm curves. The area where there are lines drawn is the operating envelope. It is best to operate the compressor within its envelope. It will still run if you go to the right of the envelope, just not well. To the left of the envelope, where it is marked "surge limit", the flow through the compressor is unstable and will go up and down and backwards unpredictably. This is surging. Do not pick a turbo that will operate in this area! It can be very damaging.
Now reading the graph
We will need work correct the flow to inches of mercury instead of psi as most graphs are displayed in that fashion, with out getting into great details here is the equation to use. I know turbonetics use 85F (545R) for their standard temps which is 29degrees so I am going assume that is the standard that all turbo manufactures use. Unless I am adviced otherwise. Pin is your inlet temperation so I am going to assume a 20c which is 70F and also I am assuming a slight vaccum of - 0.5psi in the inlet.
Corrected flow
= actual flow x (Tin/545)^0.5
--------------------------------
(Pin/13.949)
Tin = 70 + 460 = 530 deg R
Pin = -0.5 + 14.7 = 14.2 psia
So corrected flow for
2l figures first
corrected flow for 3000revs
= 18.84 x (530/545)^0.5
-----------------------------------= 18.25lb/min
(14.2/13.949)
4000revs= 24.2
5000revs= 30.42
for 2.3l
corrected flow @ 3000revs= 20.8
corrected flow @ 4000revs= 27.9
corrected flow @ 5000revs= 34.8
So lets take the gt40R that has been used on the MLR recently as an example of an evo turbo with 2.3l
So we mark that point on the bottom of the graph, and draw a straight line upward from that point
Now to get the other point of the graph the pressure ratio Pout/Pin @ 1.5bar or 22pis in the inlet manifold, you we will need to assume a 2psi pressure drop at this level which is standard again so the pressure out of the turbo will be 22psi+2=24 :
Pout/Pin
= (24 + 14.7)
-----------------= 2.72
(-0.5 + 14.7)
So then we find about where 2.72 is on the left side of the graph and draw a line horizontally from that point. Where the two lines meet is where the turbo will operate.
Look at the efficiency curves, which look like circles. Our point is just a little inside the 72% curve, so when we are running at 5000 rpm and 22 psi boost with 20 deg air outside and 45 deg air in the manifold then the compressor efficiency is a fraction over 72%.
All the above figures have assumed 100% volumetric efficiency which is not going to be correct over the whole rev range. This will vary on head design and cams design etc. but if you optimize VE at one part of the rev range you will probably loose out at another area. For the evo engine I would assume the volumetric efficiency to be around 90-95% around 5000revs and probably 100% with cams etc. so the lb/min figures are probably slightly more optimistic. You may need to multiple the figures by 0.9 or 0.95 under 5000revs.
I hope this will be of use for some people, and help understand the general principles of turbo and choice and to a certain respect engine choices as well.
maybe andy should check the calculations just in case i missed something out
afterall he deals with more complicated equations every day
i dont
Reading compressor maps and matching turbos to your engine.
A lot of people going for bigger turbos recently so I thought I will take to explain how to read a compressor map for the said turbo. This will show you how to find surge lines, max boost, max power achieved by a turbo. And when you can get (X) boost at ( Y) rpm
First we have to consider the most common engine specs that we have 2 l and 2.3l I will do a rough calucations for both
volume of air (cu ft/min)
= engine rpm x engine cid
------------------------------
(1728 x 2)
This equation is for finding the volume of air going into the engine. The displacement for a 2 l is 122 cu.in and 2.3 l is 140. cu.in . We have a four stroke engine; the intake valve on a cylinder opens once every 2 revolutions of the engine. So, for every 2 revs the engine takes in 122 cu.in for a 2 l and 140cu.in for 2.3l of air. What boost are we running to achieve that will depend depends on the pressure and temperature of the air in the intake manifold. But the volume is always the same every 2 rpm.
So for a 2 l
Vol of air at 3000revs = 3000x 122/(1728x2)= 106 CFM
Vol at 4000revs = 141 cfm
Vol oat 5000revs = 176.5 CFM
For a 2.3 same equation will give you
121 @ 3000revs, 162cfm@ 4000revs, 202cfm @ 5000revs
The Ideal Gas Law what an equation, pretty much rules everything . It relates the air pressure, temperature, volume, and mass (ie, boost in pounds) of air. If you know any three of these, you can calculate the fourth. The equation is written:
PV=nRT
where P is the absolute pressure (not the gauge pressure), V is the volume, n is related to the number of air molecules, which is an indication of the mass (or pounds) of air, R is a constant number, and T is the absolute temperature.
The Ideal Gas Law can be rearranged to calculate any of the variables. For example, if you know the pressure, temperature, and volume of air you can calculate the pounds of air:
n=PV/(RT)
That is useful, since we know the pressure (boost pressure), the volume (which we calculate as shown in the first section "Engine Volumetric Flow"), and we can make a good guess on the temperature. So we can figure out how many pounds of air the engine is moving. And the more pounds of air you move the more power you will make.
Here is the Ideal Gas Law
To get pounds of air:
n(lbs/min)
= P(psia) x V(cu.ft./min) x 29
-----------------------------------
(10.73 x T(deg R))
the reason this is handy is because most compressor maps are actually displayed in lbs/min so this will make easy reference.
Now lets work this out on a car running 1.5bar of boost, Using the figures that we have worked out earlier for volume, for both the 2 and 2.3l configurations. But we will need work out the absolute boost pressure which is the pressure you get on your gauge in psi plus 14.7. so if you are running 1.5bar which is 22psi then add 14.7 = 37.7psi absolute.
Temperature is also need to be absolute which is temperature in the intake manifold in F plus 460. so if you are running a respectable intercooler you shouldn't get more than 45c in the intake that is 113f + 460f = 573F absolute. Now we can apply the equation to get lb/min at various boost levels.
2l first
3000revs@ 1.5bar 18.84lb.min
4000revs@ 1.5bar 25lb.min
5000revs @1.5bar 31.38lbs.min
2.3l
3000revs@ 1.5bar 21.5lb.min
4000revs@ 1.5bar 28.8lb.min
5000revs @1.5bar 35.9lbs.min
I am sure I have lost most people by now but there is light at the end of the tunnel so bear with me.
Now if you look at a compressor map

There are two different sets of curves in the graph; efficiency curves and rpm curves. The area where there are lines drawn is the operating envelope. It is best to operate the compressor within its envelope. It will still run if you go to the right of the envelope, just not well. To the left of the envelope, where it is marked "surge limit", the flow through the compressor is unstable and will go up and down and backwards unpredictably. This is surging. Do not pick a turbo that will operate in this area! It can be very damaging.
Now reading the graph
We will need work correct the flow to inches of mercury instead of psi as most graphs are displayed in that fashion, with out getting into great details here is the equation to use. I know turbonetics use 85F (545R) for their standard temps which is 29degrees so I am going assume that is the standard that all turbo manufactures use. Unless I am adviced otherwise. Pin is your inlet temperation so I am going to assume a 20c which is 70F and also I am assuming a slight vaccum of - 0.5psi in the inlet.
Corrected flow
= actual flow x (Tin/545)^0.5
--------------------------------
(Pin/13.949)
Tin = 70 + 460 = 530 deg R
Pin = -0.5 + 14.7 = 14.2 psia
So corrected flow for
2l figures first
corrected flow for 3000revs
= 18.84 x (530/545)^0.5
-----------------------------------= 18.25lb/min
(14.2/13.949)
4000revs= 24.2
5000revs= 30.42
for 2.3l
corrected flow @ 3000revs= 20.8
corrected flow @ 4000revs= 27.9
corrected flow @ 5000revs= 34.8
So lets take the gt40R that has been used on the MLR recently as an example of an evo turbo with 2.3l

So we mark that point on the bottom of the graph, and draw a straight line upward from that point
Now to get the other point of the graph the pressure ratio Pout/Pin @ 1.5bar or 22pis in the inlet manifold, you we will need to assume a 2psi pressure drop at this level which is standard again so the pressure out of the turbo will be 22psi+2=24 :
Pout/Pin
= (24 + 14.7)
-----------------= 2.72
(-0.5 + 14.7)
So then we find about where 2.72 is on the left side of the graph and draw a line horizontally from that point. Where the two lines meet is where the turbo will operate.
Look at the efficiency curves, which look like circles. Our point is just a little inside the 72% curve, so when we are running at 5000 rpm and 22 psi boost with 20 deg air outside and 45 deg air in the manifold then the compressor efficiency is a fraction over 72%.
All the above figures have assumed 100% volumetric efficiency which is not going to be correct over the whole rev range. This will vary on head design and cams design etc. but if you optimize VE at one part of the rev range you will probably loose out at another area. For the evo engine I would assume the volumetric efficiency to be around 90-95% around 5000revs and probably 100% with cams etc. so the lb/min figures are probably slightly more optimistic. You may need to multiple the figures by 0.9 or 0.95 under 5000revs.
I hope this will be of use for some people, and help understand the general principles of turbo and choice and to a certain respect engine choices as well.
maybe andy should check the calculations just in case i missed something out