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In a water pipeline, at point P the flow speed is 1 m/s and the pressure is 3 x 10⁵ Pa. At point Q the cross-sectional area is half of that at P and the pressure is 5 x 10⁵ Pa. Taking g = 10 m/s² and water density 1000 kg/m³, find the difference in height between P and Q (in metres).
- 10.5
- 20.15
- 4.5
- zero
Correct answer: 20.15
Solution
Continuity: v_Q = v_P*(A_P/A_Q) = 1*2 = 2 m/s. Bernoulli: P_P + (1/2)rho v_P² + rho g h_P = P_Q + (1/2)rho v_Q² + rho g h_Q. Rearrange for height difference: rho g (h_P - h_Q) = (P_Q - P_P) + (1/2)rho(v_Q² - v_P²). Compute: (5e5-3e5) + 0.5*1000*(4-1) = 2e5 + 1500 = 201500. Then h_P - h_Q = 201500/(1000*10) = 20.15 m. So Q is 20.15 m below P; the height difference magnitude is 20.15 m.
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