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ECET 2026 MECH

Day 6 Flow Through Pipes – ECET 2026 Mechanical

Why this topic is important for ECET?

In Hydraulics, the concept of flow through pipes is crucial because it forms the base for solving real-world fluid problems like water supply systems, pumping stations, turbines, and pipelines.
In ECET 2026, this topic helps test your conceptual clarity (types of flows, losses, laws) and numerical solving skills (Darcy–Weisbach, continuity, energy equations). Many direct and application-based questions are expected.


📘 Concept Notes

🔹 Types of Flow in Pipes

  1. Laminar Flow – Fluid particles move in smooth layers (low velocity).
    • Reynolds number Re < 2000.
    • Example: Oil flowing slowly in thin tubes.
  2. Turbulent Flow – Fluid moves irregularly, with eddies (high velocity).
    • Reynolds number Re > 4000.
    • Example: Water in domestic supply pipes.
  3. Transition Flow – Intermediate, between 2000–4000.

🔹 Continuity Equation

For incompressible fluid,
 Q = A V = constant
where Q = discharge, A = area, V = velocity.


🔹 Head Loss in Pipes

Fluid flow in pipes causes loss of energy (head) due to friction and fittings.

  1. Darcy–Weisbach Equation

 h_f = \frac{f L V^2}{2 g D}

Minor Losses

  • Sudden enlargement:  h_e = \frac{(V_1 - V_2)^2}{2 g}
  • Sudden contraction:  h_c = 0.5 \frac{V^2}{2 g}
  • Bend/valve:  h_b = K \frac{V^2}{2 g}

🔹 Power Transmission through Pipe

 P = \rho g Q H_{useful}

Maximum power transmitted when

 h_f = \frac{H}{3}


🔹 Pipes in Series and Parallel

  • Series: Same discharge Q, head loss is additive.
  • Parallel: Same head loss, discharge divides among pipes.

🔹 Example (Simple)

Water flows through a 200 m long, 0.2 m diameter pipe at 3 m/s. Find head loss (take f=0.02).

 h_f = \frac{f L V^2}{2 g D} = \frac{0.02 \times 200 \times 9}{2 \times 9.81 \times 0.2} \approx 9.18 , m


⚙️ Formulas

  1. Reynolds number:

 Re = \frac{\rho V D}{\mu}

Continuity:

 Q = A V

Head loss (Darcy–Weisbach):

 h_f = \frac{f L V^2}{2 g D}

Sudden enlargement:

 h_e = \frac{(V_1 - V_2)^2}{2 g}

Sudden contraction:

 h_c = 0.5 \frac{V^2}{2 g}

Bend loss:

 h_b = K \frac{V^2}{2 g}

Power transmitted:

 P = \rho g Q (H - h_f)

Max power condition:

 h_f = \frac{H}{3}


🔟 10 MCQs

Q1. Reynolds number less than 2000 indicates:
a) Turbulent flow
b) Laminar flow
c) Transition flow
d) None

Q2. For turbulent flow in a pipe, Reynolds number is:
a) <2000
b) 2000–4000
c) >4000
d) 0

Q3. Continuity equation is valid for:
a) Compressible flow only
b) Incompressible flow only
c) Both compressible and incompressible
d) Unsteady flow only

Q4. Head loss due to friction in a pipe varies as:
a) V
b) V^2
c) 1/V
d) 1/V^2

Q5. Water flows at 3 m/s through a 0.1 m diameter pipe. Discharge is:
a) 0.003 m³/s
b) 0.0236 m³/s
c) 0.00785 m³/s
d) 0.01 m³/s

Q6. Head loss in sudden enlargement is proportional to:
a) latex^2[/latex]
b) latex^2[/latex]
c) V^2
d) 1/V

Q7. In parallel pipes, which is constant?
a) Discharge
b) Velocity
c) Head loss
d) Reynolds number

Q8. Maximum power transmitted through a pipe when head loss due to friction equals:
a) H
b) H/2
c) H/3
d) 2H/3

Q9. In Darcy–Weisbach equation, friction factor f depends on:
a) Flow type & pipe roughness
b) Pipe length
c) Velocity only
d) Pipe material only

Q10. A 0.2 m dia pipe carries water at 2 m/s. What is discharge?
a) 0.0314 m³/s
b) 0.0628 m³/s
c) 0.125 m³/s
d) 0.25 m³/s


✅ Answer Key

QAns
1b
2c
3c
4b
5c
6b
7c
8c
9a
10a

🧠 Explanations

  • Q1: Re < 2000 = laminar → (b).
  • Q2: Re > 4000 = turbulent → (c).
  • Q3: Continuity applies to all fluids (steady incompressible ideal case) → (c).
  • Q4: h_f \propto V^2 → (b).
  • Q5: Q=AV=\frac{\pi D^2}{4}V=\frac{\pi (0.1)^2}{4}\times3=0.00785 → (c).
  • Q6: Sudden enlargement loss = latex^2/2g[/latex] → (b).
  • Q7: In parallel, head loss same across each pipe → (c).
  • Q8: Max power when h_f=H/3 → (c).
  • Q9: f depends on flow (laminar/turbulent) & pipe roughness → (a).
  • Q10: Q=\pi D^2 V/4=\pi(0.2)^2/4 \times 2=0.0314 → (a).

🎯 Motivation / Why Practice Matters

Pipe flow questions check formula application + conceptual clarity.
In ECET 2026, time is short, so you need speed (quickly apply equations), accuracy (avoid silly unit mistakes), and edge (attempt extra numericals correctly).
Daily practice ensures you solve pipe flow questions in less than 2 minutes each – which gives a big lead over competitors.


📲 CTA

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