Computerized Gear System
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Transcript of Computerized Gear System
SUBJECT : ENGINEERING LABORATORY IV (BDA 2721)
CODE : BDA 2711
TOPIC : COMPUTERIZED GEAR SYSTEM
OBJECTIVES :
i. Describe the different type of gear system and some of their application
ii. Calculate gear ratios, angular velocity, input and output torque.
iii. Calculate the efficiency of the gears.
LEARNING OUTCOMES :
i. Understand the concept of gear system, types of gears and its related function and
application.
ii. Implement and analyze the required data collectively within member of group.
iii. Produce good technical report according to the required standard
THEORY :
Gears are used to transmit motion, and therefore power, between one shaft and
another shaft. The function of a gear box is to transmit rotational motion from a
driving prime mover to a driven machine. Gear also know as a toothed wheel
designed to transmit torque to another gear or toothed component. Different size gears
are often used in pairs, allowing the torque of the driving gear to produce a large
torque in the driven gear at lower speed, or a smaller torque at higher speed. The large
gear known as wheel and the smaller gears as a pinion. Consider a simple schematic
of a gear box with an input and output shaft as shown in Figure 1.
Figure 1
N1
N2
And another theory of gear is a component within a transmission device that transmits
rotational torque by applying a force to the teeth of another gear or device. A gear is different
from a pulley in that a gear is a round wheel that has linkages ("teeth" or "cogs") that mesh
with other gear teeth, allowing force to be fully transferred without slippage. Depending on
their construction and arrangement, geared devices can transmit forces at different speeds,
torques, or in a different direction, from the power source.
The gear's most important feature is that gears of unequal sizes (diameters) can be combined
to produce a mechanical advantage, so that the rotational speed and torque of the second gear
are different from those of the first. In the context of a particular machine, the term "gear"
also refers to one particular arrangement of gears among other arrangements (such as "first
gear"). Such arrangements are often given as a ratio, using the number of teeth or gear
diameter as units.
Gear Types
internal gears Worms Gear
Spur gears
Bevel gears
Gear ratio, G.R = Input Speed / Output Speed, = N1 / N2
Gear ratio, G.R = (product of driven teeth)/(product of driving teeth)
Gear ratio, G.R = Input Speed / Output Speed, = ω1 / ω2
The power transmitted by a torque,T (Nm) applied to the shaft rotating at N (rev/min) is given by :-
Power, P = TωP = [ 2πNT] / 60
In the ideal gearbox, the input and output power are the same so,[2πN2T2] / 60 = [2πN1T1] / 60
In a real gear box, power is lost through friction and the power output is smaller than the power input.
The efficiency is defined as;η = Power Output / Power Inputη = [2πN2T2 x 60] / [2πN1T1 x 60]η = T1ω1 / T1ω1
Note:- N – speed in rev/min ω – angular velocity (rad/s)
Table 1 : Technical Specification for EquipmentType
of Gear
Class of Gear
DescriptionGear
1 2 3 4
4 - Stage
Same Size
a. No of teeth
18 78 78 78
b. Pitch diameter
36 156 156 156
c. Outside diameter
40 160 160 160
4 - Stage
Different Size
a. No of teeth
18 38 78 118
b. Pitch diameter
36 76 156 232
c. Outside diameter
40 80 160 236
EQUIPMENT :
ITEM NO
Unit of Gear Set 2
Control Panel 1
Computer Set 1
Printer 1
Protective Transparent Cover 1
Equipment
Motor
Potentiometer
Gear Set Speed Set
Motor DisplayMotor Driver Computer
Figure 2 : Connection Diagram
Protective transparent cover Control Panel
Unit of Gear set
Main Table
Table
CPU Figure 3 : Overall System
Protective transparent
Control Panel
Gear Set Main Table
Computer
PROCEDURES :
A. EXPERIMENT FOR 4 STAGE GEAR WITH SAME SIZE GEAR - GEAR SET 1
1) Make sure gear set 1 is in place. If not, install the gear set 1 into the system according
to the following steps:-
i. Remove the transparent protective cover of the system.
ii. Disconnect the sensor cable. Remove the locking bolts of the gear set.
iii. Remove the gear set by lifting it using the handles.
iv. Put the removed gear set lifting on storage table.
v. Take the new gear set and put it on system. Make sure the gear set is
completely in place.
vi. Tighten the locking bolts. Connect the sensor cable.
vii. Put the transparent protective cover of the system. The new set is ready to
operate.
2) Turn on the computer. Turn on the Infix gear software.
3) Turn on the system by pressing ON button. The controller and motor displays will
ON.
4) Slowly increase the speed until it reaches 100rpm. Record the motor speed into the
lab sheet accordingly.
5) Press RECORD on the monitoring page software. Let the system run and stabilize for
an about 10 seconds.
6) Slowly increase the speed until it reaches 200rpm. Record the motor speed into the
lab sheet 1.
7) Let the system run and stabilize for about 10 seconds.
8) Follow the step 5.1.6 to 5.1.7 for the speed adjustment of 300rpm, 400rpm, 500rpm,
600rpm, and 700rpm.
9) Turn OFF the potentiometer slowly.
10) Record all values into your lab sheet.
B. EXPERIMENT FOR 4 STAGE GEAR WITH DIFFERENT SIZE GEAR –
GEAR SET 2
1) 5.2.1. Make sure gear set 2 is in place. If not, install the gear set 2 into the system
according to the steps 5.1.1. (i – vii).
2) 5.2.2. Turn on the computer. Turn on the Infix gear software.
3) 5.2.3. Turn on the system by pressing ON button. The controller and motor displays
will ON.
4) 5.2.4. Slowly increase the speed until it reaches 100rpm. Record the motor speed
into the lab sheet accordingly.
5) 5.2.5. Press RECORD on the monitoring page software. Let the system run and
stabilize for an about 10 seconds.
6) 5.2.6. Slowly increase the speed until it reaches 200rpm. Record the motor speed
into the lab sheet 1 accordingly.
7) 5.2.7. Let the system run stabilize for about 10 seconds.
8) 5.2.8. Follow the step 5.1.6 to 5.1.7 for the speed adjustment of 300rpm, 400rpm,
500rpm, 600rpm, and 700rpm.
9) 5.2.9. Turn OFF the potentiometer slowly.
10) 5.2.10. Record all values into your lab sheet.
OBSERVATIONS :
TABLE 2 : 4 Stage Gear with Same Size Gear – Gear Set
Gear 0
( Motor )Gear 1 Gear 2 Gear 3
No of teeth (t) 18 78 78 78
Pitch Diameter (D) mm 36 156 156 156
Gear Ratio (Calculation) - 0.23 0.23 0.23
Speed (rpm) 100 23 5 2
200 46 11 2
300 70 16 4
400 92 22 5
500 115 27 6
600 138 32 8
700 162 38 8
Gear Ratio (From Data) - 0.23 0.05 0.01
Gear Efficiency - 0.87 0.84 0.87
TABLE 3 : 4 Stage Gear with Same Size Gear – Gear Set 2
Gear 0
( Motor )Gear 1 Gear 2 Gear 3
No of teeth (t) 18 38 78 118
Pitch Diameter (D) mm 36 76 156 232
Gear Ratio (Calculation) - 0.47 0.23 0.15
Speed (rpm) 100 48 12 2
200 95 23 4
300 143 34 5
400 189 42 6
500 237 55 7
600 285 66 11
700 332 78 14
Gear Ratio (From Data) - 0.23 0.05 0.01
Gear Efficiency - 0.87 0.87 0.87
Gear with different size make motor speed (Rpm) increase.
The value is not accurate and very sensitive to define.
Speed motor with same size or different size with Gear 0 to 3, the value of speed
motor from result will decrease until the lowest value in the data from Gear 0 to Gear
3
DISCUSSIONS :
i. From table 1 and 2, calculate the gear ratio theoretically and experimentally and compare the result.
From the data, the value from table 1 and 2 shows that the gear ratio of calculation and the gear ratio of theory are different.
It show that data with experimentally, that motor speed is higher than data With theory.it because depending with the size and teeth that we use to
Experiment
Gear Ratio (Calculation) 0.23 0.23 0.23
Gear Ratio (From Data) 0.23 0.05 0.01
Gear Ratio (Calculation) 0.47 0.23 0.15
Gear Ratio (From Data) 0.23 0.05 0.01
ii. From the experiments, plot the graph Speed (rpm) versus Time (seconds) of the Gear set 1 and 2. Review the results.
SET 1
SET 2
iii. What are the input torque and the output torque of the gear system. Given that the input power Pin equal to 20kW and the efficiency, ƞ equal to 0.7.
The value taken from data Table 1
Calculation For () Gear 0
Formula : 2πN/60
2 x π x 100 =10.47 rad/s
60
Calculation Torque (T) ,Given data input power =20kW
T=P/
T= 20kW/10.47
= 1910.2 Nm (input Torque)
The value taken from data Table 1
Calculation For () Gear 1
Formula : 2πN/60
2 x π x 23 = 2.408 rad/s
60
Calculation Tork (T) ,Given data input power =20kW
T=P/
T= 20kW/2.408
= 8305.6Nm (Output Torque)
The value take from data Table 1
Calculation For () Gear 2
Formula : 2πN/60
2 x π x 5 = 0.524 rad/s
60
Calculation Torque (T) ,Given data input power =20kW
T=P/
T= 20kW/0.524
= 38167.9 Nm (Output Tork)
The value take from data Table 1
Calculation For () Gear 3
Formula : 2πN/60
2 x π x 2 = 0.209 rad/s
60
Calculation Torque (T) ,Given data input power =20kW
T=P/
T= 20kW/0.209
= 95693.7 Nm (Output Torque)
Gear 0 Gear 1 Gear 2 Gear 3
N
(rpm)
Rad/s
T
(Nm)
N
(rpm)
Rad/s
T
(Nm)
N
(rpm)
Rad/s
T
(Nm)
N
(rpm)
Rad/s
T
(Nm)
100 10.47 1910.2 48 5.026 3979.3 12 1.256 15915.5 2 0.209 95693.7
200 20.94 955.10 95 9.948 2010.5 23 2.408 8305.6 4 0.418 47746.5
300 31.42 6365.3 143 14.97 1336.0 34 3.560 5617.9 5 0.523 38197.1
400 41.89 476.42 189 19.79 1010.6 42 4.398 4547.5 6 0.628 31847.1
500 52.36 381.97 237 24.82 805.8 55 5.760 3472.2 7 0.733 27285.1
600 62.83 318.32 285 29.85 670.0 66 6.911 2893.9 11 1.152 17362.3
700 73.30 272.85 332 34.76 575.37 78 8.168 2448.5 14 1.466 13641.8
Table 2: 4 stage Gear with Different size Gear - Gear Set 2
The value take from data Table 2
Calculation For () Gear 0
Formula : 2πN/60
2 x π x 100 =10.47 rad/s
60
Calculation Torque (T) ,Given data input power =20kW
T=P/
T= 20kW/10.47
= 1910.2 Nm (input Torque)
The value take from data Table 1
Calculation For () Gear 1
Formula : 2πN/60
2 x π x 48 = 5.026 rad/s
60
Calculation Torque (T) ,Given data input power =20kW
T=P/
T= 20kW/2.408
= 3979.3 Nm (Output Torque)
The value take from data Table 1
Calculation For () Gear 2
Formula : 2πN/60
2 x π x 12 = 1.256 rad/s
60
Calculation Torque (T) ,Given data input power =20kW
T=P/
T= 20kW/0.524
= 15915.5 Nm (Output Torque)
The value take from data Table 1
Calculation For () Gear 3
Formula : 2πN/60
2 x π x 2 = 0.209 rad/s
60
Calculation Torque (T) ,Given data input power =20kW
T= P/
T= 20kW/0.209
= 95693.7 Nm (Output Torque)
iv. Calculate the efficiency of the gear system (gear set 1 and 2) of the following combination gear :-
a. Gear 1 to gear 0b. Gear 2 to gear 0c. Gear 3 to gear 0
Set 1
Gear 0 – Gear 1 Gear 0 – Gear 2 Gear 0 – Gear 3
N (rpm) Gear Ratio η =efficency Gear Ratioη=efficenc
yGear Ratio η=efficency
100 0.23 1.0 0.05 1.0 0.02 1.0
200 0.23 1.0 0.06 1.0 0.02 1.0
300 0.23 0.1 0.05 0.1 0.01 0.1
400 0.23 1.0 0.06 1.0 0.01 1.0
500 0.23 1.0 0.05 0.8 0.01 1.0
600 0.23 1.0 0.05 1.0 0.01 1.0
700 0.23 1.0 0.05 1.0 0.01 1.0
Average
Gear ratio0.23 - 0.05 - 0.01 -
Average
Effeciency- 0.87 - 0.84 - 0.87
Set 2
Gear 0 – Gear 1 Gear 0 – Gear 2 Gear 0 – Gear 3
N (rpm) Gear Ratio η =efficency Gear Ratioη=efficenc
yGear Ratio η=efficency
100 0.23 1.0 0.05 1.0 0.02 1.0
200 0.23 1.0 0.06 1.0 0.02 1.0
300 0.23 0.1 0.05 0.1 0.01 0.1
400 0.23 1.0 0.06 1.0 0.01 1.0
500 0.23 1.0 0.05 1.0 0.01 1.0
600 0.23 1.0 0.05 1.0 0.01 1.0
700 0.23 1.0 0.05 1.0 0.01 1.0
Average
Gear ratio0.23 - 0.05 - 0.01 -
Average
Effeciency- 0.87 - 0.87 - 0.87
CONCLUSION :
From the experiment, the objectives have been achieved which are describing the different
types of gear system, calculating the ratios, angular velocity, input and output speed and
efficiency. From what I have learn to this experiment, this experiment similarly like system
gearbox in car is to transmit rotational motional from a driving prime mover to a driven
machine.
troubleshooting the software and the speed sensor so the value given are accurate and precise.
REFERENCES :
http://en.wikipedia.org/wiki/Gear
McGraw Hill Encyclopedia of Science and Technology, "Gear"
Gear Dive Systems: Design and Application, by Peter Lynwander
http://www.codecogs.com/reference/engineering/materials/
shear_force_and_bending_moment.php