Final answer:
The discharge through the pipe is 0.28276 m^3/s. The velocity of water at section 2-2 is 2.246 m/s.
Step-by-step explanation:
To find the discharge through the pipe, we can use the equation Q = A × V, where Q is the discharge, A is the cross-sectional area, and V is the velocity.
The cross-sectional area can be calculated using the formula A = πr^2, where r is the radius of the pipe. At section 1-1, the diameter is 300mm, so the radius is 150mm or 0.15m.
Therefore, the cross-sectional area at section 1-1 is π(0.15^2) = 0.07069 m^2.
The velocity at section 1-1 is given as 4 m/s.
Plugging in the values, we get Q = 0.07069 m^2 × 4 m/s
= 0.28276 m^3/s.
To find the velocity of water at section 2-2, we can use the equation A1V1 = A2V2, where A1 and V1 are the cross-sectional area and velocity at section 1-1, and A2 and V2 are the cross-sectional area and velocity at section 2-2.
We already know A1 and V1 from the previous calculations. The cross-sectional area at section 2-2 can be calculated using the formula A = πr^2, where r is the radius of the pipe.
At section 2-2, the diameter is 400mm, so the radius is 200mm or 0.2m.
Therefore, the cross-sectional area at section 2-2 is π(0.2^2)
= 0.12567 m^2.
Plugging in the values, we get 0.07069 m^2 × 4 m/s
= 0.12567 m^2 × V2.
Solving for V2, we get:
V2 = (0.07069 m^2 × 4 m/s) / 0.12567 m^2
= 2.246 m/s.