Informe 4 Fluidos - Tuvo Pitot Datos 1=0,01 2=0,02 =4=0,1016m Solucin A= d 22=(0,1016 m)2=8,1073 x 103 m2 4 4 = 2P =1000 kg 3 m f = Hg=133280 N m3 Por

Informe 4 Fluidos - Tuvo Pitot Datos 1=0,01 2=0,02...

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Tuvo Pitot Datos: ? 1=0,01 ? ? 2=0,02 ? ? =4 ???? =0,1016m Solución: A = π 4 d 2 2 = π 4 ( 0,1016 m ) 2 = 8,1073 x 10 3 m 2 ? = 2 P ρ ρ = 1000 kg m 3 γ f = γ Hg = 133280 N m 3 Por Hidrostática ? =∆ ? · γ f ? 1=0,01 ?? 133280 N m 3 = 1332,8 ?? ? 1= 2 x 1332,8 Pa 1000 kg m 3 = 1,63 27 m s ? 1= (1,6327 m s ¿ ? ( 8,1073 x 10 3 m 2 ¿ =0,013 2 m 3 s ? 2=0,02 ?? 133280 /m ? 3 =2665,6 ?? ? 2= 2 x 2665.6 Pa 1000 kg m 3 =2,3089 ? /s
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? 2= (2,3089 m s ¿ ? ( 8,1073 x 10 3 m 2 ) = 0,0187 m 3 s = 0,0132 m 3 s + 0,0187 m 3 s 2 =0,016m 3 /s Canal de Venturi Datos: ? 1=0,222 ? ? 2 =0,0504 ? ? 2 =0,102 ? ? = ? 1 2 P ρ· ( A 1 2 A 2 2 ) γ f = γ Hg = 133280 N m 3 ρ = 1000 kg m 3 Solución: A 1 = π 4 D 2 2 = π 4 ( 0,102 m ) 2 = 8,1712 x 10 3 m 2 A 2 = π 4 d 2 2 = π 4 ( 0,0504 m ) 2 = 1 , 995 x 10 3 m 2 Por Hidrostática ? =∆ ? · γ f ? =0,222 ?? 133280 /m ? 3 = 29588,16 ??
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? = ( 8,1712 x 10 3 m 2 ) 2 29588.16 Pa ( 1000 kg m 3 )( ( 8,1712 x 10 3 m 2 ) 2 ( 1,995 x 10 3 m 2 ) 2 ) = 7.93 m/s Q=7,93 /s π 4 (0,0504 ) 2 =0,016 3 /s Placa Orificio Datos: ? =0,234 ? ? 0 =0,0614 ? ? 2 =0,102 ? ? c =0,84 ? = 2 g ∆h ∗( γ m γ 1 ) 1 ( Cc Ao A 1 ) 2 V= 2 x 9,81 m s 2 x 0,234 mx ( 133280 N m 3 9800 N m 3 1 ) 1 ( 0,84 x π 4 x ( 0,0614 m ) 2 π 4 x ( 0,102 m ) 2 ) 2 = 7,98 m s Q = 7,98 m s x π 4 x ( 0,0614 m ) 2 = 0,024 m 3 s Comparación de los caudales obtenidos anteriormente con el tanque
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