E4

MAPÚA UNIVERSITY Muralla St. Intramuros, Manila School of Mechanical and Manufacturing Engineering

EXPERIMENT NO. 4 PERFORMANCE TEST OF AN AIR COMPRESSOR

11 MAHMUD, Ali R. 2015151413 ME144L – A1 Group No. 2

Date of Performance: Sep 5, 2018 Date of Submission: Sep 26, 2018

GRADE Engr. Teodulo A. Valle Instructor

TABLE OF CONTENTS

Objectives Theory and Principle List Of Apparatus Procedure Set-up of Apparatus Final Data Sheet Sample Computations Discussion of Result Questions and Answers Conclusion Reference Preliminary Data Sheet

1 1 5 5 8 9 10 13 14 17 18 19

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OBJECTIVES 1. To be acquainted with an air compressor 2. To determine the amount of air compressed and delivered 3. To be able to determine the efficiency of the air compressor THEORY AND PRINCIPLE A compressor is a device used to increase the pressure of a compressible fluid. The inlet pressure level can be any value from a deep vacuum to a high positive pressure. The discharge pressure can range from subatmospheric levels to high values in the tens of thousands of pounds per square inch. The inlet and outlet pressures are related, corresponding with the type of compressor and its configuration. The fluid can be any compressible fluid, either gas or vapor, and can have a wide molecular weight range. Recorded molecular weights of compressed gases range from 2 for hydrogen to 352 for uranium hexafluoride. Applications of compressed gas vary from consumer products, such as home refrigerators, to large complex petrochemical plant installations. Compressors have numerous forms; their exact configurations being based on the application. For comparison, the different types of compressors can be subdivided into two broad groups based on the compression mode, either intermittent or continuous. The intermittent mode of compression is cyclic in nature: A specific quantity of gas is ingested by the compressor, acted upon, and discharged before the cycle is repeated. In the continuous compression mode, the gas is moved into the compressor, acted upon, moved through the compressor, and discharged without interruption of the flow at any point in the process. Compressors using the intermittent compression mode are referred to as positive displacement compressors, of which there are two distinct types: reciprocating and rotary. Continuous-mode compressors are also characterized by two fundamental types: dynamic and ejector. Figure 1-1 diagrams the relationship of the various compressors by type. Figure 1-2 shows the typical application range of each compressor.

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2

The reciprocating compressor is probably the best known and the most widely used of all compressors. It consists of a mechanical arrangement in which reciprocating motion is transmitted to a piston that is free to move in a cylinder. The displacing action of the piston, together with the inlet valve or valves, causes a quantity of gas to enter the cylinder where it is in turn compressed and discharged. Action of the discharge valve or valves prevents the backflow of gas from the discharge line into the compressor during the next intake cycle. When the compression takes place on one side of the piston only, the compressor is said to be single-acting. The compressor is double-acting when compression takes place on each side of the piston. Configurations consist of a single cylinder or multiple cylinders on a frame. When a single cylinder is used or when multiple cylinders on a common frame are connected in parallel, the arrangement is referred to as a single-stage compressor. When multiple cylinders on a common frame are connected in series, usually through a cooler, the arrangement is referred to as a multistage compressor. The reciprocating compressor is generally in the lower flow end of the compressor spectrum. Inlet flows range from less than 100 to approximately 10,000 cfm per cylinder. It is particularly well suited for high pressure service. One of the highest-pressure applications is at a discharge pressure of 40,000 psi. Above an approximately 1.5-to-1 pressure ratio, the reciprocating compressor is one of the most efficient of all the compressor types. In the reciprocating compressor family, there is a piece of equipment that is not as well-known as the piston type reciprocating compressor. This compressor is the diaphragm compressor-used primarily in lower flow applications—but is capable of high pressure services. It consists of a crankcase assembly very much like the piston compressor. The piston moves oil rather than gas, using the oil to move a diaphragm pack that in turn moves the gas. By the nature of the head construction, the diaphragm does not have any leakage to the atmosphere. Also, there is no leakage toward the inside to contaminate the process gas. Because relatively little power is used, most units are belt-driven by a motor. The reciprocating compressor is the patriarch of the compressor family. In the process industry, the reciprocating compressor is probably the oldest of the compressors with wide application ranging from consumer to industrial usage. This compressor is manufactured in a broad range of configurations and its pressure range is the broadest in the compressor family extending from vacuum to 40,000 psig. The reciprocating compressor declined in popularity from the late 1950s through the mid-1970s. Higher maintenance cost and lower capacity, when compared to the centrifugal compressor, contributed to this decline. However, recent rises in energy cost and the advent of new specialty process plants have given the more flexible, higher efficiency, though lower capacity, reciprocating compressor a more prominent role in new plant design. 3

The reciprocating compressor is a positive displacement, intermittent flow machine and operates at a fixed volume in its basic configuration. The few single-acting crosshead compressors are normally single stage machines with vertical cylinders. The more common double-acting type, when used as a single-stage, commonly has a horizontal cylinder. The double-acting cylinder compressor is built in both the horizontal and the vertical arrangement. There is generally a design trade-off to be made in this group of compressors regarding cylinder orientation. From a ring wear consideration, the more logical orientation is vertical; however, considering size and the ensuing physical location as well as maintenance problems, most installations normally favor the horizontal arrangement. A feature of reciprocating compressors that is somewhat unique when compared to the rest of the compressor family is the number of available drive arrangements, which is almost as complex as the cylinder arrangements. In single and multistage arrangement small compressors, particularly the trunk type, are usually V-belt driven by electric motors. The single-acting crosshead type and the small, double-acting, singlestage compressor are also driven in a similar manner. The figure on the right shows the complete crankshaft revolution and encompasses a complete compression cycle. To begin the cycle, refer to the figure at (a) the location where the piston is at the lower end of the stroke (bottom dead center) and is at path point 1 on the indicator diagram. At this point, the cylinder has filled with gas at intake pressure 𝑃1 . Note that the valves are both closed. At (b), the piston has started to move to the left. This is the compression portion of the cycle and is illustrated by Path 1-2. When the piston reaches point 2 on the indicator diagram, the exhaust valve starts to open. The discharge portion of the cycle is shown at (C). This is shown on the indicator diagram Path 2-3. Note that the discharge valve is open during this period while the intake valve is closed. The gas is discharged at the discharge line pressure 𝑃2 . When the piston reaches point 3, it has traveled to the upper end of its stroke (top dead center). Physically, at this point in the stroke, there is a space between the piston face and the head. This space results in a trapped volume and is called the clearance volume. Next in the cycle, the piston reverses direction and starts the expansion portion of the cycle, as illustrated at (d) in the figure. Path 3-4 shows this portion of the cycle. Here the gas trapped in 4

the clearance volume is re-expanded to the intake pressure. Note that the discharge valve has closed, and the intake valve is still closed. At point 4, the expansion is complete, and the intake valve opens. The intake portion of the cycle is shown at (e). This is indicated by Path 4-1 on the indicator diagram. The cylinder fills with gas at intake line pressure Pl. When the piston reaches point 1, the cycle is complete and starts to repeat. The formulas involved in the experiment were projected below: Solving for compressor discharge temperature: T2 / T1 = [P2 / P1] n – 1 / n

Equation 1

V1 = Vs + V c

Equation 2

Solving for cylinder volume:

Where V1 = cylinder volume, m3 Vs = swept volume or stroke volume, m3 =  / 4 (d) 2 x L d = cylinder bore, m L = stoke length, m V c = clearance volume, m3 Solving for mass of air in the cylinder at the beginning of compression: m1 = P1V1 / R T1

Equation 3

Solving for mass of air remaining in the cylinder at the end of discharge: Mc = p c V c / R Tc = p2Vc / R T2

Equation 4

Solving for the mass of air delivered per cycle: m del = m1 – mc

Equation 5

m f = mdel (n cyl ) (n speed)

Equation 6

Solving for the capacity of compressor:

Where mf = mass flow rate of air, kg/s n cyl = number of cylinders n speed = compressor speed, rps m del = mass of air delivered per cycle, kg. Solving for indicated power (single-stage compressor): 5

Indicated Power = (n/n – 1) [mf R T1][(P2/P1)n – 1 / n – 1 ]

Equation 7

Solving for indicated power (multi-stage compressor): Indicated Power = (sn/n – 1) [mf R T1][(P2/P1)n – 1 /s n – 1 ]

Equation 8

Indicated Power = s(n/n – 1) [mf R T1][(Pi/P1)n – 1 / n – 1 ]

Equation 9

Or

Where Pi = (P1s-1 P2)1/s S = no. of stages Solving for mechanical efficiency: m = Indicated power/Compressor shaft power

Equation 10

Solving for transmission efficiency: trans = compressor shaft power/motor shaft power

Equation 11

Solving for motor efficiency: motor= motor shaft power/electrical power input

6

Equation 12

LIST OF APPARATUS 1. Air Compressor

2. Electric motor 3. Water Manometer

4. Pitot Tube

5. Pressure gage

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6. Tachometer

7. Amprobe

8. Stop Watch

8

9. Digital thermometer

9

PROCEDURES 1. Determine the diameter of the nozzle.

2. Assume that the atmospheric pressure will be the intake pressure. 3. Turn on the motor of the compressor with the discharge valve closed first so that the compressed air will remain inside the tank receiver.

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4. Open the valve and throttle it. Observe the compressor discharge pressure by looking on the pressure gage.

5. For 3 minutes, determine the air intake temperature.

6. Use the Pitot tube to determine the velocity head and place it in the nozzle of the compressor.

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7. Use the tachometer to determine the compressor speed. Obtain 3 sample data and get its average.

8. Determine the line current of the compressor under operation using an Amprobe. Gather at least 3 sample data and get its average. The power factor within the campus is 0.8.

9. Repeat the procedure for each item for the second trial. Record all data in the data sheet. 10. Calculate the remaining data needed for the experiment.

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SET-UP OF APPARATUS

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FINAL DATA SHEET

Duration, min Compressor Discharge Pressure, psi (P2) Air Intake Temperature, ℉, (T1) Compressor Discharge Temperature, ℉ (T2) Compressor Speed, rpm Size of Nozzle, inches Velocity Head, inH2O (hv) Velocity of Air, ft./sec Free Air Delivered, ft3/min Line Current, Amperes Power Input, HP Mass Flow Rate, kg/sec (mf) Indicated Power, HP Mechanical (Compressor) Efficiency, %

1 3 75

2 3 65

89.6

90.68

374.28

353.42

393

399 7/8

1 1⁄4 76.14 19.02 8.02 3.28 0.0051709 1.22 48.62

14

1 1⁄8 72.30 18.11 7.535 3.08 0.0052671 1.16 49.25

SAMPLE COMPUTATIONS Trial 1 Given: 3

P2 = 75 psi + 14.7 psi = 89.7 psi

Bore = 4 16 𝑖𝑛 = 4.1875 𝑖𝑛

T1 = 89.6 ℉ = 549.6 R

Stroke = 3 2 𝑖𝑛 = 3.5 𝑖𝑛

N = nspeed = 393 rpm

Diameter of Nozzle = 8 𝑖𝑛 = 0.875 𝑖𝑛

hv = 1 inH2O

c = 5%

I = 8.02 A

Motor Efficiency = 85%

P1 = 14.7 psi

Transmission Efficiency = 90%

1

7

Solution: •

Solving for Compressor Discharge Temperature: 𝑃2 𝑛−1

𝑇2

=( )

𝑇1

𝑇2 549.6 𝑅

•

𝑛

𝑃1

=(

89.7 𝑝𝑠𝑖 1.3−1

14.7 𝑝𝑠𝑖

)

1.3

; T2 = 834.28 R = 374.28 ℉

Solving for Mass Flow Rate delivered to the tank: 𝜋

V1 = Vs + Vc; Vs = 4 (𝑏𝑜𝑟𝑒)2 (𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠𝑡𝑟𝑜𝑘𝑒) = c=

Vc Vd

𝜋 4

(4.1875 𝑖𝑛. )2 (3.5 𝑖𝑛. ) = 48.20 𝑖𝑛3

→ V𝑐 = c𝑉𝑑 = (0.05)(48.20 𝑖𝑛3 ) = 2.41 𝑖𝑛3

V1 = 48.20 𝑖𝑛3 + 2.41 𝑖𝑛3 = 50.61 𝑖𝑛3

𝑚1 =

𝑚𝑐 =

𝑃1 𝑉1 𝑅𝑇1

𝑃𝑐 𝑉𝑐 𝑅𝑇𝑐

𝑙𝑏𝑓

=

(14.7 2 )(50.61 𝑖𝑛3 ) 𝑖𝑛 𝑙𝑏𝑓 −𝑓𝑡

12𝑖𝑛 (53.34 𝑙𝑏 −𝑅)(549.6 𝑅)( 1𝑓𝑡 ) 𝑚

= 2.1148 × 10−3 𝑙𝑏

𝑙𝑏𝑓

=

(89.7 2 )(2.41 𝑖𝑛3 ) 𝑖𝑛 𝑙𝑏𝑓 −𝑓𝑡

12𝑖𝑛

(53.34 𝑙𝑏 −𝑅)(834.28𝑅)( 1𝑓𝑡 ) 𝑚

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= 4.0482 × 10−4 𝑙𝑏

𝑚𝑑𝑒𝑙 = 𝑚1 − 𝑚𝑐 = 2.1148 × 10 1.70998 × 10−3 𝑙𝑏

−3

𝑙𝑏 − 4.0482 × 10−4 𝑙𝑏 =

𝑚𝑓 = 𝑚𝑑𝑒𝑙 𝑛𝑠𝑝𝑒𝑒𝑑 𝑛𝑐𝑦𝑐𝑙𝑒 = (1.70998 × 10−3 𝑙𝑏)(400 𝑟𝑝𝑚) (1 𝑚𝑓 = 0.6840 •

𝑙𝑏𝑚 𝑚𝑖𝑛

×

1 𝑘𝑔

×

2.2046 𝑙𝑏𝑚

1 𝑚𝑖𝑛 60 𝑠

𝑐𝑦𝑐𝑙𝑒 𝑟𝑒𝑣

)

= 5.1709 × 10−3 𝑘𝑔/𝑠

Solving for velocity of air:

{𝜌𝑔ℎ}𝑎𝑖𝑟 = {𝜌𝑔ℎ}𝑤𝑎𝑡𝑒𝑟

ℎ𝑎𝑖𝑟 =

ℎ𝑤𝑎𝑡𝑒𝑟 𝜌𝑤𝑎𝑡𝑒𝑟 𝜌𝑎𝑖𝑟

Hence, ℎ𝑎𝑖𝑟

=

; but 𝜌𝑎𝑖𝑟

ℎ𝑤𝑎𝑡𝑒𝑟 𝜌𝑤𝑎𝑡𝑒𝑟 𝑃1 𝑅𝑇1

𝑣2

ℎ𝑎𝑖𝑟 = 2𝑔 → 90.02 𝑓𝑡 = •

𝑣2 𝑓𝑡 𝑠

2(32.2 2 )

𝑚𝑎𝑖𝑟

=

𝑉𝑎𝑖𝑟

=

𝑃1 𝑅𝑇1

𝑙𝑏 1𝑓𝑡 (1.25 𝑖𝑛)(62.4 𝑚 ) )( 𝑓𝑡3 12𝑖𝑛

𝑙𝑏𝑓 14.7 𝑖𝑛2 𝑙𝑏𝑓−𝑓𝑡 (53.34 )(549.6 𝑅) 𝑙𝑏𝑚 −𝑅

= 90.02 𝑓𝑡

; 𝑣 = 76.14 𝑓𝑡/𝑠

Solving for Free Air Delivered: 𝑓𝑡

•

=

1 𝑓𝑡 2

𝜋

6𝑠

𝐹𝐴𝐷 = 𝑣𝐴𝑛𝑜𝑧𝑧𝑙𝑒 = (76.14 ) ( ) (0.875 𝑖𝑛)2 ( ) (1min) = 19.08 𝑐𝑓𝑚 𝑠 4 12 𝑖𝑛 Solving for Indicated Power:

𝐼. 𝑃. = 2(

𝑛

𝑛−1 𝑛

𝑃𝑖

)(𝑚𝑓 𝑅𝑇1 )(( ) 𝑃

𝑛−1

1

− 1)

𝑃𝑖 = √𝑃1 𝑃2 = √(14.7 𝑝𝑠𝑖)(89.7 𝑝𝑠𝑖) = 36.31 𝑝𝑠𝑖

𝐼. 𝑃. = 2(

1.3

)(0.6840

1.3−1

𝐼. 𝑃. = 40324

𝑓𝑡−𝑙𝑏𝑓 𝑚𝑖𝑛

×

𝑙𝑏𝑚

)(53.34

𝑚𝑖𝑛 1 𝐻𝑃

33,000

𝑓𝑡−𝑙𝑏𝑓 𝑚𝑖𝑛

𝑙𝑏𝑓 −𝑓𝑡 𝑙𝑏𝑚 −𝑅

36.31 𝑝𝑠𝑖

1.3−1 1.3

)(549.6 𝑅)((14.7 𝑝𝑠𝑖 )

− 1)

= 1.22 𝐻𝑃

Another way of solving indicated power:

𝐼. 𝑃. = ( 𝐼. 𝑃. = (

2𝑛−1 𝑛

𝑃2

𝑛−1 2𝑛

)(𝑚𝑓 𝑅𝑇1 )(( ) 𝑃

2(1.3) 1.3−1

1

)(0.684

𝑙𝑏𝑚 𝑚𝑖𝑛

− 1)

)(53.34

𝑙𝑏𝑓 −𝑓𝑡 𝑙𝑏𝑚 −𝑅

89.7 𝑝𝑠𝑖

1.3−1 2(1.3)

)(549.6 𝑅)((14.7 𝑝𝑠𝑖)

16

− 1)

𝐼. 𝑃. = 40328 •

𝑓𝑡−𝑙𝑏𝑓 𝑚𝑖𝑛

1 𝐻𝑃

×

33,000

𝑓𝑡−𝑙𝑏𝑓 𝑚𝑖𝑛

= 1.22 𝐻𝑃

Solving for Input Power (3-Phase): 1 𝐻𝑃

𝑃𝑖𝑛𝑝𝑢𝑡 = √3𝐸𝐼𝑐𝑜𝑠∅ = √3(220)(8.02)(0.8)𝑊 × 746 𝑊 = 3.28 𝐻𝑃 •

Solving for Compressor Efficiency:

𝑛𝑚𝑜𝑡𝑜𝑟 = 𝑛𝑡𝑟𝑎𝑛𝑠 =

𝑀𝑜𝑡𝑜𝑟 𝑆ℎ𝑎𝑓𝑡 𝑃𝑜𝑤𝑒𝑟 𝑃𝑜𝑤𝑒𝑟 𝐼𝑛𝑝𝑢𝑡

→ 𝑀𝑜𝑡𝑜𝑟 𝑆ℎ𝑎𝑓𝑡 𝑃𝑜𝑤𝑒𝑟 = 𝑛𝑚𝑜𝑡𝑜𝑟 𝑃𝑜𝑤𝑒𝑟 𝐼𝑛𝑝𝑢𝑡

𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑜𝑟 𝑆ℎ𝑎𝑓𝑡 𝑃𝑜𝑤𝑒𝑟 𝑀𝑜𝑡𝑜𝑟 𝑆ℎ𝑎𝑓𝑡 𝑃𝑜𝑤𝑒𝑟

→ 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑜𝑟 𝑆ℎ𝑎𝑓𝑡 𝑃𝑜𝑤𝑒𝑟 =

𝑛𝑡𝑟𝑎𝑛𝑠 𝑛𝑚𝑜𝑡𝑜𝑟 𝑃𝑜𝑤𝑒𝑟 𝐼𝑛𝑝𝑢𝑡 𝑛𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑜𝑟 = 𝑛𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑜𝑟 =

𝐼𝑛𝑑𝑖𝑐𝑎𝑡𝑒𝑑 𝑃𝑜𝑤𝑒𝑟 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑜𝑟 𝑆ℎ𝑎𝑓𝑡 𝑃𝑜𝑤𝑒𝑟 1.22 𝐻𝑃 (0.85)(0.9)(3.28 𝐻𝑃)

=

𝐼𝑛𝑑𝑖𝑐𝑎𝑡𝑒𝑑 𝑃𝑜𝑤𝑒𝑟 𝑛𝑡𝑟𝑎𝑛𝑠 𝑛𝑚𝑜𝑡𝑜𝑟 𝑃𝑜𝑤𝑒𝑟 𝐼𝑛𝑝𝑢𝑡

= 48.62%

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DISCUSSION OF RESULT

Based on the data gathered, the compressor discharge temperature was increased when the air was compressed but the pressure of the air was increased. It follows the combined gas law wherein the pressure is inversely proportional to the volume but directly proportional to the temperature, considering that the mass of air delivered to the air tank is constant. The volume swept by the piston is constant as well as its clearance volume considering a percent clearance of 5%. The intake pressure was taken to be atmospheric pressure (P=14.7 psi) since the air is withdrawn within the sea level. Since the free air delivered is required, the researcher determined the velocity head by utilizing a manometer and recorded the deflection of water. For trial 1, the free air delivered to the tank is 19.08 cfm while on trial 2, it is 18.11 cfm since the velocity head did not change. The indicated power was also computed by the researcher using 2 methods: utilizing the formula of compressor power using suction and discharge pressure and the other one is to utilize the intercooler pressure and the suction pressure. Both of them when used project same value for indicated power and was proven by the researcher. The input power was calculated by the researcher by utilizing the formula for input power. The compressor is in 3-phase, so the formula will be multiplied by the power factor of the University. The efficiency of the compressor is to be calculated so the researcher also considered the transmission losses in the equipment. The mechanical efficiency of the compressor in trial 1 is 48.62% while on trial 2, it is 49.25% which was affected by some errors made by the researcher or by the condition of the equipment. Human error includes the calculation of values as well as the improper procedure performed by the researcher. The condition of the equipment also played a major role since the calculation of mechanical efficiency is an important factor to test the performance of such equipment. The higher the mechanical efficiency, the less power input required to operate the equipment at a certain output but since there is no equipment that is 100% efficient, we assume that there are some losses such as friction losses are present in the equipment.

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QUESTION AND ANSWERS 1. An air compressor has a cylinder diameter of 140 mm and a piston stroke of 200 mm. Air is taken in at 0.98 bar and delivered at 7 bar with a clearance volume of 10%. If the law of compression is PV1.28 = C, calculate the volume of air delivered per stroke. Solution: 𝜋

Stroke Volume = 4 (14 𝑐𝑚)2 (20 𝑐𝑚) = 3078 𝑐𝑚3 Clearance Volume = 0.1(3078 𝑐𝑚3 ) = 307.8 𝑐𝑚3 V1 = stroke volume +clearance volume = 3078 𝑐𝑚3+307.8 𝑐𝑚3 = 3385.8 𝑐𝑚3 P1V11.28 = P2V21.28 = (0.98 𝑏𝑎𝑟)(3385.8 𝑐𝑚3 )1.28 = (7 𝑏𝑎𝑟)(𝑉2 )1.28 V2 = 728.74 cm3 Volume delivered per stroke = 728.74 𝑐𝑚3 − 307.8 𝑐𝑚3 = 420.94 𝑐𝑚3 2. A compressor has a cylinder diameter of 80 mm and the stroke of the piston is 160 mm. The pressures at the beginning and end of compression are 1 bar and 5 bar respectively and the delivery valves open when the piston is 120 mm from the beginning of compression. Find the clearance volume if the law of compression is PV1.3 = C. Solution: 𝜋

Clearance Volume = Area x length = 4 (8 𝑐𝑚)2 (𝐶) Solving for clearance length, C: Let V1 = stroke + clearance = 160 mm + C V2 = 160 mm + C – 120 mm = 40 mm + C P1V11.3 = P2V21.3 = (1𝑏𝑎𝑟)(160 + 𝐶)1.28 = (5 𝑏𝑎𝑟)(40 + 𝐶)1.28 C = 9 mm = 0.9 cm 𝜋

Clearance Volume = 4 (8 𝑐𝑚)2 (0.9 𝑐𝑚) = 45.14 𝑐𝑚3 3. A single acting air compressor has a volumetric efficiency of 87% operates at 500 rpm. It takes in air at 100 kPa and 30 ℃ and discharges it at 600 kPa. The air handled is 6 m3/min at discharge condition. If compression is isentropic, find the mean effective pressure in kPa. Solution: W = PmVd

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Solving for Vd: 𝑚3

P1V1k = P2V2k = (100 𝑘𝑃𝑎)V11.4 = (600 𝑘𝑃𝑎)(6 𝑚𝑖𝑛)1.4 V1 = 21.58 m3/min m3

Vd = (21.58 min)/0.87 = 24.80 m3/min Solving for W: 𝑘−1 𝑃2 𝑘

𝑘𝑃1 𝑉1 𝑊= [( ) 𝑘 − 1 𝑃1

(1.4)(100 𝑘𝑃𝑎) (21.58 − 1] =

m3 𝑚𝑖𝑛)

1.4 − 1

1.4−1 1.4

600 [( ) 100

− 1] = 5049𝑘𝐽/𝑚𝑖𝑛

Hence, 5049𝑘𝐽/ min = 𝑃𝑚 ( 24.80 m3 /min) Pm = 203.59 kPa 4. A two-stage air compressor has a suction pressure of 101.325 kPa and a discharge pressure of 1,140 kPa. What is the intercooler pressure in kPa? 𝑃𝑖 = √𝑃𝑠 𝑃𝑑 = √(101.325 𝑘𝑃𝑎)(1,140 𝑘𝑃𝑎) = 339.87 𝑘𝑃𝑎 5. Air taken the atmospheric at 100 kPaa and 21 degree Celsius is delivered to the air receiver at 689 kPaa and 150 degree Celsius by means of a water jacketed reciprocating compressor. The compressor has a rated capacity of 360 m3/ hr free air. Determine the power required to compress the air.

Power =

𝑛𝑃1 𝑉1 𝑛−1

Solving for n:

𝑃2

(( )

𝑛−1 𝑛

𝑃1

𝑇2 𝑇1

− 1)

𝑃2 𝑛−1

=( ) 𝑃1

𝑛

150+273

→

= (

21+273

689 𝑘𝑃𝑎 𝑛−1

100 𝑘𝑃𝑎

)

𝑛

n = 1.23 V1 = 360 m3/ hr = 0.1 m3/ s

Power =

(1.23)(100 kPa)(0.1 m3/ s) 1.23−1

((

689 𝑘𝑃𝑎

100 𝑘𝑃𝑎

1.23−1 1.23

)

20

− 1) = 23.24 kW

CONCLUSION

Familiarization of the researcher to a compressor is essential since the researcher is studying mechanical engineering. Air compressor got its name because it compresses air to be delivered in the tank for operational purposes. The researcher was able to be familiar with the basic components of an air compressor as well as its operation and the projected PV diagram. An air compressor is important especially for a gas-powered engines since it delivers air from suction and discharge it to the combustion chamber which is needed in combustion. The amount of air compressed and delivered was also determined by the researcher using the formulas on Theory and Principles following the principles in thermodynamics. The amount of air delivered is essential to have a work done on the system. The indicated power was determined by the researcher which was affected by the following parameters: mass flow rate of air delivered, temperature, pressure and volume as well. The efficiency of the air compressor was determined in order to test its performance which factors such as losses were considered by the researcher. These losses are present in the motors, transmitting element, and the compressor itself. Having a low efficiency decreases the capacity of the equipment so proper maintenance to the equipment must be performed in order to save energy. Also, proper lubrication must be applied to the joints of the compressor when needed so that friction losses can be minimized and the save power will be increased. Lastly, follow the instructions made by the instructor and analyze the procedure carefully in order to obtain data. As much as possible, the researcher should wear PPE since the researcher is dealing with equipment and any type of accidents can occur.

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REFERENCES •

Air Compressors. (2018). Retrieved from https://www.thomasnet.com/articles/machinerytools-supplies/Air-Compressors

•

Power Calculation. (2018). Retrieved from http://users.sussex.ac.uk/~tafb8/design05_06/lec6_power_calculation2.pdf

•

Basic Thermodynamics for Reciprocating Compression. (2016). Retrieved from http://oaktrust.library.tamu.edu/bitstream/handle/1969.1/159804/01_Phillippi.pdf?sequence= 1

•

Compressor Efficiency. (2018). Retrieved from https://www.physicsforums.com/threads/calculate-compressor-efficiency.462046/

•

How Air Compressor Works? (2018). Retrieved from https://www.popularmechanics.com/home/how-to/a151/how-air-compressors-work/

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