The maximum volume flow rate of water is determined as 0.029 m³/s.
<h3>Power of the pump</h3>
The power of the pump is watt is calculated as follows;
1 hp = 745.69 W
7 hp = ?
= 7 x 745.69 W
= 5,219.83 W
<h3>Mass flow rate of water</h3>
η = mgh/P
mgh = ηP
m = ηP/gh
m = (0.82 x 5,219.83)/(9.8 x 15)
m = 29.12 kg/s
<h3>Maximum volume rate</h3>
V = m/ρ
where;
- ρ is density of water = 1000 kg/m³
V = (29.12)/(1000)
V = 0.029 m³/s
Learn more about volume flow rate here: brainly.com/question/21630019
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Answer:
Sorry it doesnt tall me anythikng
Explanation:
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Hook's law holds good up to. A elastic limit. B. plastic limit. C.yield point. D.Breaking point
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The heat of the arc melts the surface of the base metal and the end of the electrode. The electric arc has a temperature that ranges from 3,000 to 20,000 °C
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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
