MOTORS AND ENGINES

(Review)

Arsenio N. Resurreccion

 Electro-mechanical devices used to convert one form of energy to another

form.

 Motors convert electrical energy into mechanical energy. Electrical

power is manifested by voltage and current as:

P = VICosθ where P = power

V = voltage
I = current
Cos θ = phase factor

 Engines convert the heat energy of fuel into mechanical power by

burning it and allowing the burning gas to expand to produce mechanical
power.

 Mechanical power is manifested by the rotation and torque of the output

shaft as:

P = 2ΠTN/C where P = power

N = rpm
T = torque
C = constant

 Electric motors are more efficient in converting input energy into

mechanical energy compared to gasoline and diesel engines.

Table 1. Energy conversion efficiencies of major farm power units.

POWER UNIT ENERGY CONVERSION EFFICIENCY

%
Electric motor 50 - 90
Gasoline engine 25 - 30
Diesel engine 30 – 36

 Despite the low conversion efficiency, engines are still considered one of
the few inventions that have great influence on human lives.
 Electric motors are used as prime movers for stationary and indoor
equipment such as dryers, rice mills, conveyors and size reduction
equipment.

 Engines are used as prime movers for mobile and outdoor machinery
such as tractors, water pumps, sprayers, harvesters, threshers and shellers.

 Engines are also used in fishing boats to increase the mobility of fishermen
and the scope of their fishing grounds.

 In general, engines are used in applications where electric motors cannot be

used.

ELECTRIC MOTORS

 Classification of electric motors based on electrical service required:

1. Single-phase alternating current (< 5 hp) – more common

2. Three-phase alternating current (> 5 hp)
 Motor Ratings and Selection:

1. Torque requirement:

Select a motor whose torque output is always greater than the torque
required by the load.

Figure 1. Speed-vs-torque curve for general-purpose motor

2. Starting current requirement:

Determines the size and capacities of the electrical installations needed to

operate the motor (transformer capacity, wire size, breaker capacity, switch
rating).

Table 2. Motor code

CODE LETTER LOCKED ROTOR, KVA/HP
F 5.1 - 5.6
G 5.6 - 6.3
H 6.3 - 7.1
J 7.1 - 8.0
K 8.1 - 9.0
L 9.0 - 10.0
 For example, calculate the maximum locked-rotor current (starting
current) for a ½-hp, 220-volt motor with an H motor code.

7,100 VA/Hp X 0.5 Hp

Max. starting current = A = ------------------------------ = 16.14 amp.
220 V

Table 3. Starting and running power requirements for commonly used 60

cycle, single-phase electric motors

Motor Approx. Amperes Watts Required to Start Watts

Hp (Full Load) Required
Rating 120V 240V Split-phase Capacitor to run
Start
½ 9.8 4.9 2,300 575
¾ 13.8 6.9 3,345 835
1 8 4,000 1,000
1½ 10 6,000 1,500
2 12 8,000 2,000
3 17 12,000 3,000
5 28 18,000 4,500
7½ 40 15,100 28,000 7,000
10 50 81,900 36,000 9,000
Table 4. Related sizes for wires and overcurrent devices

Wire Size Wire Rating Overcurrent Device

AWG No. Amp. Amp. Rating
14 15 15
12 20 20
10 30 30
8 40 40
6 55 50 or 60
4 70 70
2 95 100
 3. Temperature rating:

Motors are built with four insulation classes that can withstand only
certain amounts of operating temperature.

Table 4. Insulation class of motors

INSULATION CLASS MAX. HOT SPOT CONT. TEMPT.
ºC ºF
A 105 221
B 130 266
F 155 311
H 180 356

For convenience, nameplate data often gives the maximum ambient

temperature that will keep the motor temperature within limits. Normal
maximum ambient temperature for motor operation is 40 ºC

4. Duty rating:

Motors are classified according to:

a. Continuous duty rating – motor is operated for more than 60

minutes without rest.

b. Intermittent duty rating – is operated only for a few minutes

at a time with a rest or cooling period between operations.

5. Enclosure:

Motors are classified according to the type of enclosure used to protect

the motor from the environment:

a. Open-type – allows easy movement of air through the inside

of the motor by providing slots or holes in the end shields or
frame where the air can pass through. Efficient.

b. Closed-type – does not allow entry of air inside the motor.

Fan blows air on the outside surface of the frame to dissipate
heat. Fins are constructed on the frame to increase the
surface area for heat dissipation. Inefficient
 6. Service factor – indicates the maximum load that can be safely
carried by the motor under continuous operation.

Fractional horsepower motor: S.F. = 1.25 to 1.4

Integral horsepower motor: S.F = 1.15

The service factor is multiplied by the horsepower rating to give the

maximum load the motor can sustain.

 Motor nameplate:

Motor nameplate carries a good deal of essential information about the

motor. Refer to it when you specify, buy, install, replace, operate and repair
motors.

Figure 2. Typical electric motor nameplate

 Sample problem 1:

Determine the minimum capacity (KW) of a standby generator if it will

be used to operate the following compost processing equipment: Belt
conveyor – 5Kw, Hammer mill – 8 KW, Siever – 3 KW, Electronic
weighing scale – 0.2 KW, lights – 1 KW, and exhaust fans – 0.6 KW.
Specify the sequence these equipment will be turned on.

Assume that the starting power is 3X the running power for each of the
equipment with electric motors and all the equipment have to be operated.
Given:

Equipment Running Power, KW Starting Power, KW

Belt conveyor 5 15
Hammer mill 8 24
Siever 3 9
Weighing scale 0.2 0.2
Lights 1 1
Exhaust fans 0.6 1.8

Requirement:

Minimum capacity (KW) of the standby generator and the sequence of

turning on the equipment.

Solution:

Arrange the equipment in the order of highest to lowest power

requirement.

Equipment Running Power, Starting Power, Available

KW Kw Power, KW
Hammer mill 8 24 24
Conveyor belt 5 15 16
Siever 3 9 11
Exhaust fans 0.6 1.8 8
Lights 1 1 7.4
Weighing scale 0.2 0.2 6.4
6.2

Therefore, a 24 KW standby generator is the minimum capacity and the

equipment are to be turned on according to the order of the list.
 Sample problem 2:

Suppose the conveyor belt has a rating of 6 KW, what would be the
minimum capacity of the standby generator?

Running Starting Available Power, KW

Equipment
Power, KW Power, KW Trial 1 Trial 2
Hammer mill 8 24 24 26
Conveyor belt 6 18 16 18
Siever 3 9 - 12
Exhaust fans 0.6 1.8 - 9
Lights 1 1 - 8.4
Weighing scale 0.2 0.2 - 7.4
7.2

Therefore, a 26 KW standby generator is the minimum is capacity and the

equipment are to be turned according to the order of the list.

ENGINES

 Construction and parts of Engines:

  The engine is composed of parts that convert reciprocating linear
motion into rotational motion.

Figure 1. Parts of an engine

 Cycle of operation of an engine:

  Cycle – series of events occurring one after the other in a definite
sequence and repeats itself after the last event has occurred.

Figure 2. Engine cycle of operation

 The events which make up the cycle of an engine are:

1. Intake – air-fuel mixture enters the combustion chamber.

2. Compression – air-fuel mixture is compressed inside the
combustion chamber.
3. Power – spark is produced and the air-fuel mixture is ignited
resulting to a very rapid expansion of the burning mixture.
This constitutes the power of the engine.
4. Exhaust – burned gases are removed from the combustion
chamber and the cycle is repeated.

Figure 3. Events in the cycle of an internal combustion engine

 Classification of internal combustion engines:
 1. Based on the number of strokes to complete the cycle:

 TDC (top dead center) – highest position attained by the

piston when it goes up.

 BDC (bottom dead center) – lowest position attained by the

piston when it goes down.

 Stroke – movement of the piston from TDC to BDC or vice

versa.

a. Four-stroke-cycle engine – engine requiring four

strokes of the piston to complete the cycle.

b. Two-stroke-cycle engine – engine requiring only two

strokes of the piston to complete the cycle.

1) Downward stroke – At the beginning of the stroke,

power event is being accomplished. Towards the end of the stroke, the piston
uncovers a port at the side of the cylinder to start the exhaust event. Still further
down, the piston uncovers the inlet port, to start the intake event.

2) Upward stroke – as the piston goes up, both the inlet

and exhaust ports will be covered to start the compression event.

2. Based on the manner of igniting the fuel:

a. Spark-ignition engine – engine uses a spark produced in the

spark plug to ignite the compressed air-fuel mixture.

 Occurrence of spark is timed before or at TDC

 Fuel used is gasoline and metering is accomplished by
a carburetor.

b. Compression-ignition engine –engine uses heat of

compressed air to ignite the fuel.

 During intake, only air is taken in.

 Air is compressed to a very high degree

 Before end of compression, fuel is injected

  Fuel used is diesel and metering is done by the fuel
injection system.

 Based on the classification, four different types of engines may be identified:

1. Four-stroke-cycle-spark-ignition engine (Fig. 4)

2. Four-stroke-cycle-compression-ignition engine (Fig. 5)

3. Two-stroke-cycle-spark-ignition engine (Fig. 6)

4. Two-stroke-cycle-compression-ignition engine (Fig. 7)

Figure 4. Operation of 4-stroke spark ignition engine

Figure 5. Operation of 4-stroke compression ignition engine

Figure 6. Operation of 2-stroke spark ignition engine

 Figure 7. Operation of 2-stroke compression ignition engine

 Engine specifications:

1. Engine type (number of strokes, manner of ignition, cylinder

number and arrangement).

2. Engine dimensions (bore, stroke, displacement, weight, length,

width, and height)

3. Output (maximum power, rated power, compression ratio,

maximum torque and rpm)

4. Capacities (fuel, lubricating oil and cooling water)

5. Auxiliary systems (valve, ignition, fuel, air cleaning, cooling,

lubrication, starting and governing)

Example of engine specifications

 CHARACTERISTIC SPECIFICATION
Engine type : 4-stroke spark-ignition vertical single cylinder
Maximum output : 6.8 hp at 4,000 rpm
Rated output : 5.0 hp at 3,600 rpm
Bore x stroke, displacement : 72 x 59 mm, 240 cc
Compression ratio : 6:1
Maximum torque : 1.25 kg-m at 3,000 rpm
Ignition system : flywheel magneto
Starting system : recoil starter
Air cleaner : semi-dry type (polyurethane)
Lubrication system : splash type
Lubrication capacity : 0.82 li
Fuel tank capacity : 6.9 li
Dry weight : 28 kg
Dimensions : 410 mm x 370 mm x 490 mm

 Basic engine terminology:

1. Bore (D) – diameter of the cylinder, inches or millimeters.

2. Stroke (L) – length of travel of the piston from TDC to BDC,

inches or millimeters
3. Piston displacement (PD) – volume displaced by the piston as it
moves from BDC to TDC, cubic inches, cubic millimeters or cubic
centimeters. For a multi-cylinder engine, PD is multiplied by the
number of cylinders.

4. Clearance volume (CV) – volume of the cylinder when the piston

is at TDC.

5. Total Volume (TV) – volume of the cylinder when the piston is at

BDC. It is equal to PD + CV.

6. Compression ratio (CR) – ratio of total volume over clearance

volume.

 Sample problem 1:
 Determine the compression ratio of the engine with the following
specifications: Total volume = 70 cc, Bore = 4 cm and Stroke = 5 cm.

Given: TV = 70 cc, D = 4 cm, L = 5 cm

Required: Find CR

Solution:
2
ΠD 2 L Π ( 4 cm ) 5 cm
PD= = =62. 83 cc
4 4

CV =TV −PD=70 cc−62. 83 cc=7 . 17 cc

TV 70 cc 9 . 76
CR= = = =9 . 76 :1
CV 7 . 17 cc 1

 Sample problem 2:

Determine the bore of the engine if its clearance volume is 7.18 cc,
compression ratio is 8:1 and stroke is 4 cm.

Given: CV = 7.18 cc, CR = 8:1. L = 4 cm

Required: Bore = D

Solution:

TV
CR=
CV TV =CRXCV =8 X 7 .18=57 . 44 cc

TV =PD +CV PD =TV −CV =57 . 44 cc−7 .18 cc =50. 26 cc

ΠD 2 L 4 PD 4 X 50.26 cc
PD=
4 cm
4
D=
√ ΠL = √ 3.14 X 4 cm = √ 16 =

 Sample problem 3:
 Determine the size of the engine given the following engine
specifications: Bore = 4. cm, stroke = 5 cm, and compression ratio = 8:1.

Given: D = 4 cm, L = 5 cm, CR = 8

Required: Find TV

Solution:

π D2
PD= ×L
4

TV
CR= →TV =CR× CV eqn. 1
CV

CV =TV −PD eqn. 2

Substituting eqn. 2 in eqn. 1:

TV =CR × ( TV −PD ) =CR × TV −CR × PD

CR ×TV −TV =CR × PD

TV ( CR−1 )=CR × PD

42
8 ×(π ×5)
CR× PD 4
TV = = =71.805 cc
CR−1 8−1

Reading Assignment:

1. PAES 116:2001 Agricultural Machinery – Small Engine –

Specifications
2. PAES 129:2002 Agricultural Machinery – Electric Motor –
Specifications