Classification of gears. Depending on the cross-sectional shape of the belt, transmissions can be: flat belt, V-belt, round belt, poly-V-belt (Fig. 69). Flat drive gears are classified as cross and semi-cross (angular), Fig. 70. In modern mechanical engineering, V-belts and poly-V belts are most widely used. Round belt transmission has limited applications ( sewing machines, desktop machines, instruments).
A type of belt drive is Toothed belt, transmitting the load by engaging the belt with the pulleys.
Rice. 70. Types of flat belt drives: a – cross, B – semi-cross (angular)
Purpose. Belt drives refer to mechanical friction transmissions with a flexible connection and are used when it is necessary to transfer the load between shafts that are located at significant distances and in the absence of strict requirements for the gear ratio. A belt drive consists of a drive and driven pulleys located at some distance from each other and connected by a belt (belts) placed on the pulleys under tension. The rotation of the driving pulley is converted into rotation of the driven pulley due to the friction developed between the belt and the pulleys. According to the cross-sectional shape they are distinguished Flat , Wedge , Polywline And Round drive belts. There are flat belt drives - Open which carry out transmission between parallel shafts rotating in the same direction; Cross, Which carry out transmission between parallel shafts when the pulleys rotate in opposite directions; V Corner (semi-cross) In flat belt drives, the pulleys are located on intersecting (usually at right angles) shafts. To ensure friction between the pulley and the belt, tension is created on the belts by preliminary elastic deformation, by moving one of the transmission pulleys or using a tension roller (pulley).
Advantages. Thanks to the elasticity of the belts, the transmissions operate smoothly, without shock and silently. They protect mechanisms from overload due to possible belt slippage. Flat drive gears are used for large center distances and operating at high belt speeds (up to 100 M/s). For small center distances, large gear ratios and transmission of rotation from one drive pulley to several driven pulleys, V-belt drives are preferable. Low cost of transfers. Easy installation and maintenance.
Flaws. Large gear dimensions. Change in gear ratio due to belt slippage. Increased loads on shaft supports with pulleys. The need for devices for tensioning belts. Low belt durability.
Areas of application. A flat drive transmission is simpler, but a V-belt drive has increased traction capacity and fits into smaller dimensions.
Poly V-belts are flat belts with longitudinal V-ribs on work surface included in the wedge grooves of the pulleys. These belts combine the advantages of flat belts - flexibility and V-belts - increased adhesion to pulleys.
Round belt drives are used in small machines, such as sewing machines and Food Industry, benchtop machines, as well as various devices.
In terms of power, belt drives are used in various machines and units at 50 HF T, (in some gears up to 5000 kW), at peripheral speed - 40 M/s, (in some programs up to 100 M/s), according to gear ratios 15, gear efficiency: flat-belt 0.93...0.98, and V-belt – 0.87...0.96.
Rice. 71 Belt drive diagram.
Power calculation . Circumferential force on drive pulley
. (12.1)
Calculation of belt drives is carried out according to the calculated circumferential force, taking into account the dynamic load coefficient and the transmission operating mode:
Where is the dynamic load coefficient, which is assumed to be =1 for a quiet load, =1.1 for moderate load fluctuations, =1.25 for significant load fluctuations, =1.5 for shock loads.
Initial belt tension force F O (pre-tension) is taken such that the belt can maintain this tension sufficiently long time, without being subjected to large stretching and without losing the required durability. Accordingly, the initial tension in the belt for flat standard belts without automatic tensioners = 1.8 MPa; with automatic tensioners = 2 MPa; for standard V-belts =1.2...1.5 MPa; for polyamide belts = 3...4 MPa.
Initial belt tension
Where A - The cross-sectional area of a flat belt drive belt or the cross-sectional area of all V-belt drive belts.
Tension forces of the driving and driven S 2 The branches of the belt in a loaded transmission can be determined from the equilibrium condition of the pulley (Fig. 72).
Rice. 72. Scheme for power transmission calculations.
From the equilibrium condition of the drive pulley
(12.4)
Taking into account (12.2), the circumferential force on the drive pulley
Leading branch tension
, (12.6)
Driven branch tension
. (12.7)
Drive pulley shaft pressure
. (12.8)
The relationship between the tension forces of the leading and driven branches is approximately determined by Euler’s formula, according to which the tensions of the ends of a flexible, weightless, inextensible thread encircling the drum are related by the dependence
Where is the coefficient of friction between the belt and the pulley, and is the angle of circumference of the pulley.
The average value of the friction coefficient for cast iron and steel pulleys can be taken: for rubber-fabric belts = 0.35, for leather belts= 0.22 and for cotton and wool belts = 0.3.
When determining the friction forces in a V-belt transmission, instead of the friction coefficient, the reduced friction coefficient for V-belts must be substituted in the formulas
, (12.10)
Where is the angle of the belt wedge.
When considering the given force relations for the belt together, we obtain the circumferential force on the drive pulley
, (12.11)
Where is the thrust coefficient, which is determined by the dependence
An increase in the circumferential force on the drive pulley can be achieved by increasing the belt pre-tension or by increasing the traction coefficient, which increases with increasing wrap angle and friction coefficient.
The tables with reference data on the characteristics of belts indicate their sizes, taking into account the required traction coefficients.
Geometric calculation . Estimated length of belts with a known center distance and pulley diameters (Fig. 71):
Where . For end belts, the length is finally agreed upon with standard lengths according to GOST. To do this, perform a geometric calculation according to the diagram shown in Fig. 73.
Fig.73. Scheme for geometric calculation of belt drive
According to the finally established length of a flat-belt or V-belt open transmission, the actual center-to-center distance of the transmission, provided that
Calculation formulas without taking into account sagging and initial deformation of the belt.
Angle of wrapping of the drive pulley by the belt in radians:
, (12.14)
In degrees 
.
The procedure for performing design calculations. For a belt drive, during design calculations based on given parameters (power, torque, angular, speed and gear ratio), the dimensions of the belt and drive pulley are determined, which provide the necessary fatigue strength of the belt and the critical traction coefficient at maximum efficiency. Based on the selected diameter of the drive pulley, the remaining dimensions are determined from a geometric calculation: 
Design calculation of flat belt transmission according to traction capacity they are produced according to the permissible useful voltage , Which is determined by slip curves. As a result of the calculation, the width of the belt is determined by the formula:
, (12.15)
Where is the circumferential force in the transmission; - permissible specific circumferential force, which corresponds to the maximum traction coefficient, which is determined at belt speed = 10 m/s and wrap angle = 1800; - gear location coefficient depending on the angle of inclination of the line of centers to horizontal line: =1.0, 0.9, 0.8 for tilt angles =0...600, 60...800, 80...900; - pulley wrap angle coefficient; - speed coefficient: ; - operating mode coefficient, which is assumed: =1.0 quiet load; =0.9 load with small changes, =0.8 – load with large fluctuations, =0.7 – shock loads.
For the calculation, the diameter of the drive pulley is first determined using empirical formulas
, (12.16)
Where is the transmitted power in kW, is the rotation speed.
The drive pulley diameter is rounded to the nearest standard.
The type of belt is adopted, according to which the permissible specific circumferential force is determined according to table 12.1.
Table 12.1
Parameters of flat drive belts
The estimated belt width is rounded to the nearest standard width according to Table 12.2.
Table 12.2 Standard width flat drive belts
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20, 25,32, 40, 50, 63, 71, 80, 90, 110, 112, 125, 140, 160, 180, 200, 224, 250, 280… |
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30, 60, 70, 115, 300… |
Table 12.3 Width of flat belt pulley rim.
Design calculation of V-belt transmission according to traction capacity, they are made according to the permissible power transmitted by one belt of the selected cross-section, which is also determined from the slip curves. As a result of the calculation, the number of belts of the selected section is determined using the formula:
, (12.17)
Where is the permissible power transmitted by one cross section; - pulley wrap angle coefficient: ; - belt length coefficient: ; - coefficient that takes into account uneven loading between belts 
.
To calculate using formula (12.17), the type of belt cross-section (Fig. 74) is first determined from empirical dependencies, and from it the diameter of the drive pulley is preliminarily determined in terms of transmitted power and rotational speed, according to Table 12.3.
Table 12.4
Power N 0, which is transmitted by one V-belt at α =180o, belt length ℓ 0 quiet loading and gear ratio U = 1
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d 1, mm |
Р0 (kW) at belt speed υ, m/s |
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l 0=1320mm |
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l 0=1700mm |
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l 0=2240mm |
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l 0=3750mm |
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l 0=6000mm |
Translation of the designation system for V-belt sections according to GOST 1284 into international standards: O – Z, A – A, B – B, V – C, G – D, D – E, E – E0
The center distance can be specified in the source data, or taken in the range
,
Where is the height of the selected belt section.
As a result of the geometric calculation of the transmission, the parameter values are specified and determined gauge length belt, which is rounded up to the nearest standard value, according to table 12.5.Table 12.5
Standard length of V-belts
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Length, mm |
Belt section |
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400; 425; 450; 475; 500; 530 |
* | |||
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560; 600; 630; 670; 710; 750 |
* | * | ||
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800; 850; 900; 950; 1000; 1060 |
* | * | * | |
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1120; 1180; 1250; 1320; 1400; 1500; 1600; 1700; 1800; 1900; 2000; 2120; 2240; 2360;2500 |
* | * | * | * |
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2650; 2800; 3000; 3150; 3350; 3550; 3750; 4000 |
* | * | * | |
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4250; 4500; 4750; 5000; 5300; 5600; 6000 |
* | * | ||
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6300; 6700; 7100; 7500; 8000; 8500; 9000; 9500; 10000; 10600 |
* | |||
The estimated number of V-belts is rounded to the nearest higher whole number.
Durability test calculation . The durability of a belt is determined by its resistance to fatigue under cyclic loading. Fatigue resistance is determined by the number of load cycles, which increases with increasing belt speed and decreasing belt length. To ensure belt durability within 1000...5000 hours of operation, the number of belt runs per second is checked, which corresponds to the number of loads per second
Table 12.7
Table 12.7
Dimensions and parameters of V-belts
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Designation |
section, mm |
F, mm2 |
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Normal section |
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08-10-2011(long ago)
Dust fan No. 6, No. 7, No. 8
Motor 11kW, 15kW, 18kW.
The engine speed is 1500 rpm.
There are NO pulleys on either the fan or the motor.
There is a TURNER and an IRON.
What size pulleys does a turner need to turn?
What speed should the fans be at?
THANK YOU
08-10-2011(long ago)
08-10-2011(long ago)
put a 240 pulley on the fan and on an engine 140-150.2 or 3 strands of profile with the volute will have 900-1000 revolutions if on engines 1500. On large fans they don’t put a high frequency because of vibrations. That’s how it is for me.
08-10-2011(long ago)
08-10-2011(long ago)
08-10-2011(long ago)
If speed is needed as for engine. then 1:1, if one and a half times more then 1:1.5, etc. how much should you increase the speed and make the difference in diameters?
08-10-2011(long ago)
there is a dependence on the belt profile
if the belt profile is “B”, then the pulley should be 125 mm or more, and the groove angle should be from 34 degrees (up to 40 degrees with a pulley diameter of 280 mm).
09-10-2011(long ago)
It’s not difficult to calculate pulleys. Convert the angular speed to linear speed through the circumference. If there is a pulley on the engine, calculate its circumference, that is, multiply the diameter by pi, which is 3.14, and get the pulley circumference. Let’s say the engine has 3000 rpm minute, multiply 3000 by the resulting circumference, this value shows how far the belt runs per minute of operation, it is constant, and now divide it by the required number of revolutions of the working shaft and by 3.14, get the diameter of the pulley on the shaft. This is the solution simple equation d1*n*n1=d2*n*n2/in short I explained it as best I could. I hope you understand.
09-10-2011(long ago)
On No. 8 there are three belts profile B (C).
Driven pulley diameter is 250mm.
Select the presenter for 18 kW
In catalogs for fans
there is data (power, fan speed)
09-10-2011(long ago)
03-08-2012(long ago)
01/28/2016(long ago)
thanks to Victor...as I understand it...if my engine has 3600 rpm...then...on the nsh-10 pump I need a maximum of 2400 rpm...from this I assume that...on the engine the pulley is 100mm...and on the pump 150mm...or 135mm? ?? in general, roughly with errors, I hope somewhere like this...
01/29/2016(long ago)
http://pnu.edu.ru/media/filer_public/2012/12/25/mu-raschetklinorem.pdf
01/29/2016(long ago)
3600:2400=1.5
This is your gear ratio. It refers to the ratio of the diameters of your pulleys on the engine and on the pump. Those. If the pulley on the engine is 100, then the pump should have 150, then there will be 2400 rpm. But here the question is different: aren’t there too many revolutions for NS?
| Time is Irkutsk time everywhere (Moscow time +5). |
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Increasing the pulley diameter improves belt durability.
Tension roller.| Tensioners.| Checking for a fracture in the joint of the split pulley. Increasing the diameter of the pulley is possible only within certain limits, determined by the gear ratio, dimensions and weight of the machine.
The coefficient cp increases with an increase in the diameter of the pulleys and the peripheral speed, as well as when using clean and well-soaked belts when working on smooth pulleys, and, conversely, falls with dirty belts and when working on rough pulleys.
According to experimental data, as the pulley diameter increases, the friction coefficient increases.
According to experimental data, with increasing pulley diameter, the friction coefficient increases.
YuOn-150, which does not entail an increase in pulley diameters.
As can be seen from the previous one, as the diameter of the pulley increases, the bending stress decreases, which has a beneficial effect on increasing the durability of the belt. At the same time, the specific pressure decreases and the friction coefficient increases, as a result of which the traction capacity of the belt increases.
With increasing pretension at the same relative load, slip increases slightly and decreases with increasing pulley diameter. When working with a reduced load, the slip decreases.
With an increase in pretension at the same relative load, the slip increases slightly and decreases with an increase in the pulley diameter.
With increasing pretension at the same relative load, slip increases slightly and decreases with increasing pulley diameter.
Most in a simple way increasing the performance of compressors is an increase in the number of their revolutions, which with a belt drive is achieved by increasing the diameter of the electric motor pulley. For example, the Type I compressor was originally rated at 100 rpm. However, during the operation of these compressors it was found that the speed can be increased to 150 per minute without violating the conditions safe work.
Formula (87) shows that for belts with the same rope diameter, the tension, which depends on the bending resistance, decreases with increasing pulley diameter.
Practice recent years indicates the feasibility of: using large ratios between the diameter of the pulley and the rope (Dm / d up to 48); increasing the diameter of pulleys; using stronger, larger diameter ropes.
A study of a transmission with pulleys without ring grooves: at speeds above 50 m/s showed that its traction capacity decreases, despite the increase in the diameter of the pulleys. The latter is explained by the appearance of air cushions in places where the belt runs over the pulleys, which cause a decrease in the belt wrap angles, the greater the higher its speed. This is most noticeable on the driven pulley because the driven belt leg is weakened, allowing penetration. air cushion into the contact area between the belt and the pulley and causes it to slip.
The diameter of the pulleys of the traveling system should be 38 - 42 times the diameter of the rope. Increasing the diameters of the pulleys helps reduce friction losses and improve the operating conditions of the rope.
Belt drives. Belt drives (Fig. 47) require round, flat and V-belts. When the diameter of the drive shaft pulley increases, the number of revolutions of the driven shaft increases, and, conversely, if the diameter of the drive shaft pulley is reduced, the number of revolutions of the driven shaft also decreases.
Technical specifications traveling blocks. The pulleys of crown blocks and traveling blocks have the same design and dimensions. The pulley diameter, profile and groove dimensions significantly affect the service life and consumption of hoisting ropes. The fatigue life of the rope increases with increasing diameter of the pulleys, since this reduces the recurrent stresses that arise in the rope when bending around the pulleys. In drilling rigs, the diameters of the pulleys are limited by the dimensions of the tower and the convenience of work associated with carrying the candles to the candle holder.
The diameter of the transmission pulley is one of the most important parameters belt operation. In the tables of power transmitted by belts, to ensure a given transmission reliability, the amount of power is indicated depending on the smaller diameter of the transmission pulley. Initially, the thrust coefficient increases sharply with increasing pulley diameter, then after reaching certain value diameter of the pulley, the traction coefficient remains virtually unchanged. Thus, further increasing the diameter of the pulley is impractical.
The cyclically changing stress that arises in a rectilinear belt traction element is largely determined by the magnitude of the bending stress that appears in the tape when it rolls over pulleys and reels. The amount of bending stress can be reduced by thickness of the belt or by increasing the diameter of the pulley. However, the thickness of the tape has a minimum limit, and an increase in the diameter of the pulley is undesirable due to a significant increase in the weight of the winding organ and the overall cost lifting installation.
From consideration of the table. 30 and slip curves the following can be seen. The traction abilities of belts with a section of 50X22 mm do not differ significantly, despite the difference in the materials of the supporting layer. These belts give a high speed loss of the driven shaft (up to 3 5% at d 200 - 204 mm, a0 0 7 MPa and f 0 6), which increases with increasing belt tension and decreases with increasing pulley diameters. Highest value t] 0 92 have belts with anide cord fabric and lavsan cord with d 240 - n250 mm.
The necessary pre-tension of the ropes is determined depending on their condition: a distinction is made between a new rope and a rope that has already been stretched under load.
As the transmission operates, the ropes gradually lengthen and their sag increases. In this case, the decrease in tension m, caused by the pre-tension of the rope, is partially replaced by an increase in tension from an increase in the weight of the sagging part of the rope, and to a greater extent, the greater the sag of the rope. More favorable conditions for rope operation, they are created by increasing the diameters of the pulleys and using elastic ropes. When installing transmission at distances of 25 - 30 m, intermediate pulleys are installed (Fig. The use of support pulleys, as already mentioned, leads to a decrease in transmission efficiency.
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03/23/2016(long ago) |
There is a 1000 rpm motor. what diameter pulleys need to be placed on the engine and shaft so that the shaft turns out to be 3000 rpm | ||||
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03/24/2016(long ago) |
???
The big one turns the small one - the speed of the latter increases and vice versa... Above joke! :) |
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03/24/2016(long ago) |
will not cut anything, change the motor to 3000 rpm. The difference in the diameters of the pulleys will be 560/190 mm. Can you imagine a 560 mm pulley??? it will cost as much as an airplane wing and there is no point in installing it. |
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03/29/2016(long ago) |
???
Arthur - the questions above (black ones) are “for nagging”... Humanity has put its activities in this dimension at 750; 1000; 1500; 3000 rpm - choose a CONSTRUCTOR!!! PS The faster the engine speed, the cheaper and more compact it is)))… |
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03/31/2016(long ago) |
Did you count correctly?
Engine 0.25 kV 2700 rpm pulley on the engine 51mm transfers to a 31mm pulley and on lap 127 I got 27-28 m/s I want to replace the 51mm pulley with 71mm then I get 38-39 m/s am I right? |
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03/31/2016(long ago) |
Your truth!!!
But!!! — by increasing the sharpening (cutting) speed you will reduce the grain feed and, as a result, the specific cutting work will increase, which will lead to an increase in power! The engine will need to be more powerful if there is no reserve in the existing one! PS There are no miracles (((, i.e.: “You can’t get anything without giving something”)))!!! |
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03/31/2016(long ago) |
"I'll give 0.25kv for 0.75kv"))
Thanks SVA. And another question is what is better to leave as is or make 38-39 m/s. |
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01-04-2016(long ago) |
For the interval:) in kW - there (from memory) between 0.25 and 0.75 there are still 0.37 and 0.55))) In short, before the speed increased, the currents shot out (at 0.25 kW - the nominal value is 0.5 A roughly), we increased the speed, again we hit the teeth and measure the current. Ilyas - as I understand it, sharpen the tape to reduce the surface roughness in the cavity of the tooth, am I interpreting it correctly? PS Right now Sergey Anatolyevich (Beaver 195) will read my writings - and explain everything both for the stones and for m/s!!!))) |
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01-04-2016(long ago) |
Thanks again SVA. I will do so. Previously, I changed the abrasive to full profile and thought that the speed was low. And the motor is connected by a star, should it be connected to a delta or left on a star? | ||||
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03-04-2016(long ago) |
Hello!
Sorry for the delay. At the same time, I checked on him to see how he was there after the holidays, whether he was alive or not... So for the grain...
Instructions1. Calculate the diameter of the drive pulley using the formula: D1 = (510/610) · ??(p1·w1) (1), where: - p1 - motor power, kW; — w1 — angular velocity of the drive shaft, radians per second. Take the motor power value from the technical data in its passport. As usual, the number of motor cycles per minute is also indicated there. 2. Convert the number of motor cycles per minute to radians per second by multiplying the starting number by the exponent 0.1047. Substitute the detected numerical values into formula (1) and calculate the diameter of the drive pulley (assembly). 3. Calculate the diameter of the driven pulley using the formula: D2= D1·u (2), where: - u - gear ratio; - D1 - calculated according to formula (1) the diameter of the leading node. Determine the gear ratio by dividing the angular velocity of the driving pulley by the desired angular velocity of the driven unit. Conversely, based on the given diameter of the driven pulley, it is possible to calculate its angular velocity. To do this, calculate the ratio of the diameter of the driven pulley to the diameter of the driving pulley, then divide the angular velocity of the driving unit by this number. 4. Find the minimum and highest distance between the axes of both nodes according to the formulas: Amin = D1+D2 (3), Amax = 2.5·(D1+D2) (4), where: - Amin - minimum distance between axes; - Amax - highest distance; - D1 and D2 - diameters of the driving and driven pulleys. The distance between the axes of the nodes should not be more than 15 meters. 5. Calculate the length of the transmission belt using the formula: L = 2A+P/2·(D1+D2)+(D2-D1)?/4A (5), where: - A is the distance between the axes of the driving and driven units, - ? — number “pi”, — D1 and D2 — diameters of the driving and driven pulleys. When calculating the length of the belt, add 10 - 30 cm to the resulting number for its stitching. It turns out that using the given formulas (1-5), you can easily calculate the optimal values of the units that make up the flat belt drive. Modern life is in constant motion: cars, trains, planes, everyone is in a hurry, running somewhere, and it is often important to calculate the speed of this movement. To calculate speed there is a formula V=S/t, where V is speed, S is distance, t is time. Let's look at an example in order to understand the algorithm of actions.
Instructions1. Interesting to know how fast you walk? Choose a path whose footage you know correctly (at a stadium, say). Record your time and walk along it at your normal pace. So, if the length of the path is 500 meters (0.5 km), and you covered it in 5 minutes, then divide 500 by 5. It turns out that your speed is 100 m/min. If you covered it on a bicycle in 3 minutes, then your speed is 167 m/min. By car in 1 minute, that means the speed is 500 m/min.
2. To convert speed from m/min to m/sec, divide the speed in m/min by 60 (the number of seconds in a minute). So, it turns out that when walking, your speed is 100 m/min / 60 = 1.67 m/sec. Bicycle: 167 m/min / 60 = 2.78 m/sec. Car: 500 m/min / 60 = 8.33 m/sec.
3. To convert speed from m/sec to km/h, divide the speed in m/sec by 1000 (the number of meters in 1 kilometer) and multiply the resulting number by 3600 (the number of seconds in 1 hour). Thus, it turns out that the walking speed is 1.67 m/sec / 1000*3600 = 6 km/h. Bicycle: 2.78 m/sec / 1000*3600 = 10 km/h. Car: 8.33 m/sec / 1000*3600 = 30 km/ h. 4. To facilitate the procedure for converting speed from m/sec to km/h, use the indicator 3.6, the one that is used further: speed in m/sec * 3.6 = speed in km/h. Walking: 1.67 m/sec *3.6 = 6 km/h. Bicycle: 2.78 m/s*3.6 = 10 km/h. Car: 8.33 m/s*3.6= 30 km/h. Apparently, that is significant It’s easier to remember the exponent 3.6 than the entire multiplication-division procedure. In this case, you will easily convert speed from one value to another.
Video on the topic |
When designing equipment, it is necessary to know the speed of the electric motor. To calculate the rotation speed, there are special formulas that are different for AC and DC motors.
Synchronous and asynchronous electric machines
Engines AC voltage There is three types: synchronous, the angular speed of the rotor coincides with the angular frequency magnetic field stator; asynchronous - in them the rotation of the rotor lags behind the rotation of the field; commutator motors, the design and operating principle of which are similar to DC motors.
Synchronous speed
Electric machine rotation speed alternating current depends on the angular frequency of the stator magnetic field. This speed is called synchronous. In synchronous motors, the shaft rotates at the same speed, which is an advantage of these electric machines.
For this purpose in the rotor of machines high power There is a winding to which a constant voltage is applied, creating a magnetic field. In low power devices, permanent magnets are inserted into the rotor, or there are pronounced poles.
Slip
In asynchronous machines, the number of shaft revolutions is less than the synchronous angular frequency. This difference is called the "S" slip. Thanks to sliding in the rotor, electricity, and the shaft rotates. The larger S, the higher the torque and the lower the speed. However, if the slip exceeds a certain value, the electric motor stops, begins to overheat and may fail. The rotation speed of such devices is calculated using the formula in the figure below, where:
- n – number of revolutions per minute,
- f – network frequency,
- p – number of pole pairs,
- s – slip.
There are two types of such devices:
- With squirrel-cage rotor. The winding in it is cast from aluminum during the manufacturing process;
- With wound rotor. The windings are made of wire and are connected to additional resistances.
Speed adjustment
During operation, it becomes necessary to adjust the speed electric machines. This is done in three ways:
- Increasing additional resistance in the rotor circuit of electric motors with a wound rotor. If it is necessary to greatly reduce the speed, it is possible to connect not three, but two resistances;
- Connecting additional resistances in the stator circuit. It is used to start high-power electrical machines and to regulate the speed of small electric motors. For example, the number of revolutions table fan can be reduced by connecting an incandescent lamp or capacitor in series with it. The same result is achieved by reducing the supply voltage;
- Changing the network frequency. Suitable for synchronous and asynchronous motors.
Attention! The rotation speed of commutator electric motors operating from an alternating current network does not depend on the frequency of the network.
DC motors
In addition to AC machines, there are electric motors connected to the network direct current. The speed of such devices is calculated using completely different formulas.
Rated rotation speed
The speed of a DC machine is calculated using the formula in the figure below, where:
- n – number of revolutions per minute,
- U – network voltage,
- Rya and Iya – armature resistance and current,
- Ce – motor constant (depending on the type of electric machine),
- Ф – stator magnetic field.
These data correspond to the nominal values of the parameters of the electric machine, the voltage on the field winding and the armature or the torque on the motor shaft. Changing them allows you to adjust the rotation speed. Define magnetic flux in a real motor it is very difficult, so for calculations they use the current flowing through the field winding or armature voltage.
The speed of commutator AC motors can be found using the same formula.
Speed adjustment
Adjustment of the speed of an electric motor operating from a DC network is possible within a wide range. It is possible in two ranges:
- Up from nominal. To do this, the magnetic flux is reduced using additional resistances or a voltage regulator;
- Down from par. To do this, it is necessary to reduce the voltage on the armature of the electric motor or connect a resistance in series with it. In addition to reducing the speed, this is done when starting the electric motor.
Knowing what formulas are used to calculate the rotation speed of an electric motor is necessary when designing and setting up equipment.
Video
A belt drive transmits torque from the drive shaft to the driven shaft. Depending on it, it can increase or decrease the speed. The gear ratio depends on the ratio of the diameters of the pulleys - drive wheels connected by a belt. When calculating the drive parameters, you must also take into account the power on the drive shaft, its rotation speed and the overall dimensions of the device.
Belt drive device, its characteristics
A belt drive consists of a pair of pulleys connected by an endless looped belt. These drive wheels are usually placed in the same plane, and the axles are made parallel, with the drive wheels rotating in the same direction. Flat (or round) belts allow you to change the direction of rotation by crossing, and mutual arrangement axes - through the use of additional passive rollers. In this case, part of the power is lost.
V-belt drives, due to the wedge-shaped cross-section of the belt, make it possible to increase the area of engagement with the belt pulley. A wedge-shaped groove is made on it.
Toothed belt drives have teeth of equal pitch and profile on inside belt and on the surface of the rim. They do not slip, allowing more power to be transmitted.
The following basic parameters are important for calculating the drive:
- number of revolutions of the drive shaft;
- power transmitted by the drive;
- required number of revolutions of the driven shaft;
- belt profile, its thickness and length;
- settlement, external, inner diameter wheels;
- groove profile (for V-belt);
- transmission pitch (for toothed belt)
- center distance;
Calculations are usually carried out in several stages.
Main diameters
To calculate the parameters of pulleys, as well as the drive as a whole, they use different meanings diameters, so for the V-belt pulley the following are used:
- calculated D calculated;
- outer D out;
- internal, or landing D int.
To calculate the gear ratio, the design diameter is used, and the outer diameter is used to calculate the dimensions of the drive when configuring the mechanism.
For a toothed belt drive, D calc differs from D plan by the height of the tooth.
The gear ratio is also calculated based on the value of D calculated.
To calculate a flat belt drive, especially when large size rim relative to the profile thickness, D calculated is often taken equal to the outer one.
Pulley diameter calculation
First, you should determine the gear ratio based on the specified rotation speed of the drive shaft n1 and the required rotation speed of the driven shaft n2/ It will be equal to:
If a ready-made engine with a drive wheel is already available, the pulley diameter according to i is calculated using the formula:
If the mechanism is designed from scratch, then theoretically any pair of drive wheels that satisfy the following conditions will be suitable:
In practice, the calculation of the drive wheel is carried out based on:
- Dimensions and design of the drive shaft. The part must be securely attached to the shaft, correspond to it in terms of the size of the internal hole, the method of fit, and fastening. The maximum minimum pulley diameter is usually taken from the ratio D calculated ≥ 2.5 D in
- Permissible transmission dimensions. When designing mechanisms, it is necessary to meet the dimensions. In this case, the interaxle distance is also taken into account. The smaller it is, the more the belt bends when flowing around the rim and the more it wears out. Too large a distance leads to the excitation of longitudinal vibrations. The distance is also specified based on the length of the belt. If production is not planned unique detail, then the length is selected from the standard range.
- Transmitted power. The material of the part must withstand angular loads. This is relevant for large capacities and torques.
The final diameter calculation is finalized based on the results of dimensional and power estimates.
Question from Messrs. Rabynin and Novikov, Nizhny Novgorod region.
Please answer as correct calculate pulley diameters so that the knife shaft of the woodworking machine rotates at a speed of 3000...3500 rpm. The rotational speed of the electric motor is 1410 rpm (the motor is three-phase, but will be connected to a single-phase network (220 V) using a capacitor system. V-belt.
First a few words about V-belt transmission- one of the most common transmission systems rotational movement using pulleys and a drive belt (this transmission is used in a wide range of loads and speeds). We produce drive belts of two types - drive belts themselves (according to GOST 1284) and for automotive engines (according to GOST 5813). Belts of both types differ slightly in size. The characteristics of some belts are given in tables 1 and 2, cross section V-belt is shown in Fig. 1. Both types of belts are wedge-shaped with a wedge apex angle of 40° with a tolerance of ± 1°. The minimum diameter of the smaller pulley is also indicated in tables 1 and 2. However, when selecting minimum diameter pulley, you should also take into account the linear speed of the belt, which should not exceed 25...30 m/s, and it is better (for greater belt durability) that this speed be within 8...12 m/s.
Note. The names of certain parameters are given in the captions to Fig. 1.
Note. The names of certain parameters are given in the figure captions to Fig. 1.
The pulley diameter, depending on the shaft rotation speed and the linear speed of the pulley, is determined by the formula:
D1=19000*V/n,
where D1 is the pulley diameter, mm; V - linear speed of the pulley, m/s; n - shaft rotation speed, rpm.
The diameter of the driven pulley is calculated using the following formula:
D2 = D1x(1 - ε)/(n1/n2),
where D1 and D2 are the diameters of the driving and driven pulleys, mm; ε - belt slip coefficient equal to 0.007...0.02; n1 and n2 - rotation speed of the drive and driven shafts, rpm.
Since the value of the slip coefficient is very small, the slip correction can be ignored, that is, the above formula will take on a simpler form:
D2 = D1*(n1/n2)
The minimum distance between pulley axes (minimum center distance) is:
Lmin = 0.5x(D1+D2)+3h,
where Lmin is the minimum center-to-center distance, mm; D1 and D2 - pulley diameters, mm; h - belt profile height.
The smaller the center-to-center distance, the more the belt bends during operation and the shorter its service life. It is advisable to take the center-to-center distance greater than the minimum value Lmin, and the closer the gear ratio is to unity, the larger it becomes. However, to avoid excessive vibration, very long belts should not be used. By the way, the maximum center-to-center distance Lmax can be easily calculated using the formula:
Lmax<= 2*(D1+D2).
But in any case, the value of the center-to-center distance L depends on the parameters of the belt used:
L = A1+√(A1 2 - A2),
where L is the calculated center-to-center distance, mm; A1 and A2 are additional quantities that will have to be calculated. Now let's look at the quantities A1 and A2. Knowing the diameters of both pulleys and the standard length of the selected belt, determining the values of A1 and A2 is not difficult at all:
A1 = /4, a
A2 = [(D2 - D1) 2 ]/8,
where L is the standard length of the selected belt, mm; D1 and D2 - pulley diameters, mm.
When marking a plate for installing an electric motor and a device driven into rotation, for example, a circular saw, it is necessary to provide for the possibility of moving the electric motor on the plate. The fact is that the calculation does not give an absolutely accurate distance between the axes of the engine and the saw. In addition, it is necessary to ensure that the belt can be tensioned and compensate for its stretching.
The configuration of the pulley groove and its dimensions are shown in Fig. 2. The dimensions indicated by letters in the figure are available in the appendices to the relevant GOST standards and in reference books. But if there are no GOSTs and reference books, all the required dimensions of the pulley groove can be approximately determined by the dimensions of the existing V-belt (see Fig. 1), assuming that
e = c + h;
b = act+2c*tg(f/2) = a;
s = a/2+(4...10).
Since the case we are interested in is associated with a belt drive, the gear ratio of which is not very large, we do not pay attention to the angle of coverage of the smaller pulley by the belt when calculating.
As a practical guide, let's say that the material for the pulleys can be any metal. We also add that in order to obtain maximum power from a three-phase electric motor connected to a single-phase network, the capacitor capacities must be as follows:
Wed = 66Рн and Sp = 2Ср = 132Рн,
where Cn is the capacitance of the starting capacitor, μF; Ср - capacity of the working capacitor, μF; Рн - rated engine power, kW.
For V-belt transmission An important circumstance that greatly affects the durability of the belt is the parallelism of the axes of rotation of the pulleys.