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A non-viscous liquid of constant density 1000 kg/m³ flows steadily (streamline flow) through a tube of variable cross section, inclined in a vertical plane. At point P (height 2 m) the cross-sectional area is 4*10⁻³ m² and the liquid speed is 1 m/s. At point Q (height 5 m) the area is 8*10⁻³ m². Find the work done per unit volume on the fluid by (a) the pressure forces and (b) the gravity forces, as the fluid flows from P to Q. (Take g = 9.8 m/s².)

1. Pressure work = +29025 J/m³; gravity work = -29400 J/m³
2. Pressure work = -29025 J/m³; gravity work = +29400 J/m³
3. Pressure work = +29400 J/m³; gravity work = -29025 J/m³
4. Pressure work = -29400 J/m³; gravity work = +29400 J/m³

Correct answer: Pressure work = +29025 J/m³; gravity work = -29400 J/m³

Solution

Continuity: AP*vP = AQ*vQ => vQ = (4*10⁻³*1)/(8*10⁻³) = 0.5 m/s. Gravity work per unit volume = -rho*g*(hQ - hP) = -1000*9.8*3 = -29400 J/m³. The change in kinetic energy per unit volume is dKE = (1/2)rho(vQ² - vP²) = (1/2)(1000)(0.25 - 1) = -375 J/m³. By the work-energy theorem W_pressure + W_gravity = dKE, so W_pressure = dKE - W_gravity = -375 + 29400 = 29025 J/m³.

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