2- Oil with S.G.= = 0.82 flows at 5 ft³/s through the cylindri...

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2- Oil with S.G.= = 0.82 flows at 5 ft³/s through the cylindrical reducing section as shown Figure 2. Calculate the force that needs to be applied to the reducer to hold it in place. 1 2 P = 60 psig P = 6 psig D = 4 in D =14 in Figure 2. Reducing section Useful Formulas: S.G. = = p = y y = pg Pwater Ywater Q = V. A V1A1 = V2A2 P1 + P2 + + 2g y [ F = pQAV, Exp.: [ Fx = ppl(V2x-Vix) TABLE F.1 FORMULAS FOR UNIT CONVERSIONS* Name, Symbol, Dimensions Conversion Formula Length L L 1 m = 3.281 ft = 1.094 yd = 39.37 in = km/1000= 106 m 1 ft = 0.3048 m = 12 in = mile/5280 = km/3281 1 mm - m / 1000 - in /25.4 - 39.37 mil - 1000 um - 107A Speed V L/T I m/s = 3.600 km/hr = 3.281 f/s = 2.237 mph = 1.944 knots I ft/s = 0.3048 m/s = 0.6818 mph = 1.097 km/hr = 0.5925 knots Mass m M 1 kg = 2.205 lbm = 1000 g = slug/14.59 = (metric ton or tonne or Mg). / 1000 1 lbm = lbf- 2/(32.17ft) = kg/2.205 = slug/32.17 = 453.6 g 1- A Venturi meter with an inlet diameter of 0.7 m is designed to handle 8 m³/s of air (Pair=1.2 kg/m³). What is the required throat diameter if this flow results the pressure difference (P1-P2) - of 884 Pa. Hint: B.E. for Venturi meter reduces to: p 2 = P2, V2 P1, V1 2 h Figure 1. Venturi meter Useful Formulas: p y S. G. = = y = pg Pwater Ywater Q = v.A V1A1=V2A2 = P1 y + + Z2 + 2g 2 [ F = pQAV, Exp.: E Fx = ppl(2xx-V1x) - TABLE F.1 FORMULAS FOR UNIT CONVERSIONS* Name, Symbol, Dimensions Conversion Formula Length L L 1 m = 3.281 ft = 1.094 yd = 39.37 in = km /1000 = 106 um 1 ft = 0.3048 m = 12 in = mile/5280 = km/3281 1 mm - m / 1000 - in/25.4 - 39.37 mil - 1000 um - 107A

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