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[Kibwe Walker]

I got this a few years ago from sky vr4

Spool vs. Turbo Size:
The Infamous Single vs. Twin Debate

When you think about spool on a turbo, there are a few considerations to observe. Firstly, you want is as low as possible, for obvious reasons. This brings up the dilemma of using a single turbocharger or twin turbochargers. There are ideas out there that may cause you to lean towards one setup or the either, but most can be dismissed with a real look at the issue. Here are some examples:

-Some people argue “one turbo gets all the cylinders’ energy, twins gets half per turbo and significant exhaust pulses, which make the twins more inefficient”. This is mostly invalid because modern turbine housings are radially split and tangential, so the exhaust pulses are calmed upon entry to the turbine, and more effectively handle unsteady energy flow. Also, inertia for the most part prevents any major transience with (heavily) pulsed exhaust energy.

-You get more exhaust pressure with a single than twins, which spools it faster. This can be eliminated because exhaust pressure is not really dependant on the turbo, moreso the manifold. Additionally, exhaust pressure does not have a direct effect on exhaust energy. While it IS an indirect effect on exhaust pressure, nothing is going to add energy to the exhaust flow, especially making the engine pressurize the exhaust more. This not only makes the engine and turbo less efficient, it is useless. Looking at the thermodynamic equation for energy in a fluid flow:

Q + ∑mi(hi + 1/2 (Vi)2+ gZi) = ∑me(he + 1/2(Ve)2+gZe) +[m1(u1+1/2(V1)2+gZ1)-m2(u1+1/2(V2)2+gZ2)] + W

Where Q = heat transfer rate to the control volume
mi=mass exhaust flow in
me=mass of exhaust flow out
m1=initial mass of gas in the control volume (turbine housing)
m2=mass of gas in turbine at the end of the control volume process examination
h= enthalpy (1 = initial, 2=final, I=flow in, e= flow out)
u = internal energy (same conditions stated above)
V= velocity
Z= vertical height
g=force of gravity (9.807 m/s2)
W=work rate by the control volume


This shows that exhaust pressure does not directly affect the energy of the gas; it would increase the velocity of the inflow, but also increase the outflow, which would negate the effect since the mass flow rate should not change. Wastegates do not affect this since they exit the flow before the turbine. Arguing the pressure increases the heat may also be argued because of the same reason. The inflow enthalpy and outflow enthalpy are affected the same way: more heat in means more heat out. While is not equal out perfectly because of turbine performance, it is not enough to affect the thermodynamic flow of the turbine. Examine this closer and you might notice that increasing the pressure (by the ideal gas condition) with the (required) decrease in volume of the gas under inspection will have a very minimal effect. Raising the pressure by 1 bar (a lot) in a 40mm ID pipe (primary) requires a bit smaller then a 35mm pipe to keep the temperature constant. This is a 23.4 % decrease in pipe diameter, which is a LOT. This means that you would have to decrease pipe diameter by more than 23 percent to see any effect increased heat by increasing pressure by 1 bar. At 3 bar (44psi) of exhaust pressure, a 1 bar increase in exhaust pressure at an exhaust gas temperature of 600 degrees Celsius (873K) requires a radius decrease of ~10%. At 1000C (1273K) the radius must decrease of 11%. Thusly, you can see that a mere 100-degree increase in temperature requires a much larger change in radius that desirable. 100 degrees might net you some good energy on the inflow, but it will also increase the energy of the outflow. The difference might be 2%, which is a negligible gain.

Ideal gas formula:

PV=mRT
PV/T = constant
P1V1/T1=P2V2/T2 which yields the equations:

V2(T1) =(T1+100)*P1*V1 P1, P2, V1 known

and
P2T1
P2(T1)=(T1+100)*P1*V1 P1, V1, T1 known
T1*V2

Conclusion: more exhaust pressure is not as desirable as you’d like to think. As you will read, with forced induction you fit the exhaust size to the maximum torque output of the engine. Exhaust/intake pressure rations can be observed also. The effectiveness of aggressive valve timing can be absolutely eliminated if the exhaust pressure ratio is high. More on that later.

Now back to the turbo spool comparison:

There are two reasons to argue a single vs. twin turbos. If you compare the mass moment of inertia for a solid disc, you get 1/2*m*R2 on the axial axis based at the center of the circle (fig. 1) (call this z) and 1/4*m*R2 on the radial axis (x and y). For a circular cone (or circular cone section), the mass moment of inertia is 3/10*m*R2 along the axial axis centered at the center of the base circle (fig. 2), is 3/10*m*R2. The same is true for a cone sectioned by cutting it along the z axis with a plane parallel to the XY plane. Now observe the mass. In such a cone, with a constant density and a constant trim turbine wheel, the relationship of radius to mass is:

trapezoid: A = H*(R1+R2)
trapezoidal cylinder (frustum of right circular cone): V = pi*H*(Ra2 + Ra*Rb + Rb2)/3
Trim of turbine wheel: [inducer/exducer]2 =(I/E)2 = I2/E2
for the frustum, call the inducer Rb and the exducer Ra.
mass = volume * density
so, M1/M2 = (V1*p)/(V2*p)
with equal density, M1/M2=V1/V2




Now lets double the size of the turbine wheel (since trim is equal, the exducer doubles as well as the inducer.

V1= pi*h*(E12 + E1*I1 + I12)/3
V2= pi*h*(E22 + E2*I2 + I22)/3
set I2=2*I1 and E2=2*E1 ::
V2= Pi*h*(4E12 + 2*E1*2*I1 + 4*I12)/3
so M1/M2=V1/V2=>cancel terms =>(1+1+1)(4+2*2+4) = (3/12)=1/4

Thusly, doubling the radius results in4 times the mass.



This roughly simulates a turbine wheel, as close as simple calculations will go. Now if you take 1 large wheel and two small ones of half the diameter you get 1/4 + 1/4 = 1/2 versus 1. It is half the mass. This becomes even more important when you implement the mass moment of inertia. For a frustum of a right circular cone (solid of revolution of a trapezoid), the mass moment of inertia is 2/10*m*R2

If R2 is 2* R1, and hence mass 2 (large turbo) = 4*mass 1 (smaller turbo), then

I1= 3/10 * m1 * R12 and
I2= 3/10 * 4*m1* (2R1)2 = 3/10*4*m1*4*R1 = 3/10*16*m1*R1

So I1/I2 = 1/16. That means the large single turbo has 16 times the inertia. Inertia is the property of mass that is defined as “resistance to change in movement”. Like the old saying goes: “An object in motion tends to stay in motion, and object at rest tends to stay at rest”.

This makes single turbos look slow and laggy. But wait, there’s more.

Lets talk efficiency. Tolerances in a turbo are pretty much the same regardless of size. What this means is that the error can be twice as much area, which can be twice the percentage of error. This affects smaller turbochargers a lot more than larger one. This, as expected, results in 4 times the efficiency loss due to clearances. The big single makes “better boost” then the twins.

Hence, the twins will spool faster, but once spooled, the single is more efficient. If you’re driving a street only car and spool is your first concern, then twin turbochargers will suit your application better. If you are racing and know that you will not fall out of the powerband, and thus spool, the single will be a better choice.

Also keep in mind another adage, “spool only exists in 1st gear and there is [among other methods] a simple solution in a bottle!” The use of tuning to decrease spool time will be addressed in another article, as well as actual turbine performance and efficiency.

The examples used here are simplified for ease of calculation, but will work for any scenario. You probably aren’t going to have a turbine wheel twice the size of another, but interpolation and some calculations give a result that says a turbine wheel about 15% bigger has twice the inertia.