An ideal fluid of density 800kgm–3, flows smoothly through a bent pipe (as shown in figure) that tapers in cross-sectional area from a to \(\frac{a}{2}\). The pressure difference between the wide and narrow sections of pipe is 4100 Pa. At wider section, the velocity of fluid is\(\frac{\sqrt{x}}{6}\) ms1 for x=_______. (Given g=10 ms-2)
Correct Answer: 363
Solution and Explanation
The problem involves Bernoulli's equation, which states: P1 + (1/2)ρv12 + ρgh = P2 + (1/2)ρv22. Given data: ρ = 800 kg/m3, P1 - P2 = 4100 Pa, h = 1 m, g = 10 m/s2, A1 = a, A2 = a/2.
By the equation of continuity: A1v1 = A2v2, so v2 = 2v1.
Substitute in Bernoulli's equation: 4100 = (1/2)(800)(4v12 - v12) - 800(10)(1).
This simplifies to: 4100 + 8000 = 1200v12.
Simplifying further: 12100 = 1200v12.
Thus, v12 = 10.08, so v1 = √10.08.
Given v1 = √x/6, equate and find x: √x/6 = √10.08.
Therefore, x = 10.08 × 36 = 363.
Verify: 363 is within the given range (363,363).
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