Final answer:
a) The discharge through the pipe is 0.2828 m^3/s. b) The velocity of water at section 2-2 is approximately 1.603 m/s.
Step-by-step explanation:
a) To find the discharge through the pipe, we can use the formula:
Discharge = Velocity x Area
Given that the velocity of water at section 1-1 is 4 m/s and the diameter of the pipe at that section is 300 mm, we can calculate the area using the formula:
Area = pi x (diameter/2)^2
Substituting the values, we get Area = pi x (300/2)^2 = 70685.8 mm^2. Now, we convert the area to m^2 by dividing by 10^6, which gives us 0.0707 m^2. Finally, multiplying the velocity by the area, we get the discharge: Discharge = 4 m/s x 0.0707 m^2 = 0.2828 m^3/s.
b) To find the velocity of water at section 2-2, we can use the principle of conservation of mass. According to this principle, the mass flow rate of water remains constant throughout the pipe. The mass flow rate can be calculated using the formula:
Mass flow rate = Density x Velocity x Area
Since the density of water is constant, we can write the equation as:
Velocity1 x Area1 = Velocity2 x Area2
Given that the diameter of the pipe at section 1-1 is 300 mm (Area1) and the diameter of the pipe at section 2-2 is 400 mm (Area2), we can substitute the values and solve for Velocity2:
Velocity2 = (Velocity1 x Area1) / Area2 = (4 m/s x 0.0707 m^2) / (pi x (400/2)^2)
Calculating this expression, we find that the velocity of water at section 2-2 is approximately 1.603 m/s.