Tank Flow Analysis: Steady-State Condition
Problem Statement
A water tank with constant level h receives flow through:
- Section 1: Velocity v₁ = 5 m/s (40mm diameter)
- Section 3: Discharge Q₃ = 0.012 m³/s
- Exit velocity v₂ at section 2 (60mm diameter)
Given Data
| d₁ = 40mm = 0.04m | A₁ = π/4 × (0.04)² = 0.001257 m² |
| d₂ = 60mm = 0.06m | A₂ = π/4 × (0.06)² = 0.002827 m² |
| V₁ = 5 m/s | Q₃ = 0.012 m³/s |
1. Inflow Through Section 1 (Q₁)
Q₁ = A₁ × V₁ = 0.001257 × 5 = 0.006285 m³/s
2. Total Outflow Calculation (Q₂)
Q₂ = Q₁ + Q₃ = 0.006285 + 0.012 = 0.018285 m³/s
3. Exit Velocity at Section 2 (V₂)
V₂ = Q₂/A₂ = 0.018285/0.002827 = 6.5 m/s
Physical Significance
Key observations from the solution:
- Constant water level indicates steady-state flow (ΣQin = ΣQout)
- Larger diameter at exit (60mm vs 40mm inlet) results in lower velocity despite higher flow rate
- Exit velocity (6.5 m/s) is 30% higher than inlet velocity (5 m/s) due to combined flows
- Demonstrates mass conservation in tank systems
- Highlights relationship between pipe size and velocity
- Important for designing overflow protection systems
