The rate of flow of water through a horizontal pipe is 0·25 m³/s. The diameter of the pipe which is 200 mm is suddenly enlarged to 400 mm. The pressure intensity in the smaller pipe is 11·772 N/cm². The approximate power lost due to enlargement is : (Take g = 10 m/s²)
- A4·50 kWCorrect
- B9·00 kW
- C2·25 kW
- D1·80 kW
Solution
Concept
Head loss in a sudden enlargement
Formula
hL = (V₁ − V₂)² / 2g; Power = ρ g Q hL
Step-by-step solution
1. V₁ = 0.25 / (π × 0.2² / 4) = 7.96 m/s; V₂ = 0.25 / (π × 0.4² / 4) = 1.99 m/s. 2. hL = (7.96 − 1.99)² / (2 × 10) = 1.78 m. 3. Power lost = 1000 × 10 × 0.25 × 1.78 = 4 452 W ≈ 4.5 kW.
Explanation
Energy is lost in the turbulence at a sudden enlargement. The loss of head is (V₁ − V₂)²/2g, and multiplying by the weight flow rate gives the power. The pressure intensity is not needed.
Why the other options are wrong
9.00 kW is twice, 2.25 kW half and 1.80 kW is about 40% of the loss calculated.
Shortcut
hL = (V₁ − V₂)² / 2g.
Common mistake
Using (V₁² − V₂²)/2g.
Correct answer: A) 4·50 kW