1. What is the volumetric flow rate?

The volumetric flow rate Q is the volume of fluid passing through a cross-section per unit time: Q = A·v.

2. What is the mass flow rate?

The mass flow rate ṁ is the mass of fluid passing per unit time: ṁ = ρ·Q = ρ·A·v.

3. State the continuity equation for incompressible flow.

For incompressible fluids, ρ A v = constant, so A₁v₁ = A₂v₂ along a streamline.

4. How do streamlines represent fluid flow?

Streamlines are curves tangent to the instantaneous velocity vector at every point; no fluid crosses a streamline in steady flow.

5. How does velocity change when area decreases?

If area decreases, velocity increases to conserve Q = A·v (for incompressible flow).

6. Define laminar flow.

Laminar flow is smooth, orderly flow in parallel layers, typically at low Reynolds number (Re ≲ 2000).

7. Define turbulent flow.

Turbulent flow is chaotic, mixing flow with eddies, occurring at high Reynolds number (Re ≳ 4000).

8. What is the Reynolds number?

The Reynolds number Re = ρ·v·D/μ characterizes the ratio of inertial to viscous forces.

9. What does Re < 2000 imply?

It generally indicates laminar flow in a circular pipe.

10. What does Re > 4000 imply?

It generally indicates turbulent flow in a circular pipe.

11. What is the unit of volumetric flow rate?

SI unit for Q is cubic meters per second (m³/s).

12. What is the unit of mass flow rate?

SI unit for ṁ is kilograms per second (kg/s).

13. How do you compute Q for a pipe of known diameter and velocity?

Compute A = πD²/4, then Q = A·v.

14. How do you compute velocity if Q and area are known?

Solve v = Q/A.

15. Why is ρ·A·v constant?

Mass conservation: for steady incompressible flow, mass entering equals mass leaving each section.

16. What distinguishes steady from unsteady flow?

In steady flow, properties at a point do not change with time; in unsteady flow, they do.

17. What is a flow rate of 10 L/min in m³/s?

Convert: 10 L/min = 10×10⁻³ m³ / 60 s ≈ 1.67×10⁻⁴ m³/s.

18. How does fluid density affect mass flow rate?

Higher density increases ṁ = ρ·Q for the same Q.

19. What is the significance of streamtubes?

Streamtubes are bundles of streamlines that enclose constant Q; no fluid crosses their walls in steady flow.

20. How is continuity modified for compressible flow?

Continuity becomes ∂ρ/∂t + ∇·(ρv) = 0, accounting for density variations.

Problem 1: Water (ρ = 1000 kg/m³) flows at 2 m/s in a 0.10 m diameter pipe. Find Q and ṁ.

Solution:
A = π(0.10)²/4 ≈ 7.85×10⁻³ m²;
Q = A·v ≈ 7.85×10⁻³×2 = 1.57×10⁻² m³/s;
ṁ = ρ·Q = 1000×1.57×10⁻² ≈ 15.7 kg/s.

Problem 2: Oil (ρ = 850 kg/m³) flows at 1.5 m/s in a 0.08 m diameter pipe. Compute Q and ṁ.

Solution:
A = π(0.08)²/4 ≈ 5.03×10⁻³ m²;
Q = 5.03×10⁻³×1.5 ≈ 7.54×10⁻³ m³/s;
ṁ = 850×7.54×10⁻³ ≈ 6.41 kg/s.

Problem 3: A pipe tap delivers 10 L/min. If its diameter is 1 cm, find the exit velocity.

Solution:
Q = 10×10⁻³/60 ≈ 1.67×10⁻⁴ m³/s;
A = π(0.01)²/4 ≈ 7.85×10⁻⁵ m²;
v = Q/A ≈ 1.67×10⁻⁴/7.85×10⁻⁵ ≈ 2.13 m/s.

Problem 4: Flow in a pipe of area 0.02 m² decreases to 0.005 m². If initial velocity is 3 m/s, find new velocity.

Solution:
A₁v₁ = A₂v₂ → v₂ = (A₁/A₂)·v₁ = (0.02/0.005)×3 = 4×3 = 12 m/s.

Problem 5: A mass flow rate of 20 kg/s passes through a duct of area 0.1 m². If ρ = 1.2 kg/m³, find velocity.

Solution:
Q = ṁ/ρ = 20/1.2 ≈ 16.67 m³/s;
v = Q/A = 16.67/0.1 = 166.7 m/s.

Problem 6: Compute the velocity if Q = 0.05 m³/s through a circular pipe of diameter 0.2 m.

Solution:
A = π(0.2)²/4 ≈ 0.0314 m²;
v = Q/A = 0.05/0.0314 ≈ 1.59 m/s.

Problem 7: Find the Reynolds number for water (ρ=1000 kg/m³, μ=1.0×10⁻³ Pa·s) flowing at 1 m/s in a 0.05 m pipe.

Solution:
Re = ρvD/μ = 1000×1×0.05/1.0×10⁻³ = 5.0×10³ (turbulent).

Problem 8: Classify the flow in Problem 7 as laminar or turbulent.

Solution:
Re = 5000 > 4000, so the flow is turbulent.

Problem 9: A streamtube narrows from area 0.03 m² to 0.01 m². If Q is constant at 0.06 m³/s, find velocities in both sections.

Solution:
v₁ = Q/A₁ = 0.06/0.03 = 2 m/s;
v₂ = Q/A₂ = 0.06/0.01 = 6 m/s.

Problem 10: If gasoline (ρ=740 kg/m³) flows at 4 m/s in a 0.1 m² conduit, find ṁ.

Solution:
Q = A·v = 0.1×4 = 0.4 m³/s;
ṁ = ρ·Q = 740×0.4 = 296 kg/s.

Problem 11: Water flows through a nozzle of diameter 0.02 m at Q = 0.001 m³/s. Find exit velocity.

Solution:
A = π(0.02)²/4 ≈ 3.14×10⁻⁴ m²;
v = Q/A ≈ 0.001/3.14×10⁻⁴ ≈ 3.18 m/s.

Problem 12: A fluid with ṁ = 50 kg/s flows in a pipe of area 0.2 m²; density ρ = 1200 kg/m³. Find velocity.

Solution:
Q = ṁ/ρ = 50/1200 ≈ 0.0417 m³/s;
v = Q/A = 0.0417/0.2 ≈ 0.208 m/s.

Problem 13: A constriction halves the pipe diameter. By what factor does velocity change?

Solution:
v₂/v₁ = (D₁/D₂)² = (1/0.5)² = 4.

Problem 14: Convert a flow of 500 L/min in a 0.05 m² channel to mass flow rate for water.

Solution:
Q = 500×10⁻³/60 ≈ 8.33×10⁻³ m³/s;
ṁ = 1000×8.33×10⁻³ ≈ 8.33 kg/s.

Problem 15: A ρ = 950 kg/m³ fluid has Q = 0.02 m³/s in D = 0.1 m pipe. Compute Re (μ=1.5×10⁻³ Pa·s).

Solution:
v = Q/A = 0.02/(π×0.1²/4) ≈ 2.55 m/s;
Re = 950×2.55×0.1/1.5×10⁻³ ≈ 1.62×10⁵ (turbulent).

Problem 16: Explain why mass flow rate is constant in a closed conduit for steady incompressible flow.

Solution:
Mass conservation: what enters per unit time must exit, so ṁ is constant along the conduit.

Problem 17: In a T-junction, one branch area is 0.02 m² at 3 m/s, the other 0.03 m². Find velocity in outlet given main inlet 0.05 m².

Solution:
Inlet Q = 0.05×v₁ (unknown). But Q splits: 0.02×3 + 0.03×v₂ = 0.05×v₁. If v₁ known, solve. (More data needed).

Problem 18: Water flows steadily in a pipe that narrows; is Re constant before and after?

Solution:
Re = ρvD/μ changes because v and D change; not constant.

Problem 19: Describe the difference between streamlines and pathlines in unsteady flow.

Solution:
Streamlines are instantaneous tangent curves; pathlines trace actual particle trajectories over time.

Problem 20: A pipe carries two fluids in series: water then oil. How does continuity apply at the interface?

Solution:
Mass flow ṁ is continuous: ρ_wA_wv_w = ρ_oA_ov_o at the interface.