A Formula for Minor Losses in Air Duct Elements

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Yuri B. Power-User Join Date: Jun 2009 Posts: 262 Good Answers: 1	A Formula for Minor Losses in Air Duct Elements 03/08/2012 2:35 PM
	"Engineering toolbox" says that: Minor Loss - Head or Pressure Loss in Air Duct Components - can be expressed as h_{minor_loss} = ξ v²/ 2 g (1) where h_{minor_loss} = minor head loss (m, ft) ξ = minor loss coefficient v = flow velocity (m/s, ft/s) g = acceleration of gravity (m/s², ft/s²) But how this formula can be used for the compressible fluid - air - not taking into account its density ?!
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Codemaster Guru Join Date: Jul 2005 Location: Stoke-on-Trent, UK Posts: 4496 Good Answers: 137	#1 Re: A Formula for Minor Losses in Air Duct Elements 03/08/2012 4:17 PM
	For air in ducts it sounds like a H & V type application, and pressure and temperature do not differ from standard P & T enough to have a significant effect. In h_{minor_loss} = ξ v²/ 2 g (1), what does the (1) mean? Is it just an equation number from a book? This formula is OK for liquid flow, but not much use for gases. It gives loss in m or ft of gas, which isn't very helpful! That's for regular values of ξ, eg 0.3 for a 90° bend. If you have values of ξ with conversion factor built in, could be OK, but you don't give figures. For gases, better to use h_{minor_loss} = ξ ρ v²/ 2 where ρ = density kg/m³ etc. If P & T are much different from standard, eg compressed air system, formula works with ρ and v referred to actual conditions. You usually have these at standard conditions, ρ from a table or calculated from molecular weight, and v from standard flowrate. In that case, you need to apply the gas laws and some math, and remember ρ v is constant, to convert to a formula with ρ and v at standard conditions, and P & T in the formula. __________________ Give masochists a fair crack of the whip
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Mikerho Guru Join Date: Jan 2007 Location: Moses Lake, WA, USA, Thulcandra - The Silent Planet (C.S. Lewis) Posts: 4216 Good Answers: 194	#2 In reply to #1 Re: A Formula for Minor Losses in Air Duct Elements 03/08/2012 11:59 PM
	Very good information with practical application guidance. Along with that, you spelled the plural of gas correctly! For that, you get a GA from me! __________________ "Reason is not automatic. Those who deny it cannot be conquered by it. Do not count on them. Leave them alone." - Ayn Rand
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Codemaster Guru Join Date: Jul 2005 Location: Stoke-on-Trent, UK Posts: 4496 Good Answers: 137	#5 In reply to #2 Re: A Formula for Minor Losses in Air Duct Elements 03/09/2012 4:32 PM
	Thanks for the GA. Did you notice I also spelled math to suit a mainly US audience? __________________ Give masochists a fair crack of the whip
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Mikerho Guru Join Date: Jan 2007 Location: Moses Lake, WA, USA, Thulcandra - The Silent Planet (C.S. Lewis) Posts: 4216 Good Answers: 194	#7 In reply to #5 Re: A Formula for Minor Losses in Air Duct Elements 03/10/2012 8:11 PM
	Now I have. That has always been a puzzle to me. US people think of mathematics as a single-entity field of endeavour (UK sp). The UK use of the word maths would seem to indicate a plethora of topics in that same field. Either way, it means the same thing! __________________ "Reason is not automatic. Those who deny it cannot be conquered by it. Do not count on them. Leave them alone." - Ayn Rand
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Yuri B. Power-User Join Date: Jun 2009 Posts: 262 Good Answers: 1	#3 In reply to #1 Re: A Formula for Minor Losses in Air Duct Elements 03/09/2012 12:12 PM
	(You baffled me at first with your "ρ v is constant" thinking ρ meant density - later got it was pressure) I knew as more appropriate the formula you have given and am not very happy with the fact engineertoolbox may put out in internet inappropriate calculations. Thanks for your answer !
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Codemaster Guru Join Date: Jul 2005 Location: Stoke-on-Trent, UK Posts: 4496 Good Answers: 137	#4 In reply to #3 Re: A Formula for Minor Losses in Air Duct Elements 03/09/2012 4:28 PM
	Sorry Yuri B, that's not right. ρ v is constant where ρ means density. This is exact, and follows from the fact for a given mass, if the volume decreases, the density increases. Ideal gas law says P v = RT so is not constant, and real gases don't follow it exactly, though for e.g. air in many applications it's close enough in practice. To elaborate on post #1 h_{minor_loss} = ξ ρ_a v_a²/ 2 where ρ_a= density kg/m³, v_a= velocity m/s at actual conditions. ρ_s , v_s at standard conditions. P_a P_s and T_a, T_s are pressure and temperature at actual and standard conditions, both absolute. ξ ρ_a v_a²/ 2 = ξ ρ_s v_sv_a/ 2 (as ρ v is constant) = ξ ρ_s v_s/2(v_s*P_s/P_aT_a/T_s) = ξ ρ_s v_s²(P_s/P_aT_a/T_s) __________________ Give masochists a fair crack of the whip
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Yuri B. Power-User Join Date: Jun 2009 Posts: 262 Good Answers: 1	#6 In reply to #4 Re: A Formula for Minor Losses in Air Duct Elements 03/10/2012 1:53 AM
	Thank you Codemaster !
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Codemaster (3); Mikerho (2); Yuri B. (2)

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