Design project to twin turbocharge an inline six BMW M3 engine using mathematical computations and engine simulations to increase the power output upwards of 500 horsepower.

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Twin Turbocharged Inline Six Internal Combustion Engine Design
ME 4680
Engine Design
Robert Beneteau and Sawyer Caley
April 18th, 2019

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Abstract
In this report, the engine from a 2003 BMW M3 will be redesigned to include a pair of
turbochargers with the goal of increasing its power output. The factory naturally aspired inline
six-cylinder engine, code named S54B32, has a maximum power output of 333 horsepower at
7900 rpm and a maximum torque value of 262 ft-lbs at 4900 rpm. In order to meet all three
design goals laid out in this report of a 0-60 mph time of 4.0 seconds, a quarter mile requirement
of 110 mph at 12.5 seconds, and a 20mph-40mph passing requirement of 1.8 seconds, a
minimum horsepower value of 500.6 was needed. Several of the key parameters of this modified
engine including the bore, stroke, connecting rod length, number of valves and their sizes, and
the turbochargers will be discussed to meet this goal. To conclude this report, an engine
performance evaluation was performed using the software Engine Analyzer Pro to show the final
maximum horsepower value of this newly designed engine to be 543.5 horsepower at 6000 rpm
and the maximum brake torque value to be 563 ft-lbs at 4000 rpm. Thus, the newly designed
engine will meet all the requirements set forth in this report.

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Table of Contents
Abstract ........................................................................................................................................... 2
Introduction..................................................................................................................................... 4
Mission Statements ......................................................................................................................... 4
Design Goals................................................................................................................................... 4
Vehicle Steady State Requirements ................................................................................................ 5
Vehicle Transient Power Requirements.......................................................................................... 6
0-60 mph..................................................................................................................................... 7
Quarter Mile Requirements......................................................................................................... 8
Passing Requirement................................................................................................................... 9
Power Plant Selection and Sizing ................................................................................................. 10
Engine Component Specifications............................................................................................ 10
Engine Schematic...................................................................................................................... 11
Fuel and Air Flow Rates and Equivalence Ratio ...................................................................... 12
Number of Valves and Sizing ................................................................................................... 14
Preliminary Performance Evaluations .......................................................................................... 16
Vehicle Energy and Thermal Loads.............................................................................................. 17
Cooling System Design................................................................................................................. 18
Exhaust/Turbocharging System Design........................................................................................ 20
Final Engine Performance Evaluation .......................................................................................... 23
Summary and Conclusions ........................................................................................................... 24
References..................................................................................................................................... 25

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Introduction
The objective of this project is to increase the performance of an inline six-cylinder
internal combustion engine from a 2003 BMW M3 by adding twin turbochargers which will see
daily street driving and can achieve successful performance figures in a track environment. The
BMW M3 can be described as a front-engine, rear-wheel-drive high performance vehicle. This
project will consist of three primary segments: vehicular requirements, power plant and energy
consumption, and technology innovations in this design. This report will only include the second
segment, power plant and energy consumption, with vehicle requirements being covered in the
first report, and technology innovations being covered in the final report. Power plant
development will be focused around providing an efficient everyday vehicle that is capable of
handling forced air induction from the two turbochargers that are introduced to the system. The
result of this will be an engine that is more powerful and fuel efficient than that of the original
design.
Mission Statements
The goal of this design is to create an engine for those who seek the extraordinary
performance from the vehicle they pilot. To most people, commuting to and from their
destinations in an automobile which doesn’t spark joy is not an issue. But the enthusiasts who
seek a car that will throw them back in their seat in a milliseconds notice are a rare breed that
won’t settle on their performance. Therefore, a complete engine will be designed around the
basis of pleasing these customers.
Acceleration: This engine will provide intense acceleration for the user who wishes to get
up to speed quickly and overtake another vehicle in its path, whether it be on the road or on a
track.
Track: The chassis of this automobile will be optimized to handle corners easily and stay
flat throughout curves, so the power output of the engine must match the astounding suspension
fitted to the vehicle.
Endurance: Due to the fact this engine will be placed in a production vehicle and sold at
normal dealerships with a warranty, it will need to perform reliably with little to no issues and
have a life of at least 150,000 miles.
Design Goals
Characteristics: The engine being designed for this report must be able to withstand
piston speeds of up to 8,000 RPM to provide for a thrilling experience as the needle on the
tachometer rapidly climbs to such a high number.

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Performance: A horsepower figure around 400 should launch this car from 0-60 mph in
less than 4.0 seconds and allow the vehicle to accelerate from 20-40mph in less than 1.8 seconds
to pass other automobiles which are in its path.
Efficiency: The goal with designing this engine was to make impressive performance
figures; therefore, the decision to make it forced induction was chosen. Efficiency was the
driving factor behind choosing to turbocharge the vehicle over supercharge it. Choosing to
capture wasted energy sent out the exhaust and put it back into the system via turbochargers
rather than robbing the engine of power by driving a supercharger off the crankshaft will lead to
a more efficient design.
Reliability: The engine characteristics of this vehicle need to be simplistic enough for a
mother to drive to a grocery store with her children, but at the same time need to be able to react
quickly when the throttle pedal is completely depressed and to be able to withstand numerous
runs to redline without failure.
Reproducibility: Due to the fact this engine will be placed in a street legal car which is
estimated to have production numbers north of 60,000 units, the engine must contain
components which are easily accessible and inexpensive enough to mass produce.
Vehicle Steady State Requirements
Steady state is defined as an unvarying condition in a physical process. Steady state
requirements include factors which work against the motion of a vehicle. These factors include
the aerodynamic resistance force, rolling resistance force, the force of power losses through the
drivetrain, and the force of resistance of any existing grades. The sum of these forces is the total
force required for a vehicle to overcome while operating in a steady state condition.
Frequired = Faero + Frolling + Fdrivetrain losses + Fgradient
The equation used to calculate the aerodynamic force experienced against the vehicle at a
certain speed is shown below, where ρair represents the density of air, v represents the velocity of
the vehicle, Af represents the frontal area of the vehicle, and Cd represents the coefficient of drag
the vehicle experiences.
Faero = 0.5 * ρair * v2
* Af * Cd
Knowing this engine will be placed in a BMW M3 with a frontal area of 2.18 m2
and
coefficient of drag of 0.32, and taking the density of air to be 1.225 kg/m3
, the equation can
simply down to:
Faero = 0.42728 * v2

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where to find the aerodynamic force in newtons can be found by plugging in the velocity
of the vehicle in meters/second. To find the force acting upon the vehicle in a common steady
environment such as driving down a highway at 70 mph (31.29 m/s), the equation simplifies to:
Faero = 0.42728 * v2
= 0.42728 * (31.29)2
= 418.34 N
The rolling force of the vehicle can be calculated using the equation below with a vehicle
mass of 1550 kg and a rolling coefficient of 3%:
Frolling = mass * g * Crolling = (1550 kg)(9.81 m/s2
)(0.03) = 456.17 N
Drivetrain loss includes the amount of power lost in-between the flywheel of the engine
and the wheels of the vehicle. The drivetrain is the sum of components including clutch, gearbox,
driveshaft, differential, and driveshafts. A drivetrain loss of 5% was assumed for this vehicle. In
addition to this, an accessory loss of 5% was assumed as well to account for accessories such as
the alternator, power steering pump, water pump, and air conditioning. This accounts to losses
totaling to 10%.
For the purpose of this design, the assumption that the vehicle will not be operating on a
grade is made, resulting in Fgradient = 0.
Therefore, the total force required sums to:
Frequired = (418.34 N + 456.17 N)(1.1) = 961.95 N
Thus, the power required to operate at steady state conditions while moving at a velocity
of 70 mph (31.29 m/s) is:
Power = Frequired * v = (961.95 N)(31.29 m/s) = 30099.4 W (1 hp/745.7 W) = 40.36 hp
The power required for this vehicle to maintain a steady speed of 70 mph on a freeway
without an incline is 40.36 horsepower.
Vehicle Transient Power Requirements
Transient is defined as lasting for a short time. This segment will focus on engine
requirements such as 0-60 mph times, quarter mile speed, and passing requirements to overtake
another vehicle. With the BMW M3 being a high-performance sports car, intense acceleration
values were chosen to provide impressive times in a straight line as well as on a race track.

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0-60 mph
For this twin turbocharged engine, a 0-60 mph run was selected to be 4.0 seconds. To
calculate how much power is necessary to achieve this time, the aerodynamic force acting on the
vehicle at these speeds must be calculated. The force at varying speeds can be seen in Table 1.
Table 1: 0-60 mph Velocity and Aerodynamic Force
Velocity (mph) Velocity (m/s) Faero (N)
0 0 0
10 4.47 8.54
20 8.94 34.15
30 13.41 76.85
40 17.88 136.62
50 22.35 213.46
60 26.82 307.39
The largest aerodynamic force value of 307.39 newtons was used below to calculate the
required force to ensure the power requirements were satisfied.
Frequired = Faero + Frolling + Fdrivetrain losses + Fgradient = (307.39 N + 456.17 N)(1.1) = 839.91 N
Next, the acceleration necessary to achieve this time was calculated in order to find
Facceleration.
Facceleration = mass * a = (1550 kg)(6.71 m/s2
) = 10393.7 N
The acceleration force required to launch this vehicle from 0-60 mph in 4.0 seconds is
10393.7 newtons. Finally, the transient force and power required can be calculated.
Ftransient = Frequired + Facceleration = 839.91 N + 10393.7 N = 11233.6 N
Power = Ftransient * v = (11233.6 N)(23.82 m/s) = 267585 W (1 hp/745.7 W) = 358.84 hp

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Therefore, for this BMW to achieve a 0-60 mph time of 4.0 seconds, a minimum of 358.84
horsepower is necessary for the engine to output.
Quarter Mile Requirements
The intended goal for quarter mile performance of this engine is 110 mph in 12.5
seconds. Using the sum of all forces, the power required by the engine to achieve this goal can be
calculated. To calculate the necessary power, the aerodynamic resistance must be known at 110
mph. The forces of resistance can be seen below in Table 2.
Table 2: 0-110 mph Velocity and Aerodynamic Force
Velocity (mph) Velocity (m/s) F aero (N)
0 0 0
10 4.47 8.54
20 8.94 34.15
30 13.41 76.85
40 17.88 136.62
50 22.35 213.46
60 26.82 307.39
70 31.29 418.39
80 35.76 546.46
90 40.23 691.62
100 44.70 853.85
110 49.17 1033.16
At 110 mph, the aerodynamic force value will be used to calculate the force required to
maintain that speed.
Frequired = Faero + Frolling + Fdrivetrain losses + Fgradient = (1033.16 N + 456.17 N)(1.1) = 1489.33 N
After finding the force required to maintain 110 mph, the force required to accelerate the
vehicle to 120 mph was calculated.
Facceleration = mass * a = (1550 kg)(3.9352 m/s2
) = 6099.56 N

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It was found that the force of acceleration required for the vehicle to reach a speed of 110
mph in a distance of a quarter mile is 11085.6 newtons. Using this data, the sum of the required
force and the acceleration force will result in the transient force of the vehicle.
Ftransient = Frequired + Facceleration = 1489.33 N + 6099.56 N = 7588.89 N
Using the transient force, the required horsepower that the vehicle must produce can be
calculated and converted to horse power.
Power = Ftransient * v = (7588.89N)(49.19 m/s) = 406732.93 W (1 hp/745.7 W) = 500.6 hp
To produce a 12.5 second quarter mile time at a speed trap speed of 110 mph, the engine
will have to produce at least 500.6 horsepower.
Passing Requirement
The horsepower required to overtake a vehicle at a certain speed can be calculated in a
similar fashion as how the 0-60 mph horsepower was calculated. First, the initial and end speeds
must be established with a desired time in mind. For this engine, a time of 1.8 seconds has been
specified to overtake another vehicle from 20 mph to 40 mph. The first step to calculate the
required horsepower for this action is to find the aerodynamic forces which will act upon the
vehicle.
Table 3: 20-40 mph Velocity and Aerodynamic Force
Velocity (mph) Velocity (m/s) Faero (N)
20 8.94 34.15
30 13.41 76.84
40 17.88 136.60
The largest force experienced in this case is 136.60 newtons, so that will be the value
used to calculate the acceleration force.
Facceleration = mass * a = (1550 kg)(4.97 m/s2
) = 7703.5 N
It is found that the acceleration force required for this vehicle to perform this action is
7703.5 newtons. Next, the total forces acting upon the vehicle are calculated and summed with
the acceleration force.

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Frequired = Faero + Frolling + Fdrivetrain losses + Fgradient = (136.60 N + 456.17 N)(1.1) = 652.05 N
Ftransient = Frequired + Facceleration = 652.05 N + 7703.5 N = 8355.55 N
Finally, the power required is calculated and converted to horsepower.
Power = Ftransient * v = (8355.55 N)(17.88 m/s) = 149397 W (1 hp/745.7 W) = 200.35 hp
It is necessary for this engine to produce at least 200.35 horsepower for accelerate from
20 mph – 40 mph in a time of 1.8 seconds to overtake another vehicle on the road.
Power Plant Selection and Sizing
Engine ComponentSpecifications
The following figures pertain to a factory S54B32 engine found in a 2003 BMW M3.
These figures are going to remain unchanged for the design of this updated engine.
Cylinder bore 87 mm
Piston stroke 91 mm
Connecting rod length 139 mm
Crank radius 45.5 mm
Number of cylinders 6
To find the total volumetric displacement of the engine, the cylinder bore, piston stroke,
and number of cylinder can be plugged into the equation shown below. The total volumetric
displacement of this engine comes out to be just over 3.2 liters.

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Engine Schematic
A 3-dimensional model of the factory S54B32 engine was obtained online [1]. A section
view straight down the middle of the cylinders can be seen on the next page. This figure helps
illustrate the design of the straight 6 engine layout with the bore, stroke, connecting rod length,
and crank radius mentioned earlier. This illustration is advantageous in understanding the layout
of the engine and other important components can be seen as well such as the intake valves,
valve springs, and timing chain.

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Fuel and Air Flow Rates and Equivalence Ratio
Another value of the factory S54B32 engine which will remain unchanged in this new
design is the compression ratio. With the compression ratio of the M3 engine known to be
11.5:1, the efficiency of the Otto cycle can be found by using this equation:
𝜂 𝑜𝑡𝑡𝑜 = 1 −
1
𝐶𝑅 𝑘−1
= 1 −
1
11.50.4
= 0.6235 = 62.35%
where,
 k = Specific heat ratio of air = 1.4
 CR = Compression ratio of engine
The Otto cycle efficiency equation above shows the ideal maximum efficiency of 62.35%
that this engine should be able to achieve with a compression ratio of 11.5:1. A more realistic
value for engine efficiency is the brake efficiency which takes into account losses such as
drivetrain, friction, and accessories. This value is normally between 40-60% of the Otto cycle
efficiency of the engine. Thus 50% of the Otto cycle will be chosen for this design to
approximate the brake efficiency:
𝜂 𝑏𝑟𝑎𝑘𝑒 = 𝜂 𝑜𝑡𝑡𝑜 ∗ 0.5 = 62.35% ∗ 0.5 = 31.175%
To calculate the mass flow of gasoline entering the engine, the following equation is
used:
𝑚̇ 𝑓𝑢𝑒𝑙 =
𝑊 𝑛𝑒𝑡,𝑏𝑟𝑎𝑘𝑒
𝜂 𝑏𝑟𝑎𝑘𝑒∗𝐿𝐻𝑉 𝑓𝑢𝑒𝑙
The engine from the M3 must be able to be daily driven, thus must be regularly filled
from local gas stations with premium 91 octane fuel which contains 47 MJ/kg from its lower
heating value (LHV). Also, from the first report, it is stated that the desired power output is 500
horsepower. This power unit was converted to kilowatts for the equation above which amounts
to 372.85 kW. Using these known values, the mass flow of fuel entering the system at peak
power can be found:
𝑚̇ 𝑓𝑢𝑒𝑙 =
372.85 𝑘𝑊
0.31175 ∗ 47,000
𝑘𝐽
𝑘𝑔
= 0.02545
𝑘𝑔
𝑠

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Mass flow of air into the system can also be found using the volumetric efficiency,
density of atmospheric air, and the operating speed at peak performance. The volumetric
efficiency of the system is chosen as 98% and the density of air is selected to be 1.801 kg/m3 due
to the fact this will be a forced induction engine and the air forced into the engine will be more
dense than that of ambient air. The operating speed at peak performance is 8000 RPM.
Therefore, using the equation below, the mass flow of air into the system is calculated:
𝑚̇ 𝑎𝑖𝑟 =
𝜌 ∗ 𝑉𝑒 ∗ 𝑉𝑑 ∗ 𝑁𝑟𝑒𝑣
2
=
(1.801
𝑘𝑔
𝑚3) ∗ (0.98) ∗ (0.003246 𝑚3)(133.33
𝑟𝑒𝑣
𝑠
)
2
= 0.382
𝑘𝑔
𝑠
To then find the air-to-fuel ratio of the engine, the mass flow of the air was simply
divided by the mass flow of the fuel as seen below.
𝐴 𝐹 =
𝑚̇ 𝑎𝑖𝑟
𝑚̇ 𝑓𝑢𝑒𝑙
=
0.382
𝑘𝑔
𝑠
0.02545
𝑘𝑔
𝑠
= 15.0098
From the calculated air-to-fuel ratio of the engine, the equivalence ratio can be
determined by dividing the stoichiometric air-to-fuel ratio. To find the stoichiometric air-to-fuel
ratio, a balanced combustion chemical equation was used utilizing octane as the fuel. Octane is
used as the fuel source as it makes up at least 91% of the fuel in this scenario. The combustion
chemical equation that was used for balancing is shown below.
𝐶8 𝐻18 + 𝑋 ( 𝑂2 + 3.76𝑁2) → 𝑎𝐶𝑂2 + 𝑏𝐻2 𝑂 + 𝑐𝑁2
Balancing the chemical equation for combustion results in the following:
𝐶8 𝐻18 + 12.5( 𝑂2 + 3.76𝑁2) → 8𝐶𝑂2 + 9𝐻2 𝑂 + 47𝑁2
From the balanced chemical combustion equation above, the stoichiometric air-to-fuel
ratio can be determined using the molecular weights of both air and fuel in the equation below.

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(
𝐴
𝐹
) 𝑠𝑡𝑜𝑖𝑐ℎ =
12.5[32 + (3.76 ∗ 28)]
(8 ∗ 12) + (18 ∗ 1)
= 15.0526
Using the stoichiometric air-to-fuel ratio above, the equivalence ratio can found for
parameters stated earlier.
Φ =
𝐴𝐹𝑠𝑡𝑜𝑖𝑐ℎ
𝐴𝐹𝑎𝑐𝑡𝑢𝑎𝑙
=
15.0526
15.0098
= 1.003
Number of Valves and Sizing
A factory S54B32 engine uses a dual overhead camshaft setup with four valves per
cylinder. This aspect of the design will be kept for the newly designed engine. Having two intake
and two exhaust valves per cylinder allows for larger total valve areas compared to a design
having one intake valve and one exhaust valve per cylinder [2]. With a four valve setup, the
valves are smaller and lighter which causes less valve float at higher RPMs than that of a two
valve setup. This design is also more efficient and is optimal for high revving engines. A bottom
view of a factory S54B32 with its intake and exhaust valves can be seen on the next page.

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A number of equations were used to calculate the proper size of both the intake and
exhaust valves. These equations and results can be seen below.
Air Speed
𝑎 = √𝐾 ∗ 𝑅 ∗ 𝑇 = √1.4 ∗ 287
𝐽
𝑘𝑔 ∗ 𝐾
∗ 298 𝐾 = 346.029
𝑚
𝑠
Piston Speed
𝑆 𝑝 = 2 ∗ 𝑁 ∗ 𝑠 = 2 ∗
8,000 𝑟𝑒𝑣
60 𝑠𝑒𝑐
∗ .091 𝑚 = 24.267
𝑚
𝑠
Inlet Area
𝐼𝑛𝑙𝑒𝑡 𝑎𝑟𝑒𝑎 = 1.3 ∗ 𝑏2
∗
𝑆 𝑝
𝑎
= 1.3 ∗ (0.087 𝑚)2
∗
24.267
𝑚
𝑠
346.029
𝑚
𝑠
= 0.00069 𝑚2
Valve Area
𝑉𝑎𝑙𝑣𝑒 𝑎𝑟𝑒𝑎 =
𝐼𝑛𝑡𝑎𝑘𝑒 𝑎𝑟𝑒𝑎
𝐹𝑙𝑜𝑤 𝐼𝑛𝑡𝑎𝑘𝑒 𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡
=
0.00069 𝑚2
.5
= .00138 𝑚2
Valve Diameter
𝑉𝑎𝑙𝑣𝑒 𝑎𝑟𝑒𝑎 =
𝜋
4
∗ 𝑣𝑎𝑙𝑣𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟2
. 00138 𝑚2
=
𝜋
4
∗ 𝑣𝑎𝑙𝑣𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟2
0.04192 𝑚 = 𝑣𝑎𝑙𝑣𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟
Thus, the valve diameter for the intake valves was calculated to be 41.92 mm and
rounded up to 42 mm. The exhaust valve size was chosen to be 80% of the intake valve size at a
diameter of 32 mm.

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Preliminary Performance Evaluations
With the specifications already outlined in this report thus far, important preliminary
performance evaluations can be calculated such as brake specific fuel consumption, thermal
efficiency, torque, and brake mean effective pressure. These can be seen below.
𝑏𝑠𝑓𝑐 =
𝑚̇ 𝑓𝑢𝑒𝑙
𝑊̇ 𝑏
=
0.02545 ∗ 1000
372.85
= .068
𝑔
𝑘𝑊 ∗ 𝑠
When compared to the average gasoline engine bsfc (0.089g/kW*s), the calculated bsfc
of the engine that is being used for this project is much lower, making this engine more fuel
efficient. From the calculated bsfc, the thermal efficiency can also be calculated.
𝑛𝑡 =
1
𝑏𝑠𝑓𝑐 ∗ 𝐿𝐻𝑉
=
1
. 068 ∗ (
47,000
1,000
)
= 31.29%
Based on an average thermal efficiency of about 20 percent for internal combustion
engines, with this engine reaching low 30 percent range, the thermal efficiency of this engine is
ideal.
𝑇𝑜𝑟𝑞𝑢𝑒 =
𝐻𝑜𝑟𝑠𝑒𝑝𝑜𝑤𝑒𝑟 ∗ 5252
𝑅𝑃𝑀
=
500 ∗ 5252
8,000
= 328.25 𝑓𝑡 ∗ 𝑙𝑏𝑠
At the maximum engine speed of 8,000 RPM, this engine will be producing 500
horsepower and 328.25 foot-pounds of torque (445.047 newton-meters).
Using this torque data, the brake mean effective pressure is calculated.
𝐵𝑀𝐸𝑃 =
2 ∗ 𝜋 ∗ 𝑛 𝑟 ∗ 𝑇
𝑉𝑑
=
2 ∗ 𝜋 ∗ 2 ∗ 445.047
0.0032
= 17.47 𝑏𝑎𝑟

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Vehicle Energy and Thermal Loads
In order to calculate the overall average heat transfer correlation (Taylor Correlation), the
Reynolds number must be calculated to determine if the flow is turbulent. This is shown below:
𝑅𝑒𝑦𝑛𝑜𝑙𝑑𝑠 𝑛𝑢𝑚𝑏𝑒𝑟 =
(𝑚̇ 𝑎 + 𝑚̇ 𝑓) ∗ 𝑏𝑜𝑟𝑒
𝐴 𝑝 ∗ μ 𝑔
=
(. 408
𝑘𝑔
𝑠
) ∗ (.087m)
(
𝜋
4
) ∗ (. 087𝑚)2 ∗ (20 ∗ 10−6 𝑁𝑠
𝑚2 )
= 74640.4
From there, the Nusselt number can be calculated:
𝑁𝑢 = 10.4 ∗ 𝑅𝑒.75
= 10.4 ∗ 74540.4.75
= 46916.7
Using this Nusselt number and assuming k = 0.06 W/m*k, the heat transfer coefficient
can be calculated:
ℎ 𝑔 =
𝑁𝑢 ∗ 𝑘
𝑏𝑜𝑟𝑒
=
46916.7 ∗ (0.06)
0.087
= 28150
𝑊
𝑚2 ∗ 𝑘
This heat transfer coefficient is necessary to calculate the heat flux of the engine. The
heat flux value is calculated using a combustion gas temperature of 1000 K and a coolant
temperature of 380 K. This yields a heat flux of:
𝑞" = ℎ 𝑔 ∗ (𝑇 𝑔 − 𝑇𝑐) = 28150∗ (1000− 380) = 17.45
𝑀𝑊
𝑚2
The fuel conversion efficiency for a power output of 500 horsepower (372.9 kW) at the
maximum speed of 8000 rpm was also calculated:
𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =
100
( 𝐵𝑆𝐹𝐶 ∗ 𝑄𝐿𝐻𝑉 )
=
100
(.068 ∗ 44,000)
= 0.033
Finally, the heat transfer rate of the exhaust was calculated:
𝑄̇ 𝑒𝑥ℎ = 𝑚̇ 𝑓 ∗ 𝐶𝑝 𝑓𝑢𝑒𝑙 ∗ (Δ𝑇) = (.02545
𝑘𝑔
𝑠
) (1.008
𝑘𝐽
𝑘𝑔 ∗ 𝐾
)(1000 𝐾) = 25.65
𝑘𝐽
𝑠

18. 18 | P a g e
Cooling System Design
For the cooling system design, the key aspect was the mass flow of coolant moving
through the system to cause the engine to operate at safe temperatures. To do this, the heat loss
rate from the engine to the coolant had to be calculated using a set of two equations shown
below.
𝑄̇ 𝑐𝑜𝑜𝑙 = −𝑘 ∗ 𝐴 ∗ (
𝑇 𝑤𝑐 − 𝑇 𝑤ℎ
𝑋
)
𝑄̇ 𝑐𝑜𝑜𝑙 = 𝑚̇ 𝑐 ∗ 𝐶𝑝 𝑐𝑜𝑜𝑙𝑎𝑛𝑡 ∗ (Δ𝑇)
With a cylinder wall thickness of 3.5mm, represented by the variable ‘X’, a thermal
conductivity value for a cast iron engine block being 70 watts/meter*K, a cold coolant temp of
80 degrees Celsius, and a hot coolant temp of 104 degrees Celsius and an assumed coolant pipe
sizing of 2 cm2, the heat loss to the coolant that is necessary for this engine can be calculated.
𝑄̇ 𝑐𝑜𝑜𝑙 = −70
𝑊
𝑚 ∗ 𝐾
∗ 0.0002 𝑚2
(
80 𝐶 − 104 𝐶
0.0035𝑚
) = 96 𝑘𝑊
From the heat loss value above, the mass flow of coolant needed to keep the engine
within its normal operating range can be found using the second equation listed above. For this
equation, the Cp value for coolant is 3.14 kJ/(kg*C) assuming that it is a 1:1 mix of water and
ethyl glycol.
96 𝑘𝑊 = 𝑚̇ 𝑐 ∗ 3.14
𝑘𝐽
𝑘𝑔 ∗ 𝐶
∗ (104 𝐶 − 80 𝐶)
𝑚̇ 𝑐 = 1.273
𝑘𝑔
𝑠

19. 19 | P a g e
For the sizing of the turbocharger intercooler, the frontal area exposed to ambient air
must be calculated. First it is necessary to have the internal flow volume from the outlet of the
turbocharger to the inlet of the intercooler. This internal flow volume will then be the same for
the exit of the intercooler to the intake manifold of the engine. The internal flow volume is
calculated by:
𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝐹𝑙𝑜𝑤 𝐴𝑟𝑒𝑎 (𝑖𝑛2) =
6 ∗ (𝑑𝑒𝑠𝑖𝑟𝑒𝑑 ℎ𝑜𝑟𝑠𝑒𝑝𝑜𝑤𝑒𝑟)
100
6 ∗ 500 ℎ𝑝
100
= 30 𝑖𝑛2
= 193.55 𝑐𝑚2
From this, the area of the intercooler charge face can be calculated using the following
equation.
𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝐹𝑙𝑜𝑤 𝐴𝑟𝑒𝑎
0.45
= 𝐶ℎ𝑎𝑟𝑔𝑒 𝐹𝑎𝑐𝑒 𝐴𝑟𝑒𝑎
193.55 𝑐𝑚2
0.45
= 430 𝑐𝑚2

20. 20 | P a g e
Exhaust/Turbocharging System Design
Most aftermarket applications of turbocharging call upon one turbocharger to create the
large increase in mass flow of air to allow for the required power increase. In this case, under
high loads, a single turbocharger system can become very hot, and very inefficient. Due to this, a
parallel twin turbocharging system will be used.
To find the size of the two parallel turbochargers that would be used, the pressure ratio
and air mass flow into the engine to satisfy the 500 horsepower requirement must be known. In
the air flow rate calculations from earlier, it was found that the density ratio needed to keep an
equivalence ratio of 1 was about 1.47. From the density ratio, the pressure ratio needed to be
calculated. This was done by combining the equations shown below.
𝜌2
𝜌1
=
𝑃2
𝑃1
∗
𝑇1
𝑇2
𝑃2
𝑃1
= (
𝑇2𝑠
𝑇1
)3.5
𝜂𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑜𝑟 =
𝑇2𝑠 − 𝑇1
𝑇2 − 𝑇1
The implementation of the isentropic process equation and the compressor efficiency
equation into the ideal gas ratio equation allows for the removal of the unknowns of P2 and T2.
The result of manipulating these three equations leaves the equation shown below.
𝜌2
𝜌1
= (
𝑇2𝑠
𝑇1
)3.5
∗
𝑇1
(
𝑇2𝑠 − 𝑇1
𝜂𝑐𝑜𝑚𝑝
) + 𝑇1
The equation shown above leaves T2s as the only unknown variable. Once having solved
for T2s, the necessary pressure ratio can be found from the isentropic process equation above.
1.801
𝑘𝑔
𝑚3
1.225
𝑘𝑔
𝑚3
= (
𝑇2𝑠
300 𝐾
)3.5
∗
300 𝐾
(
𝑇2𝑠 − 300 𝐾
0.75
)+ 300 𝐾
Solving for T2s gives a value of 356 Kelvin. Using this value, the pressure ratio is solved
for.
𝑃2
𝑃1
= (
356
300
)3.5
= 1.88

21. 21 | P a g e
After solving for the pressure ratio needed for the engine, the volumetric flow of air into
the engine was calculated using the volumetric displacement of the engine, and the speed of the
engine.
𝑉̇ 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 = 𝑉𝑑 ∗
𝑟𝑝𝑚
60 ∗ 𝑘1
For this equation, the volumetric displacement of the engine was 3.246 liters which was
converted to cubic feet, the engine rpm was 8000, and k1 was 2 as it was for a 4-stroke engine.
𝑉̇ 𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 = 0.11463 𝑓𝑡3
∗
8000
60 ∗ 2
= 7.642 𝑐𝑓𝑚
The cfm value calculated from the theoretical volumetric flow rate, is what would be
needed to enter into the engine. This value is significantly lower that what can be seen on the
turbocharger map below. However, the air leaving the turbocharger must make its way through
an intercooler where the velocity of the air decreases greatly allowing for the volumetric flow
into the engine that was calculated above to accurate.
From the calculations that were done in this section, it was determined that the best
turbocharger to add in a parallel system was the Mitsubishi TD04-13G as it allowed for the
power output that was required of the engine.

22. 22 | P a g e
In an attempt to minimize the power loss of the exhaust, two 200 cell high flowing
performance catalytic converters will be utilized in the first section of the exhaust underneath the
car. This is the only modification which will be chosen for the standard exhaust found on a 2003
M3. The factory original catalytic converters hinder the flow of exhaust gases and will hurt top
end power on an engine such as the one designed in this report. A high performance catalytic
converter allows for a more efficient flow of the exhaust gas of the engine while the honeycomb
structure of the Platinum, Palladium, and Rhodium eliminate the harmful oxides which are
produced through the combustion process. Because this engine is being designed to make much
more power than stock, it only makes sense to equip it with high performance catalytic
converters to increase the flow over stock.

23. 23 | P a g e
Final Engine Performance Evaluation
To analyze the final performance values of the engine designed in this report and identify
whether or not it met the standards set forth, the computer software Engine Analyzer Pro was
downloaded and utilized. This software allows the user to input nearly every single characteristic
of an engine to simulate the results it will provide when tested on a dynamometer. Using the
values calculated in this report, such as bore, stroke, compression ratio, number of cylinders,
connecting rod length, number of valves and their sizes, air density, number of turbochargers and
their efficiency, and boost pressure, a horsepower and torque vs rpm graph was able to be
produced. The results showed a maximum brake horsepower of 543.5 horsepower at 6000 rpm
and a maximum brake torque of 563 ft-lbs at 4000 rpm. This exceeds the goals which were set
forth at the beginning of this report. Moreover, the power and torque values which were
calculated earlier in this report were very close to the data Engine Analyzer Pro provided. The
power desired at 8000 rpm was 500 rpm and the simulation stated a power output of 490.6
horsepower at this engine speed. The torque which was calculated at this speed earlier in the
report was 328.3 ft-lbs, and the simulation stated the torque output at this speed to be 322 ft-lbs.
Overall, the results of this simulation justify the success of this internal combustion engine and
demonstrate achieved goals for this project.

24. 24 | P a g e
Summary and Conclusions
In conclusion, the engine designed in this report achieved and exceeded the power
requirements set forth. The twin turbochargers which were chosen to be paired with this
normally naturally aspirated engine pushed the maximum power output from 333 horsepower to
544 horsepower and the maximum torque output from 262 ft-lbs to 563 ft-lbs. With these new
values, the M3 which this engine powers will exceed the steady state and transient power
requirements stated at the be beginning of this report. This report demonstrates the capabilities of
a twin turbocharger setup and how much power it can add to an engine when properly sized. The
maximum power of this new engine when the vehicle remains under speeds of 3000 rpm is 275
horsepower, which shows if the speed of the engine is kept down that this vehicle can serve as a
normal daily driver which was one of the design goals of this report. However, once this vehicle
is taken to a track environment and the engine is allowed to stay in its power band of 4000 rpm
and higher, it will prove to be a very competent competitor against other vehicles it will be
racing against.

25. 25 | P a g e
References
Ferguson, Colin R., and Allan Kirkpatrick. Internal Combustion Engines: Applied
Thermosciences. John Wiley & Sons, Inc., 2016.