Question and answers

2

What are the limitations of shoe brake ?

Limitations of a shoe brake :

1. Heavy side thrust causes bending of the shaft.

2. More wear & tear as the contact surface is large.

4

Explain the working of internal expanding brake with neat sketch.

Internal Expanding shoe brake:

An internal expanding brake consists of two shoes S1 and S2. The outer surface of the shoes are lined with some friction material (usually with Ferodo) to increase the coefficient of friction and to prevent wearing away of the metal. Each shoe is pivoted at one end about a fixed fulcrum O1and O2 and made to contact a cam at the other end. When the cam rotates, the shoes are pushed outwards against the rim of the drum. The friction between the shoes and the drum produces the braking torque and hence reduces the speed of the drum. The shoes are normally held in off position by a spring . The drum encloses the entire mechanism to keep out dust and moisture. This type of brake is commonly used in motor cars and light trucks.

4

Explain with neat sketch construction and working of eddy current dynamometer.

Eddy Current Dynamometer : It consists of a stator on which are fitted a number of electromagnets and a rotor disc made of copper or steel and coupled to the output shaft of the engine. When the rotor rotates, eddy currents are produced in the stator due to magnetic flux set up by the passage of field current in the electromagnets. These eddy currents oppose the motion of the rotor thus loading the engine. The eddy currents are dissipated in producing heat so that this type of dynamometer also requires some cooling arrangements. The torque is measured similar to absorption dynamometers i.e. with the help of moment arm. The load is controlled by regulating the current in the electromagnets.

4

Explain construction and working of eddy current dynamometer.

Construction and Working of Eddy current dynamometer :

8

A conical pivot with angle of cone as 100o, supports a load of 18 kN. The external radius is 2.5 times the internal radius. The shaft rotates at 150 rpm. If the intensity of pressure is to be 300 kN/m2 and coefficient of friction as 0.05, what is the power lost in working against the friction ?

4

Discuss the working of Rope brake dynamometer with the help of a neat sketch.

8

8

Determine the power lost in a footstep bearing due to friction if a load of 15 kN is supported and the shaft is rotating at 100 r.p.m. The diameter of bearing is 15cm and coefficient of friction is 0.05. Assume : (i) Uniform wear condition (ii) Uniform pressure condition.

4

Explain working of hydraulic brake dynamometer with sketch.

Hydraulic dynamometer is also called as water brake absorber. Invented by British engineer William Froude in 1877 in response to a request by the Admiralty to produce a machine capable of absorbing and measuring the power of large naval engines, water brake absorbers are relatively common today. The schematic shows the most common type of water brake, known as the "variable level" type. Water is added until the engine is held at a steady RPM against the load, with the water then kept at that level and replaced by constant draining and refilling (which is needed to carry away the heat created by absorbing the horsepower). The housing attempts to rotate in response to the torque produced, but are restrained by the scale or torque metering cell that measures the torque.

8

A simple band brake is operated by lever 40 cm long. The brake drum diameter is 40 cm and brake band embrance 5/8 of its circumference. One end of band is attached to a fulcrum of lever while other end attached to pin 8 cm from fulcrum. The coefficient of friction 0.25. The effort applied at the end of lever is 500 N. Find braking torque applied if drum rotates anticlockwise and acts downwards.

4

A casting weighing 9 kN hangs freely from a rope which makes 2.5 turns round a drum of 300 mm diameter revolving at 20 rpm. The other end of the rope is pulled by a man. Taking μ = 0.25, determine (i) the force required by the man (ii) the power to raise the casting.  

Given: W= T1= 9 kN =9000N, d= 0.3 m, N = 20 rpm , µ= 0.25

(i) Force reqd. by a man 

T2- force reqd. by man

As rope makes 2.5 turns,

Therefore angle of contact ,

θ =2.5x2π = 5 π rad.

We know that,

2.3 log {T1/T2} = µ θ = 0.25 x 5 π = 3.9275

log {T1/T2} = 3.9275/2.3 = 1.71 or T1/T2 = 51

T2 = 9000/51 = 176.47 N

(ii) Power to raise casting : 

As velocity of rope, v = πdN/60 = 3.14x0.3x20/60 = 0.3142 m/s

Power to raise casting = (T1-T2) x v = (9000-176.47) x 0.3142

                                     = 2.772 kW.

8

In a simple band brake, the band acts on the 3/4th of circumference of a drum of 450 mm diameter which is keyed to the shaft. The band brake provides a braking torque of 225 N.m. One end of the band is attached to a fulcrum pin of the lever and the other end to a pin 100 mm from the fulcrum. It the operating force is applied at 500 mm from the fulcrum and the coefficient of friction is 0.25, find the operating force when the drum rotates in the (i) anticlockwise direction and ii) clockwise direction