NAZARIN B. NORDIN
nazarin@icam.edu.my
What you will learn:
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Simple machines
Mechanical advantage (force ratio)
Movement ratio (velocity ratio)
Machine efficiency (in %)
Gears: gear box ratio, gear efficiency
Directions of rotations
Machines
• Is a device that can change magnitude or line of action, or
both magnitude and line of action of a force.
• A simple machine usually amplifies an input force, call the
effort , to give a larger output force, called the load.
• Examples; pulley systems, screw jacks, gear systems
and lever systems.
Machines
Machines
• A machine is a device that receives energy in
some available form and converts it to a useable
form.
• For example, a person may wish to lift a weight of
2 tonnes; this is not possible unaided but the use
of a jack or a hoist permits the person to achieve
their objective. A lever such as a pry bar is an
example of a simple machine. A relatively small
manual force can be converted into a large force
in order to lift or move an object.
Levers
• Leverage and the use of levers occurs in the
use of tools such as spanners, pry bars, pliers
etc., and in many vehicle mechanisms such as
clutch and brake pedals, throttle linkages and
suspension units.
Mechanical advantage
(force ratio)
• The mechanical advantage of a machine is the
ratio of the load to the effort:
mechanical advantage, MA = load
effort
• Lifting machines such as jacks and cranes have a
large mechanical advantage so that relatively small
manual forces can be used to raise heavy weights.
Velocity ratio (movement ratio)
• The velocity ratio, VR, of a machine, or
movement ratio, MR, is the ratio of the
distance moved by the effort to the distance
moved by the load:
velocity ratio, VR = distance moved by effort
distance moved by load
or
movement ratio, MR = distance moved by effort
distance moved by load
Efficiency of a machine
• The efficiency of a machine = (energy
output/energy input) in the same time and
this also
Efficiency of machine = MA ×100%
VR
Example problem
• In the trolley jack example shown before an
effort of 250 newtons is lifting a load of 2
tonnes. In lifting the load through a distance
of 15 cm the operator performs 40 pumping
strokes of the handle each of which is 50 cm
long. Calculate the mechanical advantage,
velocity ratio and efficiency of the jack.
• MA = 78.5
• VR =133.3
• The efficiency = 59%
Gears
• Leverage and gears
The gear ratio = revolutions of input gear
revolutions of output gear.
• The gear ratio in this case is 4:1
Gearbox
• Gear trains
• Ratio of gear set
= driven × driven
driver driver
= 35×28 = 1.225
25×32
Problem 1
A simple machine raises a load of 160 kg through
a distance of 1.6 m. The effort applied to the
machine is 200 N and moves through a distance
of 16 m. Taking g as 9.8 m/s2, determine
i) the mechanical advantage (force ratio),
ii) the velocity ratio (movement ratio) and
iii) the efficiency
of the machine.
Solution 1
• Mechanical advantage (force ratio) = Load/effort
= 160 kg/200 N
= (160 x 9.8)N/200 N
= 7.84
• Velocity ratio (movement ratio) = distance moved by
effort/distance moved by the load
= 16 m/1.6 m
= 10
• Efficiency = (force ratio/movement ratio) x 100%
= (7.84/10) x 100
= 78.4%
Problem 2
For the simple machine of Problem 1, determine:
(a) the distance moved by the effort to move the
load through a distance of 0.9 m,
(b) the effort which would be required to raise a
load of 200 kg, assuming the same efficiency,
(c) the efficiency if, due to lubrication, the effort
to raise the 160 kg load is reduced to 180N.
Solution 2
• Distance moved by the effort
= 10 x distance moved by the load
= 10 x 0.9
mechanical advantage = 9 m
• Since the mechanical advantage is 7.84 ,
effort = load/7.84
= (200 x 9.8) / 7.84
= 250 N
Solution 2
• The new mechanical advantage (force ratio)
load/effort = (160 x 9.8)/180
= 8.711
• The new efficiency after lubrication
= (8.711/10) x 100
= 87.11%
Problem 3
•A driver gear on a shaft of a motor has 35 teeth
and meshes with a follower having 98 teeth. If
the speed of the motor is 1400 revolutions per
minute, find the speed of rotation of the
follower.
Solution 3
Speed of driver
speed of follower
=
teeth on follower
teeth on driver
1400/speed of follower = 98/35
Speed of follower = (1400 x 35)/98
= 500 rev/min
Problem 4
•A compound gear train similar to that shown in
last figure consists of a driver gear A, having 40
teeth, engaging with gear B, having 160 teeth.
Attached to the same shaft as B, gear C has 48
teeth and meshes with gear D on the output
shaft, having 96 teeth. Determine;
(a) the movement ratio of this gear system
(b) the efficiency ratio when the force ratio is 6
Solution 4
(a) The speed of D = speed of A x (TA/TB)x (TC/TD )
movement ratio =(NA/ND )=(TB/TA )x (TD/TC)
= (160/40) x (96/48)
=8
(b) The efficiency = (force ratio/movement ratio) x 100%
= (6/8 ) x 100
= 75%
NAZARIN B. NORDIN
nazarin@icam.edu.my
What you will learn:
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•
•
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Torque and power
Indicated power and brake power
Volumetric efficiency
Thermal efficiency
Mechanical efficiency
Torque & power
Indicated power
• Indicated power (ip) is the power that is
developed inside the engine cylinders. It is
determined by measuring the pressures inside
the cylinders while the engine is on test, on a
dynamometer.
• Because the pressure varies greatly
throughout one cycle of operation of the
engine, the pressure that is used to calculate
indicated power is the mean effective
pressure.
Indicated mean effective pressure
(Imep)
• Because the pressure varies greatly
throughout one cycle of operation of the
engine, the pressure that is used to calculate
indicated power is the mean effective
pressure.
Imep
Example 6.1
• An indicator diagram taken from a singlecylinder 4-stroke engine has an effective area
of 600 mm2. If the base length of the indicator
diagram is 60 mm and the constant is 80
kPa/mm calculate the indicated mean
effective pressure.
Indicated mean effective pressure
(i.m.e.p)
Solution
• Indicated mean effective pressure or
(imep)= (effective area of indicator
diagram/base length of the diagram)×constant
imep = (600/60)×80 kPa/mm
imep = 800 kPa = 8 bar
Calculation of indicated power
• Indicated power is calculated from the
formula
• Indicated power (ip) = (P ×l×a×N)/1000
• Unit as kW
where P = mean effective pressure in N/m2, l =
length of engine stroke in m, a = cross sectional
area of cylinder bore in m2, N = number of
working (power) strokes per second.
Calculation indicated power
Example 6.2
• In a test, a certain single-cylinder 4-stroke
engine develops a mean effective pressure of
5 bar at a speed of 3000 rev/min. The length
of the engine stroke is 0.12 m and the crosssectional area of the cylinder bore is 0.008 m2.
Calculate the indicated power of the engine in
kW.
Calculation of indicated power
Solution
• The engine is single-cylinder 4-stroke, so there is
one working stroke for every two revolutions.
N, the number of working strokes per second
N = (3000÷60)/2 = 25;
P = 5 bar= 500 kPa=500,000 Pa = 500,000 N/m2;
l = 0.12 m;
a = 0.008 m2.
• Substituting these values in the formula gives
ip = [(500 000×0.12×0.008×25)/1000] kW
= 12 kW
Number of working (power) strokes
• As mentioned, each cylinder of a 4-stroke engine
produces one power stroke for every two revolutions
of the crankshaft. In a multi-cylinder engine, each
cylinder will produce one power stroke in every two
revolutions of the crankshaft. A formula to
determine the number of power strokes per minute
for a multi-cylinder, 4-stroke engine is:
number of power strokes per minute
= (number of cylinders/2)×rev/min
Indicated power
Example 6.3
• A 4-cylinder, 4-stroke engine develops an
indicated mean effective pressure of 8 bar at
2800 rev/min. The cross-sectional area of the
cylinder bore is 0.01 m2 and the length of the
stroke is 120 mm. Calculate the indicated
power of the engine in kW.
• Solution
Brake power
• The engine power that actually reaches the
output shaft or flywheel of an engine is known
as the brake power. It is the power that is
measured by a dynamometer and a
dynamometer is also known as a brake, hence
the term, brake power.
• The simplest form of dynamometer is shown
in the Figure 14.1 on next slide:
Simple dynamometer
Simple dynamometer
Work done per revolution of the flywheel
= force × distance
= (W−S)×2R joules
(W−S×R also = T the torque that the engine is
exerting. By substituting T for W−S×R the)
Work done per revolution becomes
= T×2π joules
Brake power
Brake power = work done per second
= 2π × Torque× number of revolutions per second
=2π x T x n ie. 2π.T.N
This is normally written as brake power
(bp = 2πTN where N = number of revolutions per second.)
When the torque is in newton metres (N m), the formula
bp = 2πTN gives the power in watts. As engine power is
given in thousands of watts, i.e. kilowatts, the formula for
brake power normally appears as:
bp = 2π.T.N/1000 kW
Brake power
Example 6.4:
Volumetric efficiency
Volumetric efficiency
Example 6.5
• A 4-cylinder 4-stroke petrol engine with a bore
diameter of 100 mm and a stroke of 110mm
has a volumetric efficiency of 74% at an
engine speed of 4000 rev/min. Determine the
actual volume of air at STP that flows into the
engine in 1 minute.
• Solution
Thermal efficiency
• The thermal efficiency of an engine is a term
that is used to express the effectiveness of an
engine’s ability to convert heat energy into
useful work. Thermal efficiency is the ratio of
energy output of the engine to energy
supplied to the engine in the fuel.
Thermal efficiency
Example 6.6
• During a 10-minute dynamometer test on a petrol
engine, the engine develops a brake power of 45 kW
and uses 3 kg of petrol. The petrol has a calorific value
of 43 MJ/kg. Calculate the brake thermal efficiency.
Mechanical efficiency
The mechanical efficiency of an
engine is defined as
brake power, b.p.
indicated power, i.p. X 100 %
An engine develops a brake
power of 120 kW at a speed of
3000 revs/min. At this speed,
the indicated power is 140 kW.
Calculate the mechanical
efficiency of the engine at this
speed.