Q. Using this information is it possible to determine a theoretical fuel consumption rate? Happy to list some assumptions to simplify things, such as constant speed of 100km/h (zero acceleration) and zero wind speed at 0km/h.
Just to confirm John's numbers. I'm using slightly different values for BSFC, VE, turbo efficiency and intercooler effectiveness, so our numbers differ by a few percent but generally agree well.
For the 3000rpm point from the plots much earlier in this thread I've got
140cc max fuel.
176kw.
25psi boost.
75C into the engine
18:1 A/F ratio.
28.3 lb/min airflow.
17.3kw of heat shed by the intercooler.
43.5kw required to drive the turbo compressor (turbo shaft power).
Now driving this using a T25 turbine with 0.49 A/R housing (as I currently do) this is what we can expect for turbo drive pressures and temps.
700C EGT (I set this temp)
65% efficient.
Turbine expansion ratio 2.59.
Exhaust backpressure of 116kPa (1.9psi).
Drive pressure of 28.9psi (including that 1.9psi back pressure).
Wastegate is open enough to divert 12% of the exhaust flow around the turbine.
EGT in the downpipe is 468C, the temp drop across the turbine is 232C.
But what is most interesting, I ran these same figures for a GT2256 turbo (wastegated, not VNT).
The 56mm compressor can't flow enough air for 180kw. But it can do 170kw.
It has a big enough compressor efficiency improvement that a noticeable drop in boost can deliver the same density ratio and get the same air in each cylinder.
It has a big enough turbine efficiency improvement that a noticeable drop in drive pressure results.
23psi boost at 3000rpm.
27.2lb/min
170kw flat from 2800rpm to 3000rpm and dropping after that.
70C air into the engine.
GT22 Turbine, 0.56 A/R.
37.6kw shaft power.
70% efficient
700C EGT.
115kPa exhaust backpressure (1.9psi)
Expansion ratio 2.4
Drive pressure 24.9psi (including 1.9psi back pressure).
EGT after turbine, 485C.
Temp drop across turbine 215C.
Wastegating 15% of the exhaust flow.
These figures are just for one rpm point (3,000rpm). What it doesn't show is the huge low rpm gain possible with the GT22 turbine over the T25.
In short the GT22 turbine is a more modern design that is better everywhere. It provides more low end boost, it provides more top end flow (more shaft power) and is more efficient so it produces more boost for the same drive pressure.
The upshot of all this. I now consider even a 0.64 A/R turbine housing to be too big for optimal performance on our engines. There are gains to be made everywhere by going to a smaller but better turbine.
The only turbines in the current garrett range which feature the better design in our size range are the GT22 turbines and the largest trim GT28 turbine which Ben (Isuzurover) is running on his nissan ricer turbo. That very efficient turbine is the reason he can run a 0.86 A/R housing and still have a drivable result. If you put a 0.86 A/R housing on the other T25/GT25/T28/T28 turbines the result will be very disappointing.
Q. Using this information is it possible to determine a theoretical fuel consumption rate? Happy to list some assumptions to simplify things, such as constant speed of 100km/h (zero acceleration) and zero wind speed at 0km/h.
The biggest factor in fuel consumption is engine load, which is your engine.
Some coast-down tests with a bit of data (weight etc) can give you a rough idea of the power required for steady state cruising.
The other part is estimating or calculating engine efficiency at part load. That is the hard part.
Nevermind. Found it.
So far we have determined the density ratio (DR) for the air mass flow needed for effective combustion of the fuel required for the desired performance, and made an estimate of the pressure ratio (PR) to achieve that density ratio. That estimate was used to read the approximate adiabatic efficiency from a compressor map.
Now we want to determine the required PR to achieve the DR.
Recall for the compressor, density ratio was defined as DR = density of air at outlet / density of air at inlet
If an intercooler is used then use density at intercooler outlet in place of compressor outlet.
Also recall that PR = absolute pressure at outlet / absolute pressure at inlet
In the calculation for DR we use the following formula for both the outlet and inlet density:
Density of air = (Pa x M) / (R x Ta)Then, both M and R are cancelled out and density ratio simplifies to: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 temperature in degrees K (Kelvin) = degrees Celcius + 273
DR = POUT / PIN x TIN / TOUTRearranging to find PR we get:
Where:
POUT is absolute pressure at compressor, or intercooler outlet
PIN is absolute pressure at compressor inlet
TIN is absolute temperature at compressor inlet
TOUT is absolute temperature at compressor, or intercooler outlet
PR = POUT / PIN = DR x TOUT / TINNow the difficulty that arises is we need to know the PR before we can find the outlet temperature
TOUT = (TIN x (PR0.288 - 1) / adiabatic efficiency) + TATMTo use the methods that Garrett and other some others give for selecting a turbo they assume we know TOUT but in reality the best that can be done is make an educated guess.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)
We could use the outlet temperature we found in the last stage in the equation given above for PR. Then repeat the process if the calculated PR differed too far from the approximate PR used to find the adiabatic efficiency and outlet temperature.
In the equation for outlet temperature the temperature increase created by the compressor is the left part of the expression:
(TIN x (PR0.288 - 1) / adiabatic efficiency)And for the intercooler:
TMAN = TOUT – [(TOUT – TATM) x effectiveness]Here the temperature reduction created by the intercooler is the left part of the expression:
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 between 0.6 and 0.7)
[(TOUT – TATM) x effectiveness]There is another way!
It is easy to find DR for a given PR if we know the adiabatic efficiency of the compressor and the effectiveness of the intercooler (if fitted). This is useful in a spreadsheet for our calculations, by allowing us is to construct a 'look-up' table (or graph) of DR vs PR
In the look-up table, we can have a row (or column) for:
PR over the range we might be interested inThen it is simply a matter of finding the required DR (intercooled or non-intercooled) in the table and looking up the corresponding PR.
TOUT (from the equation above) for each PR value
TMAN (from the equation above) for each TOUT value
DR from DR = PR x TIN / TOUT if no intercooler
DR from DR = PR x TIN / TMAN if an intercooler is used
Then plot the points for PR vs air flow on the compressor map to find the adiabatic efficiency. If the new adiabatic efficiency is close to the value we used above, then we have a valid PR.
If it is too far out, we need to use the new adiabatic efficiency to re-calculate the look-up table.
Once we have the above values, the next stage is to determine the power needed to drive the compressor and move on to the turbine.
I should point out here, that most methods for choosing a turbo take a different approach than I have followed in this thread, and start with a target boost pressure at each engine rpm of interest. Then the sequence of calculation becomes:
PRIf the target power is not achieved, then guess another boost pressure and repeat the process.
Air flow
Adiabatic Efficiency
Density
Air mass flow
Fuel rate from A/F ratio
Power from fuel rate and SFC
This procedure is useful if you have already installed your turbocharger and can monitor the boost pressure at the different rpm points. It allows you to see the approximate power and conduct 'what if' calculations to see what can happen if the boost pressure is increased or decreased.
However remember that with a diesel engine the fuel rate is not going to change, simply because you have 'such and such' an air flow or boost pressure. Governor adjustment is required, and the full load adjustment will be for a particular rpm point. With our mechanical injection pump we don't have the ability to MAP the fuel rate to air flow over the full range of engine revs.
The best we can do is, over the range of rpm and load, use a setting that doesn't create smoke or egt's that are too high. So at some sections of the range, we may have surplus air, but we can't increase the fuel rate (torque/power) there because of consequences elsewhere. I will discuss this further in another post.
For our examples at 3000 rpm, we arrived at the following requirements for density ratio:
(Ex a) for 90 kW:
DR = 0.108 kg/sec / 0.0887 kg/sec = 1.218 (for A/F = 18:1)
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)
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)
DR = 0.240 kg/sec / 0.0887 kg/sec = 2.706 (for A/F = 20:1)
And for these examples we estimated an adiabatic efficiency of:
(Ex a) for 90 kW:
adiabatic efficiency = 0.70 (for A/F = 18:1)
adiabatic efficiency = 0.72 (for A/F = 20:1)
(Ex b) for 135 kW:
adiabatic efficiency = 0.74 (for A/F = 18:1)
adiabatic efficiency = 0.74 (for A/F = 20:1)
(Ex c) for 180 kW:
adiabatic efficiency = 0.72 (for A/F = 18:1)
adiabatic efficiency = 0.72 (for A/F = 20:1)
Also for these examples we used a local ambient air temperature of 303 K (approx 30 C) and an intercooler effectiveness of 0.65
Using the look up table in a spreadsheet the following PR's are found:
(Ex a) for 90 kW (no intercooler):
PR = 1.39 (for A/F = 18:1)
PR = 1.64 (for A/F = 20:1)
(Ex b) for 135 kW (with intercooler):
PR = 2.02 (for A/F = 18:1)
PR = 2.29 (for A/F = 20:1)
(Ex c) for 180 kW (with intercooler):
PR = 2.85 (for A/F = 18:1)
PR = 3.24 (for A/F = 20:1)
And the greatest difference in adiabatic efficiencies was for (Ex a) and A/F = 18.1 where the new adiabatic efficiency is 0.68 (was 0.70). This changes the PR to 1.4 from 1.39.
I will make the spreadsheet available later.
Specific Work
Is the work per unit mass evaluated for conventional turbo machinery such as pumps compressors and turbines. The theory for specific work for compressible fluids is useful for matching a turbocharger's compressor and turbine.
The work from the turbine = work from compressor + turbo losses (friction).
Specific work has SI units:
N m/kg = J/kg = m2/s2
N (Newton) is the derived SI unit for force (kg m/s2)
J (Joule) is the derived SI unit for work and energy (N m)
For Isentropic Compression(or Expansion)
p1 v1κ = p2 v2κ
where:
p = absolute pressure (Pa)
v = volume (m3)
κ = cp / cv = ratio of specific heats (J/kg K)
cp = specific heat at constant pressure
cv = specific heat at constant volume
K (Kelvin) is SI unit for absolute temperature
Specific Work of Compressor
A compressor increases the air pressure from p1to p2 and the specific work can be expressed as:
w = κ / (κ -1) x R x T1 x [( p2 / p1)^((κ-1)/κ) - 1]
where:
R = individual gas constant (J/kg K)
T = absolute temperature (K)
Specific Work of Turbine
A turbine expands the exhaust gas from p1 to p2 and the specific work can be expressed as:
w = κ / (κ -1) x R x T1 x [1 - ( p2 / p1)^((κ-1)/κ)]
Power from Specific Work
Power = specific work x mass flow
Head
With turbo machines it can be convenient to express specific work in terms of head (of the fluid).
w = g h
where
h = head (m)
g = acceleration of gravity = 9.81 (m/s2) approximately
Then head is given by:
h = w / g
Last edited by Bush65; 23rd September 2013 at 04:46 PM. Reason: fixed equations
Just a quick clarification for anyone going through the equations - in the equation given for temperature increase during compression
i.e. TOUT = (TIN x (PR0.288 - 1) / adiabatic efficiency) + TATM
PR0.288 should say PR^0.288 (or PR to the power of 0.288).
HTH
1/3.5 is a quick and easy way to get that expression to whatever accuracy you require.
(1.4-1)/1.4 = 1/3.5 = 0.2857142..........
I have almost finished running through two turbo examples on the 4BD1T. I have run these on a compressor and turbine match. This is where I have to match the power generated by the turbine to the power requirements of the compressor.
1. GT2256V VNT used on the merc ML270 (I have one of these turbos).
This turbo has a 0.64 A/R turbine housing and I've assumed the VNT can pull that down to about half the A/R.
2. GT2259 wastegated (used on Hino and Iveco diesels).
This turbo has a 0.56 A/R turbine housing and the compressor can flow about 10% more.
The results are interesting.
Low End.
Essentially the GT2259 turbo still spools just as quick. Both these turbos can provide enough boost to burn 140cc of fuel and produce over 600Nm of torque by 1400rpm.
And the wastegated turbo does with with less drive pressure.
Call it even.
Mid-range.
They are pretty much equal at 2000rpm, the wastegated turbo is running less drive pressure (because the turbine is more efficient without the extra vanes).
They can both provide more boost than needed through the mid-range, for the wastegated turbo that is no problem. You simply use the wastegate to cap boost and this lets the turbine breathe a bit easier.
But the VNT turbo still has to push that exhaust through the same turbine. The only way to get less power from the turbine is lower exhaust temp. The only way to do this is to run extra boost which costs power and spikes the exhaust drive pressure.
By 2,500rpm the VNT is running 5psi more drive pressure than the wastegated turbo.
Winner, the wastegated turbo.
Top End.
The VNT has a smaller compressor, it can only do about 27lb/min which starts to cap power at the 2,700rpm point. Now we've already had to drop off max fuel with this turbo as the turbine starts to spike and not be able to flow all the exhaust at max EGT.
So from here we're continually droppping fuel and boost to stay within the limits of the turbine and compressor.
In fact, we passed peak power (~150kw) at about 2,500rpm and have been losing ever since.
Drive pressures also keep spiking and are over 30psi by 3000rpm. Basically the party is over and it's time to change gear.
The wastegated turbo keeps producing more boost and power to beat the engines dropping VE. The larger compressor doesn't start to cap power until around 3,200rpm. At which point 140cc of diesel will be delivering around 180kw.
Drive pressures are comparable to the VNT, but the boost and power are well head.
Winner. Wastegated GT2259.
Summary GT2256V
Torque over 600Nm from 1400-2400rpm
Max power approx 150kw at 2,500rpm. 27psi here.
Limited by compressor and turbine flow limits.
Summary GT2259
Torque over 600Nm from 1400-2600rpm.
Max power approx 180kw at 3,200rpm. 26psi here.
Limited by compressor flow.
So this size VNT turbo has no real advantage unless you are already limited to lower power through other driveline concerns. We need bigger VNT's for these engines, a GT2359V might do it, but I have no info on those.
Clearly Hino, Iveco and the others running GT2259's on 4-5 litre diesels have already done this work. It's always good to agree with the big boys.![]()
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