THE COPPERBELT UNIVERSITY
SCHOOL OF ENGINEERING
DEPWARTMENT OF MECHANICAL ENGINEERING
Heat Engines & Thermodynamics II - EM 440
Assignment Two
NAME:
SIN:
DUE DATE:
CHISENGA CHAMA MISENGO
19137459
25th, February 2023
LECTURER:
Mr. BENNET SIYINGWA
QUESTION ONE:
(A) Give an example of open system ICE and a closed system ICE
.
1. Open System ICE: An example of an open system ICE is a typical gasoline engine
used in a car. In an open system, mass and energy can flow freely across the
system's boundary. In the case of a gasoline engine, air flows into the engine's
intake system, fuel is added to the air, and combustion occurs to produce energy.
Exhaust gases are then expelled from the engine through the exhaust system. The
air, fuel, and exhaust gases all flow freely into and out of the system, making it an
open system.
2. Closed System ICE: An example of a closed system ICE is a rocket engine. In a
closed system, no mass can cross the system's boundary, but energy can be
exchanged between the system and its surroundings. In the case of a rocket engine,
the combustion of fuel produces hot gases that are trapped inside the engine's
combustion chamber, and the energy from this combustion is used to power the
rocket. No mass enters or exits the system during operation, making it a closed
system.
b. Explain the three types of flows associated with an open system ICE.
Air Flow: The air flow is the flow of air into the engine's intake system, which is
then mixed with fuel to create combustion. Air flow is an important factor in the
engine's efficiency and power output. The more air that can be drawn into the
engine, the more fuel that can be burned, resulting in more power. The air flow is
controlled by the throttle, which regulates the amount of air entering the engine.
Fuel Flow: The fuel flow is the flow of fuel into the engine's combustion chamber,
where it is mixed with air and burned to produce power. The fuel flow is controlled
by the fuel injectors, which release precise amounts of fuel into the combustion
chamber based on the engine's needs. The amount of fuel injected into the engine is
determined by the engine control unit (ECU), which uses various sensors to
monitor the engine's operating conditions.
Exhaust Flow: The exhaust flow is the flow of exhaust gases out of the engine's
exhaust system. After the fuel and air have been burned, the exhaust gases are
expelled from the engine through the exhaust system. The exhaust flow is
controlled by the exhaust valves, which open and close to allow the exhaust gases
to escape from the engine. The exhaust gases are typically treated with a catalytic
converter to reduce harmful emissions before they are released into the
environment
QUESTION TWO:
(A) The first law of Thermodynamics can be stated as dQ = dW + dU. Explain
each term of this law and how each term applies to IC Engines
1. dQ (Heat Transfer): This term represents the transfer of heat energy into or out of
the system. In an internal combustion engine, heat is transferred into the system
during the combustion process, where fuel and air are burned to create a highpressure and high-temperature mixture. This heat energy is then converted into
mechanical work to move the piston and generate power.
2. dW (Work Done): This term represents the amount of work done by the system. In
an internal combustion engine, work is done by the expanding gases produced
during the combustion process. This work is transmitted through the piston to the
crankshaft and then to the wheels of the vehicle, propelling it forward.
3. dU (Change in Internal Energy): This term represents the change in the internal
energy of the system. In an internal combustion engine, the internal energy of the
system is increased during the combustion process, as the fuel and air mixture is
converted into high-pressure and high-temperature gases. This increase in internal
energy is then converted into mechanical work as the gases expand and move the
piston.
QUESTION THREE (14 Marks)
a. What is an air standard cycle? What assumptions are made in the analysis of the
Ideal Otto Cycle?
b. Sketch the ideal and real Otto Cycle and the ideal and real Diesel Cycle
c. Compare and contrast between the ideal and real Otto cycle
d. Compare and contrast between the ideal and real Diesel cycle
a. What is an air standard cycle? What assumptions are made in the analysis of the
Ideal Otto Cycle?
a. What is an air standard cycle? What assumptions are made in the analysis of the
Ideal Otto Cycle?
1.
2.
3.
4.
An air standard cycle is a theoretical cycle used to analyze and model the
thermodynamic processes that occur within an internal combustion engine. It is a
simplified model that makes several assumptions to simplify the analysis and
calculation of engine performance.
The Ideal Otto Cycle is a specific type of air standard cycle that describes the
idealized thermodynamic cycle of a gasoline engine. It consists of four processes:
Isentropic Compression: The air-fuel mixture is compressed adiabatically and
reversibly from point 1 to point 2. This process is assumed to be isentropic,
meaning there is no heat transfer and the process is reversible.
Constant Volume Heat Addition: The compressed air-fuel mixture is ignited and
the combustion takes place at a constant volume from point 2 to point 3. This
process is also assumed to be adiabatic.
Isentropic Expansion: The combustion products expand adiabatically and
reversibly from point 3 to point 4. This process is also assumed to be isentropic.
Constant Volume Heat Rejection:The combustion products are then cooled at
constant volume from point 4 to point 1, and the cycle starts over again.
Assumptions made in the Ideal Otto Cycle analysis include:
Air Standard Assumption: The air inside the cylinder is considered to be the
working fluid and is assumed to behave like an ideal gas. This simplifies the
calculation of thermodynamic properties such as specific heat and enthalpy.
Neglecting Friction and Heat Transfer: The analysis assumes that there is no heat
transfer or frictional losses during the combustion and expansion processes. This
simplifies the calculation of engine efficiency but does not reflect real-world
conditions.
Perfect Combustion: The combustion process is assumed to be perfect, with
complete and instantaneous combustion of the fuel-air mixture. This assumption
ignores the effects of imperfect combustion, such as incomplete combustion and
emissions.
Constant Specific Heat: The specific heat of the working fluid is assumed to be
constant throughout the cycle. In reality, the specific heat of the working fluid
changes as it undergoes different thermodynamic processes.
(B) Sketch the ideal and real Otto Cycle and the ideal and real Diesel Cycle
Ideal and real Otto Cycle
Ideal and real Diesel Cycle
(C) Compare and contrast between the ideal and real Otto cycle
1. Efficiency: The ideal Otto cycle is more efficient than the real Otto cycle. This is
because the ideal cycle assumes that there is no heat loss during the combustion
process, while the real cycle takes into account heat loss due to conduction,
convection, and radiation.
2. Pressure-volume diagram: The pressure-volume (PV) diagram for the ideal Otto
cycle is a perfect rectangle, while the PV diagram for the real Otto cycle is more
rounded due to the slower combustion process and heat loss.
3. Compression ratio: The ideal Otto cycle assumes that the compression ratio is
infinitely high, while the real cycle has a finite compression ratio. This is because
real engines have limitations on how much they can compress the fuel-air mixture
before it detonates prematurely.
4. Combustion: In the ideal Otto cycle, combustion occurs instantaneously, while in
the real cycle, combustion takes a finite amount of time. This results in a slower
pressure rise in the real cycle.
5. Exhaust: The ideal Otto cycle assumes that the exhaust process is instantaneous,
while in the real cycle, it takes a finite amount of time for the exhaust gases to
leave the cylinder.
d. Compare and contrast between the ideal and real Diesel cycle
1. Efficiency: The ideal Diesel cycle is more efficient than the real Diesel cycle. This
is because the ideal cycle assumes that there is no heat loss during the combustion
process, while the real cycle takes into account heat loss due to conduction,
convection, and radiation.
2. Pressure-volume diagram: The pressure-volume (PV) diagram for the ideal Diesel
cycle is a perfect rectangle, while the PV diagram for the real Diesel cycle is more
rounded due to the slower combustion process and heat loss.
3. Combustion: In the ideal Diesel cycle, combustion occurs instantaneously after
fuel injection, while in the real cycle, combustion takes a finite amount of time due
to fuel-air mixing and ignition delay.
4. Heat loss: The ideal Diesel cycle assumes no heat loss during combustion, while in
the real cycle, heat is lost due to conduction, convection, and radiation from the
cylinder walls.
5. Compression ratio: The ideal Diesel cycle assumes that the compression ratio is
infinitely high, while the real cycle has a finite compression ratio. This is because
real engines have limitations on how much they can compress the fuel-air mixture
before it detonates prematurely.
QUESTION FOUR (4 Marks)
a. What is a crank-slider mechanism and why is it important when analyzing
engine characteristics and performance? Does it apply to rotary engines?
A crank-slider mechanism, also known as a slider-crank mechanism, is a
mechanism that converts rotary motion to reciprocating motion or vice versa. It
consists of a rotating crankshaft, a sliding connecting rod, and a reciprocating
piston. When the crankshaft rotates, the connecting rod slides back and forth,
which in turn causes the piston to move in a reciprocating motion.
Crank-slider mechanisms are important when analyzing engine characteristics and
performance because they are the basic mechanism used in most internal
combustion engines, including reciprocating engines, such as gasoline and diesel
engines. The motion of the piston is directly related to the motion of the
crankshaft, which affects the engine's power output, efficiency, and other
performance characteristics.
The crank-slider mechanism also plays an important role in engine dynamics, such
as balancing and vibration. Engine designers need to carefully consider the mass
and distribution of the components in the mechanism to ensure smooth operation
and minimize unwanted vibrations.
In contrast, rotary engines, such as the Wankel engine, do not use a crank-slider
mechanism. Instead, they use a rotating rotor with an epitrochoidal profile that
revolves around an eccentric shaft. The rotary engine produces a different type of
motion, and its performance characteristics and design considerations are different
from those of reciprocating engines.
b. Define curb weight and how you can reduce this during engine design
Curb weight is the total weight of a vehicle including all the standard equipment,
necessary operating fluids, and a full tank of fuel, but without any passengers or
cargo. It is essentially the weight of the vehicle as it sits on the curb or stands still
on the ground.
Reducing the curb weight of a vehicle is important for improving its performance,
fuel efficiency, and overall handling. Here are some ways to reduce the curb
weight during engine design:
1. Material Selection: By choosing lightweight materials such as aluminum, carbon
fiber, or high-strength steel, the overall weight of the vehicle can be reduced
without compromising on strength and durability.
2. Component Optimization: Engineers can optimize the design of each component of
the engine, including the block, cylinder heads, pistons, connecting rods, and
crankshaft, to minimize their weight.
3. Downsizing: Reducing the size of the engine can also help reduce its weight. By
using smaller and more efficient engines, less weight will be needed for supporting
structures and components.
4. Innovative design: Employing innovative design methods and manufacturing
techniques like additive manufacturing or 3D printing can produce components
with intricate shapes and reduced weight.
5. Aerodynamics: Aerodynamic design of the engine compartment can be improved
to reduce the weight and resistance offered by the air flow and improve overall fuel
efficiency.
By implementing these measures, engine designers can reduce the curb weight of a
vehicle and achieve better performance and fuel efficiency.
QUESTION FIVE (11 Marks)
a) Explain what an atmospheric engine is and why it is no longer used today.
[2] Marks
The atmospheric engine worked by heating water in a boiler to create steam, which
was then directed into a cylinder. As the steam entered the cylinder, it pushed a
piston upward, which was connected to a pump or other machinery. After the
steam was used, cold water was then injected into the cylinder, which caused the
steam to condense and created a partial vacuum. The weight of the atmosphere
then pushed the piston back down, and the cycle was repeated.
However, atmospheric engines had several limitations that made them less efficient
than later steam engines. They required a significant amount of fuel to heat the
water and generate the steam, which was expensive and time-consuming.
Additionally, the condensation of steam in the cylinder cooled it down, requiring
additional fuel to reheat it and making the engine less efficient. Finally, the slow
and jerky motion of the engine made it unsuitable for many industrial applications.
As a result of these limitations, atmospheric engines were gradually replaced by
more efficient types of steam engines, such as the Watt steam engine and the highpressure steam engine
b) Explain FOUR differences between a SI engine and a CI engine [4] Marks.
Ignition system: The main difference between SI and CI engines is the way they
ignite the fuel. In SI engines, a spark plug is used to ignite a mixture of air and
fuel, while in CI engines, fuel is ignited by the heat generated from the
compression of air in the cylinder.
Fuel type: Another significant difference between the two engines is the type of
fuel they use. SI engines typically run on gasoline or other fuels with lower
viscosity and volatility, while CI engines use heavier and more viscous fuels such
as diesel or biodiesel.
Compression ratio: The compression ratio in SI engines is typically lower than in
CI engines, due to the need to avoid pre-ignition and knock. This is because the
fuel is ignited by a spark, and a high compression ratio can cause it to ignite
prematurely. In contrast, CI engines can operate with higher compression ratios
since the fuel is ignited by compression.
Emissions: SI engines tend to produce more carbon monoxide (CO) and unburned
hydrocarbons (HC) due to incomplete combustion, while CI engines tend to
produce more nitrogen oxides (NOx) due to the high temperatures and pressures
generated during combustion. As a result, different emissions control systems are
used for each type of engine. For example, SI engines often use catalytic
converters to reduce CO and HC emissions, while CI engines use selective
catalytic reduction (SCR) to reduce NOx emissions.
c) Compare and contrast between a Four stroke and a Two stoke Engine (2
comparisons and 2 contrasts) [4] Marks
Comparisons:
Combustion process: Both four-stroke and two-stroke engines use a similar
combustion process, where a fuel-air mixture is ignited in the cylinder to produce
mechanical energy.
Components: Both types of engines have similar components, including a piston,
crankshaft, and connecting rod, and operate on the same basic principles of intake,
compression, combustion, and exhaust.
Contrasts:
1. Cycle: The main difference between four-stroke and two-stroke engines is the
number of cycles required to complete one combustion cycle. Four-stroke engines
require four separate strokes of the piston to complete one cycle (intake,
compression, combustion, and exhaust), while two-stroke engines require only two
strokes (compression/ignition and exhaust).
2. Efficiency: Four-stroke engines are typically more efficient than two-stroke
engines, as they provide more power with less fuel consumption. This is due to the
greater time available for fuel combustion in a four-stroke engine, which allows for
a more complete burn of the fuel-air mixture
d) Explain the difference between opposed piston and opposed cylinder
Engine. Explain how a radial engine operated and how it is different from a
rotary engine.
Opposed piston and opposed cylinder engines are both types of internal
combustion engines that have pistons moving towards each other in opposite
directions. In an opposed piston engine, two pistons share a single combustion
chamber, and they are connected to a common crankshaft. On the other hand, in an
opposed cylinder engine, each cylinder has its own piston, and the two pistons are
connected to a separate crankshaft.
A radial engine is a type of internal combustion engine where the cylinders are
arranged in a circular pattern around a central crankshaft, like the spokes of a
wheel. The pistons move in a reciprocating motion, and are connected to a master
rod, which is then connected to the crankshaft. The radial engine is commonly used
in aircraft, and is known for its high power-to-weight ratio and smooth operation.
A rotary engine, also known as a Wankel engine, is a type of internal combustion
engine where a triangular rotor rotates within a housing. The rotor has three sides,
each of which acts like a piston, compressing and expanding air and fuel as it
rotates. The rotary engine is known for its high power output and compact size, but
it also has a reputation for high fuel consumption and emissions.
e) Explain what Cross Scavenging, BMEP, Mechanical Efficiency and
Thermal efficiency are in heat engine.
Cross scavenging is a technique used in two-stroke engines where fresh air/fuel
mixture is forced into the cylinder through ports on one side, while the exhaust
gases are expelled through ports on the opposite side of the cylinder. This
technique helps to improve the efficiency of the engine by reducing the amount of
unburned fuel in the exhaust gases. It is commonly used in small engines, such as
those found in motorcycles, boats, and lawn equipment.
BMEP:
BMEP, or Brake Mean Effective Pressure, is a measure of the average pressure
exerted on the piston by the combustion process in an engine. It is calculated by
dividing the engine's brake horsepower by its displacement volume, and is
typically expressed in units of pounds per square inch (psi)
Mechanical Efficiency:
Mechanical efficiency refers to the efficiency with which a heat engine converts
the energy of the fuel into useful work. It is the ratio of the engine's brake
horsepower to its indicated horsepower, and takes into account losses due to
friction and other mechanical inefficiencies within the engine.
Thermal Efficiency:
Thermal efficiency is a measure of how effectively a heat engine converts the
energy of the fuel into useful work. It is the ratio of the engine's output work to the
input heat energy, and is typically expressed as a percentage. The higher the
thermal efficiency, the more efficient the engine is at converting fuel into work.
QUESTION SIX (25 Marks)
a) Explain why ethanol is blended with Petrol. [5 Marks]
Oxygenate: Ethanol is a type of alcohol that contains oxygen, which makes it an
effective oxygenate additive for petrol. Adding ethanol to petrol increases its
oxygen content, which improves combustion and reduces harmful emissions. This
makes the fuel burn cleaner and more efficiently, resulting in better fuel economy.
Renewable Energy Source: Ethanol is derived from renewable sources such as
corn, sugarcane, and other plants. It is a biofuel that can be produced domestically,
which helps reduce dependence on foreign oil.
Lower Greenhouse Gas Emissions: Ethanol has lower greenhouse gas emissions
compared to petrol. Ethanol production emits fewer greenhouse gases than the
production of petrol. When blended with petrol, it can help reduce overall
greenhouse gas emissions from vehicles.
Octane Boost: Ethanol has a higher octane rating than petrol, which means it has a
higher resistance to knocking or detonation. When blended with petrol, ethanol can
increase the octane rating of the fuel, which improves engine performance and
reduces engine wear and tear.
Cost-effective: Ethanol is generally less expensive than petrol.
b) A former CBU student is producing bio-fuel and blending it with Petrol. The
fuel has some impurities from the fermentation process but he has determined that
a blend of the bio-fuel with petrol burns well in most ICEs. An analysis of a
canister of one mole of the fuel revealed the following constituents; a. C2H6 - 45%
b. C8H18 - 33%
c. Unburnable Ash - 10%
d. O2 - 12%
Calculate;
1) The stoichiometric A/F ratio [8] Marks
2) with 74 % excess air (Where X = fifth and sixth digits of your student number).
Find the A/F ratio [7] Marks
3) Find the Lamda ration ( )? [5] Marks
Weight of Carbon atom = 12, Weight of Hydrogen atom = 1, Weight of Oxygen
atom = 16, Weight of Nitrogen atom = 14. You may assume that air contains 79%
Nitrogen and 21% Oxygen, Assume the weight of 1 mole of unburnable Ash is
40% molar weight of C2H6
1) The stoichiometric A/F ratio [8] Marks
The stoichiometric A/F ratio:
To calculate the stoichiometric air-to-fuel ratio (A/F), we need to determine the
balanced chemical equation for the combustion of the bio-fuel. Assuming that the
bio-fuel is composed of only C2H6 and C8H18, the balanced chemical equation
for the combustion of one mole of the bio-fuel is:
C2H6 + 7(O2 + 3.76N2) → 2CO2 + 3H2O + 26.32N2
C8H18 + 25(O2 + 3.76N2) → 8CO2 + 9H2O + 94N2
To calculate the stoichiometric A/F ratio, we need to determine the amount of air
required for complete combustion of one mole of the bio-fuel. Using the balanced
chemical equation, we find that one mole of C2H6 requires 7 moles of O2 and one
mole of C8H18 requires 25 moles of O2. Therefore, the total amount of O2
required for complete combustion of one mole of the bio-fuel is:
7 mol O2/mol C2H6 × 0.45 mol C2H6 + 25 mol O2/mol C8H18 × 0.33 mol
C8H18 + 0.12 mol O2 = 6.12 mol O2
The total amount of air required is then:
6.12 mol O2 × (1 mol air/0.21 mol O2) = 29.14 mol air
Therefore, the stoichiometric A/F ratio is:
29.14 mol air/ (0.45 mol C2H6 + 0.33 mol C8H18) = 10.0
So, the stoichiometric A/F ratio is 10.0.
2) with x % excess air (Where X = fifth and sixth digits of your student
number). Find the A/F ratio [7] Marks
19137457
The A/F ratio with 74% excess air:
To calculate the A/F ratio with 74% excess air, we need to determine the amount
of air required for 74% excess air over the stoichiometric amount. The
stoichiometric amount of air required is 29.14 mol, so the amount of excess air
required is:
0.74 × 29.14 mol = 21.56 mol
The total amount of air required is then:
29.14 mol + 21.56 mol = 50.70 mol
Therefore, the A/F ratio with 74% excess air is:
50.70 mol air/ (0.45 mol C2H6 + 0.33 mol C8H18) = 14.1
So, the A/F ratio with 74% excess air is 14.1.
3) Find the Lamda ration ( )? [5] Marks
The lambda ratio (λ):
The lambda ratio is defined as the actual A/F ratio divided by the stoichiometric
A/F ratio. Therefore, the lambda ratio can be calculated as follows:
λ = actual A/F ratio/stoichiometric A/F ratio
For the actual A/F ratio, we will use the A/F ratio with 74% excess air that we
calculated in part (2), which is 14.1. For the stoichiometric A/F ratio, we calculated
in part (1) which is 10.0.
λ = 14.1/10.0 = 1.41
So, the lambda ratio is 1.41.
QUESTION SEVEN (18 Marks)
a) Given the following data on a Spark Ignition Engine, i) Number of Cylinders =
10
ii) Engine Type = 2 Stroke
iii) Indicated Mean Effective Pressure =459 (Last 3 SIN digits) KPa
iv) Stroke = 228mm
v) Bore = 210mm
vi) Fuel Mass Flow Rate = 22.85g/s
vii) Engine Speed = 6500 revolutions per minute.
viii) Engine Friction Power = 12.80 kW
ix) Pumping losses = 9 kW
Calculate the following;
a) Mean Piston Velocity- (m/s) [2Marks]
b) Swept Volume (Vd) (m3 ) [2 Marks]
c) Indicated Power - (Watts) [4 Marks]
d) Brake Power (Watts) [4 Marks]
e) Mechanical Efficiency % [2 Marks]
f) Brake Mean Effective Pressure- BMEP (Pa.) [2 Marks]
g) Brake Specific Fuel Consumption(bsfc) g/kW.h [2 marks ]
a) Mean Piston Velocity:
The mean piston velocity can be calculated using the following formula:
V = 2 * S * N / 60
Where
S is the stroke length in meters,
N is the engine speed in revolutions per minute.
Using the given data:
S = 228mm = 0.228m
N = 6500 rpm
V = 2 * 0.228 * 6500 / 60 = 49.4 m/s
Therefore, the mean piston velocity is 49.4 m/s.
b) Swept Volume:
The swept volume can be calculated using the following formula:
Vd = (π / 4) * B^2 * S
Where
B is the bore diameter in meters
S is the stroke length in meters
Using the given data:
B = 210mm = 0.21m
S = 228mm = 0.228m
Vd = (π / 4) * 0.21^2 * 0.228 = 0.007897m3
Therefore, the swept volume is 7.897x10^-3 m3
c) Indicated Power:
The indicated power can be calculated using the following formula:
P_ind = P_imep * Vd * number of cylinders*N / 60
Where P_imep is the indicated mean effective pressure in Pa, Vd is the swept
volume in m^3, and N is the engine speed in revolutions per minute.
Using the given data:
P_imep = Last 3 SIN digits =19137 459 kPa
Vd = 7.897x10^-3 m^3
N = 6500 rpm
n = number of cylinders
P_ind = 459000 * 7.897 x10^-3 * 6500*10 / 60 = 3926783.25W
Therefore, the indicated power is 3.927 MW.
d) Brake Power:
The brake power can be calculated using the following formula:
P_b = P_i - P_f - P_p
Brake power= indicated power –friction power –pumping losses
Where
P_i is the indicated power,
P_f is the friction power
P_p is the pumping losses.
Using the given data:
P_i = 3.927 MW
P_f = 12.80 kW
P_p = 9 kW
P_b = (3.927 * 10^6) – (12.80 * 10^3 - 9 * 10^3) = 3923200W
Therefore, the brake power is 3.923 MW.
e) Mechanical Efficiency:
The mechanical efficiency can be calculated using the following formula:
η_mech = P_b / P_i
Using the given data:
P_b =3.923 MW
P_i = 3.927 MW
η_mech = 3.923 * 10^6 / 3.927 * 10^6 = 0.9989 or 99.89%
Therefore, the mechanical efficiency is 99.89%.
f) Brake Mean Effective Pressure:
The brake mean effective pressure can be calculated using the following formula:
BMEP = P_b / (Vd * N / 2)
Using the given data:
P_b = 3.923 MW
Vd 7.897x10^-3 m3
N = 6500 rpm
BMEP = 3.923 * 10^6 / (7.897x10^-3 * 6500 / 2) = 152852.59 Pa
Therefore, the Brake Mean Effective Pressure is 152.852 kPa.
g) Brake Specific Fuel Consumption:
The brake specific fuel consumption can be calculated using the
following formula:
Brake Specific Fuel Consumption (BSFC) = Fuel Mass Flow Rate / BP
where Fuel Mass Flow Rate is in g/s and BP is in W.
Given, Fuel Mass Flow Rate = 22.85 g/s and BP = 3.923 MW
=22.85/3.923 x 10^6
=0.000005824624 = 5.824
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