A uniform tapering pipe has 100mm and 80mm diameter at its end If the velocity of water at larger end is 2m/sec, find the discharge at larger end and velocity head at smaller end

Given data :
D1 = 100 mm = 0.1 m D2 = 80 mm = 0.8 m V1 = 2 m / sec
Area of section 1, A1 = π / 4 ( D1 )² = π / 4 ( 0.1 ) ² = 0.00785 m²
Area of section 2, A2 = π / 4 ( D2 )² = π / 4 ( 0.08 ) ² = 0.005 m²
Now, velocity head at larger end = V1² / 2g = ( 2 ) ² / 2 × 9.81 = 0.203 m
velocity head at smaller end = V2² / 2g
From continuity equation ;
A1 × V1 = A2 × V2
V2 = ( A1 × V1 ) / A2
V2 = ( 0.00785 × 2 ) / 0.005 = 3.14 m / sec
therefore,
Velocity head at smaller end = V2² / 2g = ( 3.14 ) ² / 2 × 9.81 = 0.50 m
Also, Discharge at larger end = A1 × V1 = 0.00785 × 2 = 0.0157 m³/sec = 15.7 litres / sec