We Have Numbers Of Free Samples

Question 1:

(a) Draw a pV diagram to illustrate the operation of a two-stage compressor with intercooling

(b) List the advantages to be gained by the use of a multi-stage design compressor.

Solution 1:

 (a)

(b) Advantages of Multi stage design of Compressors:

With the help of multi staging the work done to compress the air is reduced and a considerable amount of power can be saved.

Temperature control is possible as intercooling can be implemented.

Mechanical failure due to overheating is reduced.

Size of machine is made compact.

Question 2:

(a) With the aid of a sketch, describe briefly the operating principle of a rotary vane compressor.

(b) State the reasons why oil must be injected into this machine for its efficient operation.

Solution 2:

(a) Rotary vane compressor:

Fig1. Illustrated the cross section of rotary vane compressor.

The rotary vane compressor consists of inlet, outlet, casing, rotor and vanes. Initially air is allowed inside the inlet, where the air is sucked due to partial vacuum created inside the casing. The rotor is located eccentrically inside the cylindrical casing. A set of spring loaded vanes are arranged in the rotor slots.

The atmospheric air is entrapped between the two vanes. When the rotor starts to rotate the air which is entrapped is further compressed by the action of vanes varying cross section and then finally discharged through outlet ports.

(b) Why oil must be injected into machined:

The oil must be injected into machines for the purpose of lubrication which will resist friction and thus provides long life and higher efficiencies. Reduction of friction leads to less wear and tear, reduction of heat developed in machine parts, etc.

Question 3:

(a) A mass of 400 kg is to be raised by the actuation of two identical hydraulic cylinders with a piston diameter of 120 mm. Calculate the required system pressure to just raise the load.

Solution 3:

(a)

Given:

Mass: M = 400 Kg

F = ma = 4000 N

Diameter: 120 mm

P=F/A

Where,

F is force in N

A is area in m²

Area of the cylinder:

A=πr^2=0.0113097 m^2

P=4000/0.01131=3.53 bar

The pressure required is found to be 3.53 bar

(b) If the load is to be raised 600 mm in 10 seconds, what will be the required flowrate (Q) in ℓ min–1?

Given:

Raised by: 600mm

Piston Diameter: 120 mm (From previous problem)

Time, t = 10 sec

Q =?

Q=V/t

Volume:

V=πr²h=6.78e-3 m³

Q=v/t=6.7858e-4 m³/s

Q = 40.7148 L/min.

Question 4:

Describe the operation of a variable displacement axial piston (swash plate) pump, indicating the means by which its output is varied.

Solution 4:

Swash plate axial piston pump:

The swash plate axial piston pump consists of a swash plate through which the displacement of liquid can be controlled by simply adjusting the angle of the swash plate. The piston inside the cylinder changes its stroke as per the swash position, the inlet fluid is pumped to the certain limit and the expelled at the outlet.

Question 5:

With reference to basic constructional features, contrast the methods of compression employed by positive displacement and dynamic air compressors.

Solution 5:

The positive displacement compressors sucks the air in atmosphere by creating a partial vacuum inside the cylinder during the suction stroke then the compression stroke occurs in which the air is being compressed, finally the compressed air is allowed to the storage tank.

Dynamic air compressors:

In case of dynamic air compressors the fluid compression is achieved by inertia of the motion of the fluid. The fluid is compressed with the help of blade vanes attached to the rotor, it is of rotating type.

Question 6:

An item of plant requires 2 m3 min–1 of pulsation-free and oil-free compressed air supplied at a pressure of 7 bar. Select and size a suitable type of machine with regard to output FAD (free air delivered) and quality of air supply.

Solution 6:

Q = 2 m³/min

Pressure, P = 7 bar

For this application Piston air compressor is selected which will deliver quality air at 7 bar also the discharge rate is achieved quite good.

Question 7:

Air is drawn into a compressor at normal temperature and pressure (N.T.P.) and compressed to a pressure of 6 bar gauge. After compression the air is delivered at 1.2 m3 min–1 and cooled to a temperature of 30°C, at which point condensate is collected at the rate of 2 litres per hour. Estimate the FAD (N.T.P.) of the compressor, and the relative humidity of the air entering the compressor.

Solution 7:

P = 6 bar

Q = 1.2 m3/min

T = 30 ⁰ C

Condensate: 2 L/min

FAD: P – (Pv(RH))

Pv: Vapour pressure

RH- Relative humidity.

 

The relative humidity of air at 30 degree C is 0.0055

The vapour pressure of air at 30 C is 0.04199122 bar

FAD: 5.99 m³/min

Question 8:

Describe the difference between regenerative absorption and chemical absorption air drying.

In regenerative absorption the water molecules are being absorbed by the material and kept by it until it is all vaporized and dried, in case of chemical absorption the fluid molecules are instantly vaporized with the help of chemicals.

[Download not found]

Download