Lab IV: Internal Combustion
Engine
14:650:431:03
Max Tenorio
Objective
The four-stroke internal combustion engine undergoes the Otto
cycle, which is the most commonly used thermodynamic cycle in
combustion engines.
The objective of this lab is to run an engine test, and to learn how to
use engine performance evaluation parameters and procedures. We
can use these parameters and procedures to gain insight on the
general characteristics of the cycle as well as the performance of a
particular engine. This can in turn help us increase the efficiency of
an engine and determine at what throttle levels and speed will be
most efficient for a specific purpose.
Setup
5.5 hp Honda GX 160 engine connected to a dynamometer
Piston engine: geometry and
components
Measurements
•
•
•
•
•
•
•
•
Engine RPM
Dynamometer Oil Pressure
Dynamometer Oil Flow Rate
Dynamometer Oil Temperature
Engine Cylinder Head Temperature
Fuel difference
Throttle level
Engine Run Time
Recorded Data
Some Conversion Factors
revolution s 1 min
RPM revolution s


min
60 sec
60
sec
lbs 0.45359237
kg 39.37007872 in 2


 9.80667m 2  PSI  6894.77Pa
2
2
s
1lbs
in
1m
3
Gal 3.78541178
L 0.001m 3 1 min Gal
5 m




6.309019633 10
min
1Gal
1L
60sec min
s
1 Watt = 0.00134102209 horsepower
Nm 
0.22480894 3lbf
1 ft

 Nm  0.737562149 lbf  ft
1N
0.3048 m
Engine Power/Brake Power
Power is a measure of torque per unit time. This unit can be
converted into either torque or force output by the engine
and through the crankshaft.

Power  p  Q
p  Q
BrakePower
p
Power is the calculated power from the engine.
Brake Power is the engine power taking the
 p  0.8
pump efficiency into account.
From the graph, we see that the highest amount of power
comes from a combination of full throttle at its highest
recorded load output. It is also worth noting that at low
throttle, the power peaks at a lower RPM and has a lower
magnitude at higher RPMs. This shows that with a low
throttle setting, increasing the load is not a good idea.
Torque
Engine Torque is calculated by taking the Power factor and
using the revolutions per minute to remove the time factor.
Pb
 
2N
Pb  BrakePower
N  rev / s
From the graph, we can see that the highest torque is
obtained from a full full throttle with a low load setting.
Fuel Consumption
Fuel Consumption is the rate at which fuel is drawn into the
engine. Through conversions in the recorded fuel used in a
span of 30 seconds obtains this data for us. The unit is output
in lbs of gasoline per hour.
From the graph, we see that the overall highest level of fuel
consumption happened naturally at full throttle and full load.
However, at 2600 RPM, three-quarter load actually had a
higher rate of fuel consumption than full throttle. Because of
this we cannot draw a clear conclusion as to what yields the
highest level of fuel consumption as more data is needed.
Brake Specific Fuel Consumption
Brake Specific Fuel Consumption is essentially the
measure of fuel used per unit of torque output by the
engine. It can be seen as a measure of efficiency.

BSFC 
mf
Pb

m f  FuelMassFlowRate
Pb  BrakePower
Assuming 1 Nm of torque at constant load, from the
graph we see that low throttle and high load tends to
use up the most amount of fuel, whereas full throttle
and high load uses up considerably less fuel.
Essentially, one quarter throttle gives you less torque
per gallon of fuel than full throttle at a constant load.
Cycle Efficiency
Cycle efficiency is a measure of how efficient the engine is at
turning fuel into useable energy.
f 
Pb

m f H HV
Pb  BrakePower

m f  FuelMassFlowRate
H HV  44 106 J
Kg
HHV is called the Higher Heating Value of the Fuel, given by
the lecture manual.
From the graph, Cycle efficiency is highest with full throttle
and low load. Efficiency for all throttle values tend to decline
as load is increased.
Brake Mean Effective Pressure
Brake Mean Effective Pressure is the mean effective
pressure inside the cylinder at the moment of combustion
that creates power.
BMEP 
Pb n p
Vd N
Pb  BrakePower
np 
revolutions/power stroke
Vd 
displacement volume
N
revolutions/unit time
From the graph we see that the pressure inside the
cylinder is the highest at full throttle and low load. This
graph has almost the same figure as the Torque graph.
This is most likely because the Mean Effective Pressure
and Torque are almost directly related. It also makes
sense because the highest torque corresponds to the
highest effective pressure.
Otto Cycle
As defined in the lecture manual, thermal efficiency of the Otto cycle is:
where
is defined as the Compression ratio
and
is the ratio of specific heat
capacities for constant pressure and constant
volume, respectively
We are using regular gasoline which has an octane rating of about 87. From the Lecture
Manual, this has a compression ratio of 7:1
I was not able to find a ratio of specific heat capacities for regular gasoline, however
substituting a similar gas, propane (C3H8) we use γ = 1.13
This yields a maximum thermal efficiency of ηth= 22.35%
A higher compression ratio would yield a high maximum thermal efficiency. This calculated
thermal efficiency is quite low in comparison to methane, with a Compression Ratio of
about 11:1, octane rating of 107, and γ = 1.32 which yields ηth= 53.57%
More accurate numbers would yield a more accurate thermal efficiency for gasoline.
2900 RPM
We are finally asked to calculate theoretical values for an engine at 2900 RPM.
a)
The Maximum Piston Speed
We can use the equation from the Lecture Manual
Where Sp = Piston Speed
L = stroke; N = rotational speed; θ = crank angle; R = connecting rod length
L=2 , N=2900*2pi, and R= 0.105m, leaving us with θ as our independent variable.
Optimization yields θ = 323° which gives a value of 0.56m/s
b)
The number of times per second that the spark plug fires
2900RPM=48.33 RPS
This is a 2 stroke engine, which fires twice every 4 revolutions
48.33RPS*2 fires / 4 Rev = 24.1667 strikes per second
c)
The approximate tensile stress induced on a connecting rod with a 1-cm2 cross-section
as a result of the deceleration of a 0.3-kg piston during each cycle
Downward Force = 0.3kg*9.8m/s2 = 2.94N
1 cm2 cross-section = 0.0001 m2
Tensile Stress = Downward Force / Cross Sectional Area = 2.94/0.0001=29.4kPa
References
• Lab Manual
• Lecture Manual
• Wikipedia: Heat capacity ratio