Apump and its horizontal intake pipe are located 82 m beneath the surface of a large reservoir. the speed of the water in the intake pipe causes the pressure there to decrease, in accord with bernoulli's principle. assuming nonviscous flow, what is the maximum speed with which water can flow through the intake pipe? (assume atmospheric pressure is 1.01 105 pa.)

Ans : Bernoulli's principle states for incompressible non-viscous flow that p/Ď + gâ™h + (1/2)â™v² = constant Evaluate the equation along a stream line from liquid surface of the reservoir (1) to the inlet of the pipe pâ‚/Ď + gâ™hâ‚ + (1/2)â™v₲ = pâ‚‚/Ď + gâ™hâ‚‚ + (1/2)â™v₂² => vâ‚‚ = âš[ 2â™(pâ‚-pâ‚‚)/Ď + 2â™gâ™(hâ‚-hâ‚‚) + v₲ ] lets make some assumptions: - the pressure at the liquid surface is equal to the atmospheric pressure pâ‚ = 1atm = 101325Pa - the velocity of the liquid at the surface (that is the speed at which the liquid level in reservoir decreases) is quite small, so it may be ignored: v₠≠0 So vâ‚‚ = âš[ 2â™(pâ‚-pâ‚‚)/Ď + 2â™gâ™(hâ‚-hâ‚‚) ] The height difference is fixed. So the only variable remaining is the pressure in the pipe. As higher it is as lower the velocity in the pipe is. So you get the maximum velocity for the minimum pressure. Since pressure cannot drop below zero this is pâ‚‚ = 0 Therefore vâ‚‚max = âš[ pâ‚/Ď + gâ™(hâ‚-hâ‚‚) ] = âš[ 2â™101325Pa/1000kg/mÂł + 2â™9.81m/s²â™12m ] = 20.93m/s

Look it up you feel me

Ithink the answer would be a