3/6/2024 Republic of the Philippines Tarlac State University COLLEGE OF ENGINEERING Department of Mechanical Engineering MEF343: FLUID MACHINERIES Engr. Marc Florenz P. Arnaldo Faculty, Department of Mechanical Engineering 1 Republic of the Philippines Tarlac State University COLLEGE OF ENGINEERING Department of Mechanical Engineering LECTURE 3: FUNDAMENTAL PRINCIPLES OF FANS AND BLOWERS Engr. Marc Florenz P. Arnaldo Faculty, Department of Mechanical Engineering 2 1 3/6/2024 LEARNING OUTCOMES After completion of the discussion, the students should be able to: • Explain the types, concepts, and principles related to fans and blowers • Identify the different types and components of fans and blowers • Perform energy analysis to solve related fans and blowers problems For more inquiries, message: mfarnaldo@tsu.edu.ph 3 FUNDAMENTAL PRINCIPLES OF FANS AND BLOWERS FAN A fan is a machine used to apply power to a gas to increase its energy content thereby causing it to flow or move. BLOWER A blower is a fan used to force air under pressure, that is, the resistance to gas flow is imposed primarily upon the discharge. Tube Axial Fan Centrifugal Blower Source: Industrial Plant Engineering by Capote, R. and Mandawe, J. 4 2 3/6/2024 FUNDAMENTAL PRINCIPLES OF FANS AND BLOWERS EXHAUSTER An exhauster is a fan used to withdraw air under pressure which means resistance to gas flow is imposed upon suction. BLOWER A blower is a fan used to force air under pressure, that is, the resistance to gas flow is imposed primarily upon the discharge. Source: Industrial Plant Engineering by Capote, R. and Mandawe, J. 5 FUNDAMENTAL PRINCIPLES OF FANS AND BLOWERS DIFFERENCE BETWEEN FANS AND BLOWERS The American Society of Mechanical Engineers (ASME) has defined fans and blowers based on their discharge pressure and suction pressure ratio. This way they have also defined compressors, and in a way distinguished between all these three devices. According to ASME, a fan is a device with a pressure ratio of up to 1.11. Source: Fans by American Society of Mechanical Engineers (ASME) 6 3 3/6/2024 FUNDAMENTAL PRINCIPLES OF FANS AND BLOWERS TYPES OF FANS: 1. Centrifugal Fan It is a machine which pumping action is accomplished by imparting kinetic energy to the fluid by a high speed revolving impeller with vanes and subsequently converting this kinetic energy into pressure energy either by passing through a volute casing or through a diffuser vanes. Source: Industrial Plant Engineering by Capote, R. and Mandawe, J. 7 FUNDAMENTAL PRINCIPLES OF FANS AND BLOWERS TYPES OF FANS: 2. Axial Fan It is a machine which pumping action is accomplished by imparting kinetic energy to the fluid by a high speed revolving impeller with vanes and subsequently converting this kinetic energy into pressure energy either by passing through a volute casing or through a diffuser vanes. Source: Industrial Plant Engineering by Capote, R. and Mandawe, J. 8 4 3/6/2024 FUNDAMENTAL PRINCIPLES OF FANS AND BLOWERS TYPES OF FANS: 2. Axial Fan Source: Industrial Plant Engineering by Capote, R. and Mandawe, J. 9 FUNDAMENTAL PRINCIPLES OF FANS AND BLOWERS BLOWERS: Blower is equipment or a device which increases the velocity of air or gas when it is passed through equipped impellers. They are mainly used for flow of air/gas required for exhausting, aspirating, cooling, ventilating, conveying etc. Blower is also commonly known as Centrifugal Fans in industry. In a blower, the inlet pressure is low and is higher at the outlet. The kinetic energy of the blades increases the pressure of the air at the outlet. Blowers are mainly used in industries for moderate pressure requirements where the pressure is more than the fan and less than the compressor. Source: Fans and Blowers by Power Zone Equipment, Inc. 10 5 3/6/2024 FUNDAMENTAL PRINCIPLES OF FANS AND BLOWERS TYPES OF BLOWERS: Blowers can also be classified as Dynamic and Positive Displacement Blowers. Like fans, blowers use blades in various designs such as backward curved, forward curved and radial. They are mostly driven by electric motor. They can be single or multistage units and use high speed impellers to create velocity to air or other gases. Positive displacement blowers are similar to PDP pumps, which squeezes fluid that in turn increases pressure. This kind of blower is preferred over a centrifugal blower where high pressure is required in a process. Source: Fans and Blowers by Power Zone Equipment, Inc. 11 FUNDAMENTAL PRINCIPLES OF FANS AND BLOWERS APPLICATIONS OF FANS AND BLOWERS: Fans and Blowers are mostly used for processes such as Air Ventilation, Material Handling, Air Drying etc. Source: Fans and Blowers by Power Zone Equipment, Inc. 12 6 3/6/2024 FUNDAMENTAL PRINCIPLES OF FANS AND BLOWERS APPLICATIONS OF FANS AND BLOWERS: Industrial fans are also used in a variety of applications such as chemical, medical, automotive, agricultural, mining, food processing, and construction industries, which can each utilize industrial fans for their respective processes. They are mainly used in many cooling and drying applications. Source: Fans and Blowers by Power Zone Equipment, Inc. 13 FUNDAMENTAL PRINCIPLES OF FANS AND BLOWERS APPLICATIONS OF FANS AND BLOWERS: Centrifugal blowers are routinely used for applications such as dust control, combustion air supplies, on cooling, drying systems, for fluid bed aerators with air conveyor systems etc. Positive displacement blowers are often 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. Source: Fans and Blowers by Power Zone Equipment, Inc. 14 7 3/6/2024 FUNDAMENTAL PRINCIPLES OF FANS AND BLOWERS APPLICATIONS OF FANS AND BLOWERS: Therefore, fans, and blowers, largely cover Municipal, Manufacturing, Oil & Gas, Mining, Agriculture Industry for their various applications, simple or complex in nature. These are available from various manufacturers in different designs. Few of the well-known brands are Yashiba, Hoffman, Bosch, Sutterbuilt, Sullair and Joy. Source: Fans and Blowers by Power Zone Equipment, Inc. 15 FUNDAMENTAL PRINCIPLES OF FANS AND BLOWERS APPLICATIONS OF FANS AND BLOWERS: A detailed study of all the designs and specifications is required to buy an appropriate fan or a blower that is available in the market so that it can match the requirements of your process and ensure reliability and durability at the same time. Source: Fans and Blowers by Power Zone Equipment, Inc. 16 8 3/6/2024 FUNDAMENTAL PRINCIPLES OF FANS AND BLOWERS BASIC DIFFERENCES COMPRESSORS: OF FANS & BLOWERS, AND EQUIPMENT SPECIFIC RATIO PRESSURE RAISE (mmHg) Fans Up to 1.11 1136 Blowers 1.11 – 1.20 1136 – 2066 Compressors More than 1.20 More than 2066 Source: Centrifugal Pumps by Michael Smith Engineers Co. UK 17 FUNDAMENTAL PRINCIPLES OF FANS AND BLOWERS ANALYSIS OF FANS AND BLOWERS: 1. Static Head, ππ Static Head is the height of the surface of the fluid above the gauge point. This can also be interpreted as the pressure head in the case of fans and blowers, since change in elevation is not a factor in the analysis. β =β = πΎ β πΎ 2. Velocity Head, ππ Velocity Head is the head required to produce the flow of fluid. β = π −π 2π Source: Industrial Plant Engineering by Capote, R. and Mandawe, J. 18 9 3/6/2024 FUNDAMENTAL PRINCIPLES OF FANS AND BLOWERS ANALYSIS OF FANS AND BLOWERS: 3. Total Head, π― Total Head is the sum of the static head and velocity head. π» =β +β π −π π −π + πΎ 2π πΎ (β , − β , ) π − π π»= + πΎ 2π π»= 4. Capacity of the Fan, πΈ The capacity of the fan is the volume flow rate measured at the fan outlet. π = π΄π Source: Industrial Plant Engineering by Capote, R. and Mandawe, J. 19 FUNDAMENTAL PRINCIPLES OF FANS AND BLOWERS ANALYSIS OF FANS AND BLOWERS: 5. Power Output of Fan (Air Power), π·π The power output of a fan or air power is based on fan volume and the fan total pressure. π = πΎ ππ» 6. Power Output of the Fan (Brake Power), π·π© The power output of a fan, or brake power, is the power delivered to the fan shaft. π = π π 7. Power Input to the Fan (Input Power), π·π The power input to the fan, or input power, is the power supplied to the fan motor. π = π π Source: Industrial Plant Engineering by Capote, R. and Mandawe, J. 20 10 3/6/2024 FUNDAMENTAL PRINCIPLES OF FANS AND BLOWERS ANALYSIS OF FANS AND BLOWERS: 8. Static Efficiency of Fan, ππ The static efficiency of a fan is the product of the fan/blower efficiency and the ratio of the static pressure to the total pressure. π =π β β = π , π 9. Overall Efficiency of Fan, ππ The overall efficiency of a fan is the product of the fan/blower efficiency and the mechanical/motor efficiency. π =π π Source: Industrial Plant Engineering by Capote, R. and Mandawe, J. 21 FUNDAMENTAL PRINCIPLES OF FANS AND BLOWERS ANALYSIS OF FANS AND BLOWERS: 10. Fan Affinity Laws There are three (3) basic fan laws which encompass all fan functional principles. Variation in Fan Speed (Constant Diameter & Constant Density) π π = π π π» π = π» π π π = π π Variation in Fan Diameter (Constant Speed & Constant Density) π π· = π π· π» π· = π» π· π π· = π π· Source: Industrial Plant Engineering by Capote, R. and Mandawe, J. 22 11 3/6/2024 FUNDAMENTAL PRINCIPLES OF FANS AND BLOWERS ANALYSIS OF FANS AND BLOWERS: 10. Fan Affinity Laws There are three (3) basic fan laws which encompass all fan functional principles. Variation in Gas Density (Constant Diameter & Constant Speed) π =π π» π = π» π π π = π π Source: Industrial Plant Engineering by Capote, R. and Mandawe, J. 23 FUNDAMENTRAL PRINCIPLES OF PUMPS ANALYSIS OF FANS AND BLOWERS: Source: Industrial Plant Engineering by Capote, R. and Mandawe, J. 24 12 3/6/2024 FUNDAMENTRAL PRINCIPLES OF PUMPS ANALYSIS OF FANS AND BLOWERS: Source: Industrial Plant Engineering by Capote, R. and Mandawe, J. 25 FUNDAMENTRAL PRINCIPLES OF PUMPS ANALYSIS OF FANS AND BLOWERS: π π = π π π» π = π» π π π = π π Source: Industrial Plant Engineering by Capote, R. and Mandawe, J. 26 13 3/6/2024 FUNDAMENTRAL PRINCIPLES OF PUMPS ANALYSIS OF FANS AND BLOWERS: π π· = π π· π» π· = π» π· π π· = π π· Source: Industrial Plant Engineering by Capote, R. and Mandawe, J. 27 PROBLEM SOLVING: 1. Determine the power required to move the air at 20 fpm through a 2’×3’ duct under 3 πππ» π. Given: A Fan used to move air in a duct π = 20 ππ‘/πππ β = 2 ππ‘, π€ = 3ππ‘ β = 3 ππ Required: Power supplied to the Air, π Solution: Solving for the power supplied to the air, Assumption, π =π πΎ β =πΎ β Note: β = β For the static head, πΎ β β = πΎ For the air power, π = πΎ ππ» Note: π» =β +β =β + For more inquiries, message me at: mfarnaldo@tsu.edu.ph βπ 2π 28 14 3/6/2024 PROBLEM SOLVING: 1. Determine the power required to move the air at 20 fpm through a 2’×3’ duct under 3 πππ» π. Solution: For the fan capacity, π = π΄π π = 2ππ‘ 3ππ‘ 20 ππ‘ πππ 1 πππ 60 π π = 2 ππ‘ /π For the static head, πΎ β β =β = πΎ ππ 62.40 ππ‘ β =β = πΎ 3 ππ For more inquiries, message me at: mfarnaldo@tsu.edu.ph 29 PROBLEM SOLVING: 1. Determine the power required to move the air at 20 fpm through a 2’×3’ duct under 3 πππ» π. Solution: Finally, solving for the power supplied to the air, π = πΎ ππ» 62.40 π =πΎ π π = 2 ππ‘ π ππ ππ‘ 3 ππ πΎ 62.40 ππ ππ‘ 3 ππ 1 ππ‘ 12 ππ 1 π»π − π 550 ππ − ππ‘ π·π = π. ππππ π―π· (ans) For more inquiries, message me at: mfarnaldo@tsu.edu.ph 30 15 3/6/2024 PROBLEM SOLVING: 2. A fan whose static efficiency is 40% has a capacity of 60,000 ππ‘ /βπ @ 60β and barometer of 30 inHg and gives a static pressure of 2 inπ» π on full delivery. Determine the size of the motor required to operate the fan. Given: A Fan π = 40% π = 60,000 ππ‘ /π π = 30 πππ»π β , = 2 ππ Required: Size of the Motor required to operate the fan, π Solution: Solving for the brake power using the static efficiency, π , π = π For the static head, πΎ β For the static air power, β = π , = πΎ ππ» πΎ π» =β +β Note: β = β For more inquiries, message me at: mfarnaldo@tsu.edu.ph 31 PROBLEM SOLVING: 2. A fan whose static efficiency is 40% has a capacity of 60,000 ππ‘ /βπ @ 60β and barometer of 30 inHg and gives a static pressure of 2 inπ» π on full delivery. Determine the size of the motor required to operate the fan. Solution: Hence, the static air power π , = πΎ ππ» πΎ β π , =πΎ π πΎ π , = ππΎ β ππ ππ‘ π , = 60000 62.40 βπ ππ‘ π , = 0.3152 π»π 2 ππ 1 βπ 3600π 1 ππ‘ 12 ππ 1 π»π − π 550 ππ − ππ‘ For more inquiries, message me at: mfarnaldo@tsu.edu.ph 32 16 3/6/2024 PROBLEM SOLVING: 2. A fan whose static efficiency is 40% has a capacity of 60,000 ππ‘ /βπ @ 60β and barometer of 30 inHg and gives a static pressure of 2 inπ» π on full delivery. Determine the size of the motor required to operate the fan. Solution: Finally, solving for the size of the motor required to operate the fan Using the static efficiency, π π , π = π = π π π π , π = π = π π Assume: π = 100% 0.3152 π»π π = π = 0.7880 π»π π = 0.40 Use a motor that has the rating of π·π ≈ π. ππ ππ (ans) π = 0.7880 π»π For more inquiries, message me at: mfarnaldo@tsu.edu.ph 33 PROBLEM SOLVING: 3. Air enters a fan through a duct at a velocity of 6.30 m/s and an inlet static pressure of 2.50 cm of water lower than the atmospheric pressure. The air leaves the fan through a duct at a velocity of 11.25 m/s and a discharge static pressure of 7.62 cm of water above the atmospheric pressure. If the density of the air is 1.20 ππ/π and the fan delivers 9.45 π /π of air. Determine the fan efficiency when the power delivered to the fan is 13.75 kW. Given: A Fan through a Duct (Pressurized Air = Blower) π = 6.30 π/π β , = −2.50 πππ» π π = 11.25 π/π β , = 7.62 πππ» π π = 1.20 ππ/π π = 9.45 π /π π = 13.75 ππ Required: Fan Efficiency, π For more inquiries, message me at: mfarnaldo@tsu.edu.ph 34 17 3/6/2024 PROBLEM SOLVING: 3. Air enters a fan through a duct at a velocity of 6.30 m/s and an inlet static pressure of 2.50 cm of water lower than the atmospheric pressure. The air leaves the fan through a duct at a velocity of 11.25 m/s and a discharge static pressure of 7.62 cm of water above the atmospheric pressure. If the density of the air is 1.20 ππ/π and the fan delivers 9.45 π /π of air. Determine the fan efficiency when the power delivered to the fan is 13.75 kW. Solution Solving for the fan efficiency, π π = π For the air power, π = πΎ ππ» For the total dynamic head, π» =β +β βπ βπ πΎ ββ βπ π»= + = + πΎ 2π πΎ 2π πΎ (β , − β , ) π − π π»= + πΎ 2π For more inquiries, message me at: mfarnaldo@tsu.edu.ph 35 PROBLEM SOLVING: 3. Air enters a fan through a duct at a velocity of 6.30 m/s and an inlet static pressure of 2.50 cm of water lower than the atmospheric pressure. The air leaves the fan through a duct at a velocity of 11.25 m/s and a discharge static pressure of 7.62 cm of water above the atmospheric pressure. If the density of the air is 1.20 ππ/π and the fan delivers 9.45 π /π of air. Determine the fan efficiency when the power delivered to the fan is 13.75 kW. Solution Solving for the fan efficiency, πΎ (β , − β , ) π − π π»= + ; πΎ = ππ πΎ 2π ππ 1π 1000 7.62ππ − −2.50ππ 11.25 − 6.30 100ππ π π»= + π ππ 2 9.81 1.20 π π π» = 88.76 π π π π = πΎ ππ» For more inquiries, message me at: mfarnaldo@tsu.edu.ph 36 18 3/6/2024 PROBLEM SOLVING: 3. Air enters a fan through a duct at a velocity of 6.30 m/s and an inlet static pressure of 2.50 cm of water lower than the atmospheric pressure. The air leaves the fan through a duct at a velocity of 11.25 m/s and a discharge static pressure of 7.62 cm of water above the atmospheric pressure. If the density of the air is 1.20 ππ/π and the fan delivers 9.45 π /π of air. Determine the fan efficiency when the power delivered to the fan is 13.75 kW. Solution π = πΎ ππ» ππ π π = 9.87 ππ π = 1.20 9.81 π π 9.45 π π 88.76 π 1 π−π ππ − π 1 ππ 1000 π 1 ππ − π ππ − π Finally, solving for the fan efficiency π π = π 9.87 π = 13.75 ππ = π. ππππ = ππ. ππ% (ans) For more inquiries, message me at: mfarnaldo@tsu.edu.ph 37 PROBLEM SOLVING: 4. A fan delivers 4.70 π /π of air at a static pressure of 5.08 cm of water when operating at a speed of 400 rpm. The power required to operate the fan is 2.963 kW. If 7.05 π /π of air are desired using the same fan and installation, determine the pressure in cm of water. Given: A Fan Required: The Static Pressure from the New Fan Solution: Solving for the static pressure from the new fan, Using the fans laws (@ Constand Fan Diameter) π π = π π π» π = π» π π π = π π Note: It is assumed that π» = β / π π» = → π―π = ππ. ππ ππ ππ πππππ (πππ) π π» For more inquiries, message me at: mfarnaldo@tsu.edu.ph 38 19 3/6/2024 PROBLEM SOLVING: 5. A fan running at 2000 rpm delivers 16,000 ππ‘ /πππ against 3in of water static pressure, thereby consuming 15 BHP. If the fan speed is increased to 2200 rpm so that the pressure ratio is 1:1.10, determine the new capacity of the fan. Given: A Fan π = 2000 πππ, π = 16,000 ππ‘ πππ β , = 3 ππ π , = 15 π»π π = 2200 πππ Required: New Capacity, π Solution: Solving for the new capacity of the fan, Using the fans laws (@ Constant Fan Diameter) π π = π π πΈπ = ππ, πππ πππ /πππ (ans) For more inquiries, message me at: mfarnaldo@tsu.edu.ph 39 PROBLEM SOLVING: 6. A blower draws 3000 ππ‘ /πππ of air through a duct 12in in diameter with a suction of 3in of water. The air is discharged through a duct 10in in diameter against a pressure of 2in of water. The air is measured at 70°F and 30.20 inHg. Determine the amount of air horsepower produced. Use 62.34 ππ /ππ‘ for water. Given: A Blower π = 3000 ππ‘ /πππ π· = 12 ππ = 1 ππ‘ β , = −3ππ π· = 10 ππ β , = 2ππ π = 70°πΉ; π = 30.20 πππ»π π = 62.34 ππ /ππ‘ Required: Amount of Air Horsepower, π For more inquiries, message me at: mfarnaldo@tsu.edu.ph 40 20 3/6/2024 PROBLEM SOLVING: 6. A blower draws 3000 ππ‘ /πππ of air through a duct 12in in diameter with a suction of 3in of water. The air is discharged through a duct 10in in diameter against a pressure of 2in of water. The air is measured at 70°F and 30.20 inHg. Determine the amount of air horsepower produced. Use 62.34 ππ /ππ‘ for water. Solution: Solving for the amount of air horsepower produced, π = πΎ ππ» For the total dynamic head, π» =β +β π»= π»= πΎ ββ βπ + πΎ 2π πΎ (β −β , , ) πΎ + π −π 2π For the air velocities, 4π π = π΄π → π = ππ· π = 63.66 ππ‘/π π = 91.67 ππ‘/π For more inquiries, message me at: mfarnaldo@tsu.edu.ph 41 PROBLEM SOLVING: 6. A blower draws 3000 ππ‘ /πππ of air through a duct 12in in diameter with a suction of 3in of water. The air is discharged through a duct 10in in diameter against a pressure of 2in of water. The air is measured at 70°F and 30.20 inHg. Determine the amount of air horsepower produced. Use 62.34 ππ /ππ‘ for water. Solution: For the total dynamic head, π»= πΎ (β −β , , ) , ) πΎ + π −π 2π + π −π 2π Note: πΎ = ππ π»= π (β , π −β Using the ideal gas equation, π = π π π 14.70 ππ ππ 30.20 πππ»π 144 29.92 πππ»π − ππ ππ‘ π ππ π = = = 0.0756 π π ππ‘ ππ − ππ‘ 53.34 70βinquiries, + 460 message me at: ππ − π For more mfarnaldo@tsu.edu.ph 42 21 3/6/2024 PROBLEM SOLVING: 6. A blower draws 3000 ππ‘ /πππ of air through a duct 12in in diameter with a suction of 3in of water. The air is discharged through a duct 10in in diameter against a pressure of 2in of water. The air is measured at 70°F and 30.20 inHg. Determine the amount of air horsepower produced. Use 62.34 ππ /ππ‘ for water. Solution: For the total dynamic head, π»= π»= πΎ (β , −β πΎ ππ 62.34 ππ‘ , ) + π −π 2π 2ππ − −3ππ 0.0756 1 ππ‘ 12 ππ ππ ππ‘ 91.67 − 63.66 + 2 32.20 ππ‘ π ππ‘ π π» = 411.14 ππ‘ Solving for the amount of air horsepower produced, π = πΎ ππ» ππ ππ‘ ππ‘ π = 0.0756 32.20 3000 411.14 ππ‘ ππ‘ π πππ π·π = π. ππ π―π· (ans) 1 ππ − π 32.20 ππ − ππ‘ 1 π»π − 33000 ππ − ππ‘ For more inquiries, message me at: mfarnaldo@tsu.edu.ph 43 PROBLEM SOLVING: 7. A fan delivers 12,000 ππ‘ /πππ at a static pressure of 1in of water gauge at a speed of 400 rpm and requires and input of 4 HP. If the same installation 15,000 ππ‘ /πππ are desired, determine the new fan speed and the new power input. Given: A Blower π = 12, 000 ππ‘ /πππ β , = 1ππ π = 400 πππ π , = 4 π»π π = 15,000 ππ‘ /πππ Required: New Fan Speed, N and New Power Input, π , For more inquiries, message me at: mfarnaldo@tsu.edu.ph 44 22 3/6/2024 PROBLEM SOLVING: 7. A fan delivers 12,000 ππ‘ /πππ at a static pressure of 1in of water gauge at a speed of 400 rpm and requires and input of 4 HP. If the same installation 15,000 ππ‘ /πππ are desired, determine the new fan speed and the new power input. Solution: Solving for the new fan speed, Using the fan affinity laws π π = π π π =π π π π΅π = πππ πππ (ans) Solving for the new input power, Using the fan affinity laws π π = π π π, = π, π π π , = 7.81 π»π π·π,π ≈ π. π π―π· (ans) Standard Motor Size For more inquiries, message me at: mfarnaldo@tsu.edu.ph 45 46 23 3/6/2024 Questions? Clarifications? Pleasefeelfree to contactme, Engr.MarcFlorenzArnaldo,throughemail or MS Teams. Email Address MS Teams Consultation Hours mfarnaldo@tsu.edu.ph Marc Florenz Arnaldo Tuesday - Friday 1 PM to 4 PM 49 For more inquiries, message me at: mfarnaldo@tsu.edu.ph 50 25
