Basic Concepts of Ventilation Design - GHDonline

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1 Bench Grinder Exhaust Ventilation • If there are no losses from the grinder hood entry then: SP 1 + VP 1 = SP 2 + VP 2 but: SP 1 = 0 and VP 1 we then have: 0 = SP 2 + VP 2 or SP 2 = (-VP 2 ) 0 Duct diameter = 3 inches Area = 0.0668 ft 2 2 3 V = 4490 fpm • from V = 4005 (VP) 0.5 • VP 2 = (4490/4005) 2 • VP 2 = 1.26 in w.g. • then: SP 2 = (-VP 2 ) SP 2 = -1.26 in w.g.
1 Bench Grinder Exhaust Ventilation SP h Duct diameter = 3 inches Area = 0.0668 ft 2 2 3 • However there are losses thru the grinder hood entry SP 2 = - (VP 2 + h e ) where h e is the energy loss of the hood entry • Static pressure (SP) must decrease due to acceleration of air up to the duct velocity • F h is defined as the energy loss factor (for that hood design) • Energy losses will be measured as a function of the velocity pressure in the system h e = (F h ) (VP) • Now we define the static pressure at the hood as SP h V = 4490 fpm • SP h is also called the hood static suction and is the absolute value of SP 2
Page 1 and 2: Building Design and Engineering App
Page 3 and 4: General Principles of Ventilation
Page 5 and 6: • Supply • Exhaust Types of Sys
Page 8 and 9: • Temperature • Pressure • Ai
Page 10 and 11: Air Conditioning System Water and R
Page 12 and 13: Conversion Factors Quantity To Conv
Page 14 and 15: Q = V . A Conservation of Mass 8 in
Page 16 and 17: Types of Pressure Measurements •
Page 18 and 19: Conservation of Energy • TP = SP
Page 20 and 21: Velocity Pressure & Velocity • V
Page 24 and 25: 1 Bench Grinder Exhaust Ventilation
Page 26 and 27: Flanged Inlet with F d = 0.49
Page 28 and 29: Hood Entry Coefficients C e = (VP/S
Page 30 and 31: Example Problem • V = 4000 fpm Q
Page 32 and 33: Air Flow Characteristics Exhaust Ho
Page 34 and 35: Questions ?
Page 36 and 37: HVAC Systems
Page 43: Questions ?

Basic Concepts of Ventilation Design - GHDonline

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