Download pdf - Modul 10 Clutches, Brakes, Couplings and Flywheel Part II

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Teknik Mesin - FTMD ITB MS3111 - Elemen Mesin

Modul 10Clutches, Brakes, Couplings

and Flywheel Part II


Modul 10 Clutches, Brakes, Couplings and Flywheel Part II

Energy ConsiderationSegment 1

Temperature RiseSegment 2

Friction MaterialSegment 3

Miscellaneous Clutches and CouplingsSegment 4

FlywheelSegment 5


Teknik Mesin - FTMD ITB

10.01. Energy Consideration

MS3111 - Elemen Mesin



𝜔2𝜔1

𝐼1𝐼2

Brakes or Clutch


• In the case of braking action, the kinetic energy must be absorbed by the brake system.

• For the clutching action, slipping must occur in the clutch until the driven parts have the same speed as the driver.

• Kinetic energy is absorbed during slippage of either a clutch or brake, and this energy appears as heat.

• The capacity of a clutch or a brake is limited by two factors: characteristic of material and ability of clutch/brake to dissipate heat.

• If the heat generated faster than it is dissipated, we have the temperature-rise problem.

8.8. Energy Considerations


To study what happens during a simple clutching/braking operation, let see the Fig. 16-1 .

𝜔2𝜔1

𝐼1𝐼2

Brakes or Clutch

Figure 16–1

Two basic assumptions:

✓ The two shafts are rigid

✓ The torque is constant.

➢ I1 and I2 have initial angular velocity of 1 and 2

(generaly 1 ≠ 2).

➢ During the clutch operation, both angular velocity change and eventually become equal.


➢ Equation of motion for inertia 1 and 2 can be written:

where ሷ𝜃 is angular acceleration and 𝑇 is clutch torque

➢ The instantaneous angular velocities for both inertia can be determined as follows

ሶ𝜃1 =−𝑇

𝐼1𝑡 + 𝜔1 (c) ሶ𝜃2 =

𝑇

𝐼2𝑡 + 𝜔2 (d)

(a)𝐼1 ሷ𝜃1 = −𝑇

(b)𝐼2 ሷ𝜃2 = 𝑇

➢ The difference in velocities (relative velocity), is

ሶ𝜃 = ሶ𝜃1 − ሶ𝜃2 =−𝑇

𝐼1𝑡 + 𝜔1 −

𝑇

𝐼2𝑡 + 𝜔2

(50)ሶ𝜃 = 𝜔1 − 𝜔2 − 𝑇𝐼1 + 𝐼2𝐼1𝐼2

𝑡


The clutching operation is completed at the instant when both angular velocity become equal. Let the time required for entire operation be t1. Then the Eq. (50) gives the time as

ሶ𝜃1 = ሶ𝜃2 ⇒ ሶ𝜃 = 0 ⇒ 𝑡1 =𝐼1𝐼2(𝜔1 −𝜔2)

𝑇(𝐼1 + 𝐼2)

Using Eq. (50), the rate of energy-dissipation during the clucthing operation is

𝑢 = 𝑇 ሶ𝜃 = 𝑇 (𝜔1 − 𝜔2) − 𝑇𝐼1 + 𝐼2𝐼1𝐼2

𝑡 (e)

This equation shows that the energy-dissipation rate is greatest(maximum) at the start of clutching, when t = 0.


➢ The total energy dissipated during the clutching operation or braking cycle is obtained by integrating Eq. (e) from t = 0to t = t1. The result is found to be

𝐸 = න

0

𝑡1

𝑢𝑑𝑡 = 𝑇න

0

𝑡1

(𝜔1 − 𝜔2) − 𝑇(𝐼1 + 𝐼2)

𝐼1𝐼2𝑡 𝑑𝑡

𝐸 =1

2

𝐼1𝐼2𝐼1 + 𝐼2

(𝜔1 − 𝜔2)2 (52)

Notes:

This is the energy that must be absorbed by clutch or brake.


➢ In USCS units, if the inertias are expressed in (lbf · in · s2), then the energy absorbed is in (in · lbf).Using this units, the heat generated in Btu is.

➢ In SI unit,

▪ inertias are expressed in [kg-m2]

▪ angular velocities are in [rad/s]

▪ energy dissipated is expressed in [Joules].

𝐻 =𝐸

9336(53)



10.02. Temperature Rise





➢ The temperature rise of the clutch or brake assembly can be approximated by the classic expression.

8.9. Temperature Rise

Δ𝑇 =𝐸

𝐶𝑝 ×𝑊(54)

where𝛥𝑇 = temperature rise, oF𝐶𝑝 = specific heat capacity, [Btu/(lbm ⋅ oF]; use 0.12 for steel or cast iron𝑊 = mass of clutch or brake parts, [lbm]


➢ A similar equation can be written for SI units. It is

Δ𝑇 =𝐸

𝐶𝑝 ×𝑚(55)

where𝛥𝑇 = temperature rise, oC𝐶𝑝 = specific heat capacity, [J/kg ⋅ oC]; use 500 for steel or cast iron𝑚 = mass of clutch or brake parts, [kg]

Notes:▪ The temperature rise equations above can be used to explain what

happens when a clutch or brake is operated.▪ However, there are so many variables involved that it would be most

unlikely that such an analysis would even approximate experimental results.


➢ If an object is at initial temperature T1 in an environment to temperature T, Newton’s cooling model is expressed as

𝑇 − 𝑇∞𝑇1 − 𝑇∞

= exp −ℎ𝐶𝑅𝐴

𝑊𝐶𝑝𝑡 (56)

where𝑇 = temperature at time 𝑡, oF𝑇1 = initial temperature, oF𝑇∞ = environmental temperature, oFℎ𝐶𝑅 = overall coefficient of heat−transfer,Btu/(in2 ⋅ s ⋅ oF)𝐴 = lateral surface area, in2.𝑊 = mass of the object, lbm𝐶𝑝 = specific heat capacity of the objecy, Btu/(lbm ⋅o F)

➢ Figure 23 shows an application of Eq. (56). The curve ABC is the exponential decline of temperature given by Eq. (56).


Figure 16–23The effect of clutching or braking operations on temperature.

▪ At time tB a second application of the brake occurs.

▪ The temperature quickly rises to temperature T2, and a new cooling curve is started.

▪ For repetitive brake applications, subsequent temperature peaks T3, T4,…,occur.

▪ Please read the reference book for a more detail explanation of the graph.


➢ When a disk brake has a rhythm such as discussed above, then the rate of heat transfer is described by another Newtonian equation:

𝐻𝑙𝑜𝑠𝑠 = ℎ𝐶𝑅𝐴 𝑇 − 𝑇∞ = ℎ𝑟 + 𝑓𝑣ℎ𝑐 𝐴 𝑇 − 𝑇∞ (57)

where

𝐻𝑙𝑜𝑠𝑠 = rate of energy loss, Btu/s

ℎ𝐶𝑅 = overall coefficient of heat−transfer,Btu/(in2 ⋅ s ⋅ oF)

ℎ𝑟 = radiation component of ℎ𝐶𝑅 ,Btu/(in2 ⋅ s ⋅ oF), Fig. 24(a)

ℎ𝑐 = convective component of ℎ𝐶𝑅 ,Btu/(in2 ⋅ s ⋅ oF), Fig. 24(a)

𝑓𝑣 = ventilation factor, Fig. 24(b)

𝑇 = disk temperature, oF𝑇∞ = ambient temperature, oF)


Figure 16–24(a) Heat-transfer coefficient in still air. (b) Ventilation factors.


➢ The energy E absorbed by the brake stopping an equivalent rotary inertia I in terms of original and final angular velocities ωo and ωf is given by Eq. (50) and Eq. (49) with I1 = I and I2 =0

➢ The temperature rise T due to a single stop is

Tmax has to be high enough to transfer E Btu energy in t1 seconds.

𝐸 =1

2

𝐼

9336(𝜔0

2 − 𝜔𝑓2) in Btu (58)

Δ𝑇 =𝐸

𝑊𝐶(59)


➢ For steady state, rearrange Eq. (53) as

➢ Cross-multiply, add Tmax to both sides of Eq. (57) and letting

Tmax – Tmin = T, and rearrange, we obtain

𝑇∞Δ𝑇

1 − exp −𝛽𝑡1 max

(60)

with A = lateral area (heat transfer area)

𝑇𝑚𝑖𝑛 − 𝑇∞𝑇𝑚𝑎𝑥 − 𝑇∞

= 𝑒𝑥𝑝 −𝛽𝑡1

𝛽 =ℎ𝐶𝑅𝐴

𝑊𝐶𝑝


Example 5

Solution Example 5


From Fig. 24- (b) with air speed of 25 ft/s we find fv = 4.8, and so we have


Answer

Answer



10.03. Friction Material





8.10. Friction MaterialA brake or friction clutch should have the following lining material characteristics to a degree that is dependent on the severity of service:

• A high and uniform coefficient of friction.

• Imperviousness to environmental conditions, such as moisture.

• The ability to withstand high temperatures, together with good thermal conductivity and diffusivity, as well as high specific heat capacity

• Good resiliency.

• High resistance to wear, scoring, and galling.

• Compatible with the environment

• Flexibility.


➢ Table 16–2 gives area of friction surface required for several braking powers.

➢ Table 16–3 gives important characteristics of some friction materials for brakes and clutches.

➢ The manufacture of friction materials is a highly specialized process, and it is advisable to consult manufacturers’ catalogs and handbooks, as well as manufacturers directly, in selecting friction materials for specific applications.

➢ Selection involves a consideration of the many characteristics as well as the standard sizes available.

➢ Table 16–4 lists properties of typical brake linings.➢ Table 16–5 includes a wider variety of clutch friction materials, together

with some of their properties.


Material Fabrication Field of Application

Woven-cottonlining

As a fabric belt which impregnated with resin and polymerized

Mostly in heavy machinery; usually in rolls up to 50 ft length. Thickess frm 1/8 in to 1 in; width up to 12 in.

Woven-asbestos lining

Similar to cotton lining and may contain metal particles. Not as flexible as cottong lining.

Widely used as a brake material in heavy machinery. Comes in smaller range of sizes.

Molded-asbestos lining

Contain asbestos fiber and friction modifiers; thermoset polymer is used, with heat, to form a rigid/semi0rigid molding.

Principal use is in drum brakes.

Molded-asbestos pads

Similar to molded lining but have no flexibility.

Used both for clutches and brakes.

Sintered-metal pads

Made of a mixture of copper and/or iron particles with friction modifiers; molded under high pressure and the heated to a high temperature to fuse the material.

Pads are used in both brakes and clutches for heavy-duty applications.

Cermet pads Similar to the sintered-metal pads and have a substantial ceramic content.

Pads are used in both brakes and clutches.

Table of several common friction material







10.04. Miscellaneous Clutches and Couplings





8.11. Miscellaneous Clutches and Coupling Square-jaw clutch as shown in Fig. 25 (a) in one form of positive-contact clutch.

These clutches have the following characteristics:

1. They do not slip

2. No heat is generated

3. They cannot be engaged at high speeds

4. Sometimes they cannot be engaged when both shafts are at rest

5. Engagement at any speed is accompanied by shock.

The greatest differences among the various types of positive clutches are

connected with the design of the jaws (ractched-shaped, spriral-shaped, gear-

tooth-shaped).

Although positive clutches are not used to the extent of the frictional-contact

types, they do have important applications where synchronous operation is

required.


Figure 16–25(a) Square-jaw clutch; (b) overload release clutch using a detent.

Square-jaw Overload-release clutch

../../Videos/Jaw clutch.MP4

../../Videos/Safety clutch 6 (spring arm).MKV


Devices such as linear drives or motor-operated screwdrivers must run

to a definite limit and then come to a stop.

An overload-release type of clutch required for theses application. Fig.

25 (b) is a schematic drawing illustrating the principle of the operation

of such clutch.

This clutches are usually spring-loaded so as to release at a

predetermined torque.

The clicking sound which is heard when the overload point is reached is

considered to be a desible signal.

Both fatigue and shock must be considered in obtaining the stress and

deflections of the various portions of postive clutches. In addition, wear

must generally be considered.


An overrunning clutch or coupling

permits the driven member of a

machine to “freewheel” or

“overrun” because the driver is

stopped or because another source

of power increases the speed of the

driven mechanism.

The driving action is obtained by wedging the rollers between the

sleeve and the cam flats.

There are many varieties of overrunning clutches available, and

they are built in capacities up to hundreds HP.

Since no slippage is involved, the only power loss is that due to

bearing friction and windage.

../../Videos/Roller overrunning clutch 2.MP4



Plain shaft coupling Light duty toothed coupling

BOST-FLEX through-bore coupling Three-jaw coupling




10.05. Flywheel





8.12. Flywheel A flywheel is an inertial energy-storage device. It absorbs mechanical

energy by increasing its angular velocity and delivers energy by decreasing its velocity.

A flywheel is used to smooth out variations in the speed of a shaft caused by torque fluctuations.

Piston comp., punch presses, rock crusher, etc. all have time-varying loads. The prime mover (ex.: internal combustion engine) introduce torque

oscillations to the transmission shaft. Other systems may have both smooth torque sources and loads such as an

electrical generator driven by a steam turbine ➔ no need flywheel. If the source of driving torque or the load torque have a fluctuating

nature (see Fig. 16-28) then a flywheel is usually needed.


Figure 16–28

Relation between torque and crank angle for a one-cylinder, four-stroke–cycle internal combustion engine.



Note:

➢Ti is considered positive and To negative.

➢Values of Ti and To may depend on the angular displacement and/or their angular velocities.

➢ In many case (ex. induction motor), its characteristic depends upon only the speed of the motor. Generally, motor manufacturers publish the torque-speed characteristics of their various motors.

The equation of motion for the flywheel represented in Fig. 16-1 is

𝑇i, 𝜃i

Flywheel

𝐼, 𝜃

𝑇o, 𝜃o

𝑇 = torque; 𝜃 = angular displacement

Figure 16–1

𝑀 = 𝑇𝑖(𝜃𝑖 , ሶ𝜃𝑖) − 𝑇𝑜(𝜃𝑜, ሶ𝜃𝑜) − 𝐼 ሷ𝜃 = 0

or𝐼 ሷ𝜃 = 𝑇𝑖(𝜃𝑖 , 𝜔𝑖) − 𝑇𝑜(𝜃𝑜, 𝜔𝑜) (a)


When the input and output torque functions are given, Eq. (a) can be solved for

the motion of the flywheel using well-known techniques for solving linear or non-

linear differential equation.

We can simplify this by assuming a rigid shaft, giving i = = o .

Thus, Eq. (a) becomes

𝐼 ሷ𝜃 = 𝑇𝑖(𝜃, 𝜔𝑖) − 𝑇𝑜(𝜃, 𝜔𝑜) (b)

When the two torque functions are known and the starting values of the

displacement and velocity are given, Eq. (b) can be solved.

Here, we are not really interested in the instantaneous values of these terms at all.

Primary, we want to know the overall performance of the flywheel (what should its

moment of inertia?, how do we match the power sources to the load?, what are the

resulting performance?, etc.)


To gain insight into the problem, a hypothetical situation is diagrammed in Fig. 24.

𝑈𝑖

Energy is given to and store by flywheel

1 cycle

𝜃1 𝜃2

𝜃3 𝜃4

𝜔1

𝑇i, 𝜃i

𝜔4

𝜔2

𝜔𝑇𝑖

𝑇𝑜

𝑇,𝜔

𝜃𝑈𝑜

Energy released back to system

The work input to the flywheel is the area of the rectangle between 1

and 2 , or.

𝑈𝑖 = 𝑇𝑖 × (𝜃2 − 𝜃1) (c)

Figure 16–27


The work output of the flywheel is the area of the rectangle

between 3 and 4 , or.

𝑈𝑜 = 𝑇𝑜 × (𝜃4 − 𝜃3) (d)

Three conditions possible (assume: no friction losses)

If 𝑈𝑜 > 𝑈𝑖 ⇒ 𝜔4 < 𝜔1; load uses more energy than has been delivered to the flywheelIf 𝑈𝑜 = 𝑈𝑖 ⇒ 𝜔4 = 𝜔1; the gains and the losses are equalIf 𝑈𝑜 < 𝑈𝑖 ⇒ 𝜔4 > 𝜔1; load uses less energy than has been delivered to the flywheel

These relation can also be written in terms of kinetic energy.

At = 1 the flywheel has a velocity of 1 rad/s, and so its kinetic energy is

𝐸1 =1

2𝐼𝜔1

2 (e)


At = 2 the velocity is 2 rad/s, and so its kinetic energy is

𝐸2 =1

2𝐼𝜔2

2 (f)

Δ𝐸 = 𝐸2 − 𝐸1 =1

2𝐼 𝜔2

2 −𝜔12 (61)

Thus the change in kinetic energy is

Notes:

➢ Many of the torque displacement functions encountered on practical engineering situations are so complicated that they must be integrated by approximate methods.

➢ One simplest integration routine is Simpson’s rule. This approximation can be handled on any computer and is short enough to use on the smallest programmable calculators.


Example 1.

The crankshaft of a punch press rotates at 60 rpm, causing holes to be punched in a steel part at the rate of 60 punches per minute.

The crankshaft torque requirement is shown in Fig. 25. The press is driven (through suitable speed reducers) by a 1200-rpm motor.

Neglecting any “flywheel effect”, what motor power is required to accommodate the peak crankshaft torque? Figure 25


Assumption:1. Friction losses are negligible2. No energy is stored as rotational kinetic energy (no flywheel)3. The motor delivers maximum torque continuously

Solution:1. Neglecting friction losses ➔ motor power = crankshaft power out, and the 20 : 1

speed reduction (1200 rpm/60 rpm) is associated with a 20 : 1 torque increase.Hence, the motor torque required is 10 kN·m/20 = 500 N·m.

2. Motor delivers the torque continuously. The work capacity corresponding to 1 revolution of the shaft is 2 x 500 N·m = 1000 J.

3. In 1 second, the motor shaft turns 20 revolutions (1200 rpm), the work capacity is 20x1000 J = 20 kJ.

This is equivalent to 20 kW or 62.8 kW or 84.2 hp (a big motor ??)


Example 2.

For the punch press in Example 1determine the motor power capacity required if we use a flywheel.

The energy required for the press is represented by the area under the actual crank torque versus the crank angle curve of Fig. 26, which

is = 2 kN·m or 6283 Joules.

Figure 26


Assumption: Friction losses are negligible

Solution:

1. Fig. 26 shows that average torque requirement during the actual punch stroke is 6

kN and that the punch stroke lasts for /3 rad. (The energy involved is this area

under the curve is 2 kN·m or 6283 J).

By using the flywheel that permits the motor to deliver a constant torque over the

entire 2 rad, we can reduce the torque requirement to 1 kN·m➔ resulting a total

energy of 2 kN.m (kJ)

This is shown in Fig. 26 as “uniform torque supplying equal energy”.

2. Since, at the same shaft speed, motor torque and motor power are proportional,

thus the 10:1 reduction (20➔ 2) in motor torque corresponds to a like reduction

in power rating required.

Hence, the motor power required is 62.8 kW/10 = 6.28 kW or equal to 8.4 hp (a

small motor).


Example 3.

Continuing with the previous problem, we choose to design a flywheel that rotates at 1/3 motor speed and that limits motor speed fluctuation to the range of 900 rpm to 1200 rpm. The flywheel is to be made of steel and have the geometric proportions as shown in Fig. 27.

Figure 27

To simplify the calculation, assume that the inertia contributed by the hub and arms is negligible. Determine the required flywheel polar moment of inertia, I, and the diameter, d.


Assumption and decisions:

Assumptions:

1. Friction losses are negligible.

2. The inertia contributed by the hub and arms is negligible.

Decisions:

1. The flywheel rotates at 1/3 motor speed.

2. Motor speed fluctuation is limited to 900 to 1200 rpm.

3. The flywheel is made of steel.

4. Geometric proportions for the flywheel are shown in Fig. 27.


Solution:

1. Fig. 26 shows that during the actual punch stroke, energy provided by the motor is represented by an approximate rectangle 1 kN·m x /3 rad. Thus, the motor provides 1000 N x /3 = 1047 J of the total of 6283 J required.The flywheel must provide the remaining (6283 – 1047) = 5236 J.

2. The flywheel inertia must be such that

5236 =1

2𝐼 𝜔𝑚𝑎𝑥

2 −𝜔𝑚𝑖𝑛2 (*)

Figure 26


2. The moment of inertia for a hollow cylinder is

𝜔1

31200 rpm

40𝜋

3

rad

s

rad

s max

𝜔1

3900 rpm

30𝜋

3

rad

s

rad

s min

Substitution these values to the Eq. (∗) gives

5236 =1

2𝐼 13.33𝜋 2 − 10𝜋 2 ⇒ 𝐼 = 13.80 kg ⋅ m2

𝐼 =𝜋

32𝑑𝑜4 − 𝑑𝑖

4 × 𝐿 × 𝜌

13.80 =𝜋

32𝑑4 − (0.8𝑑)4 0.2𝑑 7700

From wich we find 𝑑 = 688 mmComment:

If the inertia contributed by the hub and arms is included in the analysis, we would find that a smaller d is required.


Thank YouLecturers

Faculty of Mechanical and Aerospace Engineering

Institut Teknologi Bandung

Modul 10 Clutches, Brakes, Couplings and Flywheel Part II

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