undefined | unit 6 compressor

Unit – 6

Compressor

Q1) A two stage air compressor with perfect intercooling takes in air at 1 bar & 270c. The law of compression in both stages is pv1.3 = c. The compressed air is delivered at 9 bar from HP cylinder to an air receiver. Calculate per kg of air.

a). Minimum work of compression

b). Heat rejected in intercooler

c). Work required for single stage compression to the same delivery pressure.

A1)

Given:

P1 = 1 x 105 N/m2

T3 = T1 = 270c = 27 +-273 = 300 K

n = 1.3

P3 = 9 x 105 N/m2

Minimum Work = mRT1 [ ()(n-1)/2n - 1 ]

required to compress

= x1 x 287 x 300 [ ()(1.3 -1)/(2 x 1.3) - 1]

= 215.32 x 103N.m or Joule

For Perfect intercooling

P2=

= 3 bar = 3 x 105 N/m2

() = ()(n -1)/n

() = ()(1.3 -1)/1.3

T2 = 386.57 K

Heat rejected = mCp(T2 – T3)

in intercooler

= 1 x 1.005 x (386.57 - 300)

= 87 KJ/kg

Work required for = mRT1[ ()(n-1)/n - 1 ]

Single stage compression

= x1 x 287 x 300 [ ()(1.3 -1)/( 1.3) - 1]

= 246.39 x 103N.m

Q2) A two stage single reciprocating air compressor takes in air 1 bar & 200c. it discharges air at 40 bar. The temperature rise in intercooler is 220c. Assume polytrophic compression & expansion with n = 1.33, perfect intercooling with ideal intermediate pressure. Calculate work of compression, % saving in work of compression as compared to single stage. C). Mass of cooling water required in intercooler per kg of air. Assume Cp = 1.005 & 4.187 kj/kg k for air & water respectively.

A2)

Given

P1 = 1 bar = 1 x 105 N/m2

T3 = T1 = 200c = 20 + 273 = 293 K

P3 = 40 bar = 40 x 105 N/m2

Δtw = 220c

n = 1.33

Work required = mRT1 [ ()(n-1)/2n - 1 ]

for compression

= x1 x 287 x 293 [ ()(1.33 -1)/(2 x 1.33) - 1]

W1= 393.37 x 103N.m or Joule

Work required for = mRT1 [ ()(n-1)/n - 1 ]

single stage compression

= x1 x 287 x 293 [ ()(1.33 -1)/( 1.33) - 1]

W2= 507.51 x 103 Joule

% Saving in work = (W2 – W1)/W2 x 100

= (507.51 – 393.37)/507.51 x 100

= 22.49 %

P2 =

= 6.32 bar

() = ()(n -1)/n

() = ()(1.33 -1)/1.33

T2 = 462.960c

Heat rejected = Heat gain

in intercooler by water

mCp(T2 – T3) = mwCpwΔtw

1 x 1.005 x (462.96 - 293) = mw x 4.18 x 22

mw = 1.85 kg

Q3) A single cylinder double acting air compressor sucks air at the rate of 5 m3/min at a pressure of 100 kPa & 250c. it delivers air to the receiver at a pressure 6 bar. The speed of the compressor is 250 rpm & stroke is 1.5 times cylinder diameter. Neglect the effect of clearance & if law of compression is PV1.25 = C.

Find cylinder dimensions, indicated power, shaft power in the mechanical is 90%.

A3)

Given:

V1 = 5 m3/min = 5/60 = 0.0833 m3/sec

P1 = 100 x 103 N/m2

T1 = 250c = 25 + 273 = 298 k

P2 = 6 bar = 6 x 105 N/m2

N = 250 m

L = 1.5 D

n = 1.25

Indicated power = P1V1 [ ( )n-1/n - 1 ]

= x 100 x 103 x 0.0833 [ ()(1.25-1)/( 1.25) - 1]

= 17.95 x 103 watts

ɳmech= x 100

90 = x 100

Shaft Power = 19.94 x 103 watts

V1 = Vs = π/4 x D2 x L

0.0833 = π/4 x D2 x 1.5 D

D = 0.41 m

L = 1.5 D = 0.619 m

Q4) A single stage reciprocating compressor has a swept volume 2000 cm3, runs at 800 rpm & takes in air 1.013 bar & 150c. It operates on a pressure ratio of 8, with a clearance of 5% of the swept volume. Assume polytrophic compression & expansion with n = 1.25. Calculate volumetric efficiency, indicated power, isothermal efficiency, actual power needed to drive the compressor, if mechanical efficiency is 85%.

A4)

Given:

Vs = 2000 cm3 = 2 x 10-3 m3

N = 800 rpm

P1 = 1.013 bar = 1.013 x 105 N/m2

T1 = 150c = 15 + 273 = 288 k

P3/P4 = P2/P1 – 8

Vc = 5/100 Vs = 5/100 x 2 x 10-3

= 1 x 10-4 m3

ɳ = 1.25

ɳmech = 85%

V1 = Vs + Vc = 2.1 x 10-3 m3

V3 = Vc = 1 x 10-4

() = ()1/n

() = (8)1/1.25

V4 = 5.27 x 10-4 m3

Work done = P1(V1 – V4) [ ()n-1/n - 1 ]

W = x 1.013 x 105 x (2.1 x 10-3- 5.27 x 10-4)[ (8)(1.25-1)/( 1.25) - 1]

= 410.88 J

Indicated Power = W x N/60 = 410 x 800/60

= 5478.45 watts

ɳv =x 100

= 78.60 %

Isothermal work = P1(V1 – V4) x ln()

= 1.013 x 105 (2.1 x 10-3 – 5.27 x 10-4) x ln(8)

= 331.34 J

Isothermal power = WI x N/60

= 331. 34 x 800/60

= 4417.97 watts

ɳiso= x 100

= x 100

= 80.64 %

ɳmech = x 100

85 = x 100

Shaft Power = 6445.23 Watts

= 6.44 kw

Q5) Explain the construction and working of single stage compressor.

A5)

Construction:

Single stage single acting air compressor consists of a piston, which reciprocates inside a cylinder having connecting rod and crank mechanism.

There are inlet and delivery valves mounted in the head of cylinder.

The inlet and delivery valves are of pressure differential type i.e., they operate as a result of pressure difference across the valves.

Working:

Working of single stage reciprocating air compressor is completed in two strokes.

a). Suction Stroke:

When the piston moves in the downward direction, from (T.D.C.) to (B.D.C), the air compressed in the previous compressions and entrapped in the clearance space begins to expand.

Due to expansion, the pressure of air inside the cylinder starts to decrease.

After some travel of piston from T.D.C. towards B.D.C, the pressure of air inside the cylinder falls below the atmospheric pressure.

Thus, due to pressure difference, the inlet valve gets opened. Now, the atmospheric air is sucked into the cylinder.

b). Compression Stroke:

When the piston starts moving from BDC to TDC, the pressure of sucked air inside the cylinder begins to increase.

When the pressure of air inside the cylinder increases above the atmospheric pressure, the inlet valve gets closed.

Further movement of piston towards TDC causes compression of air sucked during suction stroke.

Due to this, pressure of air inside the cylinder goes on increasing.

As soon as, the pressure reaches upto desired discharge pressure, the delivery valve gets opened and compressed air is delivered into the receiver.

Now again the piston starts moving from TDC to BDC, and the cycle is repeated again and again.

Q6) Derive the expression for work done on air in a reciprocatingcompressor without clearance.

A6)

Expression for work done on air in a reciprocating compressor without clearance.

A P-V diagram of a reciprocating compressor without clearance has been shown in fig.

Process 0-1 Constant pressure suction

Process 1-2 Polytrophic compression of air

Process 2-3 Constant pressure delivery.

Work transfer will be given by area under the curve on P-V diagram.

Work transfer per cycle = ---- (1)

For process 1-2

PVn = C

Vn = C/P

V = where K = C1/n

:. Put in eqn (1)

Work transfer per cycle = -

= - K[ P-1/(n + 1) / -1/n + 1 ]P2P1

= - K * [ P2-1/(n + 1) – P1-1/(n + 1) ]

= [ V2P21/n.P2-1/n. P2 – V1P11/n . P1-1/n. P1 ]

= [ P2V2 – P1V1 ]

Hence,

Work done on the airper cycle= [ P2V2 – P1V1 ]

= P1V1[]

= P1V1[ ()-1/n– 1. . . () = ()-1/n

=P1V1[- 1]

= mRT1[- 1]

Q7) Define and Derive the Volumetric efficiency and Isentropic efficiency.

A7)

Volumetric efficiency:

It is defined as the ratio of actual volume of air sucked in per cycle to the swept volume per cycle.

ɳv =

= 1-

= 1- ……… k = clearance ratio.

=1- k[()1/n -1] () = ()1/n

Isothermal efficiency:

It is defined as ratio of isothermal power to the indicated power. It is also called as compressor efficiency.

Isothermal efficiency =

ɳiso = .. Single stage compressor

ɳiso = .. Double stage compressor

Q8) Describe the Actual indicator diagram for air compressor.

A8)

Actual indicator diagram for air compressor:

During suction stroke, the charge of air is drawn along 4-1 at constant pressure P1, which is slightly below Patm.

At point 1 piston completes suction stroke and starts compression stroke.

During starting both the valves being closed air will be compressed along 1-2. At point 2 pressure P2 is reached which is slightly higher than the pressure in the receiver.

The delivery valve at this point opens and the air is delivered along 2-3 into the receiver.

The network input required for compression and delivery of air per cycle is given by the area 1-2-3-4 on the P-V diagram.

Theoretical Indicator Diagram Actual Indicator Diagram

In actual air compressor, some air is present in the clearance volume. This air expands during suction stroke.

At point 4, the inlet valve will not open in actual practice because of valve inertia and pressure differential required to open the valve.

Thus, the pressure drops away until the valve is forced to open. Some valve bounce will take place and slowly intake will become nearly steady.

This negative pressure is known as Intake depression.

A similar situation occurs at point 2 i.e., at point 2 pressure has to increase otherwise valve will not open.

Actual work required to compress and deliver the air will be greater than theoretical work.

Q9) Explain the construction and working of Roots blower.

A9)

Construction of Roots blower:

Lobe type or Roots blower type rotary compressor consists of two rotor lobes rotating in an air-tight casing, which has inlet and outlet ports.

The lobes are so designed that, they provide an air-tight sealing at the point of their contact.

Fig. Roots Blower

Working:

Mechanical energy is provided to one of the rotors from external source (an electric motor) and the second rotor is gear driven from the first. Thus, both rotors rotate at the same speed.

The rotation of rotors creates space in the casing at the entry port.

The air is drawn into the casing to fill the space.

With further movement of lobed rotor, the air is trapped between casing and one rotor.

At this condition, we can observe that tip of rotor touches the casing.

This part of blower is not open to suction port.

But, the air flows into the space created by rotation of the other rotor.

The second rotor also carries out the same cycle as the first rotor after 900.

The trapped volume of air is not internally compressed, it is only displaced at high speed from suction side to delivery side.

The continued rotation of lobes opens the discharge port.

The compressed air at higher pressure is present at the delivery side.

Therefore, when the rotor lobe uncovers the discharge port, some pressurised air enters into the space between the rotor and casing of the compressor.

This flow of air is called as backflow of air.

This backflow of air continues until the pressure in the blower gets equalized.

After the backflow, the air is compressed irreversibly at constant volume.

Finally, at higher pressure, the air is delivered from the blower to receiver.

Q10) Explain the construction and working of vane type rotary compressor.

A10)

Construction:

A vane type rotary compressor consists of a rotor rotating eccentrically in an air-tight casing with inlet and outlet ports.

The rotor has a number of slots containing vanes, which are spring loaded.

The disc rotates in anticlockwise direction. During this rotation, the vane’s remain in contact with the casing, due to spring forces.

Fig. Vane type rotary compressor

Working:

When the rotor rotates, the vanes are pressed against the casing, due to centrifugal force and form air– tight pockets.

The mechanical energy is provided to the rotor from some external source like electric motor.

As the rotor rotates, the air is trapped in the pockets formed between the vanes and casing.

As the rotor further rotates, compression takes place due to decrease in volume provided for the trapped air.

Further compression is obtained by backflow of air from the receiver, i.e. when the rotating vane uncovers the exit port and compressed air enters into the receiver, then at the same time, some air flows back into the pocket.

Thus, the pressure of air, entrapped in the pocket, is increased first by decreasing the volume and then by back flow of air as shown in Fig.

Now, the compressed air is delivered to the receiver by the rotation of the vanes. Finally, the air at a high pressure gets stored inside the receiver.

Q11) Explain the construction and working of screw compressor.

A11)

Construction:

Screw compressor consists of two mutually engaged helically grooved rotors suitable housed in a casing.

The male rotor consists of four helical lobes and is normally the driven rotor.

The female rotor has six flutes or gullies and is the driven rotor.

The four helical lobes of male rotor are engaged in corresponding gullies (flutes) of female rotor.

Screw compressor is directly coupled to the prime mover (electric motor) and require low starting torque.

Fig. Screw compressor

Working:

When the driving rotor rotates, an inter lobe space between a pair of lobes of male-female rotors and the housing nearest to suction end gets opened and is filled with air.

Now, the air so trapped is moved both axially and radially with the rotation of rotors and gets compressed due to reduction in volume, until it reaches the discharge end.

This compressed air is finally delivered through discharge port.

As the number of lobes of male rotor and number of flutes of female rotor are different, therefore the male rotor and female rotor rotate at different speed.

A four-lobe male rotor will drive six gully (flutes) female rotor at two third of its speed.

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