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

EXPERIMENT NO. 6 ROTARY BLOWER

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

Date of Performance: Sep 19, 2018 Date of Submission: Oct 3, 2018

GRADE Engr. Teodulo A. Valle Instructor

TABLE OF CONTENTS

Objectives

Page 1

Theory and Principle

Page 1

List Of Apparatus

Page 6

Procedure

Page 9

Set-up of Apparatus

Page 12

Final Data Sheet

Page 12

Sample Computations

Page 13

Discussion of Result

Page 15

Questions and Answers

Page 16

Conclusion

Page 18

Reference

Page 19

Preliminary Data Sheet

Page 20

i

OBJECTIVES 1. To be familiar with the operating procedure and principle; and 2. To be able to determine the efficiency of the rotary air blower THEORY AND PRINCIPLE Fans and blowers provide air for ventilation and industrial process requirements. Fans generate a pressure to move air (or gases) against a resistance caused by ducts, dampers, or other components in a fan system. The fan rotor receives energy from a rotating shaft and transmits it to the air. Fans, blowers and compressors are differentiated by the method used to move the air, and by the system pressure they must operate against. As per American Society of Mechanical Engineers (ASME) the specific ratio - the ratio of the discharge pressure over the suction pressure - is used for defining the fans, blowers and compressors.

Blowers can achieve much higher pressures than fans, as high as 1.20 kg/cm2. They are also used to produce negative pressures for industrial vacuum systems. Major types are: centrifugal blower and positive-displacement blower. Centrifugal blowers look more like centrifugal pumps than fans. The impeller is typically gear-driven and rotates as fast as 15,000 rpm. In multi-stage blowers, air is accelerated as it passes through each impeller. In single-stage blower, air does not take many turns, and hence it is more efficient. Centrifugal blowers typically operate against pressures of 0.35 to 0.70 kg/cm2 but can achieve higher pressures. One characteristic is that airflow tends to drop drastically as system pressure increases, which can be a disadvantage in material conveying systems that depend on a steady air volume. Because of this, they are most often used in applications that are not prone to clogging.

1

Positive-displacement blowers have rotors, which "trap" air and push it through housing. Positive-displacement blowers provide a constant volume of air even if the system pressure varies. They are especially suitable for applications prone to clogging, since they can produce enough pressure - typically up to 1.25 kg/cm2 - to blow clogged materials free. They turn much slower than centrifugal blowers (e.g. 3,600 rpm) and are often belt driven to facilitate speed changes.

2

Blowers are mechanical or electro-mechanical devices used to induce gas flow through ducting, electronics chassis, process stacks, etc.--wherever flow is needed for exhausting, aspirating, cooling, ventilating, conveying, and so on. Key specifications include intended application, blower type, port design, as well as the parameters of flow capacity, electrical ratings, and dimensions. Blowers cool electronic enclosures, induce drafts in boilers, increase airflow on engines, and are configured in a variety of designs such as centrifugal flow or rotary lobe styles. Motors usually drive blowers, though they can be powered by other means such as engines. Often used interchangeably with “Fans,” blowers are defined by the ASME as having a ratio of discharge pressure over suction pressure between 1.11 and 1.2, while fans are defined as anything below this ratio and compressors are defined as anything above it. Some makers of portable fans refer to their units as blowers even if they do not necessarily conform to the ASME distinction, which applies to permanently install industrial process equipment. Another kind of blower is the mobile or hand-held device used for moving fallen leaves. Centrifugal blowers are routinely used for combustion air supplies, on cooling and drying systems, for fluid bed aerators, with air conveyor systems, for dust control, etc. Positive displacement blowers are also used in pneumatic conveying, and for sewage aeration, filter flushing, and gas boosting, as well as for moving gases of all kinds in the petrochemical industries. Centrifugal blowers are often built as close-coupled units, meaning that the impeller wheel is not supported by independent bearings but is cantilevered on an extension of the motor shaft and relies on the motor bearings for support. Close coupled mounting dispenses with the need for shaft couplings. Other arrangements cantilever the wheel off pillow block bearings, such as designs that use belt drives. Blowers are sometimes stepped up from motor speed, but are just as often stepped down or 1:1 ratios. The centrifugal blower outlet is usually arranged tangentially to impeller rotation and can be specified usually in one of eight angular orientations with respect to the direction of the blower wheel rotation, making for sixteen possible arrangements of rotation and discharge orientation in 45 degree increments. Industry practice specifies impeller rotation either as CW or CCW as viewed from the drive end—usually the motor end—of the unit. On smaller centrifugal blowers the housings can often be rotated through a full circle to permit any angle of discharge. Rotary lobe blowers usually orient the input and output ports in line, due to the design of the blower.

3

Centrifugal Blower (left) and Positive Displacement Blower (right) Blowers are specified on the basis of pressure and flow rate. As mentioned, the ratio of system inlet pressure and outlet pressure determines whether a fan or blower should be picked according to strict definition, although the terms are sometimes used synonymously. Where higher pressures are needed, a designer may have to select a positive displacement machine over a centrifugal type. Manufacturers often publish fan performance curves or similar charts which help the designer to narrow his choice to one or several models that match requirements. The chart at right is fairly common among blower makers. System designers decide the flow rate and pressure needed and add additional capacity to overcome frictional losses in the systems due to ducting, piping, etc. They can select materials or coatings that combat the effects of corrosive media. Most blower capacity charts are based on standard temperature and pressure, ie, 70 degree F air at sea level. Where design conditions are different, designers can apply correction factors which size the blowers based on actual conditions. In the experiment, the researcher focused on the performance of the rotary blower. Below are the formulas needed to compute for the unknown data. Solving for the density of air: 𝑃𝑉 = 𝑚𝑅𝑇 →

𝑚 𝑃 =𝜌= 𝑉 𝑅𝑇

Equation 1

Solving for the velocity head of air: 𝜌𝑎𝑖𝑟 𝑔ℎ𝑎𝑖𝑟 = 𝜌𝐻𝑔 𝑔ℎ𝐻𝑔 → ℎ𝑎𝑖𝑟 =

𝜌𝐻𝑔 ℎ𝐻𝑔 𝜌𝑎𝑖𝑟

Equation 2

Solving for the velocity of air: ℎ𝑎𝑖𝑟 =

𝑣𝑎𝑖𝑟 2 → 𝑣𝑎𝑖𝑟 = √2𝑔ℎ𝑎𝑖𝑟 2𝑔

4

Equation 3

Solving for the volume flow rate of air: 𝑄 = 𝐴𝑜𝑟𝑖𝑓𝑖𝑐𝑒 𝑣𝑎𝑖𝑟

Equation 4

ℎ𝑇𝐻𝑔 = ℎ𝑠𝑡𝑎𝑡𝑖𝑐 + ℎ𝑣

Equation 5

Solving for the total head of mercury:

Solving for the total head of air: ℎ𝑇𝑎𝑖𝑟 =

𝜌𝐻𝑔 ℎ𝐻𝑔 𝜌𝑎𝑖𝑟

Equation 6

Solving for the air power/ output power: 𝑃𝑜𝑢𝑡𝑝𝑢𝑡 = 𝛾𝑎𝑖𝑟 𝑄ℎ𝑇𝑎𝑖𝑟 = 𝜌𝑎𝑖𝑟 𝑔𝑄ℎ𝑇𝑎𝑖𝑟

Equation 7

𝑃𝑜𝑢𝑡𝑝𝑢𝑡 = 𝐸𝐼𝑐𝑜𝑠(∅)

Equation 8

𝑃𝑜𝑢𝑡𝑝𝑢𝑡 × 100% 𝑛𝑡𝑟𝑎𝑛𝑠 𝑛𝑚𝑜𝑡𝑜𝑟 𝑃𝑖𝑛𝑝𝑢𝑡

Equation 9

Solving for the input power:

Solving for the mechanical efficiency: 𝑛𝑚𝑒𝑐ℎ =

5

LIST OF APPARATUS 1. Rotary Air blower

2. Electric motor 3. Mercury Manometer

4. Pitot tube

6

5. Tachometer

6. Amprobe

7. Set of Orifices

7

8. Stopwatch 9. Thermometer (Digital)

8

PROCEDURES 1. Prepare the apparatus needed for the experiment.

2. Insert the orifice with diameter 3/4 inch into the discharge air tunnel of the rotary blower.

3. Start the motor so that the rotary blower will now able to intake air from the atmosphere. 4. Obtain the suction temperature by placing the thermometer directly to the suction side of the rotary blower. Perform this procedure for at least 3 minutes. 5. Determine the static and velocity head of the air using a manometer. In getting the velocity head, use a pitot tube and place it inside the discharge air tunnel of the rotary blower while for the static head, insert the tube of the manometer in the static section. Observe and record the deflections.

9

6. Use a tachometer to determine the speed of the rotary blower. Make at least 3 trials and obtain the average which will be recorded in the data sheet.

7. Determine the current flowing into the motor using an amprobe.

8. Use the thermometer to determine the discharge temperature and do it at least 3 mins.

10

9. Make 2 more trials by replacing the orifice diameter into 1 inch and 1 1/4 inches. 10. Record all the data in the data sheet and calculate for the necessary data. Return the materials after the experiment.

11

SET-UP OF APPARATUS

FINAL DATA SHEET

Trial

1 2 3

Orifice Orifice Pressure Temperature Current Input Output Eff. Diameter Static Velocity Suction Discharge (Amperes) (HP) (HP) (%) (inches) (in (in Hg) (°𝑅) (°𝑅) Hg) ¾ 34/8 41/8 549.24 564 12.375 2.792 2.098 75.155 549.96 1 18/8 22/8 561.3 10.84 3.1968 1.4635 59.84 551.22 1¼ 7/8 6/8 559.5 10.645 2.4025 0.3926 16.34

12

SAMPLE COMPUTATIONS Trial 2 •

Solving for the density of air: 𝑙𝑏𝑓 12𝑖𝑛 2 )( ) 𝑚 𝑃 𝑙𝑏𝑚 𝑖𝑛2 1𝑓𝑡 𝑃𝑉 = 𝑚𝑅𝑇 → = 𝜌 = = = 0.07216 3 𝑓𝑡 − 𝑙𝑏𝑓 𝑉 𝑅𝑇 𝑓𝑡 (53.34 ) (549.96 𝑅) 𝑙𝑏𝑚 − 𝑅 (14.7



Solving for the velocity head of air:

𝜌𝑎𝑖𝑟 𝑔ℎ𝑎𝑖𝑟 = 𝜌𝐻𝑔 𝑔ℎ𝐻𝑔 → ℎ𝑎𝑖𝑟 =

𝜌𝐻𝑔 ℎ𝐻𝑔 𝜌𝑎𝑖𝑟

1𝑓𝑡 𝑘𝑔 2.2046𝑙𝑏𝑚 1𝑚 3 (2.75𝑖𝑛)(12𝑖𝑛)(13600 3 )( )( ) 1𝑘𝑔 3.28𝑓𝑡 𝑚 = = 2698.3737 𝑓𝑡 𝑙𝑏𝑚 0.07216 3 𝑓𝑡 •

Solving for the velocity of air: ℎ𝑎𝑖𝑟 =



𝑣𝑎𝑖𝑟 2 𝑓𝑡 𝑓𝑡 → 𝑣𝑎𝑖𝑟 = √2𝑔ℎ𝑎𝑖𝑟 = √2 (32.2 2 ) (2698.37372 𝑓𝑡) = 416.8636 2𝑔 𝑠 𝑠

Solving for the air flow rate: 𝑄 = 𝐴𝑜𝑟𝑖𝑓𝑖𝑐𝑒 𝑣𝑎𝑖𝑟



𝜋 1𝑓𝑡 2 𝑓𝑡 𝑓𝑡 3 = (7𝑖𝑛 × ) (416.8636 ) = 2.2736 4 12𝑖𝑛 𝑠 𝑠

Solving for the total head of air: ℎ𝑇𝐻𝑔 = ℎ𝑠𝑡𝑎𝑡𝑖𝑐 + ℎ𝑣 = 18/8 𝑖𝑛 + 22/8 𝑖𝑛 = 5 𝑖𝑛

ℎ𝑇𝑎𝑖𝑟 •

𝑘𝑔 2.2046𝑙𝑏𝑚 1𝑓𝑡 1𝑚 3 𝜌𝐻𝑔 ℎ𝑇𝐻𝑔 (13600 𝑚3 ) ( 1𝑘𝑔 )(3.28𝑓𝑡) (5𝑖𝑛) (12𝑖𝑛) = = = 4906.1339 𝑓𝑡 𝑙𝑏𝑚 𝜌𝑎𝑖𝑟 0.07216 3 𝑓𝑡

Solving for the Output Power: 𝑃𝑜𝑢𝑡𝑝𝑢𝑡 = 𝛾𝑎𝑖𝑟 𝑄ℎ𝑇𝑎𝑖𝑟 = 𝜌𝑎𝑖𝑟 𝑔𝑄ℎ𝑇𝑎𝑖𝑟 𝑓𝑡 2.2736𝑓𝑡 3 𝑙𝑏𝑚 (0.07216 3 ) (32.2 2 ) ( ) (4906.1339 𝑓𝑡) 𝑠 𝑓𝑡 𝑠 = 𝑙𝑏𝑚 − 𝑓𝑡 (32.2 ) 𝑙𝑏𝑓 − 𝑠 2

13

𝑃𝑜𝑢𝑡𝑝𝑢𝑡 = (804.9149 •

𝑓𝑡 − 𝑙𝑏𝑓 1𝐻𝑃 )( ) = 1.4635 𝐻𝑃 550𝑓𝑡 − 𝑙𝑏𝑓 𝑠 𝑠

Solving for Input Power: 1𝐻𝑃 𝑃𝑜𝑢𝑡𝑝𝑢𝑡 = 𝐸𝐼𝑐𝑜𝑠(∅) = (220𝑉)(10.84 𝐴)(1.00) ( ) = 3.1968 𝐻𝑃 746𝑊



Solving for mechanical efficiency: 𝑛𝑚𝑜𝑡𝑜𝑟 = 𝑛𝑡𝑟𝑎𝑛𝑠 =

𝑀𝑜𝑡𝑜𝑟 𝐵𝑒𝑙𝑡 𝑃𝑜𝑤𝑒𝑟 → 𝑀𝑜𝑡𝑜𝑟 𝐵𝑒𝑙𝑡 𝑃𝑜𝑤𝑒𝑟 = 𝑛𝑚𝑜𝑡𝑜𝑟 𝑃𝑖𝑛𝑝𝑢𝑡 𝑃𝑖𝑛𝑝𝑢𝑡

𝐵𝑙𝑜𝑤𝑒𝑟 𝐵𝑒𝑙𝑡 𝑃𝑜𝑤𝑒𝑟 → 𝐵𝑙𝑜𝑤𝑒𝑟 𝐵𝑒𝑙𝑡 𝑃𝑜𝑤𝑒𝑟 = 𝑛𝑡𝑟𝑎𝑛𝑠 𝑀𝑜𝑡𝑜𝑟 𝐵𝑒𝑙𝑡 𝑃𝑜𝑤𝑒𝑟 𝑀𝑜𝑡𝑜𝑟 𝐵𝑒𝑙𝑡 𝑃𝑜𝑤𝑒𝑟 𝑛𝑚𝑒𝑐ℎ =

=

𝑃𝑜𝑢𝑡𝑝𝑢𝑡 𝑃𝑜𝑢𝑡𝑝𝑢𝑡 = 𝐵𝑙𝑜𝑤𝑒𝑟 𝐵𝑒𝑙𝑡 𝑃𝑜𝑤𝑒𝑟 𝑛𝑡𝑟𝑎𝑛𝑠 𝑀𝑜𝑡𝑜𝑟 𝐵𝑒𝑙𝑡 𝑃𝑜𝑤𝑒𝑟

𝑃𝑜𝑢𝑡𝑝𝑢𝑡 1.4635 𝐻𝑃 = × 100% = 59.84% 𝑛𝑡𝑟𝑎𝑛𝑠 𝑛𝑚𝑜𝑡𝑜𝑟 𝑃𝑖𝑛𝑝𝑢𝑡 (0.9)(0.85)(3.1968 𝐻𝑃)

14

DISCUSSION OF RESULT

Based on the data gathered by the students, three trials were performed with different orifice diameter ( ¾”, 1”, 1 ¼” ). As the orifice diameter increases, the speed also increases. Taking an observation at the head, the liquid used in the manometer was mercury (Hg) because the pressure in the discharge of the blower is too much high and if water will be used, the water will just jet out from the manometer. The static head decreases when the orifice diameter was increased while the velocity head increases when the orifice diameter was increased based on the data. The researcher obtained some errors on recording the velocity head because the expected trend of the data when the orifice diameter was increased is that the velocity head decreases. It can be the insertion of the Pitot tube at the discharge of air or the improper measurement of deflection made by the researcher affected the data. For the temperature, the suction temperature is constant in trial 1 and 2 at 549.24 Rankine while trial 3 had a temperature of 551.22 Rankine. However, the discharge, the temperature decreases by approximately 3 Rankine when the orifice diameter was increased. The current in the motor also decreases and as a result. The output power was computed by the students and projects a varying trend because of the data gathered by the students. It must project an increasing output power as the orifice diameter increases since it also increase the volume flow rate of air. The output power was affected also by the total head of air which decreases when the orifice diameter was increased, but the volume flow rate is significantly larger which affects the value of output power. Thus, the volume flow rate shows direct proportionality with the output power. Also, the output power was affected by the efficiency of the motor due to some losses and the resulting efficiency of the rotary blower were computed and ranging from 16%80%. The data gathered were affected by some errors that were made by the researcher like rounding off values in the calculations or the condition of the apparatus used in the experiment for the assumption that losses are present in the system.

15

QUESTION AND ANSWERS 1. A small blower handles 43.33 m3 of air per minute whose density is 1.169 kg/m3. The static and velocity heads are 16.38 and 1.22 cm WG (at 15.6 degrees Celsius) respectively. Local gravity acceleration is 9.741 m/s2. Find the power input to the air from the blower. 𝑃 = 𝛾𝑄ℎ; ℎ = 16.38 + 1.22 = 17.6 𝑐𝑚 = 0.176 𝑚 43.33𝑚3 1𝑚𝑖𝑛 𝑚3 𝑄=( )( ) = 0.72 min 60𝑠 𝑠 𝑃 = 9.741(0.72)(0.176) = 1.24 𝑘𝑊 2. A blower operating at 15000 rpm compresses air from 20 degrees Celsius and 1 atm to 1.68 atm. The design flow is 38 m3/min and at this point the power input is 60 kW. Determine the blower efficiency at the design flow. 𝐵𝑙𝑜𝑤𝑒𝑟 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =

𝐼𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐 𝑃𝑜𝑤𝑒𝑟 𝑃𝑜𝑤𝑒𝑟 𝐼𝑛𝑝𝑢𝑡

𝑘 𝑃2 (𝑃𝑉) (( ) 𝐼𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐 𝑃𝑜𝑤𝑒𝑟 = 𝑘−1 𝑃1 1.4 38 1.68 (101.325) ( ) (( = ) 1.4 − 1 60 1 𝐵𝑙𝑜𝑤𝑒𝑟 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =

1.4−1 1.4

𝑘−1 𝑘

− 1)

− 1) = 35.89 𝑘𝑊

35.89 × 100% = 59.81% 60

3. A sewerage aeration blower rotating at 3500 rpm is designed to deliver 567 m3/min of air from 20 degrees Celsius and 1 atm to a discharge of 158 kPaa with an adiabatic efficiency of 65%. During summer, the atmospheric temperature rises to 43 degrees Celsius but the barometric pressure does not change. It is desired to vary the blower speed to maintain the same discharge pressure. Determine the discharge volume of standard air with the new speed. ℎ1 𝑄1 2 ℎ1 𝑇2 43 + 273 =( ) ; = = = 1.08; ℎ2 𝑄2 ℎ2 𝑇1 20 + 273 567 2 𝑚3 1.08 = ( ) → 𝑄2 = 545.98 𝑄2 𝑚𝑖𝑛 4. A blower with the inlet open to the atmosphere delivers 3000 cfm of air at a pressure of 2 in. WG through a duct 11in. in diameter, the manometer being attached to the discharge duct at the blower. Air temperature is 70 degrees Fahrenheit, and the barometer pressure is 30.2 in Hg. Calculate the horsepower.

16

𝑃𝑎𝑖𝑟 = 𝛾𝑄ℎ; 𝑄 = 3000 𝑐𝑓𝑚 𝐴𝑖𝑟 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑎𝑡 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 𝑏𝑦 𝑐𝑜𝑟𝑟𝑒𝑐𝑡𝑖𝑛𝑔 𝑡ℎ𝑒 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑛𝑠𝑖𝑡𝑦. 30.2 𝑙𝑏 𝜌𝑎𝑖𝑟 = 0.075 ( ) = 0.076 3 29.92 𝑓𝑡 ℎ𝑠 =

2 62.4 ( ) = 136.84 𝑓𝑡 𝑜𝑓 𝑎𝑖𝑟. 12 0.076 2

3000 ( 60 2 ) 𝜋 11 𝑣2 4 (12) ℎ𝑣 = = = 89.13 𝑓𝑡 2𝑔 2(32.2) 3000 1 𝑃 = 0.076 ( ) (136.84 + 89.13) ( ) = 1.56 𝐻𝑃 60 550 5. A blower draws 3000 cfm of air through a duct 12 in. in diameter with a suction of 3 in. of water. The air is discharged through a duct 10 in. in diameter against a pressure of 2 in. of water. The air is measured at 70 degrees Fahrenheit and 30.2 in. Hg. Calculate the air horsepower. Use specific weight of 62.34 lb/ft3. 𝑃𝑎𝑖𝑟 = 𝛾𝑄ℎ 3000 3000 𝑃𝑑 − 𝑃𝑠 𝑣𝑑2 − 𝑣𝑠2 𝑓𝑡 𝑓𝑡 60 ℎ = 𝑧𝑑 − 𝑧𝑠 + + ; 𝑣𝑠 = ; 𝑣𝑑 = 60 2 = 63.66 2 = 91.67 𝛾 2𝑔 𝑠 𝑠 𝜋 10 𝜋 12 ( ) ( ) 4 12 4 12 𝑃𝑠 = 𝛾ℎ = 62.34 (

2 −3 ) = 0.072 𝑝𝑠𝑖; 𝑃𝑑 = 62.34 ( ) = −0.108 𝑝𝑠𝑖 12 12

𝑧𝑑 = 𝑧𝑠 ; 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑎𝑖𝑟 𝑎𝑡 70℉ 𝑎𝑛𝑑 30.2 𝑖𝑛. 𝐻𝑔 14.7 30.2 (29.92) (144) 𝑃 𝑙𝑏 𝛾= = = 0.0756 3 𝑅𝑇 53.34(70 + 460) 𝑓𝑡 (0.072 − (−0.108))(144) (91.67)2 − (63.66)2 ℎ= + = 410.42 𝑓𝑡 𝑜𝑓 𝑎𝑖𝑟 0.0756 2(32.2)

𝑃𝑎𝑖𝑟

3000 0.0756 ( 60 ) (410.42) = = 2.82 𝐻𝑃 550

17

CONCLUSION The researcher was able to be familiar with the components and operating procedure of a rotary blower located at the ME Laboratory Room at North Building of Mapua University. Blowers usually force air under pressure just like fans, the difference is that blowers increases the pressure of air higher than the fans. The resistance to gas flow is imposed primarily upon the discharge. Just like fans, there are several types of blowers but the researcher focused only on the performance test of a rotary blower. The researcher was also able to calculate the mechanical efficiency of the air blower and found out that the blower assessed has mechanical efficiency of 16% - 80%. The air power/ output power was affected by the variables volume flow rate and total dynamic head of air. The efficiency of a blower is essential especially in industrial applications since it also affects the costs of the company/ industry. Factors such as losses were considered by the researcher and these losses are present in the motors, transmitting element, and the rotary blower itself. A lower efficiency decreases the capacity of the air blower so proper maintenance must be applied to the equipment so that there will be a considerable amount of energy that can be saved by the facility. Also, proper lubrication must be applied to the joints of the rotary blower 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 •

Fans and Blowers (2018). Retrieved from https://www.saylor.org/site/wpcontent/uploads/2011/09/Chapter-3.5-Fans-Blowers.pdf



Introduction to fans and blowers. (2018). Retrieved from https://www.bchydro.com/content/dam/hydro/medialib/internet/documents/psbusiness/pd f/fans_blowers_guide.pdf



Types of Blowers. (2018). Retrieved from https://www.thomasnet.com/articles/pumpsvalves-accessories/types-of-blowers-industrial-fans



Fans and Blowers – Applications and Industrial Use. (2018). Retrieved from https://www.saylor.org/site/wp-content/uploads/2011/09/Chapter-3.5-Fans-Blowers.pdf



Industrial Fans and Blowers. (2018). Retrieved from https://continentalfan.com/productcategory/industrial/



Blower Efficiencies. (2018). Retrieved from http://www.machinedesign.com/archive/improving-energy-efficiency-low-pressureblowers

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