ME 444 - Internal Combustion Engines

Exam 2 Take Home

/(){) 1. A Waukesha spark ignition engine, with various versions manufactured from 1928 to

1961 has operating characteristics shown in attached brochure. Calculate the

efficiency of this engine at 7 4 HP in percent assuming that the gasoline used has an

energy of 44.0 MJ/kg.

Given Parameters:

bsfc at 74 HP = 0.61 lb/HP-hr [Ref: 1]

Energy of gasoline = 44.0 MJ/kg

In order to match the units of the given energy parameter, the bsfc was

converted from lb/HP-hr to kg/kWh.

0 61 lb X GOB.27? g/kWh

- 371 04897 J_ - 0 37104897 £' HP-hr l lb!HPh - ' kWh - ' kWh

Next, kilowatt hours was converted to joules.

0.37104897 /.;h X 3_16�'u;�J = J.03069 X 10-?'

Finally, the thermal efficiency was calculated by taking the inverse of the brake

specific fuel consumption times the energy of the fuel used (gasoline).

Efficiency:

TJ - bsfc•ener�y of fuel = (44x10\�)(l.�3069x10-7 �) =

0.2205 :::: 22.05%

�

ME 444 - Internal Combustion Engines

Exam 2 Take Home

(Problems 7 and 8 Redo - Assigned on November 20th, 2018)

7. How are emissions from an automotive spark-ignition gasoline engine sold in the United

States in 2018 controlled? What are the primary emissions?

The main mechanical method used to control emissions in a spark-ignition engine in

the US is a catalytic convertor. Three-way catalysts are installed in every vehicle with an SI

engine produced in the United States and are capable of removing 98% of the non-ideal

combustion products from the vehicle's exhaust gases. [Ref: 13] These non-ideal, regulated

combustion products include NMOG + NOx (nitrous oxides), CO (carbon monoxide), PM

(particulate matter/soot), and HCHO (hydrocarbons). [Ref: 3]

Three way catalyst is constructed using a ceramic, honeycomb patterned lining that is

coated in two different catalysts. These catalysts speed up the chemical reaction process in

order to break down the harmful combustion products into ideal emissions (i.e. water, carbon

dioxide, and nitrogen). One of the two catalysts breaks down the nitrous oxides into their

individual components, oxygen and nitrogen, in a reduction process (removing oxygen). The

other catalyst adds oxygen to the unwanted carbon monoxide to turn it into carbon dioxide.

[Ref: 14] These two catalysts together essentially perform three different chemical reactions.

[Ref: 15]

CO + 102 --+ CO2

H4 C

2 + 30

2 --+ 2C02 + 2H

20

CO + NOx --+ CO2

+ N2

These three reactions are what gives the three-way catalyst its name.

There are a few important notes about three-way catalysts in order to use them

effectively as an aftertreatment system. These catalysts do not work well in cold conditions. A

cat prefers to operate with exhaust temperatures between 250-300 degrees Celsius. During

cold-start conditions, it takes about 2-3 minutes for a modern catalyst to warm up to these

temperatures. Therefore, during this time, a large number of emissions will come out of the

tailpipe. This is why cold-start emissions tests are the most troublesome for engineers to

pass. [Ref: 13] (Aside: the equivalence ratio of the air/fuel mixture can be adjusted in order to

produce exhaust gases that are at a higher temperature. This can be done to warm up the

three-way catalyst faster, therefore producing less harmful emissions).

Another important factor to note is that catalytic converters work best at stoichiometric

conditions. This is because at near stoichiometric conditions, the exhaust temperatures are

high, allowing the catalysts to perform to their best ability. [Ref: 15]

ME 444 Exam2 Submitted

to: Dr H.Schock

- .

. ,

ME444

11/27/18

Exam 2 Revised Problems 7 & 8

7.) According to the "Natural Gas Vehicle Emissions" in the Alternative Fuels Data Center on the US Department of Energy's website, the emissions of primary concern are regulated emissions of hydrocarbons, oxides of nitrogen, carbon monoxide, as well as carbon dioxide. The EPA requires all fuels and vehicles to meet new standards (trending towards zero) for tailpipe emissions of particulate matter and pollutants. Gasoline engine emissions are reduced by catalytic converters, which convert toxic pollutants into less harmful ones. Additionally, from the ME 444 lecture slides, "Trends & Challenges for Advanced IC Engines (2016)" Closed loop combustion control can be used to reduce emissions in real time a�d d�t injection can reduceparticulate matter. V

http ://afdc.energy.go /vehicles/natural gas ernissions.html

8.) According to the United States Environmental Protection Agency, the primary regulated emissions from diesel-fuel engines are HC, NOx, CO, and PM. The EPA has standards in place for all of these pollutants and diesel engines must pass EPA regulations to be put into production. These regulations help control emissions from over-the-road diesel engines. From the ME 444 Lecture slides, "Trends & Challenges for Advanced IC Engines (2016)" Diesel particulate filters can help reduce emissions while maintaining high efficiency, and high rate of cooled EGR can be used to reduce NOx. Additionally, catalytic converters are used in th exhaust to convert toxic pollutants into less harmful ones before exiting the vehicle.

https://nepis.epa.gov/ xe/ZyPDF.cgi?Dockey=Pl 0009ZZ.pdf

ME444

Exam 2 Take Home, Fall 2018

November 12, 1020 am (before class starts)

Organization and clarity count for 30�o

Prose should be typed, figures copied referenced properly

Numerical answers must be In a box with the proper units.

1. A Waukesha spark ignition engine, with various versions manufactured from 1928 to

1961 has operating characteristics shown in attached brochure. Calculate the efficiency of

this engine at 74 HP in percent assuming that the gasoline used has an energy of 44.0

MJ/kg.

W out 'fl 111 =

Q ill

Where

p OU.1pU1

p i11pu1

bhp=74hp=55.I8kW

bhp (3600,:;;-

)

FCxCV

Fuel Consumption (FC)= 20.14 kg/hr Calorific Value (CV)= 44,000 kJ/kg

1 n=21.7% j Using the calorific value and the fuel consumption it may also be done with the following equation:

CV engme bsfc(MJ/kg) '1,1t = CV fuel CVfuel(MJ/kg)

By converting the bsfc to MJ/kg it is found that the engine uses 9.545 MJ/kg and with the calorific value of the fuel at 44 MJ/kg. This equation also reached the conclusion that n=21. 7%.

ME444

Exam 2 Take Home, Fall 2018

November 12, 1020 am (before class starts)

Organization and clarity count for 30%

Prose should be typed, figures copied referenced properly

Numerical answers must be in a box with the proper units.

1. A Waukesha spark ignition engine, with various versions manufactured from 1928 to

1961 has operating characteristics shown in attached brochure. Calculate the efficiency of

this engine at 74 HP in percent assuming that the gasoline used has an energy of 44.0

MJ/kg.

W out p au.tpuJ

Yf rh=

Q in p illput

Where "bhp=74hp=55.18kW

bhp (3600-)hr

FCXCV

Fuel Consumption (FC)= 20.14 kg/hr Calorific Value (CV)= 44,000 kJ/kg

1 n=21.7% j

Using the calorific value and the fuel consumption it may also be done with the following equation:

CVengine bf (MJ!kg) 1'/ih =

CV fuel CV fuel (Ml/kg)

By converting the bsfc to MJ/kg it is found that the engine uses 9.545 MJ/kg and with the calorific value of the fuel at 44 MJ/kg. This equation also reached the conclusion that n=21.7%.

ME 4�4 Take Home Exam 2

1. Givens [1]:

Tuesday- November 13th 2018

Power=74HP

Energy for Gasoline =44MJ/Kg

At 74 HP and graph, N=l 700 RPM

With 1700 RPM and table, Fuel Consumption= 0.61 _lb_

HP-Hr

output 11 =

input

(74HP) • (. 7457 ��) • ( 3.6 f J;) 1J

= (

Mf) lb

( 1 kg ) * 100 = 22.004%

44Kg *(o.61HP - Hr)*(74 HP)* 2.2lb /

2. Givens [1]:

I 11 = 22.004% I V

Power=80 HP At 80 HP and graph, N=l950 RMP

4 Stroke = nr = 2

vd = 404in3

lb 1 bar-14.5038�

396000 * P(HP) * n,. BMEP = mep = [2]

Vd (in3) * N(�v) mm

396000 * 80 * 2 lb BMEP =

404 * 1950 = 80.43 in2

[ 80.43 _lb3

1 1 bar mlb

= 5.55 bar 14.503873

/ m

IBMEP = 5.55 bar I

ME 444 Take Home Exam 2

3. Givens [3]:Power=74HP Energy=43.2 MJ/Kg At 74 HP and graph, N=2500 RPM With 2500 RPM and table, Fuel Consumption= 0.40 _lb_

HP-Hr output

rJ = input

(74HP) • (. 7457 t�) * (3.6Fi) r,

= (

Mf) lb (

l kg )

* 100=

34.065% 43.2 Kg * ( 0.40 HP_ Hr)* (74 HP)* 2.Z lb

1 rJ = 34.065% 1 v7

4. Givens [1,3): /

John Deere Weight=853lb John Deere Length=33.9 in, Width=24.1 in , Height=33.7 in John Deere Rated Power=74 HP Waukesha Weight=860lb Waukesha Length=43.6875 in, Width=20.5in , Height=31 in Waukesha Rated Power=85 HP

V,ohnDeere = 33.9 * 24.1 * 33.7 = 27532.563 in3

Vwaukesha = 43.6875 * 20.5 * 31 = 27763.40625 in3

. . .

W(lb) Specific Weight=

d (HP.) rate power

853 lb Specific Weight

fohnDeere = 74 = 11.53 HP

860 lb Specific Weightwaukesha = 85 = 10.12 HP

. . V(in3)Specific Volume=

d (HP) rate power

27532.563 in3

Specific VolumefohnDeere = 74 = 372.06

HP

27763.40625 in3

Specific Volumewaukesha = 85 = 326.63 HP

ME 444 Take Home Exam 2

rated power (HP) HP per cubic inch=

V(in3)

74 HP HP per cubic inchfohnDeere = 27532_563 = .0027

in3

85 HP HP per cubic inchwaukesha = 27763_40625 =

.0031 in3

When looking at the John Deere diesel engine and the Waukesha spark ignitionengine, the specific weights are very close. The John Deere engine specific weight andvolume is higher than Waukesha. Deere engine has a weight and volume of 11.53 lb/HPand 372.06 in/\3/HP. As the Waukesha engines has a weight and volume of 10.12 lb/HPand 326.63 in/\3/HP. This is because the rated power for the Waukesha engine is higher at85 HP as the Deere engine rated power is only 74 HP. However, as for the Horsepowerper cubic inch the Waukesha engine is higher due to the volume of the engine beinglarger. The Waukesha horsepower per cubic inch is .0031, and John Deere's is .0027. Tofigure out these values information from the Power Tech 4045DF270 Diesel Engine JohnDeere handout [3] and the Waukesha Engines bulletin No. 798-S [l].

5. Givens[3]:Power=74 HPAt 74 HP, N=2500 RMP4 Stroke = nr = 2vd = 275in2

lb 1 bar=14.5038"'."""z in

396000 * P(HP) * 17.rBMEP = mep = rev [2] Vd (in3) * N (min)

396000 * 74 * 2 lb BMEP = 275 * 2500 = 85.248

inz

[

lb

l

85.248 -:---z1 bar i�b = 5.88 bar

14.5038 Tri!

/ I BMEP = 5.88 barl J

ME 444 Take Home Exam 2

6. Givens [2]:

Gasolines specific energy content=44MJ/Kg Gasoline Density=6.2-1

b-gallonDiesel specific energy content=43.2 Diesel Density=7 .12-1

b-gallon. M] 1kg 6.2 lb M] Energy Content on M] /Gallon Basis9asoline = 44 Kg* 2_2 lb* 1 gallon= 124 Gallon

. M] 1kg 7.12 lb M] Energy Content on M] /Gallon BasisDiesel = 43.2- * -2 lb* 1 ll = 139.81 G ll Kg 2. ga on a on

7.

8.

(139 81 M] ) - (124 M] ) Percent Increase= · Gallon M]

Ga.lli2!!__/"

100 = 12.75% ( 124 Gallon) \_..../"

The energy content of diesel is 12.75% greater than the energy content of gasoline.

Emissions for an automotive spark-ignition gasoline engine sold in the United States in 2018 are controlled by federal standards. The United States Environmental Protection Agency (EPA) establishes these standards. The EPA authority is based on the clean air act. For new vehicles, there must be an EPA-issued certificate of conformity demonstrating that the engine/vehicles conformed to the emission requirements established. The primary emissions for an automotive spark-ignition gasoline engine is N2 , H2 0, CO2 , CO, HC, and NOx. Three of the emissions are okay for the atmosphere, these areN2, H2 0, and CO2 . However, CO, HC, and NOx are the pollutant e This information was gathered from the EPA website [4].

Emissions for an automotive spark-ignition gasoline engin States in 2018 are controlled by federal standards. The United States Environmental Protection Agency (EPA) establishes these standards. For new vehicles, there must be an EPA-issued certificate of conformity demonstrating that the engine/vehicles conformed to the emission requirements established. The primary emissions for an over-the-road diesel engine truck are PM and NOx. About half of the emissions are nitrogen oxides and two-thirds are particulate matter. This information was found from the EPA website [4]

and the ucsusa website [5].

ME 444 Take Home Exam 2

9.

The figure 1 below is from Dr. Todd Fansler's presentation, Trends and

Challenges for Advanced Internal Combustion Engines [ 6]. The sketched dashed line

shown on figure 1 is where an engineer would like to have an engine operate. This at an

equivalence ratio and temperature too low for soot and NOx formation.

sootzone Conventional

Diesel

Combustion

Figure 1: Soot vs Nox Tradeoff

/

ME 444 Take Home Exam 2

10. Givens [7,8,9]:1 acre can sequester 2.5 tons of carbon 766 million acres of forest land in US 2.3 billion acres of land in the US 16.65% arable land 1540 million metric tons of CO2 1 US ton=.907 metric ton

Amount of arable land= acres of land*% of arable land

Amount of arable land= (2.3 * 109) * .1665 = 382.95 million acres

Total amount of land available= acres of f ore st+ acres of arable

Total amount of land available = (382.95 * 106) * (766 * 106) = 1148.95 million acres

Metric tons squestered by arable and forest land in the US

( tons

) (· 907metric ton

)= 2.5 * (1148.95 * 106 acres)* year - acre 1 ton

= 2605244125 potenital metric tons of CO2 squestered per year

= Approximately 2.6 billion metric tons of CO2

Metric tons squestered by forest land in the US

( tons

) (· 907metric ton

)= 2.5 * (766 * 106 acres)* year - acre 1 ton

= 1736.905 million potenital metric tons of CO2 squestered by the forest per year

= Approximately 1.74 billion metric tons of CO2

1736.905 million metric tons Percent CO2 emissions Forest can squester = * 100

1540 million metric tons 1 = 112.786% 1

If all the arable and forest land in the United States were utilized, potentially 2.6 billion metric tons of CO2 would be sequestered. If approximately 1.5 billion metric tons of CO2 were produced annually from transportation, then the United States has enough land to sequester these emissions. When looking at only the forest land, the United Sates forests are capable of sequestering approximately 1.7 billion metric tons of CO2 annually. This makes up for 1 !,2.786% of the CO2 emitted from transportation per year. To derive the values,A info_;)llation was gathered from two websites and one brochure:Western Water hedef P:t;'World Bank [8], and FIA [9].

ME 444 Take Home Exam 2

11. Givens [7,10,11,12]:2.3 billion acres of land in the US

16.65% arable land in the US

1.24 million mi 2 land in India

52.62 % arable land in India

3. 7 million mi 2 land in China

12.68% arable land in China

6.6 million mi2 land in Russia

7.52 % arable land in Russia

37 billion acres ofland in the World

10.991 % arable land in the World

I mi 2

=640 acres

US Population=326.766 million

World Population=7.662 billion

Amount of arable land = acres of land * % of arable land

Amount of arable land united States = (2.3 * 109 acres)* .1665 = 382.95 million acres

(640 acres)Amount of arable land India = (1.24 * 106 mi2) *

.2 * .52621mi

= 417.59 million acres

(640 acres)Amount of arable land China = (3.7 * 106 mi2 ) *

.2 * .12681 mi

= 300.26 million acres

Amount of arable land world = (37 * 109 acres )* .10991 = 4.07 billion acres

382.95 * 106acres Fraction of arable land in US VS World =

4_07 * 109 acres* 100 = 9.41 %

417.59 * 106acres Fraction of arable land in India VS World= 4_07 * 109 acres * 100 = 10.26%

300.26 * 106acres Fraction of arable land in China VS World= 4_07 * 109 acres * 100 = 7.38%

317.64 * 106acres Fraction of arable land in Russia VS World=

4_07 * 109 acre

s * 100 = 7.8%

ME 444 Take Home Exam 2

Fraction of arable land in US, India, China, and Russia VS World

(382.95 * 106acres) + (417.59 * 106acres) + (300.26 * 106acres) + (317.64 * 106acres)

* 100 =134.85% 14.07 * 109 acres

326.766 * 106

Fraction of World Population is in US= 7_662 * 109 * 100 �4.26% I

The United States has a roximatel 2.3 billion acres ofland d the earth has approximately 37

billion acres ofland. In all of the land on earth, about 10.991 % of the land is arable. This means

that there is about 4.07 billion acres of arable land on earth. The United States, India, China, and

Russia make up approximately 34.85% of arable land on earth. In addition, the United States

contains 4.26% of the world's population. To obtain this data four websites were used:

Westemwatersheds [7], Nationmaster [10], Newstatesman [11), and Worldometer [12].

ME 444 Take Home Exam 2

References:

[l]Waukesha. "Six Cylinder Waukesha Engines For Tractors, Trucks, Pumps, Electrical and

Industrial Machinery." Waukesha Motor Company Engine Bulletin NO. 798-S.

[2] Schock, Harold. "Exam 1 Equation Sheet." 2018

[3]John Deere. "PowerTech 4045DF270 Diesel Engine, Industrial Engine Specification." John

Deere Power System. 2016.

[4] EPA, Environmental Protection Agency, 2018, wvvw.epa.2.ov/.

[5] "Diesel Engines and Public Health." Union of Concerned Scientists, Union of Concerned

Scientists, 2018,,.; ww.ucsllsa.org/clean-v hicl /v hicles-air-pollution-and-human-health/di�sel­

engines#.W-NR6cRReUk.

[6] Fansler, Todd. "Trends and Challenges for Advanced Internal Combustion Engines."

University of Wisconsin-Madison,2018

https://www.cgr.ms11.edu/ares/ME444/lecnu-es/Fans1er%20lC%20EnQine%?ff

rend %20&%20

Challenges%202016.J df.

[7] Wuerthner, George. "The Truth About Land Use in the United States." The Truth About Land

Use in the United States, 2002,

www.we ternwatersheds.org/watmess/\i ·atmes 2002/2002html summer/article6.htm.

[8] Food and Agriculture Orgnaization. "Arable Land(% of Land Area)." The World Bank,

2015, data.worldbank:.org/indicator/ AG. LND .ARBL.ZS ?locations=US.

[9] US Forest Resource Facts and Historical Trend,;. US. Forest Resource Facts and

Historical Trends, United States Department of Agriculture, 2014.

[10] "Countries Compared by Geography > Land Area > Square Miles. International

Statistics." NationA1aster.com, NationMaster, 2013, www.nati nmasl r.com/ ountrv­

info/ tars/Geography/Land-area/Sguar -miles.

[11] Cahill, Kevin. "Who Owns the World?" New Statesman, 2011,

www.newstatesman.com/g.lobal-is ues/20 l l/03/!and-gneen-world-australia.

[12] "CUITent World Population." United Arab Emirates Population (2018) - Worldometers,

Worldometer, 2018, v,rww.\vorldometcrs.info/world-population/.

7.

ME 444 Exam 2 Revision

Tuesday- November 20th, 2018

Emissions from an automotive spark-ignition gasoline engine sold in the United States in 2018 are controlled by complex emission standards which focus on the primary emissions of carbon monoxide, NOx, and hydrocarbons. A popular met ti of meeting these standards include a catalytic converter that is attached to the exhaust pipe. catalytic converter reduces the amount of pollutants in the exhaust by utilizing chem i I reactions to create less harmful products. The converter consists of precious meta s or example, platinum and rhodium) that react with the exhaust to form carbon dioxide and water. There are different types of catalytic converters including oxidation catalyst, duel-bed, and three-way catalytic converter. Even though their designs are different, their end goal is the same. They reduce the harmful exhaust gases though chemical reactions. [1]

8.

Emissions from an over-the-road truck diesel engine sold in the United States in 2018 are controlled by complex emission standards as well. These focus on the primary emissions of soot and NOx. Aftertreatment systems are utilized to meet these standards. A specific example of an aftertreatment system is the Detroit I-Box. The L-Box combines the use of selective catalytic reduction (SCR), diesel oxidation catalyst (DOC) diesel particl!,).a e filter (DPF), and diesel exhaust fluid (DEF). The overalJ goal. of the system is� res;hrte the NOx and soot. The NOx is reduced by a small amount of DEF being injected into Wexhaust. The DEF reacts with the NOx in the SCR catalyst and forms nitrogen and water which can be released into the environment. The soot is reduced by DOC and DPF. This captures he soot and regenerates (burns) it while stationary or active. The stationary regeneration requires a high-idle and can take up to 45 minutes. The active regeneration can occur during a normal operation cycle. Diesel fuel is injected into the exhaust system to increase the aftertreatment system temperature until the regeneration is complete. [2]

References

[1] "Catalytic Converter: Design, Types And Working Explained with Diagrams." CarBikeTech,Ca, 10 Jan.2018, carbiketech.com/catalytic-converter-types-working/.

[2] "Detroit Emissions." Demand a Superior Powertrain I Demand Detroit, Detroit, 2018,demanddetroit.com/why-detroit/emissions/.