ME 444
Assignment 11
Nov. 21, 2013
Due Nov. 26, 2013
1. Explain what is understood by MBT timing in an engine.
MBT spark timing is the optimal timing of the spark so as to obtain maximum brake torque.
If the spark timing is too early, the piston does too much work on the gases, decreasing the
engine output. If it is too late, the peak cylinder pressure reduces and the work transfer
from the gases to the piston during expansion reduces. Near MBT, the relative torque
output of the engine is somewhat insensitive to spark timing that allows the engine
calibrator to retard the spark slightly resulting in reduced unburned hydrocarbons and nitric
oxides without much loss in engine efficiency.
2. Explain how the burn rate of fuel in an engine is typically described. Give an example.
Ignition delay is the 0-10% burn time while rapid burning rate is the 10-90% burn time. The ignition
delay term encompasses the phenomena from the start of the ignition process to about 30% volume
fraction burned. During this period, the flame transitions from a small ignition kernel to a fully
developed turbulent flame. The rapid burn period of the combustion process is governed by the
turbulent fluid mechanics. The oncoming flame front entrains fresh charge into the flame reaction
zone where it is ignited and eventually consumed. At the end of turbulent burning process, the
flame front has touched the chamber walls and a turbulent diffusion burning occurs, i.e.
entrainment ends since all the gases are entrained and the unburned islands of fuel burn up in a
laminar fashion. Note that at 90% burned state, over 95% of the volume is burned.
The following bar chart illustrates the effect of burn rate of fuel based on different combustion
chamber geometries:
A lower ignition delay means that the 0-10% value is least among the available sources. Faster
burning rate implies that the charge is completely burnt quicker i.e. low 10-90% burn duration.
3. Explain why a fuel might be hydrotreated.
Generally speaking, light crude oils yield more gasoline than heavier crude oils. Heavy crude oils
have a high density and are “sour” (have a high sulfur content), require more complex and
expensive processing steps than light, sweet crude oils, and are less desirable. Hydrotreating
removes sulfur and nitrogen compounds using high pressure hydrogen in the presence of a catalyst.
Sweetening converts foul smelling mercaptans to disulfides using a catalyst.
4. Where is ethanol added to gasoline? Why?
Pipelines prohibit shipment of gasoline containing ethanol because of its hydroscopic nature. Thus,
ethanol is blended into gasoline at the terminal. Ethanol is mainly added to gasoline to increase the
octane rating of the latter. E85 has (R+M)/2 octane value of about 96-98 whereas regular grade
gasoline has an average value of 87. In addition, ethanol is added to gasoline to reduce air pollutant
emissions.
5. The vapor pressure of a fuel in the US is seasonally adjusted. In addition, in the winter E85
might be 51% ethanol by volume and in the summer 85% by volume. Prepare the following
graphs. On the ordinate $ per gallon of E85. On the abscissa $ per gallon of gasoline ranging
from $3.00, 3.50 and 4.00. Calculate the amount one should be willing to pay for E85 in the
summer and winter for the given gasoline prices. State you assumptions.
In terms of Gasoline energy equivalent:
Heating value of Gasoline = 18,500 BTU/lb
Heating value of Ethanol = 12,500 BTU/lb
Density of Gasoline = 6.073 lb/gal
Density of Ethanol = 6.584 lb/gal
When Gas price = $ 3.0/gal:
Cost of ‘fuel’ per BTU = ($ 3/gal)*(gal/6.073 lb)*(lb/18500 BTU) = 0.00267 cents/BTU.
Appropriate cost of ethanol per lb = (0.00267 cents/BTU)*(12500 BTU/lb) = 33.38 cents/lb.
Appropriate cost of ethanol per gal = (33.38 cents/lb)*(6.584 lb/gal) = $ 2.20/gal.
In winter, the E85 fuel contains 51% ethanol and 49% gasoline  E85 cost = (2.20*0.51)+(3.00*0.49)
= $ 2.59/gal
In summer, the E85 fuel contains 85% ethanol and 15% gasoline  E85 cost =
(2.20*0.85)+(3.00*0.15) = $ 2.32/gal
Performing similar analysis (using spreadsheet) for the other two costs of gasoline i.e. $ 3.50/gal and
$ 4.00/gal, the following graph can be plotted.
4
$ per gallon of E85
3.75
3.5
Winter
3.25
Summer
3
Winter_GEE
Summer_GEE
2.75
2.5
2.25
3
3.5
4
$ per gallon of gasoline
Table: Cost of E85 fuel ($/gal) in winter and summer weather conditions
Gasoline
per gal
3
3.5
4
Winter
2.64
3.08
3.52
Summer
2.45
2.87
3.28
Winter_GEE
2.59
3.02
3.45
Summer_GEE
2.32
2.7
3.09
GEE: Gasoline Energy Equivalent
Here, the pump prices for E85 for winter and summer are included based on comparative prices for
gasoline ($3.65) and E85 ($3.23) as shown in Table 1 of Slide#43 of the lecture “Road Vehicle Fuels.”
Notice that E85 fuel is about 11.5% lower than gasoline. This was further validated using the price
chart in www.e85prices.com.
6. Explain how octane number is determined and the antiknock index on the gas pump means.
An octane number measurement is based on comparing the test gasoline to primary reference fuels
in standardized, single cylinder, variable compression laboratory engines. To determine octane
number, a technician first runs the test gasoline and adjusts the compression ratio of the engine to
produce a knock of standardized intensity. Then the technician runs reference fuels until he finds
one whose knock intensity matches that of the test fuel at the set compression ratio. The octane
number of the matched reference fuel is the measured octane number for the test fuel. Primary
reference fuels are mixtures of the pure chemicals isooctane and heptane. The octane numbers of
isooctane and heptane have been defined as 100 and 0, respectively. Reference fuel blends are
made using the two chemicals to provide a range of octane numbers. For instance, a blend of 90%
isooctane and 10% heptane makes a fuel with an octane number of 90.
In general, two types of engines are used to determine a gasoline’s antiknock index. These result in
two separate octane measurements: Research octane number (RON) correlates with low speed,
mild-knocking conditions; Motor octane (MON) correlates with high speed, high temperature
knocking conditions and with part-throttle operation. The antiknock index posted on gasoline
pumps represents (R+M)/2 or the average of RON and MON values.
7. Explain how the cetane number of a fuel is determined.
The cetane number is measured by a rating procedure using a standardized laboratory single
cylinder test engine. Alternatively, a predictive number called cetane index has been developed to
approximate cetane number based on a fuel’s density and distillation properties.
8. Estimate how much forest land would be needed in the US to convert the CO2 produced by
automobiles used in personal transportation to Carbon and Oxygen. Assume it takes 240 trees
to manage the CO2 produced when a car burns 600 gallons of fuel in a year. State other
assumptions.
The assumptions are that the vehicle is driven 15,000 miles per year and achieves 25 mpg.
The 600 gallons of gasoline consumed by this vehicle per year requires 240 trees to convert
corresponding CO2 into C and O2.
In 2011, the United States consumed about 134 billion gallons
(http://www.eia.gov/tools/faqs/faq.cfm?id=23&t=10)
Hence number of trees required = 134*10e+09/600 = 22.34e+07 trees.
Assuming that there are 4 trees in 10x10 ft2 land, it requires 22.34e+09/4 = 5.58e+09
ft2*(acres/43560 ft2) = 0.13 million acres of land.