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The velocity potential for a certain inviscid, incompressible flow field is given by the equation where φ has the units of m2/s when x and y are in meters. Determine the pressure at the point x = 1.7 m, y = 1.7 m if the pressure at x = 1 m, y = 1 m is 2600 kPa. Elevation changes can be neglected, and the fluid is water.

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Answer:

P₂ = 2541.24  kPa

Explanation:

The pressure can be find using the law of gases and fluids of the thermodynamic so:

Δ P = ¹/₂  *  γ / g * ( V₂² - V₁² )

V² = u² + v²

Solving using the equation

φ = 2 x² * y  - ²/₃ * y ³  

At the point (1.1)

u = 4 * x *y  ⇒  u₁ = 4 * 1 * 1 = 4 m/s

v₁ = 2x² - 2y²   ⇒  v₁ = 2 * 1² - 2 * 1² = 0

V₁² = 4² + 0² = 16 m² / s²

At the point (1,7 . 1,7)

u = 4 * x *y  ⇒ u₁ = 4 * 1.7 * 1.7 = 11.56 m/s

v₂ = 2x² - 2y²  ⇒ v₂ = 2 * 1.7² - 2 * 1.7² = 0

V₂² = 11.56² + 0² = 133.6336 m² / s²

Replacing

Δ P = ¹/₂  *  γ / g * ( V₂² - V₁² ) ⇒ P₁ - P₂ = ¹/₂  *  γ / g * ( V₂² - V₁² )

P₂ = P₁ + 0.5 *  γ / g * ( V₂² - V₁² ) ⇒ P₂ = 2600 kPa + 0.5 *  9.8 x 10³ / 9.8 m/s₂ * ( 16 - 133.6336 )

P₂ = 2541.24 x 10³ Pa

P₂ = 2541.24  kPa

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