16. Two stage air compressor final lab report PDF

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UNIVERSITI SAINS MALAYSIA SCHOOL OF CHEMICAL ENGINEERING EKC 291 - CHEMICAL ENGINEERING LABORATORY I EXPERIMENT 16: TWO-STAGE AIR COMPRESSOR GROUP NUMBER : 16 GROUP MEMBERS : 1. CARANI A/P P (133613) 2. CHONG CI EN (133616) 3. CHONG SOON XIN (133617) 4. MUHAMMAD FARIS AFIQ BIN ZAINAL AZMAN (133649) ...

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UNIVERSITI SAINS MALAYSIA SCHOOL OF CHEMICAL ENGINEERING

EKC 291 - CHEMIC CHEMICAL AL ENGI ENGINEERING NEERING L LABORATORY ABORATORY I EXPERIM EXPERIMENT ENT 16: TWO TWO-STAGE -STAGE A AIR IR COMPR COMPRESSO ESSO ESSOR R

GROUP NUMBER

: 16

GROUP MEMBERS : 1. CARANI A/P P.NARAYANASAMY (133613) 2. CHONG CI EN (133616) 3. CHONG SOON XIN (133617) 4. MUHAMMAD FARIS AFIQ BIN ZAINAL AZMAN (133649) 5. SITI ZULAIKHA BINTI AZMI (133681) LECTURE IN CHARGE

: ASSOC. PROF. LOW SIEW CHUN/ DR. POOYA LAHIJANI AMIRI

TECH. IN CHARGE

:

EN. SHAMSUL HIDAYAT SHAHARAN

DATE OF EXPERIMENT : 27 FEBRUARY 2018 DATE OF SUBMISSION

: 5 MARCH 2018

TABLE OF CONTENTS

CONTENTS

PAGES

1.0

ABSTRACT ………………………………….………………………........

1

2.0

INTRODUCTION ……………………………………………………........

2

3.0

THEORY …………………………………………………………..............

4

4.0

OBJECTIVE ………………………………………………………….........

6

5.0

EXPERIMENTAL PROCEDURE …………………………………..........

6

6.0

RESULTS …………………………………………………………….........

7

7.0

DISCUSSION ………………………………………………………..........

10

8.0

CONCLUSION ………………………………………………………........

15

9.0

REFERENCE ………………………………………………………...........

16

10.0 APPENDIX ………………………………………………..…………........

17

1.0 ABSTRACT An experiment is carried out to study on a two-stage air compressor to understand how it works and its performance characteristic. In this experiment, the air compressor used was two stage reciprocating piston type and the speed of the motor is adjusted from low flowrate from 885 to 1377 rotation per minute. The readings of the pressure and temperature are recorded after the constant pressure (5 bar) is achieved by adjusting the valve 8. The calculations are performed to determine the free air delivery at suction side, polytropic index, indicated power input and the compressor efficiency. The ideal condition is assumed during the calculation. From the calculation made, graphs of suction flow rate versus compression ratio for each stage is plotted to make comparison between first and second stages of compressor. The graphs contains a straight line which shows that the flow rate increases with the compression ratio.

Based on the result, when the flow rate increases by increasing the motor speed, the less power input is generated. Moreover, more power input is needed when the suction flow rate increases. This indicates that the efficiency of the compressor depends on the suction flow rate and motor speed. An optimum motor speed and suction flow rate should be determined to ensure the compressor operates effectively and efficiently so that cost and power can be saved. Determining the compressor performance is important to verify the model’s goodness.

1

2.0 INTRODUCTION A two-stage air compressor is a machine that increases the pressure from lower pressure to a higher desired pressure level by reducing the volume of gas and consequently making the gas pressure and temperature increase (heat is resulted from the compression process) [1]

. A two-stage air compressor normally has two air storage chambers and two pistons

for pumping air. At the first stage, air is drawn in and compressed to an intermediate pressure. After being compressed in the first stage, the air is piped through an intercooler which is also called heat exchanger where the intermediate air is allowed to cool down, and then to be compressed in the second stage. In addition, there is an aftercooler in the machine and it works to reduce the temperature of the air and also to remove the moisture from the air as well. This is crucial as during the compression process the will be no way to remove the moisture formed in the air. The effective cooling system is important for a compression system to work at high efficiency and reduce the power input.

There are two types of compressors which are positive displacement and centrifugal compressor. The positive displacement compressor can be further divided into reciprocating piston and rotary screw or vane types of compressor. These type of compressors are suitable for relatively high pressure and low flow rates. Apart from that, centrifugal compressor is designed for higher flow rates but lower pressures [1].

Two-stage compressors are more efficient in pumping compared to single-stage compressor. The two-stage compressors are normally good and efficient for high pressures because the air is cooled between the stages. Less heat is generated when more air power is used. The compressor and facility do not wear easily due to low heat is 2

released. A two-stage air compressor generates more power than traditional electricpower generators, and they are safer than other power generation systems which causes industrial jobs get done faster. These are the reasons why the two-stage compressors are widely used in nowadays industries. There are many industrial applications for two-stage air compressor. It plays a key role in industries such as agriculture (farming) [2]. This is because compressed air provides long-lasting performance and low cost of ownership. The principal of air compressor are used in spraying the crops, glasshouse ventilation systems and also powering dairy machines and to operate pneumatic material handling equipment manufacture of chemicals and fertilizers. There are two types of sprays being used for spraying crops which are hydraulic sprayer and low-volume sprayer. Besides, a two-stage air compressor is also served as a total air-power supply system to commercial establishments and industries which can supply power to all equipment in small industries and as an efficient backup power system in large industries.

A two-stage air compressor is very efficient in most cases than other compressors because it can be used continuously for long period of time depends on the desired period. It is important to understand the characteristic of the system and the suitability of the system to certain conditions that will cause ineffectiveness. Thus, two-stage air compressor should be used based on the situation to maximise the efficiency.

3

3.0 THEORY The work of compression for an ideal (frictionless) compression is given by Bernoulli equation for compressible fluids, which can be reduced to below equation: 𝑃2

W=∫

𝑃1

𝑑𝑃 ρ

(1)

Assuming ideal gas properties, PV = nRT. The ideal gas equation is modified in term of density, ρ is shown below: ρ=

𝑃𝑀 𝑅𝑇

(2)

The path followed by the gas during compression in large compressors can be described by the polytropic process: 𝑃1 𝑃2 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 𝜌1𝑛 𝜌𝑛2

(3)

Substituting ρ2 from (3) into (1). Then, integrate and substitute ρ1 from (2) will give the following equation in term of work done, W. W=

( ) RT1 n P2 n−1 ∕n − 1] [( ) M(n − 1) P1

(4)

where P2/P1 is known as the compression ratio, r. The temperature at the end of the compression can be determined by substituting equation (2) into (3). The equation can be rearranged as: ρ2 n−1 P2 (n−1)/n T2 =( ) =( ) = r (n−1)/n P1 T1 ρ1

(5)

Free air delivery (F.A.D) can be known as compressor capacity which is the volumetric flow rate of air delivered at actual atmospheric condition. The F.A.D is calculated from the general ideal gas equation as follows: P2 T1 Q1 = Q2 ( ) ( ) 𝑃1 T2 where Q2 is the flow rate of air after compression.

4

(6)

By using equation (4) and (5), the theoretical power required by the compressor which is known as Indicated power input, Pf can be calculated. Pf =

( ) P1 Q1 n P1 Q1 n P2 𝑛−1 /𝑛 (𝑇 − 𝑇1 ) − 1] = [( ) T1 (n − 1) 2 (n − 1) 𝑃1

(7)

Compressor efficiency is the ratio of theoretical power input (or fluid power) to the actual power input (or shaft power input): Compressor efficiency, ŋ =

𝑃𝑓

𝑃𝐵

x100%

(8)

where PB= brake power of shaft power input Multi-stage Compression There is an optimum compression ratio for each stage which will decrease the total compression work in multistage operation. For a two-stage compressor, if the gas enters stage 1 at (P1, T1), leaves stage 1 at (P2, T2) was cooled to T1 then enters stage 2 at (P2, T1), and finally left at P3, the optimum interstage pressure can be calculated as 1

𝑃2 = (𝑃1 𝑃3 )2

5

(9)

4.0 OBJECTIVE This experiment was carried out to define the performance characteristics of a two-stage air compressor by calculating the value of polytropic index (ŋ), indicated power input, compressor efficiency and free air delivery from the obtained results.

5.0 EXPERIMENTAL PROCEDURE All the procedure was fully read and understood before the experiment started. Any questions regarding on how the machines that were used in the experiment were asked to the authorized personnel lab assistant so that more understanding on how the machines operates can be obtained. Before starting the experiment, tank 1 (TN1) and all flow lines had been made sure that were clear of pressures heads where they must be at atmospheric pressure. All valves were ensured to be closed except for valves V9 while regulator RG1 was ensured to be set at minimum. The cooling water was next allowed to flow to heat exchanger HX1 and HX2 by opening valves V11 and V12. The cooling water flow rates F11 and F12 was set to 5 LPM. The motor was started, and the speed was set up to 1150 rpm. Once the pressure P3 of the tank TN1 reached 5.5 bar, the valve V10 was opened. Regulator RG1 was then regulated to 3.5 bar. The pressure P3 was ensured to maintain at 5 bar by adjusting valve V8. The valves V11 was necessarily adjusted to lower the inter stage temperature as low as the initial suction temperature T1. As soon as the system reached the steady state, the flow rates F11 to F13, temperatures T1 to T9, pressure P4 to P5, torque T, rotational speed N, and electric power input were recorded. After all the results were recorded, the motor was shut off. The compressed air tank was emptied by regulating RG1 to 1.5 bar and slightly opening valve V1. The pressure heads were released at the pressure measuring points. The valves V11 and V12 were closed to shut off water supply to cooling systems. The main switch was shut off.

6

6.0 RESULTS Table 1: Data collected during experiment. Run No. 1

2

3

4

5

Set 1 2 Avg 1 2 Avg 1 2 Avg 1 2 Avg 1 2 Avg

Flow Rate (LPM) Fl1 Fl2 Fl3 5 5 160 5 5 140 5 5 150 5 5 80 5 5 120 5 5 100 5 5 110 5 5 150 5 5 130 5 5 100 5 5 60 5 5 80 5 5 60 5 5 60 5 5 60

Temperature (◦C) T1 33.80 33.70 33.75 42.20 45.00 43.60 37.50 37.20 37.35 38.60 37.40 38.00 44.80 40.10 42.45

T2 67.10 67.00 67.05 66.80 67.90 67.35 66.60 69.70 68.15 72.10 68.60 70.35 69.00 70.50 69.75

T3 39.40 39.50 39.45 66.50 54.80 60.65 60.20 58.00 59.10 65.40 61.40 63.40 69.30 61.10 65.20

T4 74.80 74.10 74.45 105.30 93.90 99.60 99.60 99.50 99.55 90.70 102.00 96.35 111.70 106.60 109.15

T5 50.50 52.60 51.55 50.60 59.10 54.85 68.70 63.50 66.10 54.00 65.10 59.55 74.20 61.30 67.75

T6 76.80 86.60 81.70 82.30 106.50 94.40 116.90 116.10 116.50 114.50 114.10 114.30 93.10 102.80 97.95

7

T7 33.80 34.50 34.15 40.60 48.60 44.60 58.90 52.70 55.80 43.10 54.10 48.60 63.60 49.80 56.70

T8 34.80 35.90 35.35 41.50 49.40 45.45 57.10 53.00 55.05 43.80 54.70 49.25 61.40 50.80 56.10

T9 33.60 34.70 34.15 40.80 48.90 44.85 56.50 52.50 54.50 43.20 54.00 48.60 60.90 50.30 55.60

Pressure (bar) P4 P5 7.00 0.5 7.00 0.5 7.00 0.50 7.00 0.5 7.00 0.5 7.00 0.50 7.00 0.5 7.00 0.5 7.00 0.50 6.90 0.2 6.80 0.2 6.85 0.20 6.80 0.2 6.90 0.2 6.85 0.20

Torque (Nm)

N (rpm)

17 16 16.5 17 17 17 17 17 17 17 16 16.5 16 16 16

885.0 885.0 885.0 950.0 952.0 951.0 1085.0 1089.0 1087.0 1169.0 1170.0 1169.5 1376.0 1378.0 1377.0

Power, PB (kW) 1.48 1.46 1.47 1.56 1.58 1.57 1.72 1.71 1.715 1.8 1.79 1.795 2 2.01 2.005

Table 2: Calculation of Polytropic Index, ŋ, Indicated Power Input, Compression Efficiency, and Free Air Delivery (FAD) for first stage.

Run

N (rpm)

Input temperature T1 (K)

Output temperature T2 (K)

Input Pressure P1 (Pa)

Output Pressure P5 (Pa)

r= (P5/P1)

log r

T2/T1

log (T2/T1)

Polytropic Index, n

F13 (LPM)

Q2 (m3/s)

F.A.D (m3/s)

PB (kW)

P f, W

Compressor efficiency, ŋ (%)

1

885.0

306.90

340.20

101325

151325

1.493

0.174

1.109

0.045

1.34558

150

0.0025

0.0034

1.47

144.19

9.81

2

951.0

316.75

340.50

101325

151325

1.493

0.174

1.075

0.031

1.2199

100

0.0017

0.0023

1.57

97.589

6.22

3

1087.0

310.50

341.30

101325

151325

1.493

0.174

1.099

0.041

1.30856

130

0.0022

0.0029

1.715

125.48

7.32

4 5

1169.5 1377.0

311.15 315.60

343.50 342.90

101325 101325

121325 121325

1.197 1.197

0.078 0.078

1.104 1.087

0.043 0.036

2.21771 1.85374

80 60

0.0013 0.001

0.0014 0.0011

1.795 2.005

27.746 20.973

1.55 1.05

Table 3: Calculation of Polytropic Index, ŋ, Indicated Power Input, Compression Efficiency, and Free Air Delivery (FAD) for second stage.

Run

N (rpm)

Input temperature T5 (K)

Output temperature T4 (K)

Input Pressure P5 (Pa)

Output Pressure P4 (Pa)

r= (P4/P5)

log r

T4/T5

1

885.0

324.70

347.60

151325

801325

5.295

0.724

2

951.0

328.00

372.75

151325

801325

5.295

0.724

3

1087.0

339.25

372.70

151325

801325

5.295

4 5

1169.5 1377.0

332.70 340.90

369.50 382.30

121325 121325

786325 786325

6.481 6.481

P f, W

Compressor efficiency, ŋ (%)

1.47

3228

219.59

1.57

2089.7

133.10

0.0104

1.715

2762.1

161.05

0.0078 0.0058

1.795 2.005

1860.1 1388.5

103.63 69.25

log (T4/T5)

Polytropic Index, n

F13 (LPM)

Q2 (m3/s)

F.A.D (m3/s)

PB (kW)

1.071

0.03

1.04263

150

0.0025

0.0124

1.136

0.056

1.08311

100

0.0017

0.0078

0.724

1.099

0.041

1.05979

130

0.0022

0.812 0.812

1.111 1.121

0.046 0.05

1.05947 1.06534

80 60

0.0013 0.001

8

Free Air Delivery (F.A.D), Q1, (m3/s)

Graph of Free Air Delivery (F.A.D), Q1 versus Compression Ratio for First Stage 0.004 0.0035 0.003 0.0025 0.002 0.0015 0.001 0.0005 0 1.1

1.15

1.2

1.25

1.3

1.35

1.4

1.45

1.5

1.55

Compression Ratio, r

Graph 1: Graph of Free Air Delivery (F.A.D) against Compression Ratio, r for 1st Stage.

Free Air Delivery (F.A.D), Q1, (m3/s)

Graph of Free Air Delivery (F.A.D), Q1 versus Compression Ratio for Second Stage 0.014 0.012 0.01 0.008 0.006 0.004 0.002 0 5

5.2

5.4

5.6

5.8

6

6.2

6.4

6.6

Compression Ratio, r

Graph 2: Graph of Free Air Delivery (F.A.D) against Compression Ratio, r for 2nd Stage.

Free Air Delivery (F.A.D), Q1, (m3/s)

Graph of Free Air Delivery (F.A.D), Q1 versus Compression Ratio for Both Stages 0.014 0.012 First Stage

0.01 0.008

Second Stage

0.006 0.004 0.002 0 0

1

2

3

4

5

6

7

Compression Ratio, r

Graph 3: Graph of Free Air Delivery (F.A.D) against Compression Ratio, r for both Stages. 9

7.0 DISCUSSION 1. Draw a schematic diagram of the air compression system.

10

2. Give your comments on the characteristics curves. A straight-line which is increased linearly is obtained from the graph for first stage. This shows that free air delivery (F.A.D) increases with the compression ratio. According to the equation 6, free air delivery (F.A.D), Q1 will increases when compression ratio increases. P

T

Q1 = Q2 ( 2 ) ( 1 ) 𝑃 1

T2

……… (5)

However, for the graph of second stage, free air delivery (F.A.D) decreased with the compression ratio. This may due to some error and deviation in the experiment which cause the air compressor showing inaccurate readings. By comparing first stage and second stage, we found out that compression ratio of second stage is bigger than of first stage. The suction flow rate for second stage also higher than of first stage. According to the theory, we should obtain a vertical line graph for the first stage and a curve identical to the alphabet ‘C’ for the second stage of air compressor which is different from the graphs that we have plotted in this experiment. The errors occurred in this experiment has caused inaccuracy in the results. 3. What will be the overall efficiency if the motor efficiency is taken into account? Motor efficiency is defined as output power divided by the input power by the motor, where ŋ=

𝑂𝑢𝑡𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟 × 100% 𝐼𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟

The motor efficiency is usually less than 1 because there is no 100% efficiency. Energy will be loss in the form of heat, sound energy, the turbulence of air and friction to the surrounding. These factors cause more power is generated to overcome the resistances. The equation for overall efficiency is:

Overall efficiency, ŋ𝑜 = ŋ × ŋ𝑚

where ŋ = compressor efficiency ŋm= motor efficiency

Hence, the overall efficiency of the system will be lower compared to the efficiency calculated in the results part if t...