TWO-STAGE AIR COMPRESSOR

COURSE TITLE

:

THERMODYNAMICS LABORATORY

TOPIC

:

TWO-STAGE AIR COMPRESSOR

1.0

OBJECTIVE  Understanding the polytropic process in the two-stage air compressor  Determination of the polytropic efficiency of a compressor.

2.0

EXPERIMENTAL EQUIPMENT Air compressor Model PCB 100.

1

3.0

THEORY An air compressor uses an electric motor to convert the electrical energy into mechanical energy which is converted to thermodynamic energy in form of compressed air. Air compressors are utilized to raise the pressure of a volume air. All air compressors designs utilize at least the basic principles: Staging, Intercooling, Compressor displacement and volumetric efficiency. Compressor is staged to reduce the temperature rise and improve the compression efficiency. The temperature of the air leaving each stage is cooled prior to entering the next stage (Intercooling). An after-cooling should also be used to reduce the temperature of the air before it enters the receiver (storage tank) and to discharge the cooling medium to the surrounding at a low temperature (see figure 1).

1

4

Compressor first stage

After-cooler

Compressor second stage Storage tank

2

3

Intercooler

3.1

Fig.1: Two-stage air compressor schematically Compressor Outline Referring to figure 2at page 9, the filtered free air is induced at the first stage compressor intake (P0, T0) for the first compression. Before going into the second stage compressor, this air is cooled by the heat exchanger to reduce the high temperature after the first compression P1, T1). The intake condition of the air for the second compressor is at P1 and T2. Before this last compressed air is delivered into the storage tank, it is again cooled by the second heat exchanger (after cooler) to reduce its temperature. The temperature of the cooling medium also decreases after cooling process (T9).Water in use hereby as cooling

2

medium for both the heat exchangers. The mass flow rate shall be regulated in the same level during the operation to prevent from unnecessary losses. 3.2

Theory and Equations Air compression in the compressor usually undergoes a polytropic process in form pv n  const , where n is the polytropic coefficient. With assumption that the real air is considered as ideal gas with its characteristic equation pV / T  const , then we can obtain a useful relationship between the temperatures and pressure, or temperatures and volume for any point during the process as follows: pV n  C

=>

(1)

p1V n 1  p 2V2n

And the property of the ideal gas follows the equation: =>

pV C T

(2)

p1 V1 pV  2 2 T1 T2

From (2) it yields:

T2 pV  2 2 T1 p1V1

And together with equation (1) we obtain:

(3)

 p2  T2    T1  p1 

n 1 n

(4)

(Note: index 1 and 2 refers to the air condition at inlet and outlet of the compressor, respectively). The value for the thermodynamic properties p and T at the intake and the outlet are measured. 3.2.1

Polytropic Compression a. Work and efficiency The work required to compress the air from the initial pressure p1 to the delivery pressure p2 is given by the equation (5):

3

p2

Windicated 

 vdp

(5)

p1

Applying the equations (1) into (5) and using the ideal gas equation, we get the equation for the indicated work: 

Win

n  p1v1   n 1 

 p2   p1

n 1 n

  

  1  

For the two-stage compressor the total work done on the system is the sum of the work done by each compressor. Between Work and polytropic Power (Pp) is given by the relation: .

(7)

Pp  m Win

Equation (6) applied in (7) and using the ideal gas equation we get: 

. n Pp  m RT1   n 1



 p2   p1

n 1 n







  1  

(8)

The polytropic efficiency of the compressor  p :

p 

Pp Ps

(9)

Where Ps is the motor shaft power. To produce Ps, the compressor requires an electric power Pe:

m 

Ps Pe

(10)

Where  m is the motor efficiency. The value for  p  98% , m  80% can be used. b. Heat Loss in the heat exchanger Heat produced in the air compressors is rejected into the heat exchangers. The total heat absorbed by the cooling medium (water): Q  Q1  Q2

(11) 4

.

.

Q  m1 (hout ,1  hin,1 )  m2 ( hout , 2  hin , 2 )

Applied this equation to our compressor it becomes: .

.

Q  m1 (h8  h7 )  m 2 (h9  h7 ) .

(12)

.

For m we use m   W ρ and W is the density and the mass flow rate of the water. c. Heat Loss in the compressor From the first law of thermodynamics the energy balance per mass in the   compressor   with   an   assumption   that   the   kinetic   and   potential energy change is negligible it follows the equation: q  w  h

(13)

Rearranging this equation we get the heat loss per unit mass in each compressor: q  h  w         

   (14)

                                                                                      

4.0

ADDITIONAL THEORY

5

The original air compressor is the lung. When you take a deep breath to blow out your birthday candles, for example, you're increasing the pressure of the air in your lungs, which effectively is an air compressor containing pressurized air. You then use the force of the air in your lungs to blow out the flames of your birthday candles. Eventually, man-made air compressors were created. These falls generally fall into one of two types, positive displacement or dynamic, defined by its mode of operation. A positive displacement air compressor works by filling and then emptying an air chamber. Three common types of positive displacement air compressors are: reciprocating, rotary screw and rotary sliding vane. A dynamic air compressor, on the other hand, uses a rotating device to accelerate and then decelerate air. This process uses the speed or velocity of the air to increase the air's pressure. Centrifugal air compressors are dynamic air compressors. Compressed air can be used in a variety of ways. It can be used to alter the chemical composition as in the case of making fertilizer or it can be used for industrial purposes like production line manufacturing processes or it can be used to maintain industrial plants. Perhaps the most well known use of the air compressor is in the case of pneumatic tools like air powered nail guns, staplers, sanders, spray guns, or ratchet wrenches. Air compressors can also be used to move debris. These tools are commonly available at hardware stores for purchase or rental.

Another way to group air compressor types is by the number of stages it has. A two-stage air compressor is usually used for heavy duty use. This type of unit 6

offers a higher level of compression than smaller, single stage air compressors. A twostage air compressor can store air for future use, and is more energy efficient since it produces more air per unit of horsepower than a single stage compressor. Also, less heat is generated in a two-stage compressor, which means that wear on the unit is reduced. Portable electric air compressors are also available for light-duty applications. Depending on the type of air compressor, operation costs can be high, as in the case of plant maintenance. While air compressors can run on manual labour, like a hand powered air compressors, most run on either electricity or natural gas. It's the natural gas air compressor that is usually more cost-effective. If the air compressor is used in a small, enclosed area, an electric model may be more desirable in order to avoid gas fumes. The American Society of Mechanical Engineers (ASME) attests to the quality and protective features of air compressors. Their rating can be considered in evaluating air compressors for purchase or rental. In some states, only ASMEcertified air compressors may be sold. Some safety features include a safety relief valve, which lets air escape if the tank's pressure exceeds the maximum. The air compressor should also have a belt guard for protection, and an enclosed air intake filtration system. 4.1

Compressor Part 4.1.1 4.1.2 4.1.3 4.1.4 4.1.5 4.1.6 4.1.7 4.1.8 4.1.9 4.1.10

4.2

Valve Assembly Suction Strainer Cool Cylinder Heads Cylinders Rings. Connecting Rods Balanced Crankshafts Crankcase Balanced Ran Type Flywheel Main Bearings

Comparison Two-Stage and Single-Stage Stage refers to the number of times intake air is compressed in a pump. A single stage compressor compresses intake air one time before sending the air into a storage tank. Single stage compressors are the more common types of compressor

7

that consumers are used to using. Two stage compressors compress air two times before sending air into a storage tank.

5.0

EXPERIMENT PROCEDURES a) Make sure that the flow rate of cooling water is sufficient, approximately 100 L/H. Adjust it using the valves. Then connect the water inlet and outlet properly.

b) Plug in the flexible cable into diaphragm D1 for compressing air at p4 = 2.5 bar absolute.

8

c) Start the air compressor by turning on the main switch on the control panel.

d) Record the data for suction conditions into the table data sheet then wait for about 10 minutes to get a steady-state condition. e) Record the value of temperature and pressure as indicated in the column ‘first stage compression’ of the data sheet and wait again for 5 minutes.

9

f) Fill up the values in the column ‘second state of compression’ and wait 5 minutes for final compression.

g) Record rest of the data in the data sheet. h) Repeat step a until g for p4= 4 bar and p4 = 7 bar absolute by replacing the cable flexible into diaphragm D2 and D3.

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6.0

DATA SHEET : AIR COMPRESSOR

Pressure

Ambient Air

First State of Compression

Final Compression

Second State of Compression

p4

t0

p0

t1

p1

t1-t0

p1-p0

T1/T0

β1

t2

t3

p2

t3-t2

p2-p1

T3/T2

t4

P3

2.5

28.6

0.05

133.3

2.1

104.7

2.05

4.661

42

35.5

79.9

2.3

44.4

0.2

2.251

32.1

1.5

4

28.1

0.05

133.3

2.1

105.2

2.05

4.744

42

51.1

85.8

2.6

34.7

0.5

1.679

30.2

2.4

7

29.6

0.05

136.8

2.1

107.2

2.05

4.621

42

38.4

125.9

6.0

87.5

3.9

3.279

30.7

5.8

Air flow rate

( m 3 / h)

Input power (kW)

Air Supply

Exchange water inlet

Exchange 1 water outlet

Exchange 2 water outlet

W2

V

Pe

t6

t7

t8

t9

78

78

6.5

3.4

30.8

36.0

37.0

32.0

30.0

90

86

2.0

3.6

31.8

28.7

35.0

39.6

44.2

100

90

0.5

4.0

32.2

29.4

38.0

39.2

Pressure

Air Tank

p4

t5

W1

2.5

31.3

4 7

P0 P1 P2

= = =

T0 = T1 = T2 = T3 =

Air Inlet Exchanger 1 Air Inlet Exchanger 1 Air Outlet Exchanger 2 Air Inlet

Water circuit

Aspiration 1st Step of Compression 2nd Step of Compression

P3 P4 T4 = T5 = T6 = T7 =

= =

Exchanger 1 Air Outlet Tank Temperature Air Supply ExchangersWaterInlet

Tank Air Network T8 = T9 =

Exchangers 1 Water Outlet Exchangers 2 Water Outlet

11

7.0

RESULT CALCULATION

Calculation for β1, β2, m1, m2; P4 = 2.5 Bar First state of compression T1-To = 130.9-28.8 = 102.1oc T1 T0

=

130.9  273 28.8  273

= 1.338 P1-p0 = 2.2-1.013=1.187bar P1 β1 = P0

2.2 = 1.013 = 2.17 Second state of compression T3-T2 = 77.8-34.1 = 43.7oc T3 T2

=

79.7  273.15 34.9  273.15

= 1.145 P2-p1 = 2.2-2.0 = 0.2 bar P2 β2 = P1 =

m1

2.2 = 1.10 2.0

= ρW1 = 1000 kg / m3 x (100 x 10 -3) m3 / s 3600 = 0.0277kg /s

12

m2

= ρW2 = 1000 kg / m3 x (75x 10 -3) m3 / s 3600 = 0.0208 kg /s

P4 = 4 Bar T1 T0

=

135.8  273 31  273

= 1.345 P1-p0 = 2.18-2.0 = 0.2 bar

β1

=

P1 P0

2.18 = 1.013 = 2.1520 T3 T2

=

96.3  273 34.9  273

= 1.1994

β2

=

P2 P1

=

3.5 2.18

= 1.606

m1

= ρW1 = 1000 kg / m3 x (93 x 10 -3) m3 / s 3600 = 0.0258 kg /s

13

m2

= ρW2 = 1000 kg / m3 x (95 x 10 -3) m3 / s 3600 = 0.0264 kg / s

P4 = 7 Bar T1 T0

=

139.8  273 29.9  273

= 1.352

β1

=

P1 P0

2.10 = 1.013 = 2.0731 T3 T2

=

117 .7  273 34.7  273

= 1.2697

β2

=

P2 P1

=

6.5 2.1

= 3.095

m1

= ρW1 = 1000 kg / m3 x (100 x 10 -3) m3 / s 3600 = 0.0278 kg / s 14

m2

= ρW2 = 1000 kg / m3 x (95 x 10 -3) m3 / s 3600 = 0.0292 kg / s

Calculation for Heat Loss, Q1 and Q2; P4 = 2.5 Bar w1 = 2.78 x 10-5 m3/s

m1 = 0.0277kg /s

w2 = 2.08 x 10-5 m3/s

m2 = 0.0208 kg /s

Q= m (h out – h in) From steam table, value of h is : T7

= 29.3 º C T

H

29

121.5

29.3

h

30

125.7

T8

= 35.7 º C T

H

34

142.4

35.7

h

36

150.7

; h1 , in = 122.76kJ / kg

; h1, out = 149.455kJ / kg

15

T9

= 34.2 º C

T

H

34

142.4

34.2

h

36

150.7

Q1

; h2,out = 143.23kJ / kg

= m1 (h1 out – h1 in ) = 0.0277 (149.455– 122.76) = 0.7395 kJ / s

Q2

= m2 (h2 out – h1 in ) = 0.0208 (143.23– 122.76) = 0.4258 kJ / s

P4 = 4 Bar w1 = 2.58 x 10-5 m3/s

m1 = 0.0258 kg /s

w2 = 2.64 x 10-5 m3/s

m2 = 0.0264kg / s

Q= m (h out – h in) From steam table, value of h is : T7

= 29.9 º C T

H

29

121.5

29.9

h

30

125.7

; h1 , in = 125.28kJ / kg

16

T8

T9

= 36.5 º C

; h1, out = 152.8kJ / kg

T

H

36

150.7

36.5

h

38

159.1

= 33.8 º C

; h2,out = 141.56kJ / kg

T

H

32

134.0

33.8

h

34

142.4

Q1

= m1 (h1 out – h1 in ) = 0.0306 (203.49 – 124.07) = 2.430 kJ / s

Q2

= m2 (h2 out – h1 in ) = 0.0278 (137.44 – 124.07) = 0.3717 kJ / s

P4 = 7 Bar w1 = 2.64 x 10-5 m3/s

m1 = 0.0264 kg / s

w2 = 2.58 x 10-5 m3/s

m2 = 0.0258 kg / s

Q

= m (h out – h in)

From steam table, value of h is : T7

= 29.6 º C

; h1 , in = 124.07kJ / kg

T8

= 48.6 º C

; h1, out = 203.49kJ / kg

T9

= 32.8 º C

; h2,out = 137.44kJ / kg

Q1 = m1 (h1 out – h1 in ) 17

= 0.0264 (203.49 – 124.07) = 2.097 kJ / s Q2

= m2 (h2 out – h1 in ) = 0.0258 (137.44 – 124.07) = 0.345 kJ / s

8.0

EXPERIMENT RESULT a. Draw the curve Q1 = f(β1), where β1 = P1/P0 P4 (Bar)

2.5

4

7

Q1 (kJ/s)

2.097

2.430

2.097

β1

2.073

1.974

1.1974

18

b. Draw the curve Q2 = f(β2), where β2 = P2/P1 P4 (Bar)

2.5

4

7

Q2 (kJ/s)

0.3343

0.3717

0.345

Β2

1.095

1.900

3.250

19

c. Calculate the polytropic coefficient n for this two-stage air compressor? P4 = 2.5 Bar T2 T1

p2 =( p ) 1

n-1 n

20

34.9  273.15 2.3 = ( 130.2  273.15 2.1 )

In 0.764

=

n -1 n

ln 1.095

= - 2.966 n -1 n

= - 2.966n = 0.25

P4 = 4 Bar T2 T1

p2 =( p ) 1

n-1 n

35.3  273.15 3 .8 = ( 133.1  273.15 2 .0 )

In 0.759

=

n -1 n

ln 1.9

= - 0.43 n -1 n

= - 0.43n = 0.70

P4 = 7 Bar T2 T1

p2 =( p ) 1

n-1 n

33.6  273.15 6.5 = ( 139.5  273.15 2.0 )

In 0.743

=

n -1 n

ln 3.25

= - 0.252 n -1 n

= - 0.252n = 0.80

d. Calculate the polytropic work for each P4 = 2.5, 4, 7 Bar? 21

P4 = 2.5 Bar, n = 0.25 Win = Pp m

RAir = 0.2871 kJ/kg.K

m = (m1 + m2)/2 = (0.0264 + 0.025)/2 = 0.0257 kg/s

Pp = (

P 2 (n-1/n) n ) mRT1 [ ( ) – 1] n -1 P1

= (-0.3333) x (0.0257) x (0.2871) x (130.2 + 273.15) x (1.0952(-3) – 1) = (-0.9919) x (-0.24) = 0.2368 kJ/s Win =

0.2368 0.0257

= 9.22 kJ/kg P4 = 4 Bar, n = 0.70 Win = Pp m

RAir = 0.2871 kJ/kg.K

m = (m1 + m2)/2 = (0.0306 + 0.0278)/2 = 0.0292 kg/s

Pp = (

P 2 (n-1/n) n ) mRT1 [ ( ) – 1] n -1 P1

= (-2.3333) x (0.0292) x (0.2871) x (133.1 + 273.15) x (1.9(-0.429) – 1) = (-7.947) x (-0.24) = 1.913 kJ/s Win =

1.913 0.0292

= 65.51 kJ/kg

22

P4 = 7 Bar, n = 0.80 Win = Pp m

RAir = 0.2871 kJ/kg.K

m = (m1 + m2)/2 = (0.0264 + 0.0258)/2 = 0.0261 kg/s

P 2 (n-1/n) n ) mRT1 [ ( ) – 1] n -1 P1

Pp = (

= (-4) x (0.0261) x (0.2871) x (139.5 + 273.15) x (3.25(-0.25) – 1) = (-12.37) x (-0.26) = 3.157 kJ/s Win =

3.157 0.0261

= 120.96 kJ/kg

e. What is the polytropic efficiency of the air compressor,  p ? P4 = 2.5 Bar, Pe = 3.5 kW, nm = 80% nm

=

Ps Pe

Ps = 0.8 x 3.5 = 2.8 kJ/s np

= =

Pp Ps 0.2368 2.8

= 0.085 x 100% = 8.5 %

23

P4 = 4 Bar, Pe = 3.7 kW, nm = 80% nm

=

Ps Pe

Ps = 0.8 x 3.7 = 2.96 kJ/s np

=

=

Pp Ps

1.913 2.96

= 0.646 x 100% = 64.6 % P4 = 7 Bar, Pe = 4.2 kW, nm = 80% nm

=

Ps Pe

Ps = 0.8 x 4.2 = 3.36 kJ/s np =

=

Pp Ps

3.157 3.36

= 0.94 x 100% = 94 %

24

9.0

DISCUSSION a)

Based on the value n at question 5(c) above, is it acceptable if we assume that the compression process follows an adiabatic process? Give the reasonable arguments. Value of polytrophic coefficient, n for the two stage air compressor is 0.25, 0.70 and 0.80 for the pressure of 2.5, 4 and 7 bar. According to the value of the coefficient, we can say that this process did not follow the adiabatic process because there was heat been transfer to the environment while process in running. Range for n is 1 < n < 1.4. In thermodynamics, an adiabatic process or an isocaloric process is a c process in which no heat is transferred to or from the working fluid. The term "adiabatic" literally means impassable in Greek, corresponding here to an absence of heat transfer. Conversely, a process that involves heat transfer (addition or loss of heat to the surroundings) is generally called adiabatic. One opposite extreme—allowing heat transfer with the surroundings, causing the temperature to remain constant—is known as an isothermal process. Since temperature is thermodynamically conjugate to entropy, the isothermal process is conjugate to the adiabatic process for reversible transformations. A transformation of a thermodynamic system can be considered adiabatic when it is quick enough that no significant heat is transferred between the system and the outside. At the opposite extreme, a transformation of a thermodynamic system can be considered isothermal if it is slow enough so that the system's temperature remains constant by heat exchange with the outside.

25

b)

Assume that the compression process shall be adiabatic. Calculate the adiabatic efficiency, a ? P4 = 2.5 bar Qin = 2.097 kJ/s

Qout = 0.3343 kJ/s

= 84.06 % P4 = 4 bar Qin = 2.430 kJ/s

Qout = 0.3717 kJ/s

= 84.70 % P4 = 2.5 bar Qin = 2.047 kJ/s

Qout = 0.345 kJ/s

= 83.15 %

26

c)

Mark and plot the thermodynamic process on the P-H diagram provided.

27

28

10.0

QUESTIONS a) Is air compressor considered as a closed or open system? Explain it. Compressor is considered as an open system because there is no addition or reduction of it. An open system is a state of a system, in which a system continuously interacts with its environment. Basic characteristics of an open system are environment, input, throughput and output. And some control systems with feedback. b) Describe how a diaphragm works? A diaphragm compressor consists of a gas and a hydraulic system separated by a metal diaphragm group.

During the experiment,

diaphragm that we used to control intake of air is diaphragm valve. This entire diaphragm D1, D2, and D3 has a same diameter but during the experiment we gain a different output or discharge of pressure. First, the gas compression system features three metal diaphragms, which are clamped in between two cavity plates and which regulate the movement of the gas through inlet and outlet check valves. Second, the hydraulic system utilizes a motor-driven crankshaft and piston to pressurize hydraulic fluid, which in turn moves the diaphragms through the cavity that contains the gas. Diaphragm valves can be manual or automated. Their application is generally as shut-off valves in process systems within the food and beverage, pharmaceutical 29

and biotech industries. The older generation of these valves is not suited for regulating and controlling process flows, however newer developments in this area have successfully tackled this problem.

c) What is the function of the heat exchanger and give example of two industries that use the equipment? Heat exchangers are devices where two moving fluid streams exchange without mixing. Heat exchangers are widely used in various industries and mostly their using water to cooling and reduce the heat, and they come in various designs. Heat exchanger used in chemical industries for the purpose of petrochemical processing and in food industry and example of it is pasteurization of milk. And other example used at power plant for an industry for example methanol processing, their used intercooler and heat exchanger to cooling down the methanol that go through in the pipe line before it deliver the pure methanol to the save tank. d) How can we exactly measure the rotation speed of the motor? By using tachometer and dynamometers.

30

11.0

CONCLUSION

From this experiment, the objective of the experiment to understanding the polytropic process in the two-stage air compressor and determination of the polytropic efficiency of a compressor is achieved. Besides that, we get to know the air compression operation and how to calculate the polytropic efficiency of compression. Even if had a something wrong at the experiment device, the problem can be soft by using the theory value at P1. At the pressure P4 = 2.5 Bar, the polytropic efficiency is 8.5%, P4 = 4 Bar the polytropic efficiency is 64.4% and at the pressure P4 = 7 Bar. The conclusion of the experiment is successfully and the objective is achived.

12.0

REFERENCE

1. Thermodynamics (An Engineering Approach Sixth Edition), Yunus A. Cengel, Micheal A. Boles, Mc Graw Hill, Sixth Edition, 2007. 2. Termodinamik Gunaan (Masalah Dan Contoh Penyelesaian), J. Sutanto, Dewan Bahasa Dan Pustaka KPM, 1988. 3. Internet,  http://www.maxustools.com/404.html  http://www.popularmechanics.com/home_journal/how_your_house_works/12 75131.html?page=1  http://www.sophocom.com.my/  http://www.abn-drucklufttechnik.de/

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