Thread: Comparison between 2.5" and 3" exhaust

SteveG,

Looks like your falling off, way off the efficiency peak. The pressure ratio is a bit too high for a single stage compressor. Looks like they don't like to do much more than PR = 1.8.

Having dropped the EGT it suggests that there is less energy in the exhaust, don't know if it will let you get up to 25 psi.

Well not a bad problem to have, although probably will start to get expensive to solve if your going for 25+ psi. The pump will need attention to fuel that much air.

If you set out to drop your EGT then mission accomplished. Probably got a faster spool up as well.

Originally Posted by slug_burner

SteveG,

Looks like your falling off, way off the efficiency peak. The pressure ratio is a bit too high for a single stage compressor. Looks like they don't like to do much more than PR = 1.8.

Having dropped the EGT it suggests that there is less energy in the exhaust, don't know if it will let you get up to 25 psi.

Well not a bad problem to have, although probably will start to get expensive to solve if your going for 25+ psi. The pump will need attention to fuel that much air.

If you set out to drop your EGT then mission accomplished. Probably got a faster spool up as well.

Are you referring to the map Dougal posted for his smaller compressor, or mine?

Here's the map for mine with a very rough plot of the full-load line from Dougals.



If I've understood correctly, then it should be pretty happy to ~20psi (as Dougal suggested earlier). Definitely not trying to get 25psi out of it!

I'm not sure I get what you mean about there being less energy in the exhaust now since the EGT has dropped. Temperature is only one component of the exhaust energy and since its essentially the same fuel going in I would have thought the exhaust would have the same energy.

The original purpose behind the larger exhaust was that I was sure it was choking on the 2.5" system. Others have reported getting sustained 15psi at reasonable EGT's, but I wasn't able to fuel it up enough to get that and stay below ~700degC under sustained load. I'm pretty much happy I've proved that was the case as I now have the same boost with much lower EGT.
I'm still not getting a solid 15psi, but now have the headroom in EGT to be able to fuel it to build the extra boost. Will be interesting to see how it behaves.

I'm very much a see it, touch it, learn it sort of person - so its great that there are people around here that are prepared to share their understanding of the theory and maths side of things and help me put all the pieces together in my head.

Steve

Originally Posted by steveG

Are you referring to the map Dougal posted for his smaller compressor, or mine?

Here's the map for mine with a very rough plot of the full-load line from Dougals.



If I've understood correctly, then it should be pretty happy to ~20psi (as Dougal suggested earlier). Definitely not trying to get 25psi out of it!

I'm not sure I get what you mean about there being less energy in the exhaust now since the EGT has dropped. Temperature is only one component of the exhaust energy and since its essentially the same fuel going in I would have thought the exhaust would have the same energy.

The original purpose behind the larger exhaust was that I was sure it was choking on the 2.5" system. Others have reported getting sustained 15psi at reasonable EGT's, but I wasn't able to fuel it up enough to get that and stay below ~700degC under sustained load. I'm pretty much happy I've proved that was the case as I now have the same boost with much lower EGT.
I'm still not getting a solid 15psi, but now have the headroom in EGT to be able to fuel it to build the extra boost. Will be interesting to see how it behaves.

I'm very much a see it, touch it, learn it sort of person - so its great that there are people around here that are prepared to share their understanding of the theory and maths side of things and help me put all the pieces together in my head.

Steve

Sorry Steve, I was looking at the map that Dougal posted.

The one you put up allows a higher PR.

I know about as much about turbo selection as I know about nuclear reactors, understand the basics but no real practical exposure.

My comment about the lower EGT is relating to there being more energy in a hotter gas. A better measure might be the temperature drop across the turbo.

If you have kept the same amount of fuel going in and the EGT drops then you could be using more energy in the engine to do more work. I would find it difficult to accept that the exhaust gas has the same energy if it is cooler, even though I know that temperature is not the complete picture.

Increasing the exhaust diameter after turbine (2.5"-3") is reducing back pressure and creating a greater pressure drop across the turbine making it work more efficiently which then in turn will pump air into the motor at different rpms than before which in turn is decreasing the egts on that fuel setting...the next thing is then to increase the fuel again until egts are back upto Steves set comfortable level then there's a power increase from that extra fuel...

Originally Posted by slug_burner

SteveG,

Looks like your falling off, way off the efficiency peak. The pressure ratio is a bit too high for a single stage compressor. Looks like they don't like to do much more than PR = 1.8.

Having dropped the EGT it suggests that there is less energy in the exhaust, don't know if it will let you get up to 25 psi.

Well not a bad problem to have, although probably will start to get expensive to solve if your going for 25+ psi. The pump will need attention to fuel that much air.

If you set out to drop your EGT then mission accomplished. Probably got a faster spool up as well.

I no longer run that turbo, but in cruising conditions it returned the best fuel economy of all turbos I have run. I put this down to the cruise condition (red dot all by itself) being right near the maximum efficiency point.

I originally picked that turbo for 15psi and at that boost it ran very well. I then cranked the boost up to 20 to see how it went and ran that one off the map regularly for several years. I retired it when the compressor wheel eroded due to a poor join in the intake. I've got enough parts to rebuild it, but now run bigger compressor wheels to give more tip speed and more boost at lower rpm.

Here is a video of how that T25 built boost even with an eroded compressor wheel which limited boost 16psi it was extremely fast to react.
http://users.actrix.co.nz/dougal.ell...with%20T25.wmv

Regarding energy in the exhaust. With a bigger exhaust the total energy will be comprised of less pressure, less heat and faster flow. The energy available across the turbine will generally be higher.

I thought I had my head around Bush65's good explanation of the GT2052 maps earlier so was trying to apply it to my T25G.
The 2200rpm/20psi(PR 2.4 mass flow of about 18.9 lb/min) example is near the top of the map for my compressor.
The 2600/25psi (PR 2.7 mass flow of about 24.7 lb/min) is miles outside it. I take that to mean that mine won't do 25psi.

The radial flow compressor used for out turbos create boost pressure from the velocity that the air leaves the tip of the impeller – for efficient compressor conditions the air will have attained a velocity close to the tip velocity. The air can reach the speed of sound in parts of the compressor and trying for more pressure by increasing speed over sonic speed is rather futile – in this case practical solutions, depending on the desired increase, are; improved impeller technology, larger impeller diameter, or compound compressor stages.
This dynamic pressure is like you experience by holding your hand outside while driving at speed and is proportional to velocity squared. Immediately outside the impeller tip is a diffuser, formed between the compressor back plate and part of the volute housing, to slow down the high velocity air thus converting the dynamic pressure to static pressure and temperature.
The compressor map plots curves of constant impeller speed. These are the curves that start off horizontal from the surge line at the left, then dip down sharply as they approach the choke line at the right. Exceeding the speed beyond what the manufacturer has given on the map will likely reduce the turbo life. The results will depend on how far you go. Competition diesels in the USA exceed the given speeds by a large degree, but for them, the life of the turbo is of little importance.
There are 3 other consequences of running the compressor off it's map. The compressor efficiency is poor so temperature increases at a faster rate with increasing pressure, with the result that air density increases slowly. Greater exhaust manifold pressure is required to produce the power needed by the compressor (most of the power increase is simply increasing the air temperature) – it is desirable for turbine inlet pressure (TIP) to be less than or only a little higher than boost pressure. The other problem, is our small turbos have small impeller shafts, and the greater torque required can result in shaft failure.

How do I work out the mass flow at 20psi for eg 2500 and 3200 RPM so I can plot them on the map?

There are several ways to skin this cat and it is not necessary to follow the same order I am giving here. Here I will present it in a way that shows that what the turbo does is increase the air density. I will use imperial measurements because they are used on your compressor map. We need to calculate the actual mass flow then correct it for the inlet conditions (temperature and pressure) used for the compressor map.

The actual mass flow is the volumetric flow through the engine at the rpm of interest multiplied by the air density.

To convert the 4BD1 displacement of 3.856 litres to cubic inches, multiply litres by 61.02374. Then divide in^3 by 1728 to get ft^3

Actual volumetric flow in cubic feet per minute = (rpm / 2) x VE x displacement (in^3) / 1728
where rpm is divided by 2 because we have a four stroke engine
VE is the volumetric efficiency – for a 1989 4BD1T I estimate VE is about 0.84 at 2500 rpm and about 0.77 at 3200 rpm. Because of different valve timing, not ideal for turbocharging, earlier models may be different.

Then actual volumetric flows are:
at 2500 rpm
Qv = 2500 x 0.5 x 0.84 x 3.856 x 61.02374 / 1728
i.e. Qv = 142.982 ft^3/min

at 3200 rpm
Qv = 3200 x 0.5 x 0.77 x 3.856 x 61.02374 / 1728
i.e. Qv = 167.766 ft^3/min

Now boost pressure increases air density but the compressor also increases the air temperature which will reduce density.

The values for pressure and temperature must be used in absolute units.
To convert gauge pressure to absolute pressure we add the local ambient pressure (usually 14.7 psi near sea level).

Here we need to convert local ambient temperature in degrees Fahrenheit to degrees Rankine by adding 459.67 Now 14 deg C ~ 57.2 deg F.

We need to know the temperature of the air entering the engine. If we haven't measured it we first calculate the air temperature rise from the compressor inlet to the compressor outlet. Then calculate the temperature drop through the intercooler, if fitted.

The temperature at the compressor outlet is a function of the inlet temperature, PR (pressure ratio), compressor efficiency and the air properties:
For inlet temperature = 14 deg C i.e. 57.2 deg F
PR = (boost pressure + ambient pressure) / ambient pressure
i.e. PR = (20 psi + 14.7 psi) / 14.7 psi (assuming ambient pressure = 14.7 psi
i.e. PR = 2.36

For a first pass take the adiabatic efficiency of the compressor as 0.7 (70%) from the efficiency curves on the compressor map for PR = 2.36.

Later when we have calculated the corrected mass flow we may have to use a different adiabatic efficiency and repeat the calculations.

Now calculate the compressor outlet temperature:
Tout = 57.2 def F + (57.2 + 459.67) x (2.36^0.288 - 1) / 0.7
i.e. Tout = 264.4 deg F

If the intercooler effectiveness is 0.65 (65%) the charge air temperature will be:
T = 264.4 - [(264.4 – 57.2) x 0.65]
i.e. T = 129.7 say 130 F

Now that we have PR and inlet temperature we can calculate the air density in the inlet manifold.

Inlet air density = Absolute Pressure x 2.7027 / Absolute Temperature
For no intercooler:
inlet air density = (20 + 14.7) x 2.7027 / (264.4 + 459.67)
i.e. inlet air density = 0.1295 lb/ft^3

For an intercooler and assuming 0.5 psi pressure loss:
inlet air density = (20 – 0.5 + 14.7) x 2.7027 / (129.7 + 459.67)
i.e. inlet air density = 0.1568 lb/ft^3

Then actual mass flow is:
For no intercooler at 2500 rpm:
Actual mass flow = volumetric flow x air density
i.e. Qm = 142.982 x 0.1295 = 18.5 lb/min

For no intercooler at 3200 rpm:
Qm = 167.766 x 0.1295 = 21.7 lb/min

For intercooler and 2500 rpm:
Qm = 142.982 x 0.1568 = 22.4 lb/min

For intercooler and 3200 rpm:
Qm = 167.766 x 0.1568 = 26.3 lb/min

Before we can use the compressor map we must correct the actual mass flow to the equivalent mass flow at the reference inlet conditions. For the older Garrett turbos the reference temp was 545 deg R and reference pressure 28.4 inches of mercury (13.9 psia)

Corrected mass flow = actual mass flow x sqrt ( Tin / Tref) / (Pin / Pref)

For no intercooler and 2500 rpm
Corrected Qm = 18.5 x sqrt[(57.2 + 459.67) / 545] / (14.7 / 13.9)
i.e. corrected Qm = 17 lb/min

For no intercooler and 3200 rpm
Corrected Qm = 21.7 x sqrt[(57.2 + 459.67) / 545] / (14.7 / 13.9)
i.e. corrected Qm = 20 lb/min

For intercooler and 2500 rpm
Corrected Qm = 22.4 x sqrt[(57.2 + 459.67) / 545] / (14.7 / 13.9)
i.e. corrected Qm = 21 lb/min

For intercooler and 3200 rpm
Corrected Qm = 26.3 x sqrt[(57.2 + 459.67) / 545] / (14.7 / 13.9)
i.e. corrected Qm = 24 lb/min

Now check to see if the assumed adiabatic efficiency of 0.7 is valid for the PR and corrected mass flow. If not valid, repeat the calculations for the new efficiency.

I don't have time to go into the turbine calculations. The process involves determining the power required to drive the compressor, then the pressure ratio required for the turbine efficiency, and temperature of the exhaust gas - higher EGT requires lower PR to develop the power required to drive the compressor.

It is reasonable to assume if the PR for your turbo's compressor is nearly off it's map, the PR for the turbine might also be nearly off it's map. Note that the curve for turbine mass flow vs PR is close to horizontal at about 15 lb/min from PR 2 to 3.

The mass flow of exhaust gas = mass flow of air + mass flow of fuel.

The mass flow of fuel is calculated from the air mass flow / air/fuel ratio. Use A/F ratio of about 22 for no smoke.

The engine power and torque can be estimated from the fuel flow per hour and the BSFC. For a 1989 4BD1T the BSFC is approximately 0.36 lb/HP-hr at 2500 rpm and 0.425 lb/HP-hr at 3200 rpm. The earlier 4BD1T were not so good.

For air flow of 18.5 lb/min at 2500 rpm (no intercooler, assuming A/F ratio 22:1 (no smoke) and BSFC of 0.36 lb/HP-hr then:
HP = 18.5 / (22 x 0.36 / 60) note: here BSFC is divided by 60 to convert from hr to min
i.e. HP = 140.2

For air flow of 21.7 lb/min at 3200 rpm (no intercooler, assuming A/F ratio 22:1 (no smoke) and BSFC of 0.425 lb/HP-hr then:
HP = 21.7 / (22 x 0.425 / 60)
i.e. HP = 139.2

For air flow of 22.4 lb/min at 2500 rpm (with intercooler, assuming A/F ratio 22:1 (no smoke) and BSFC of 0.36 lb/HP-hr then:
HP = 22.4 / (22 x 0.36 / 60)
i.e. HP = 169.7

For air flow of 26.3 lb/min at 3200 rpm (with intercooler, assuming A/F ratio 22:1 (no smoke) and BSFC of 0.425 lb/HP-hr then:
HP = 26.3 / (22 x 0.425 / 60)
i.e. HP = 168.8

Note the above calculations for HP assume that the fuel injection pump has been tuned to be capable of supplying the fuel at a rate equivalent to 22:1 air/fuel ratio. Simply increasing turbo boost pressure for greater air flow will not increase power - it may reduce power because of increased pumping losses.

I have editted the last part because I read incorrect values for BSFC. Note also the 4BD1T information only went to 3000 rpm so values for VE and BSFC above 3000 rpm are extrapolated.

Originally Posted by slug_burner

SteveG,

Looks like your falling off, way off the efficiency peak. The pressure ratio is a bit too high for a single stage compressor. Looks like they don't like to do much more than PR = 1.8.

Having dropped the EGT it suggests that there is less energy in the exhaust, don't know if it will let you get up to 25 psi.

Well not a bad problem to have, although probably will start to get expensive to solve if your going for 25+ psi. The pump will need attention to fuel that much air.

If you set out to drop your EGT then mission accomplished. Probably got a faster spool up as well.

Small dia compressor impellers will not be suitable for high PR. Larger impellers can be suitable for PR 3 or more in a single stage.

For a moderate tune 4BD1T a newer technology impeller around 58 to 60 mm will be a good choice and could be suitable for PR over 2.5. But as you say the pump needs to be tuned to deliver enough fuel and generate the energy in the exhaust gas for the turbine. EGT of 650C or 700C for short periods would not concern me, but I have a late 4BD1T with better piston cooling etc.

Thanks John - very much appreciated.
It would take me half a day just to type that - let alone actually doing the calcs and thinking about the explanation.

I won't pretend to understand it all after the first few readings, but already there's a couple of missing pieces that have fallen into place.

Steve