Thread: 4BD1T Turbo Sizing and Performance Prediction.

Thanks John. Really appreciate the time you put in to make these posts.

There are other shorter, ways, e.g. the equation found in the Garrett catalogue, to determine the required boost pressure, but my goal in this thread is to provide a better understanding of what turbochargers do, and how those equations were developed.


The engine is a restriction to air flow. The volumetric air flow rate for a 4 cycle engine is:

Va = VE x Displacement x rpm / (2 x 60)

where:
VE is the volumetric efficiency, approximately = 0.80 (approximately) for a 4BD1T at 3000 rpm

Displacement = 3.856 litres for a 4BD1T

rpm is engine speed in revs per minute

2 is the number of revs to complete 4 cycles (strokes)

60 is the conversion factor for revs per minute to revs per second

Then Va = 0.8 x 3.856 litres x 3000 rpm / (2 x 60) = 77.120 litre/sec


Clearly the only way to increase the volumetric air flow is to either increase the engine displacement, improve the volumetric efficiency, or increase the engine speed.


It is possible, but expensive, relative to performance gain (bang for buck), to increase VE, e.g. improve the breathing of the head (4 valves per cylinder with modern engines), or optimise the valve timing. It is counter-productive for our purposes to increase the engine speed, unless we want to race for example.


Recapping the previous stage we found the required air mass flow for the desired power. That was because engines develop torque and power from the conversion of chemical energy in the fuel to heat energy by combustion, and that chemical process dictates mass flow. The heat energy increases the pressure of the gas in the combustion chamber (combustion pressure), which expands, forcing the piston down to generate torque. Power is simply the time rate of that torque (torque x speed).


Moving right along now, we need to convert the volumetric air flow of our engine to air mass flow. To obtain mass flow we multiply volumetric flow by the density.


Now density is mass / volume, and in the case of air it depends upon the number of molecules of air and the space (volume) they take up. As temperature increases the vibration of each molecule (temperature is a measure of vibration amplitude) each molecule takes up more space (density reduces), but as pressure increases the molecules are forced closer together (density increases).

Density of air = (Pa x M) / (R x Ta)

where:
Pa is absolute pressure in Pa (Pascal)
M is molar mass of air = 0.0289644 kg/mol
R is ideal gas constant = 8.31447 J/mol K
Ta is absolute pressure in degrees K (Kelvin) = degrees C + 273

For now we are interested in the density of air entering the engine, so Pa and Ta are the absolute pressure and temperature in the inlet manifold. From the equation for air density it should be obvious that increasing the pressure by turbocharging increases the density, as does reducing the temperature by intercooling.


Now to increase engine performance, we need to burn more fuel, which requires an increase of the air mass flow, but the volumetric flow of the engine is fixed as we have shown, so we want the turbocharger compressor to increase the density of the air.


For the turbocharger compressor we define DR (Density Ratio) as:

DR = density of air at outlet / density of air at inlet

Here we can determine the required DR from:

DR = required air mass flow / air mass flow of naturally aspirated engine

where:
required air mass flow is what we found in the previous stage
air mass flow of naturally aspirated engine is the volumetric air flow we found above, i.e. Va = 77.120 litre/sec x inlet air density

Although I'm not going to use it here, an alternate method of determining required DR is:

DR = required power with turbo / power without turbo

To determine the density for the naturally aspirated engine we need to know the ambient pressure and temperature.
For this example assume:

Pa = 100 kPa = 100000 Pa
Ta = 303 K (273 + 30 C)

Then:

density of inlet air = (Pa x M) / (R x Ta)

= (100000 Pa x 0.0289644 kg/mol) / (8.31447 J/mol K x 303 K)
= 1.1497 kg/m3

Then air mass flow of naturally aspirated 4BD1 at 3000 rpm is:

Ma = Va x density
= (77.120 litre/sec / 1000 litre/m3) x 1.1497 kg/m3

= 0.0887 kg/sec

Now using the required air mass flow determined in the previous stage we can determine the required density ratio.


(Ex a) for 90 kW:

DR = 0.108 kg/sec / 0.0887 kg/sec = 1.218 (for A/F = 18:1)

or:

DR = 0.120 kg/sec / 0.0887 kg/sec = 1.353 (for A/F = 20:1)

(Ex b) for 135 kW:

DR = 0.162 kg/sec / 0.0887 kg/sec = 1.826 (for A/F = 18:1)

or:

DR = 0.180 kg/sec / 0.0887 kg/sec = 2.029 (for A/F = 20:1)

(Ex c) for 180 kW:

DR = 0.216 kg/sec / 0.0887 kg/sec = 2.435 (for A/F = 18:1)

or:

DR = 0.240 kg/sec / 0.0887 kg/sec = 2.706 (for A/F = 20:1)

The next stage is to convert the required density ratio to required pressure ratio and correct the air mass flow so that we can plot these two values onto compressor maps to select a suitable turbo.


While it is relatively easy to determine density ratio from pressure ratio, it is more difficult to do the reverse conversion. This is because the temperature increases (thus reducing density) when air is compressed.

Originally Posted by Bush65

My understanding is that turbo manufactures, develop their turbine (and maps) using "hot gas stands".

For that reason I would expect that the reference temp and pressure, would be the inlet conditions of the hot gas stand. However they could correct them to "standard conditions", but why expect them to do that for a turbine when they don't correct compressor maps to "standard conditions".

Indeed and I chased that same thought for many hours over many months. Like you I was looking for a reference temp around 1000K which might have some real world significance.
Until I accepted that it could possibly be any arbitrary temp and pressure for correcting the flow to a standardised chart.
So I calculated it using similar temperatures and pressures as their compressor inlet conditions and it all fits.

I'm currently working on plotting out real flow and turbine power vs corrected flow on a turbine map to show how this fits at realistic exhaust temps.

Originally Posted by Bush65

If the high drive pressure is not down to the waste gate, then I would suspect that the turbine size may be a little small and so require such a high expansion ratio to develop the required torque/power.

It's that old trade-off again. Tight turbine for fast spooling and drivability vs
a larger housing for lower drive pressure and more power but worse drivability.
Personally I find myself always making the choice for the parts of driving I enjoy the most. Which is the acceleration and drivability vs top end power.

Pressure ratio (PR) is the absolute pressure at the compressor outlet (P2) divided by the absolute pressure at the compressor inlet (P2). P2is often taken as the ambient/atmospheric pressure, but a more accurate value takes account of the pressure loss that occurs because of the air filter and snorkel. Absolute pressure (PABS) is gauge pressure (PG) plus ambient pressure (PATM).


Boost pressure is the gauge pressure at the compressor outlet so to convert boost pressure to PR:
PR = (PBOOST + PATM) / P1
And to convert PR to boost pressure:
PBOOST = (PR x P1 ) + PATM


We need to know the PR to be able to select a suitable turbocharger compressor from its map.


The following figure shows DR vs PR, for some values of adiabatic efficiency.






With an intercooler fitted a lower PR is required for any DR, but the improvement is more worthwhile when effeciency is lower or PR is higher. As a rough guide; for DR up to about 2.0 and no intercooler, the PR is approximately 1.2 to 1.5 times the DR (greater for higher DR). For higher values of DR and an effective intercooler, the PR is approximately 1.1 to 1.2 times the DR.


Compressor map is a chart that shows the compressor performance. The vertical axis is the non-dimensional “Pressure Ratio”, the horizontal axis is either “Mass Air Flow” or “Volumetric Air Flow” measured at the reference inlet conditions (pressure and temperature). The units for air flow may be SI or Imperial.


Now the mass flow through the engine must be identical to the mass flow through the turbocharger, unless there is a boost leak. However the volumetric flow through the engine, which we determined before, is not the same as the volumetric flow through the compressor. We have already found the required air mass flow, we will return to the conversion to volumetric flow later.


Corrected Air Mass Flow. The compressor performance was measured on a test stand and at particular inlet conditions known as the reference pressure and reference temperature. However our calculations were based on our local ambient conditions where we want the engine to perform. For the inlet pressure we should have allowed for pressure drop through the air filter and snorkel.


If our inlet pressure and temperature are different to the reference values, we need to correct the air flow using the following formula:


MCORR = MAIR x (PREF / P1) x square root (T1 / TREF)


where:
MAIR is the actual air mass flow at the local pressure and temperature
PREF is the reference absolute pressure that the map is based upon
P1 is the local absolute pressure used to determine MAIR
T1 is the local absolute temperature used to determine MAIR
TREF is the reference absolute temperature that the map is based upon


For the sake of example, I will refer to the following map for the 62mm Borg Warner EFR compressor.






Usually Borg Warner use PREF = 981 mbar (98100 Pascal) and TREF = 293 Kelvin (19.85 C) so will use these for this example. In the previous stage we used ambient conditions of Pa = 100 kPa and Ta = 303 K.
For our three examples the corrected mass flow is:


(Ex a):
MCORR = 0.108 kg/s x (98.1 kPa / 100 kPa) x square root (303 K / 293 K) = 0.1077 kg/s


(Ex b):
MCORR = 0.162 kg/s x (98.1 kPa / 100 kPa) x square root (303 K / 293 K) = 0.1616 kg/s


(Ex c):
MCORR = 0.216 kg/s x (98.1 kPa / 100 kPa) x square root (303 K / 293 K) = 0.2155 kg/s


In this case we see it was not worthwhile correcting the mass flow as there is no significant change when rounded to 3 significant figures.



Adiabatic Efficiency of the compressor is shown on the map as a series of roughly oval shape curves of equal efficiency that have a similar appearance to contour lines. The more efficient the compressor, the less heat it adds to the air during compression. A value of 1.0 is the unobtainable 100% efficiency. At best we might get above 0.75 for a very good compressor match, but most likely the efficiency will be between 0.60 and 0.75


We need to know the adiabatic efficiency in order to calculate density and density ratio from boost pressure and pressure ratio.



One option is to take a guess between 0.6 and 0.7 for a first pass at calculating pressure ratio. If our guess is too far out we will have to plot the resulting pressure ratio vs air flow on the compressor map to refine the value for efficiency and repeat the process.


If as in this example, we have a possible turbo in mind, and because we have already determined the air mass flow and density ratio, we can use the previously mentioned approximation to choose an approximate pressure ratio that will allow us to take a better stab at the adiabatic efficiency.


(Ex a) and no intercooler:
where MCORR = 0.108 kg/s, and DR = 1.218


Approx PR = DR x 1.2 = 1.218 x 1.2 = 1.5


where MCORR = 0.108 kg/s, and DR = 1.35


Approx PR = DR x 1.2 = 1.353 x 1.2 = 1.6


Then from compressor map, adiabatic efficiency is 0.69 for PR = 1.5 and 0.70 for PR = 1.6


(Ex b) with intercooler:
where MCORR = 0.162 kg/s, and DR = 1.826


Approx PR = DR x 1.1 = 1.826 x 1.1 = 2.0


where MCORR = 0.162 kg/s, and DR = 2.029


Approx PR = DR x 1.1 = 2.029 x 1.1 = 2.2


Then from compressor map, adiabatic efficiency is 0.74 for PR = 2.0 and 0.73 for PR = 2.2


(Ex c) with intercooler:
where MCORR = 0.216 kg/s, and DR = 2.435


Approx PR = DR x 1.2 = 2.435 x 1.2 = 2.9


where MCORR = 0.216 kg/s, and DR = 2.706


Approx PR = DR x 1.1 = 2.706 x 1.2 = 3.2


Then from compressor map, adiabatic efficiency is 0.71 for PR = 2.9 and 0.7 for PR = 3.2


At this stage we can draw some preliminary conclusions from examination of the map location where the air flow and PR intersect.


We see for these examples that at maximum power engine rpm, the points lay to the left of the centre region. Ideally that point would be further to the right, which indicates that the air flow capacity of this compressor is on the large side for our engine. This is mainly a function of the inducer diameter, 49.6 mm.


It is capable of supplying even more air than needed for the plus 100% power increase of example c, if the fuel injection pump was calibrated for sufficient fuel.


We can see this by drawing lines on the map connecting the relevant points found for (ex b) and (ex c). Those lines plot intermediate values between plus 50% and plus 100% power increase at 3000 rpm with an intercooler. We can extend the lines further up and down to see what happens if we later decide on more than plus 100%, or less than plus 50%.


For higher engine rpm, say 3500 rpm the points will be further to the right, for lower rpm, say for maximum torque, the points will be further left.



That is probably enough information to digest in this post. However the discussion on this topic will be continued.

Originally Posted by Bush65

OK.

Do you know, if by manually controlling the normally ECU controlled start-up timing advance, if it would affect the injection timing across the range of the auto advance curve, such that it would simulate a change to the static advance?

If so, then if you made it so the injection timing could be easily adjusted from inside the cab, it could be useful for finding the best static advance setting for the injection. You may find, there are optimum settings for torque/power and economy for different vehicle loads, different ambient conditions, such as temperature and air density changes with altitude.

Regarding your first comment in the earlier post:



I'm willing to clarify/expand any technical issues. Your not the first to express those sentiments in this thread.

So I will make a start later today. I think it may be useful to start from the beginning and spread it over a number of post for each topic. It will take some time and I won't have any this Friday or Monday.

Given the title of this thread, I trust Dougal won't see this as a hijack. If so I can take it to a new thread.

John Apology for the late response but have been trying to chase more info on the timing advance but without success,The advance is by solenoid so is all or nothing. I will wire it to a switch in the cab and with a EGT and seat of the pants gauge will observe the results.
Also many thanks to you and Dougal for trying to educate in these extensive posts

Thanks Noel

Further to the last post, the following pic shows the points marked on a printed copy of the compressor map. Please ignore my mistaken mark-ups such as the leftmost ruled line.






This shows errors in my earlier values for adiabatic efficiency determined by eyeball on the computer screen. They should read:
(Ex a) (90 kW with no intercooler), adiabatic efficiency is:

1. 0.70 for MCORR = 0.108 kg/s, PR = 1.5 and A/F = 18:1
   0.72 for MCORR = 0.120 kg/s, PR = 1.6 and A/F = 20:1

(Ex b) (135 kW with intercooler), adiabatic efficiency is:

0.74 for MCORR = 0.162 kg/s, PR = 2.0 and A/F = 18:1
0.74 for MCORR = 0.180 kg/s, PR = 2.2 and A/F = 20:1

(Ex c) (180 kW with intercooler), adiabatic efficiency is:

0.72 for MCORR = 0.216 kg/s, PR = 2.9 and A/F = 18:1
0.72 for MCORR = 0.240 kg/s, PR = 3.2 and A/F = 20:1

The two parallel ruled lines are both for 3000 rpm, the left line for A/F ratio 18:1 and the right line for A/F ratio 20:1 As engine performance (power and torque) at 3000 rpm changes, the minimum required air mass flow and PR will lie along these lines on the map, if the A/F ratio is maintained to control exhaust smoke and EGT.


Before we can conclude that this is a suitable turbocharger compressor for our needs, it will be useful to repeat the previous exercises for lower engine rpm where we want boost to start, and where we want maximum torque. For the sake of this example use 1000 rpm and 2000 rpm (note max torque is at 2200 rpm) and use A/F ratio = 20:1


Referring to the previous performance curves for the stock 1989 4BD1T


At 1000 rpm:
Power = 26 kW
SFC = 231.7 g/kW hr
Fuel mass flow rate Mf = 26 kW x 231.7 g/kW hr / 60 min/hr = 100.40 g/min
Air mass flow Ma = 100.40 g/min x 20 / 60 sec/min = 0.0335 kg/sec (at A/F = 20:1)
Volumetric Efficiency VE = 0.9 (assumed)
Volumetric air flow through engine Va = 0.9 x 3.856 litres x 1000 rpm / (2 x 60) = 28.920 litre/sec
Density of inlet air = 1.1497 kg/m3 (as calculated before)
Air mass flow of naturally aspirated 4BD1 at 1000 rpm is:
Ma = Va x density = (28.920 litre/sec / 1000 litre/m3) x 1.1497 kg/m3 = 0.0332 kg/sec
Density Ratio DR = 0.0335 kg/sec / 0.0332 kg/sec = 1.009
Approximate Pressure Ratio PR = 1.009 x 1.01 = 1.02 (no intercooler)
Approximate Adiabatic Efficiency = off map


At 1000 rpm:
Power = 26 kW x 1.5 = 39 kW
SFC = 231.7 g/kW hr
Fuel mass flow rate Mf = 39 kW x 231.7 g/kW hr / 60 min/hr = 150.60 g/min
Air mass flow Ma = 150.60 g/min x 20 / 60 sec/min = 0.0502 kg/sec (at A/F = 20:1)
Volumetric Efficiency VE = 0.9 (assumed)
Volumetric air flow through engine Va = 0.9 x 3.856 litres x 1000 rpm / (2 x 60) = 28.920 litre/sec
Density of inlet air = 1.1497 kg/m3 (as calculated before)
Air mass flow of naturally aspirated 4BD1 at 1000 rpm is:
Ma = Va x density = (28.920 litre/sec / 1000 litre/m3) x 1.1497 kg/m3 = 0.0332 kg/sec
Density Ratio DR = 0.0502 kg/sec / 0.0332 kg/sec = 1.512
Approximate Pressure Ratio PR = 1.512 x 1.05 = 1.59 (with intercooler)
Approximate Adiabatic Efficiency = off map (surge)


At 1000 rpm:
Power = 26 kW x 2.0 = 52 kW
SFC = 231.7 g/kW hr
Fuel mass flow rate Mf = 52 kW x 231.7 g/kW hr / 60 min/hr = 200.81 g/min
Air mass flow Ma = 200.81 g/min x 20 / 60 sec/min = 0.0669 kg/sec (at A/F = 20:1)
Volumetric Efficiency VE = 0.9 (assumed)
Volumetric air flow through engine Va = 0.9 x 3.856 litres x 1000 rpm / (2 x 60) = 28.920 litre/sec
Density of inlet air = 1.1497 kg/m3 (as calculated before)
Air mass flow of naturally aspirated 4BD1 at 1000 rpm is:
Ma = Va x density = (28.920 litre/sec / 1000 litre/m3) x 1.1497 kg/m3 = 0.0332 kg/sec
Density Ratio DR = 0.0669 kg/sec / 0.0332 kg/sec = 2.016
Approximate Pressure Ratio PR = 2.016 x 1.1 = 2.22 (with intercooler)
Approximate Adiabatic Efficiency = off map (surge)


At 2000 rpm
Power = 65 kW
SFC = 213.9 g/kW hr
Fuel mass flow rate Mf = 65 kW x 213.9 g/kW hr / 60 min/hr = 231.72 g/min
Air mass flow Ma = 231.72 g/min x 20 / 60 sec/min = 0.0772 kg/sec (at A/F = 20:1)
Volumetric Efficiency VE = 0.9 (assumed)
Volumetric air flow through engine Va = 0.9 x 3.856 litres x 2000 rpm / (2 x 60) = 57.840 litre/sec
Density of inlet air = 1.1497 kg/m3 (as calculated before)
Air mass flow of naturally aspirated 4BD1 at 2000 rpm is:
Ma = Va x density = (57.840 litre/sec / 1000 litre/m3) x 1.1497 kg/m3 = 0.0665 kg/sec
Density Ratio DR = 0.0772 kg/sec / 0.0665 kg/sec = 1.161
Approximate Pressure Ratio PR = 1.161 x 1.2 = 1.39 (no intercooler)
Approximate Adiabatic Efficiency = 0.64


At 2000 rpm
Power = 65 kW x 1.5 = 97.5 kW
SFC = 213.9 g/kW hr
Fuel mass flow rate Mf = 97.5 kW x 213.9 g/kW hr / 60 min/hr = 347.59 g/min
Air mass flow Ma = 347.59 g/min x 20 / 60 sec/min = 0.116 kg/sec (at A/F = 20:1)
Volumetric Efficiency VE = 0.9 (assumed)
Volumetric air flow through engine Va = 0.9 x 3.856 litres x 2000 rpm / (2 x 60) = 57.840 litre/sec
Density of inlet air = 1.1497 kg/m3 (as calculated before)
Air mass flow of naturally aspirated 4BD1 at 2000 rpm is:
Ma = Va x density = (57.840 litre/sec / 1000 litre/m3) x 1.1497 kg/m3 = 0.0665 kg/sec
Density Ratio DR = 0.116 kg/sec / 0.0665 kg/sec = 1.742
Approximate Pressure Ratio PR = 1.742 x 1.09 = 1.90 (with intercooler)
Approximate Adiabatic Efficiency = 0.69


At 2000 rpm
Power = 65 kW x 2.0 = 130 kW
SFC = 213.9 g/kW hr
Fuel mass flow rate Mf = 130 kW x 213.9 g/kW hr / 60 min/hr = 463.45 g/min
Air mass flow Ma = 463.45 g/min x 20 / 60 sec/min = 0.154 kg/sec (at A/F = 20:1)
Volumetric Efficiency VE = 0.9 (assumed)
Volumetric air flow through engine Va = 0.9 x 3.856 litres x 2000 rpm / (2 x 60) = 57.840 litre/sec
Density of inlet air = 1.1497 kg/m3 (as calculated before)
Air mass flow of naturally aspirated 4BD1 at 2000 rpm is:
Ma = Va x density = (57.840 litre/sec / 1000 litre/m3) x 1.1497 kg/m3 = 0.0665 kg/sec
Density Ratio DR = 0.154 kg/sec / 0.0665 kg/sec = 2.321
Approximate Pressure Ratio PR = 2.321 x 1.12 = 2.60 (with intercooler)
Approximate Adiabatic Efficiency = 0.66


The following pic shows these points for PR vs Air Mass Flow plotted on the compressor map.






This shows compressor surge at 1000 rpm. The compressor supplies more air than the engine can breath. It may not be noticeable if the engine accelerates quickly beyond 1000 rpm.
So far we have used approximate values for PR, and we need to calculate better values.


Before we can calculate a better value than the approximate pressure ratio (PR) used earlier, we should revue some physics for gasses.


Ideal Gas Law in equation form is:
PV=nRT
where, using SI units of measurement:
P is the absolute pressure (Pa)
V is the volume (m3)
n is the number of gas molecules, in moles, an indication of the mass
R is the ideal (or universal) gas constant 8.314 J / K. mole (Joule per Kelvin per mole)
T is the absolute temperature (K) (Kelvin = degrees Celsius + 273.15)


Specific Heat is the amount of heat in Joules to raise the temperature of one kg of substance by one degree Kelvin (or Celsius). For dry air cP is the specific heat at constant pressure, which varies with temperature, and cV is the specific heat at constant volume.
For our calculations we need to know the value for the expression {(k-1) / k} where k is the ratio of specific heats (cP / cV ) For the intake air I will use 0.288 and later for exhaust gas use 0.222 instead of writing {(k-1) / k} in equations.


Intercooler Effectiveness is the ratio (TOUT / TATM ) For a suitably sized intercooler the effectiveness is usually between 0.6 and 0.7 A value of 1.0 (100% effective) would require the ambient cooling air at temperature TATM to reduce the temperature of the charge air to the same temperature.


To be continued ...

Continuing from the last post. When the air pressure is increased in the compressor (PR) the temperature rises. In the turbine, when the exhaust pressure is reduced by the expansion ratio the temperature falls.


For the compressor:


TOUT = (TIN x PR(0.288 - 1) / adiabatic efficiency) + TATM


where:
TOUT is the outlet temperature (C)
TIN is the absolute inlet temperature (K) = TATM + 273.15
PR is the pressure ratio developed by the compressor
0.288 is {(k-1) / k} where k is the ratio of specific heats of dry air
adiabatic efficiency is the efficiency found from the compressor map
TATM is the local ambient temperature (C)


For the turbine:
TOUT = TIN – {isentropic efficiency x (TABS x [1 – (1 / ER)] 0.222 }


where:
TOUT is the outlet temperature (C)
TIN is the inlet temperature (C) = exhaust gas temperature (EGT)
isentropic efficiency is the efficiency found from the turbine map
TABS is the EGT in Kelvin = TIN + 273.15
ER is the expansion ratio across the turbine (PIN / POUT)
where PIN is the inlet pressure POUT is outlet pressure
and POUT is local ambient pressure + pressure loss in exhaust system
0.222 is {(k-1) / k} where k is the ratio of specific heats of the exhaust gas


For the air temperature in the inlet manifold when an intercooler is used:


TMAN = TOUT – [(TOUT – TATM) x effectiveness]


where:
TMAN is the inlet manifold temperature (C)
TOUT is the outlet temperature from the compressor (C)
TATM is the local ambient temperature (C)
effectiveness is the intercooler effectiveness (usually 0.6 to 0.7)

Originally Posted by Bush65

...

For the compressor:

TOUT = (TIN x PR(0.288 - 1) / adiabatic efficiency) + TATM

where:
TOUT is the outlet temperature (C)
TIN is the absolute inlet temperature (K) = TATM + 273.15
PR is the pressure ratio developed by the compressor
0.288 is {(k-1) / k} where k is the ratio of specific heats of dry air
adiabatic efficiency is the efficiency found from the compressor map
TATM is the local ambient temperature (C)

For the turbine:
TOUT = TIN – {isentropic efficiency x (TABS x [1 – (1 / ER)] 0.222 }

where:
TOUT is the outlet temperature (C)
TIN is the inlet temperature (C) = exhaust gas temperature (EGT)
isentropic efficiency is the efficiency found from the turbine map
TABS is the EGT in Kelvin = TIN + 273.15
ER is the expansion ratio across the turbine (PIN / POUT)
where PIN is the inlet pressure POUT is outlet pressure
and POUT is local ambient pressure + pressure loss in exhaust system
0.222 is {(k-1) / k} where k is the ratio of specific heats of the exhaust gas
...

Sorry, I just realised some typos occurred in those formulae. One I made and the others occurred because the postscripts vanished when I pasted them in.

For the compressor the formula should be:
TOUT = [TIN x ((PR^0.288) - 1) / adiabatic efficiency] + TATM

And for the turbine it should be:
TOUT = TIN – {isentropic efficiency x (TABS x [1 – (1 / ER)]^0.222 }