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Unformatted text preview: V 1 2 / 2 + p 1 / ρ + gz 1 = α V 2 2 /2 + p 2 / ρ + gz 2 + h Δ Total (2) 1 (0) z o = [V 2 2 /(2g)] + [V 1 2 /(2g)][f(L/D 1 ) + K ent + K diff ] 2m = (0.36V 1 ) 2 /(2g) + [V 1 2 /(2g)] (K diff + f(2/.03) + 0.5) V 2 = V 1 (3/5) 2 (1) (2) (0) 2m = (0.36V 1 ) 2 /(2g) + [V 1 2 /(2g)] (K diff + f(2/.03) + 0.5) f = (1.8 log [( ρ V 1 D/µ)/6.9])2 With proper choice of diffuser Q with diffuser can result in 33% increase Moody Plot – guess Re D and get f, calculate V 1 , check Re D – if not close use that Re D and repeat…. V 1nodiffuser = [2gz o /(1 + K entrance )] 1/2 V 1diffuser = [2gz o /{( A 1 /A 2 ) 2 + K entrance + K diffuser }] 1/2 No pipe at exit so f = 0 0.25 0.3 .55 < 1 so V 1diffuser > V 1nodiffuser Does diffuser reduce energy “losses”? Does diffuser increase V 1 ? Is V 2 faster than V 1diffuser ? Is V 2 faster than V 1nodiffuser ?...
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 Summer '09
 Rohr
 Leat, p2/ + gz2

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