To solve this problem, we can apply the principles of fluid mechanics and Bernoulli's equation.
Given:
- Diameter of the first pipe (D1): 120 mm = 0.12 m
- Diameter of the second pipe (D2): 80 mm = 0.08 m
- Head at the first pipe (H1): 3 m
- Altitude change (Δh): 12 m
- Head at the second pipe (H2): 13 m
- Density of water (ρ): 1000 kg/m³
Step 1: Calculate the velocities in the pipes using Bernoulli's equation.
Applying Bernoulli's equation between the two points in each pipe:
For the first pipe:
P1/ρ + V1²/2g + H1 = constant (1)
For the second pipe:
P2/ρ + V2²/2g + H2 = constant (2)
Since the pipes are open to the atmosphere, we can assume P1 = P2 = atmospheric pressure (approximately).
Simplifying equation (1):
V1²/2g + H1 = constant (3)
Simplifying equation (2):
V2²/2g + H2 = constant (4)
Step 2: Solve for the velocities V1 and V2.
Using equation (3) for the first pipe:
V1²/2g + 3 = constant (5)
Using equation (4) for the second pipe:
V2²/2g + 13 = constant (6)
Step 3: Solve for the velocities V1 and V2.
Since the constants in equations (5) and (6) are the same (as it is a continuous flow), we can equate the two equations:
V1²/2g + 3 = V2²/2g + 13
V1²/2g - V2²/2g = 10
(V1² - V2²)/(2g) = 10
V1² - V2² = 20g
V1² = V2² + 20g (equation 7)
Step 4: Convert the diameter to radius for each pipe.
r1 = D1/2 = 0.12/2 = 0.06 m
r2 = D2/2 = 0.08/2 = 0.04 m
Step 5: Calculate the rate of discharge (Q) using the continuity equation.
The continuity equation states that the product of the cross-sectional area (A) and the velocity (V) is constant in a flowing fluid.
Q1 = Q2 (since it is a continuous flow)
A1V1 = A2V2
πr1²V1 = πr2²V2
(r1²V1)/(r2²) = V2
Step 6: Calculate the velocity in the smaller pipe (V2).
Substitute the values in equation (7):
V1² = V2² + 20g
(V1²r2²)/(r1²) = V2²
(0.06²V2²)/(0.04²) = V2²
V2² = (0.06²V1²)/(0.04²) [Substitute V1² = 2g(3) from equation (5)]
V2² = (0.06² × 2g × 3)/(0.04²)
V2² = 0.27g
V2 = √(0.27g)
Step 7: Calculate the rate of discharge (Q) in L/s.