Internal Combustion Engines PROJE

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    2.61 Internal Combustion Engines

    Design Project Corrected Version Number 2

    Wednesday, April 14, 2004

    Due: Thursday, April 22, 2004

    Heavy duty diesel engine with EGR and particulate trap

    For this project you need to design a heavy-duty truck diesel engine. The engine is 11

    liter, 6 cylinder, with target maximum power of 360 kW and target bsfc of 185 g/kW-hr.You will also incorporate an emissions strategy to meet current emissions standards.

    1. Base Engine

    Begin by sizing the base engine (no turbocharger), which will be 11 liter, 6 cylinder.

    Assume a maximum mean piston speed (Sp) of 10 m/s. Use data on Figure 13.7 on page722 in the text for estimates of mechanical/rubbing plus auxiliary mep. Making any other

    reasonable assumptions necessary, determine the following parameters:(a) Bore and stroke

    (b) Compression ratio

    (c) Connecting rod length(d) Brake mean effective pressure, at maximum torque and maximum power

    (e) Maximum torque and maximum power

    (f) Maximum engine speed, at maximum power

    2. Boost, turbomachinery, and intercooler

    Design the required turbomachinery and intercooler for the engine by addressing thefollowing points:

    (a) Based on your calculations in part 1, calculate the amount of boost pressurerequired at maximum speed to produce the target power.

    (b) Use a turbocharger to produce this boost, and define the main operatingparameters of the required turbomachinery. For both turbine and compressor,

    provide values for mass flow rate, pressure ratio, inlet and exhaust temperatures

    and pressures. Use typical values for isentropic efficiencies (Kt=0.85, Kc =0.80)and assume the exhaust temperature is 900K.

    (c) Include an intercooler to lower the temperature of the air coming out of the

    compressor. Provide the inlet and outlet temperatures of the air, as well as the

    coolants inlet and outlet temperatures and mass flowrate. Assume a counter-flowheat exchanger with effectiveness of 0.8. (Heat exchanger effectiveness is the

    ratio of the actual heat transfer to the heat transfer that would occur if the streamwith the minimum capacity rate were heated (or cooled) from its inlet temperature

    to the inlet temperature of the other stream).(d) Draw a schematic of your system(Hint: You will need to include the effect of turbo-charging on pumping work)

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    3. Brake efficiencyEvaluate the brake fuel conversion efficiency of the design at half maximum speed and

    full load at that speed. Does it meet the target bsfc? If not, what changes in engine designwould bring it closer?

    4. Emissions NOXThe engine you are designing must comply with 2004 EPA NOx requirements which are

    set at 2.5 g/bhp-hr (assume for simplicity that the NOx requirement must be met at all

    operating points); EGR has been chosen as the technology to reduce NOx levels.Address the questions below using available data, and an appropriate safety factor:

    (a) Draw a schematic of your system, showing clearly how you will drive EGR fromexhaust to intake. There are a few possibilities for doing this (the reference paper

    on EGR systems might be helpful).

    (b) Find the required amount of EGR to run at low load (25% max torque and 1600rpm).

    (c) Find the required amount of EGR to run at maximum power.

    (d) Similar to part 2, recalculate the boost pressure at maximum power. Resize theturbomachinery and intercooler to reach the stated target or best case power

    output; use available data to determine the exhaust temperature. Also, include anEGR cooler to lower the re-circulated gas temperature before it enters the engine;

    using the same assumptions as part 2c, provide operating temperatures and

    flowrates.As a safety measure, it is common standard to reduce the amount of NOx, by anadditional 20% to 40% of the required EPA standard. For this design please use a

    safety factor of 30% (i.e., reduce NOx to 1.75 g/bhp-hr, 30% below required

    standard). Assume that the equivalence ratio based on the mass of fresh fuel andfresh air must stay below the smoke limit of 0.7; for simplicity once you have

    selected the level of EGR, assume NOx levels remain constant, in spite of additionalboosting (this is not the actual case). Also assume that beyond 8 CAD BTDC, for

    every additional CAD delay in injection you lose 0.25 percentage points in indicated

    fuel conversion efficiency.(Note: Watch the units in the emissions data)

    5. Emissions particulate

    The engine you are designing must also comply with 2004 EPA particulate requirements.Diesel Particulate Filter (DPF) has been selected as the technology to achieve the target

    PM levels of 0.05 g/bhp-hr. Assume that current DPF technology can reach 99%

    efficiency (i.e., 99% removal of particulates). Using a PM emissions safety factor of50%, answer the following questions, all for maximum power conditions

    (a) What is the approximate level of PM coming out of the engine?

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    (b) Size the trap and calculate the average pressure drop. Assume a space velocity inthe range of 10,000 28,000 hr

    -1that minimizes the physical size of the

    particulate trap (see SAE 2003-01-0047 for reference).

    (c) What impact does the particulate trap have on the performance of theturbocharged engine (be quantitative). What changes in boost pressure and

    turbomachinery operating conditions are needed to keep best case output?

    6. 2007 Emissions requirements:Below is a table showing EPA Diesel engine emissions requirements for 2007.

    NOx (g/bhp-hr) PM (g/bhp-hr)

    0.20 0.01

    (a) Based on engine out NOx levels of part 4, how efficient a NOx catalyst is needed

    to meet 2007 emissions levels? Assume that the catalyst is used in conjunctionwith EGR.

    (b) Is it possible to achieve these levels of PM with the trap described in the SAE2003-01-0047 paper?

    (Note: Keep the same safety factors as in parts 4 and 5)

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    E

    xhaustTemperaturevs.EGR(ForDifferentStartofInjection@M

    aximumPower)

    800

    850

    900

    950

    1000

    1050

    5%

    10%

    15%

    20%

    25%

    30%

    35%

    %EGR

    Temperature(K)

    0.5CADBTDC

    7.5CADBTDC

    3.5CADBTDC

    2.5CADATDC

    6.5CADBTDC

    156

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    25%Maxtorque,1600

    RPM

    bsNO

    xvs.

    StartOfMainInjection

    Low-LoadCondition

    0.0

    00

    0.5

    00

    1.0

    00

    1.5

    00

    2.0

    00

    2.5

    00

    3.0

    00

    3.5

    00

    4.0

    00

    4.5

    00

    5.0

    00

    -8

    -6

    -4

    -2

    0

    2

    4

    6

    8

    10

    S

    tartofMainInjection[CAFromT

    DC]

    bsNOx[g/kW-hr]

    Increase

    dEGR

    StockEG

    R

    Reduced

    EGR

    IncreasingEGR

    23.8

    %

    10.9

    %

    11.0

    %

    10.8

    %

    9.1

    %

    8.6

    %

    31.8

    %

    32.1

    %

    32.3

    %

    32.2

    %

    32.7

    %

    23.6

    %

    24.4

    %

    24.2

    %

    24.9

    %

    157

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    bsNOxvs.

    Star

    tOfMainInjection(Maxim

    umP

    ower)

    0.0

    00

    1.0

    00

    2.0

    00

    3.0

    00

    4.0

    00

    5.0

    00

    6.0

    00

    7.0

    00

    8.0

    00

    9.0

    00

    -10

    -8

    -6

    -4

    -2

    0

    2

    4

    6

    8

    StartofMainInjection[CAFromT

    DC]

    bsNOx[g/kW-hr]

    IncreasedEGR

    StockEGR

    Reduced

    EGR

    Increasing

    19.9

    %

    20.6

    %

    20.4

    %

    20.2

    %

    20.0

    29.5

    30.9

    29.6

    %

    29.5

    8.5

    %

    9.4

    %

    7.5

    %

    7.4

    7.6

    %

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    bsPMvs

    .StartOfMainInjection

    Lo

    w-LoadCondition

    0.0

    000

    0.0

    200

    0.0

    400

    0.0

    600

    0.0

    800

    0.1

    000

    0.1

    200

    0.1

    400

    0.1

    600

    0.1

    800

    -8

    -6

    -4

    -2

    0

    2

    4

    6

    8

    10

    Start

    ofMainInjection[CAFromT

    DC]

    bsPM[g/kW-hr]

    IncreasedEGR

    StockE

    GR

    Reduce

    dEGR

    Increasing

    20.0

    22.2

    22.9

    23.8

    23.8

    %

    31.8

    32.1

    32.3

    32.2

    32.7

    %

    10.9

    %

    11.0

    10.8

    9.1

    %

    8.6

    %

    25%Maxtorque,1600RP

    M

    159

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    bsPM

    vs

    .StartOfMainInjection(MaximumP

    ower)

    0.0

    00

    0.0

    50

    0.1

    00

    0.1

    50

    0.2

    00

    0.2

    50

    0.3

    00

    0.3

    50

    0.4

    00

    0.4

    50

    0.5

    00

    -10

    -8

    -6

    -4

    -2

    0

    2

    4

    6

    8

    StartofMainInjection[CAFromT

    D

    C]

    bsPM[g/kW-hr]

    Increase

    d

    EGR

    StockEGR

    Reduced

    EGR

    Increasing

    8.5

    %

    9.4

    %

    7.5

    7.4

    %

    7.6

    %

    29.5

    %

    30.9

    29.6

    29.5

    %

    20.0

    %

    20.2

    %

    20.4

    19.9

    %

    20.6

    %

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    Displacement (m3) 11

    Cylinders 6

    Bore (m) 0.1326

    Stroke (m) 0.1326

    Compression Ratio 18

    Connecting Rod Length (m) 0.3315

    Bmep @ max torque (kPa) 895

    Bmep @ max power (kPa) 835

    Maximum torque (N-m) 783

    Maximum power (kW) 173

    Maximum engine speed at

    maximum power(RPM)2261

    2.61 Internal Combustion Engines

    Design Project Solution

    Here is a possible solution for the design problem.

    1. Base Engine

    Table 1 below summarizes the main parameters of the base engine

    Table 1 Base Engine Summary

    There are two possible methods to size the engine, and they should be consistent witheach other:

    Method 1:

    2

    /,, stoichoaHVdvifm AFQVNP

    IUKKK (1)

    Method 2:Assume a bmep based on practical limits and fuel-air cycle charts, and solve for the

    power output:

    2000

    NVdbmepP (2)

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    Using the first method, we must determine N,,,,, ,, IUKKK oavifm

    Calculate Engine Speed (N)

    L2

    pSN max find L (3)

    Assume B ~ L, so

    4

    L6

    4

    B6V

    32

    d

    SS

    Lcylinders (4)

    mL 1326.06

    (0.011)4

    6

    V4 31

    3

    1

    d

    SS

    so:

    RPM2260or,sec/71.37)1326.0(2

    10m/sN revs

    Determine I, and Kf,iChose rc=18 (maybe a bit high), and I=0.7 (smoke limit, maximum possible fuel we canget in per mass of air). Using Fuel-air cycle results (Fig. 5-9, Heywood p. 182), then Kf,i=0.575. Applying a correction factor of around 80%, actual Kf,i= 46%. The correctionfactor can be between 80% and 85%; For this case, I chose 80% so that Method 1, and 2,as explained above, are consistent with each other.

    Determine IMEP

    For phi=0.7, and rc=0.8, we get

    5.10imepso5.10imep

    1PPi

    , (5)

    Note that Pi is not atmospheric pressure. At WOT, there is a pressure loss in the intake

    system, due to frictional losses that scale with speed. Pi will be less than atmospheric.

    Likewise, the exhaust pressure (Pe) is not atmospheric; a higher than atmosphericpressure is needed to pump the gases through the exhaust system. Once the gases leave

    the exhaust system and reach ambient conditions, they will expand to atmospheric

    pressure. Additionally, depending on the opening timing of the exhaust valves, the gasesmight exit at a higher pressure than what is required to overcome the pumping loss in the

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    exhaust system. To get an idea, of the value of Pi, look at Figure 13-13 in the text(Heywood P. 725). For a piston speed of 10 m/s,

    (6)

    Now allocate this pumping loss between Pe and Pi. At high speeds around 18% of theloss is on the intake side, and the remaining 82% on the exhaust side. This will beconsistent with volumetric efficiency as explained below. So:

    Pi= 101 kPa 0.18(40 kPa) = 93.8 kPa

    Pe=101 kPa + 0.82(40 kPa) = 133.8 kPa

    We can now calculate an imep:

    9kPa.9845.10kPa8.93imep

    Determine Mechanical Efficiency Km

    ;1imep

    tfmep

    imep

    tfmepimepm

    K (7)

    wherepmepfmeptfmpe mep)auxiliaryandfriction(rubbingmepfrictiontotal

    From figure 13-7 (Heywood p 722), fmep for a fired engine at 2260 rpm | 140 kPa. So

    %7.81985

    1801

    and;18040140

    m

    kPakPatfmpe

    K

    Determine Volumetric Efficiency and oa,U

    Using figure 6-8 (Heywood p. 217), assume a volumetric efficiency of 90% for a piston

    speed of 10 m/s. Note that this volumetric efficiency measures the efficiency of theentire intake system. Also note that we have chosen the right pressure loss allocation for

    the intake system (as calculated in the imep section), consistent with volumetricefficiency. The air density oa ,U , is just calculated from ideal gas law, at ambient

    conditions. The value is 1.17 kg/m3

    40)10(4.0p)S(4.0Pi)-(Pe 22 xpmep

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    Fuel-to-Air Ratio & Heating Value

    From table D.4 in the text (Heywood p. 915) we get the stoichiometric Fuel-to-Air ratio

    of gasoline as 0.0697, and its heating value of 43.2 MJ/kg

    Power calculation

    With the estimates for each value, we can now calculate the power

    kWP

    mkgkgkJemmP

    173

    2

    )0697.0)(7.0(/17.1/32.43)011.0sec)(/7.37)(90.0)(46.0)(817.0( 3

    We also use method 2 to check for consistency. Rearranging equation 2.19b (Heywood

    p50), we get:

    kW

    revdmkPaNVdbmepP 167

    2000

    sec/7.37311805

    2000

    the methods are close

    For low loads, follow the same procedure, with lower pumping loss, due to lower speed

    (see figure 13-13, Heywood), and lower rubbing and auxiliary friction (see figure 13-7Heywood); additionally, the allocation of pressure losses is different, and must be

    consistent with volumetric efficiency.

    2. Boost, Turbo-machinery and Intercooler

    Boost pressure:

    To find the boost pressure required, we use equation 1, and replace the volumetricefficiency for the entire inlet system with the volumetric efficiency for the valves only

    ( vK ~ 94%). We also replace the ambient air density with the air density right before the

    valves, ia,U . This density can be determined from the ideal gas law, knowing the pressure

    (which is approximately cylinder pressure divided by volumetric efficiency), and thetemperature (about the same as the cylinder temperature). Thus, we can vary the cylinder

    pressure until we get the required power level, as defined by equation 1.

    2

    /,, stoichiaHVdvifm AFQVNP

    IUKKK

    Note that as we vary the cylinder pressure, and consequently the density, the mechanical

    efficiency (as defined by equation 7 above) will also change because the pumping loss

    will change.

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    Pmep= Pexhaust - Pintake

    Thus the solution to this problem is iterative, and can easily be done with a spreadsheet.

    After varying the cylinder pressure, determining the corresponding air density at thevalves through the ideal gas law, and calculating the mechanical efficiency, we get the

    following target power:

    kWP

    mkgkgkJemmP

    360

    2

    )0697.0)(7.0(/065.2/32.43)011.0sec)(/7.37)(944.0)(46.0)(918.0( 3

    For this case the pressure that gives a density of 2.065 is 176 kPa, as dictated by the ideal

    gas law:

    )944.0(314/97.28/314.8

    /065.2* 3_)()(, Kkmolekg

    kmoleKkJmkgRTP valvesvesbeforevalvesbeforevalviacylinder

    KU

    which gives Pcylinder=176 kPa. The pressure that must come out of the compressor is

    approximately:

    kPakPaP

    Pv

    cylinder

    comp 186944.0

    176

    K

    Thus the desired boost is 85 kPa. That is we have to compress 85 kPa aboveatmospheric. Note that to relate pressure before the valves, and after the valves, as a first

    approximation I have used the volumetric efficiency.

    Turbo-machinery

    Knowing the desired boost, the turbo-machinery can now be sized to generate the

    required pressure. This is done using the insentropic relationships for the compressor andturbine. First we must size the compressor by finding the work required to compress the

    gas to the desired pressure. Second, we must size the turbine to produce the work that

    drives the compressor.

    To determine the amount of work that is required to compress the gas we do an energybalance assuming an adiabatic compressor:

    )( 12 TTCmW apc (8)

    where,

    T2a= Actual compressor exit temperatureT1= Compressor inlet temperature (300K)

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    We calculate T2a using the compressor efficiency, and isentropic relationships:

    112

    2 TTT

    Tc

    sa

    K(9)

    and:

    J

    J1

    2

    112

    P

    PTTs (10)

    T1, P1, P2, J, and Kc are all known, so T2a, and consequently the compressor work canbe calculated.

    Knowing the compressor work, we now size the turbine using the following equations:

    m

    ct

    WW

    K (11)

    where mK is the mechanical efficiency for the turbine and compressor system. 95% isreasonable estimate for this number

    Cp

    WTT ta 45 (12)

    where:T5a=Actual turbine exhaust temperature

    T4 = Turbine inlet temperature (engine exhaust, given at 900K)

    To find the required turbine pressure ratio:

    J

    J

    1

    4

    5

    5

    4

    T

    TP

    P

    P sr (13)

    where:

    445

    5 TTT

    Tt

    as

    K(14)

    Thus, enough equations for enough unknowns. Values for the temperatures, pressures,

    and compressor work, are show in table 2.

    Intercooler

    Adding an intercooler to lower the intake temperature, will increase the density of the

    gas, and consequently decrease the required boost, as reflected in table 2. For a given

    pressure rise we get a higher change in density (due to lower gas temperatures going intothe engine). To size the intercooler you can select a coolant, and based on adequate

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    compressor turbine

    Heat Exchanger Engine

    1

    2 3

    4

    5

    67

    estimates for inlet and outlet coolant temperatures, you can determine the required massflow-rate that is needed to achieve a certain temperature change in the air. You must use

    the definition for heat exchanger to determine the allowed change in air temperature:

    For the coolant I used water (Cp=4.2 kJ/kg K), and assumed that it goes in at 300 K, andI want it to leave at 380 K.

    To find the exit temperature of the air, and to determine the required flow rate of water, Iuse the definition for heat exchanger effectiveness in conjunction with an energy balance:

    For effectiveness we have:

    )(

    )(

    max_min

    airor,

    outin

    coolantoutinairorcoolant

    TTCpm

    TTCpm

    H

    and for the energy balance we have:

    aircoolant TCpmTCpm ''

    For this case, I chose the air and water to have about the same capacitance (mCp). Using

    the effectiveness equation I can solve for Tout air. Note that the capacitances will cancelout in the equation, and the maximum change in temperature occurs when Tout air=

    Twater in, thus:

    )()(

    )(____

    __

    irr

    inwaterinairinairoutair

    inwaterairin

    aoutin TTTTTT

    TT

    HH

    Assuming, water temperature increases from 300 to 357, then Tair in is 314 K. Valuesfor the heat exchanger temperatures and flow-rates are also shown in table 2.

    Figure 1

    Schematic of Turbocharged Engine with Intercooler

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    State Temperature (K) Pressure (kPa) Temperature (K) Pressure (kPa)

    1 300 101 300 1012 401 233 372 186

    3 401 233 314 186

    4 900 223 900 188

    5 801 127 833 127

    6 N/A N/A 101 300

    7 N/A N/A 101 380

    Work Compressor 40 kW Work Compressor 29 kW

    Intercooler

    m_dot water 0..097 kg/sec

    m_dot air 0.404 kg/sec

    No intercooling With intercooling

    Table 2

    Turbocharged Engine with Intercooler: Operating Parameters

    3. Brake Efficiency

    At half maximum speed, and full load at that speed, I kept the same boost, but loweredthe fmep, per figure 13-7 in the text. The BSFC came out to be 188 g/kw-hr. This

    number is actually quite good for industry standards. Other people perhaps got lower

    (around 175), however, as I previously explained, I was more conservative in myefficiency estimate from fuel air cycle tables, to be consistent with different ways of

    calculating power. My calculation is shown below:

    hrkWgkgghrkg

    kWPower

    hrgmbsfc

    f /188193

    )/1000)(sec/3600sec(/010078.0

    )(

    )/(

    Note that if we directly use break engine efficiency, we should get the same answer:

    hrkWgQQ

    bsfcHVifmHVbf

    /188)2.43)(460.0(96.0

    1)(

    1)(

    1

    ,, KKK

    Ways to decrease bsfc include raising compression ratio, and reducing frictional losses.

    4. Emissions NOx

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    compressor turbine

    Inter-

    coolerEngine

    1

    2 3

    5

    6

    Inter-

    coolerVenturi/

    Mixer

    4

    7

    Engine ModeNOx Standard

    g/bhp-hr

    NOx safety

    target g/bkW-

    hr

    NOx Safegy

    target g/bkW-

    hr

    EGRTiming (CA

    from TDC)

    Hit in Fuel

    economy

    (percentage

    points)

    25% maxtorque, 1600

    rpm

    2.50 1.75 2.35 24% 1 2%

    Maximum

    power2.50 1.75 2.35 24% 0.5 2%

    Figure 2

    Schematic of Turbocharged Engine with Intercooler

    The schematic shows how EGR will be driven from the engine. There are a few ways of

    doing this; one way is to use a Venturi system, as shown above. Another way is tooptimize the system so that the pressures at the air and EGR intersection are about thesame. It is necessary for these pressures to be equal, otherwise there will be backflow in

    the direction of lower pressure. Overall, however, the addition of EGR will impact the

    fuel economy of the engine. This is the price that we must pay to have lower emissions.

    The first step of this problem is to define the amount of EGR that is needed to meet EPAemissions levels. The emissions requirements along with their safety levels are shown in

    table 3 below.

    Table 3

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    Must operate

    below dashed

    line

    Using the figures provided, there are various possibilities for selecting EGR, depending

    on the hit on fuel economy. Figure 3 below is an example that shows that there is a rangeof timings and EGR levels that will give the proper amount of NOx

    Figure 3

    Acceptable operating area for low load

    Once EGR has been calculated, the loss in engine efficiency can be assessed, as well as

    the required boost. Again this is an iterative process. There are many variables affectingengine power, and they are all related as well, thus at least a spreadsheet must be setup.

    For example, boost affects engine power, but it also affects mechanical efficiency, which

    in turn affects engine power, thus all these variables must be connected when solving thesystem.

    One important implication of adding EGR, is that the pressure in the cylinder chamber

    must increase if we are to maintain constant mass of fresh fuel and air; this is what we

    should desire if we are to maintain the same power output from the engine as the casewithout EGR.

    The total pressure is equal to the sum of the partial pressures of air and the EGR:

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    State Temperature (K) Pressure (kPa)

    1 300 101

    2 434 295

    3 327 295

    4 349 279

    5 990 241

    6 864 121

    7 438 241

    With intercooling

    IntercoolerMass flowrate

    (kg/sec)Tin (K) Tout (K)

    EGR 0.171 300 380

    Compressor 0.131 300 380

    EGRairT PPP

    Assuming the molecular weights of both Air and EGR are about the same, then the mole

    fraction is approximately equal to the mass fraction of each mixture, and PT can beexpressed as:

    EGR

    PP airT

    1

    Additionally, since there is a pressure loss of around 16 to 20 kPa associated with the

    venturi, a higher boost is still needed. To reach the target power output, a total boost of

    194 kPa is required, for total pressure of 295 kPa. This is a high boost, higher thanindustry standard for this size engines. Perhaps a more practical boost is 150 kPa (PT=251 kPa), or less. However this limits the maximum power to 310 kW. If yourecognized the practical limitations, this is a perfectly acceptable answer.

    Table 4

    Table 5

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    5. Emissions Particulates

    (a) The particulate emissions corresponding to the chosen EGR level, can be obtained

    from the data provided (PM levels vs injection timing for various EGR fractions). At

    24% EGR, the PM coming out of the engine is approximately 0.2 g/bkW-hr

    (b) To size the trap, use the space velocity that will minimize the volume of the trap, inthis case 28,000 hr-1

    . This is evident from the relationship for space velocity:

    velocitySpaceV

    V_

    where V is the volume flow-rate of the gases going through the trap, and V is the volume

    of the trap. For a smaller trap volume we get a higher space velocity.

    Solving for the volume of the trap we get

    velocitySpace

    VV

    _

    Using the ideal gas law to solve forV

    sec/10.1)(

    3mP

    RTmmmV

    exhaust

    exhaustegrfuelair

    Solving for Volume

    LmmV 141141.0sec3600/000,28

    sec/10.1 33

    Values used are shown in table 6 below

    Table 6

    As shown in figure 4 of SAE 2003-01-0047, The maximum pressure loss through the

    trap, at a space velocity of 28,000/hr, is 6 kPa. This is a very small percentage of thetotal exhaust pressure (~2.5%), and the effect on turbo-machinery is small.

    Mdot_air&fuel

    (kg/sec)

    Mdot_egr

    (kg/sec)Texhaust (K) Pexhaust (kPa)

    Space Velocity

    (1/sec)

    Volume

    flowrate

    (m3/sec)

    Trap Volume

    (L)

    0.43 0.10 864.74 121.00 7.78 1.10 140.87

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    6. 2007 Emissions requirements

    Using the same safety factor as in part 6, the table below shows the new emissions that

    must be met:

    Table 7

    Where catalytic converter efficiency is defined as:

    intpollu

    outtpollu

    catm

    m

    ,tan

    ,tan1

    K

    As shown in table 7, a catalytic converter with 92% efficiency will be needed to meet

    2007 NOx emissions requirements.

    The required particulate trap efficiency is fairly high (97%) but the trap presented in the

    Ford paper seems to have efficiencies of around 99%, so it should work fine for 2007emissions requirements.

    NOx

    Standard

    g/bhp-hr

    NOx safety

    target

    g/bhp-hr

    NOx

    Safegy

    target

    g/bkW-hr

    Current

    Engine out

    g/bkW-hr

    Required

    efficiency

    (catalytic)

    PM

    Standard

    g/bhp-hr

    PM safety

    target

    g/bhp-hr

    PM Safegy

    target

    g/bkW-hr

    Current

    Engine out

    g/bkW-hr

    Required

    Efficiency

    (trap)

    0.2000 0.1400 0.1879 2.3490 0.9200 0.0100 0.0050 0.0067 0.2000 0.966