Water is flowing with a velocity of 1·5 m/s in a pipe of length 2500 m and of diameter 500 mm. At the end of the pipe, a valve is provided. What is the rise in pressure, if the valve is closed in 25 seconds ? (Take the value of C as 1460 m/s)
- A15 N/cm²Correct
- B150 N/cm²
- C7·5 N/cm²
- D75 N/cm²
Solution
Concept
Gradual closure of a valve
Formula
Critical time Tc = 2L / C; for t > Tc: Δp = ρ L V / t
Step-by-step solution
1. Tc = 2L/C = 5000 / 1460 = 3.4 s. 2. Closure time 25 s > 3.4 s, so the closure is gradual. 3. Δp = ρ L V / t = 1000 × 2500 × 1.5 / 25 = 1.5 × 10⁵ N/m² = 15 N/cm².
Explanation
If the valve closes slower than the time 2L/C taken by the pressure wave to travel to the reservoir and back, the pressure rise is computed from the momentum (ρLV/t) relation, which is much smaller than the instantaneous value ρCV.
Why the other options are wrong
150 N/cm² is 10 times larger. 7.5 and 75 N/cm² correspond to closure times of 50 s and 5 s, not 25 s.
Shortcut
Slow closure: Δp = ρ L V / t. Rapid closure: Δp = ρ C V.
Common mistake
Using ρCV (instantaneous closure).
Correct answer: A) 15 N/cm²