Answer:
The maximum discharge rate of water through the pipe is 0.00545 m³/s or 5.45 L/s.
Friction head and pressure head will cause the actual flow rate to be less.
Explanation:
Considering point 1 at the free surface of the pool, and point 2 at the exit of
pipe.
Using Bernoulli equation between
these two points simplifies to
P1/(p*g) + V1²/2g + z1 = P2/(p*g) + V2²/2g + z2
Let the reference level at the pipe exit (z2 = 0). Noting that the fluid at both points is open to the atmosphere (and thus P1 = P2 = Patm) and that the fluid velocity at the free surface is very low (V1 ≅ 0),
P/(p*g) + z1 = P/(p*g) + V2²/2g
z1 = V2²/2g
Note; z1 = h
V2max = √2gh
h = 3 m
V2max = √2 * 9.81 * 3
V2max = √58.86 = 7.67 m/s
maximum discharge rate of water through the pipe Qmax = Area A * Velocity of discharge V2max
Qmax = A * V2max
Diameter d = 3 cm = 0.03 m
A = Πd²/4 = (Π * 0.03²)/4 = 0.00071m³
Qmax = 0.00071 * 7.67 = 0.00545 m³/s
Qmax = 5.45 L/s
The maximum discharge rate of water through the pipe is 0.00545 m³/s or 5.45 L/s.
Actual flow rate will be less because of heads such as friction head and pressure head.
