Answer:


Explanation:
= Area of section 1 = 
= Velocity of water at section 1 = 100 ft/min
= Specific volume at section 1 = 
= Density of fluid = 
= Area of section 2 = 
Mass flow rate is given by

The mass flow rate through the pipe is 
As the mass flowing through the pipe is conserved we know that the mass flow rate at section 2 will be the same as section 1

The speed at section 2 is
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Answer:
0.245 m^3/s
Explanation:
Flow rate through pipe a is 0.4 m3/s Parallel pipes have a diameter D = 30 cm => r = 15 cm = 0.15 m Length of Pipe a = 1000m Length of Pipe b = 2650m Temperature = 15 degrees Va = V / A = (0.4m3/s) / (3.14 (0.15m)^2) = 5.66 m/s h = (f(LV^2)) / D2g (fa(LaVa^2)) / Da2g = (fb(LbVb^2)) / Da2g and Da = Db; fa = fb LaVa^2 = LbVb^2 => La/Lb = Vb^2/Va^2 Vd^2 = Va^2(La/Lb) => Vb = Va(La/Lb)^(1/2) Vb = 5.66 (1000/2650)^(1/2) => 5.66 x 0.6143 = 3.4769 m/s Vb = 3.4769 m/s V = AVb = 3.14(0.15)^2 x 3.4769 m/s = 0.245 m^3/s
Answer:
Following the ways of dealing with incomplete questions, i was able to get the complete question, please look at the attachment for ans.