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5/13/2020
FLUID MACHINERIES
FANS AND BLOWERS PREPARED BY: ENGR. ANGIELO U. COMIA MSEUF – College of Engineering
DEFINITION OF TERMS
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FAN •Is a machine used to apply power to a gas to increase its energy content thereby causing it to flow or move.
BLOWER •Is a fan used to force air under suction, that is, the resistance to gas flow is imposed primarily upon the inlet
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EXHAUSTER •Is a fan used to withdraw air under suction, that is, the resistance to gas flow is imposed primarily upon the inlet.
TYPES OF FAN
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1. PROPELLER FAN •Consists of a propeller or disk type wheel within a mounting ring or plate and including driving mechanism supports either for belt drive or direct connection.
2. TUBEAXIAL FAN •Consists of a propeller or disk type wheel within a cylinder and including driving mechanism supports either for belt drive or direct connection.
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3. VANEAXIAL FAN •Consists of a disk type wheel within cylinder, a set of air guide vanes located either before or after the wheel and including driving mechanism supports either for belt drive or direct connection.
4. CENTRIFUGAL FAN •Consists of a fan rotor or wheel within a scroll type of housing and including driving mechanism supports either for belt drive or direct connection.
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PERFORMANCE OF FANS
FAN PERFORMANCE •Is a statement of volume, total pressures; static pressure, speed, power input, mechanical and static efficiency, at standard air density.
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1. STATIC HEAD, ℎ𝑠 • Static head is the height of the surface of the fluid above the gauge point.
𝜸𝒘 𝒉𝒘 𝝆𝒘 𝒉𝒘 𝒉𝒔 = = 𝜸𝒂 𝝆𝒂
Where: ℎ𝑠 = static head, m of air ℎ𝑤 = manometer reading, m of water 𝛾𝑤 = specific weight of water = 9.81 kN/𝑚3 𝛾𝑎 = specific weight of air = 0.011772 kN/𝑚3 = 11.772 N/𝑚3 𝜌𝑤 = density of water = 1000 kg/𝑚3 𝜌𝑎 = density of air = 1.2 kg/𝑚3 at 101.325 kPa and 21.2 degree C
2. VELOCITY HEAD • Velocity head is the head required to produce the flow of fluid.
𝑽𝒈 𝟐 𝒉𝒗 = 𝟐𝒈 Where: ℎ𝑣 = velocity head of air, m 𝑉𝑔 = outlet velocity g = gravitational acceleration = 9.81 m/𝑠 2 = 32.2 ft/𝑠 2
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3. TOTAL HEAD •Total Head is the sum of the static head and velocity head. 𝒉 = 𝒉𝒔 + 𝒉𝒗 Where: h = total head ℎ𝑠 = static head of air, m ℎ𝑣 = velocity head of air, m
4. CAPACITY OF FAN •The capacity of fan is the volume flow rate measured at fan outlet. 𝑸 = 𝑨𝑽 Where: A = fan outlet area V = fan outlet velocity
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5. POWER OUTPUT OF FAN OR AIR POWER •The power output of a fan or air is based on fan volume and the fan total pressure. 𝑷𝒂𝒊𝒓 = 𝜸𝑸𝒉 Where: 𝛾 = specific weight of air = 𝜌𝑔 = (1.2 kg/𝑚3 )(9.81 m/𝑠 2 ) Q = fan capacity (volume flow rate) h = total head
6. POWER INPUT OF FAN or BRAKE POWER •The power input of a fan or brakepower is the power delivered to the fan shaft.
𝑷𝒃𝒓𝒂𝒌𝒆 Where:
𝑷𝒂𝒊𝒓 = 𝒆𝒎
𝑃𝑎𝑖𝑟 = power output of fan 𝑒𝑚 = mechanical efficiency of fan
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7. STATIC EFFICIENCY OF FAN The static efficiency of a fan is the mechanical efficiency multiplied by the ratio of the static pressure to the total pressure.
𝒆𝒔 = 𝒆𝒎
𝑷𝒔 𝑷
Where: 𝑒𝑠 = static efficiency 𝑒𝑚 = mechanical efficiency 𝑃𝑠 = static pressure P = total pressure
FAN LAWS There are three(3) basic fan laws encompass all fan functional principles.
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1. FAN SPEED VARIATIONS (constant fan size, constant density) If the speed (rpm) is varied. Capacity / Volume Flow Rate 𝑸 𝟏 𝑵𝟏 = 𝑸 𝟐 𝑵𝟐 Static / Total Pressure or Head 𝒉𝟏 𝑵𝟏 = 𝒉𝟐 𝑵𝟐 Impeller Absorbed Power 𝑷𝟏 𝑵𝟏 = 𝑷𝟐 𝑵𝟐
𝟐
𝟑
2. FAN SIZE VARIATIONS (constant speed, constant density) • If fan wheel diameter (D) is varied. Capacity / Volume flow rate 𝑸𝟏 𝑫𝟏 = 𝑸𝟐 𝑫𝟐 Static / Total Pressure or Head 𝒉𝟏 𝑫𝟏 = 𝑷𝟐 𝑫𝟐 Impeller Absorbed Power 𝑷𝟏 𝑫𝟏 = 𝑷𝟐 𝑫𝟐
𝟑
𝟐
𝟓
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3. GAS DENSITY VARIATION (constant fan size, constant speed) • If density is varied. Capacity / Volume flow rate
𝑸𝟐 = 𝑸𝟏 Static/Total Pressure or Head
𝒉𝟏 𝝆𝟏 = 𝒉𝟐 𝝆𝟐
Impeller absorbed Power
𝑷𝟏 𝝆𝟏 = 𝑷𝟐 𝝆𝟐
BERNOULLI’S EQUATION APPLIED FOR FAN
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Total Head = Static Pressure Head + Velocity Head 𝑷𝟐 − 𝑷𝟏 𝑽 𝟐 𝟐 − 𝑽 𝟏 𝟐 𝒉= + 𝜸𝒂 𝟐𝒈 Where: 𝑃2 = 𝑑𝑖𝑠𝑐ℎ𝑎𝑟𝑔𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 = 𝛾𝑤2 ℎ2 𝑃1 = 𝑖𝑛𝑙𝑒𝑡 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 = 𝛾𝑤1 ℎ1 𝛾𝑎 = specific weight of air 𝛾𝑤 = specific weight of water 𝑉2 = discharge velocity 𝑉1 = inlet velocity g = gravitational acceleration
SAMPLE PROBLEM: 1. What Horsepower is supplied to air moving at 20 fpm through a 2 X 3 ft duct under a pressure of 3 in. water gage?
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SOLUTIONS: 𝑷𝒂𝒊𝒓 = 𝜸𝒂 𝑸𝒉 Solving for Capacity (Volume flow rate), Q: 𝑸 = 𝑨𝑽 Cross Sectional Area of the Duct: 𝐴 = 2𝑓𝑡 3𝑓𝑡 = 6𝑓𝑡 2 Air velocity: 𝑓𝑡 1 𝑚𝑖𝑛 1 𝑉 = 20 = 𝑓𝑡/𝑠𝑒𝑐 min 60 𝑠𝑒𝑐 3
𝑄 = (6𝑓𝑡 2 )
Solving for h:
1 𝑓𝑡/𝑠𝑒𝑐 3
= 𝟐𝒇𝒕𝟑 /𝒔𝒆𝒄
SOLUTIONS: (continued..)
ℎ = ℎ𝑠 + ℎ𝑣 𝑁𝑂𝑇𝐸: 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 ℎ𝑒𝑎𝑑 𝑐𝑎𝑛 𝑏𝑒 𝑛𝑒𝑔𝑙𝑒𝑐𝑡𝑒𝑑 𝑑𝑢𝑒 𝑡𝑜 𝑖𝑡𝑠 𝑠𝑚𝑎𝑙𝑙 𝑣𝑎𝑙𝑢𝑒 Static Head: 1 𝑓𝑡 𝑙𝑏 62.4 3 3𝑖𝑛 12 𝑖𝑛 𝛾𝑤 ℎ𝑤 𝑓𝑡 ℎ𝑠 = = = 𝟐𝟎𝟖𝒇𝒕 𝛾𝑎 0.075 𝑙𝑏/𝑓𝑡 3 For water: 𝛾𝑤 = 62.4 𝑙𝑏/𝑓𝑡 3 ℎ = ℎ𝑠 + ℎ𝑣 For air: 𝛾𝑎 = 0.075 𝑙𝑏/𝑓𝑡 3 −3 Velocity head:
ℎ = 208𝑓𝑡 + 1.725328𝑥10 𝒉 = 208.0017253 ft
𝑓𝑡
2 1 𝑓𝑡/𝑠𝑒𝑐 𝑉 3 ℎ𝑣 = = = 𝟏. 𝟕𝟐𝟓𝟑𝟐𝟖𝒙𝟏𝟎−𝟑 𝒇𝒕 2𝑔 2 32.2 𝑓𝑡 𝑠𝑒𝑐 2 𝑓𝑡 g = gravitational acceleration = 32.2 𝑠𝑒𝑐 2 2
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SOLUTIONS (continued).. 𝑃𝑎𝑖𝑟
𝑙𝑏 = 𝛾𝑎 𝑄ℎ = 0.075 3 𝑓𝑡
𝑃𝑎𝑖𝑟 = 31.2
NOTE: 1HP = 550
𝑓𝑡 − 𝑙𝑏 𝑠𝑒𝑐
𝑓𝑡 3 2 𝑠𝑒𝑐
208.0017253 𝑓𝑡 = 31.2
𝑓𝑡 − 𝑙𝑏 𝑠𝑒𝑐
1 𝐻𝑃 = 𝟎. 𝟎𝟓𝟔𝟕𝟑, 𝑯𝑷 𝑓𝑡 − 𝑙𝑏 550 𝑠𝑒𝑐
𝒇𝒕−𝒍𝒃 𝒔𝒆𝒄
ACTIVITY NO. 1: •Calculate the air power of a fan that delivers 1200 𝑚3 /min of air through a 1 m by 1.5 m outlet. Static pressure is 120 mm WG and density of air is 1.18. NOTE: Deadline of submission: May 18, 2020, Monday (write on any blank paper – neat. Scanned or captured output are accepted) Please upload on your class google drive. Thank you!
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