A upper tank containing water situated at 20m height from water level...
F = 0.005
Apply Bernoulli’s equation between point 1 & 2
P1 / ρg + V12 / 2g + z1 = P2 / ρg + V22 / 2g + z2+ loss of head due to friction
P1 = P2 = atmospheric pressure and V1 = V2 = 0
So, z1 – z2 = loss of head = 4fLV2 / 2dg
Here, z1 – z2 = 20 = 4×0.005×500×V2/ 2×0.2×9.81
So, V = 2.8 m/s
Pressure at summit, apply Bernoulli equation between 1 & 3
P1 / ρg + V12/2g + z1= P3 / ρg + V32/ 2g + z3 + loss of head due to friction
Take a datum line passing through 1
0 + 0 + 0 = P3 / ρg + 2.82/ 2×9.81 + 4+ 4×0.005×100×2.82 / 2×0.2×9.81
So, P3 /ρg = -8.395 m of water = 1.9 m of water absolute
We know pressure at 3 become less than 2.7 of water absolute, the dissolved air and gasses would come out from water and collect at summit, so flow of water is obstructed.