Answer to Question #160471 in Physics for kyle

Question

At one point in a pipeline, the water’s speed is 3.50 [m/s] and the gauge pressure is 5.00 × 10^4[Pa]. Find the gauge pressure at a second point in the line, 13.0 [m] lower than the first, if the pipe diameter at the second point is twice that at the first.

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Answer 1

$$A_1v_1=A_2v_2\to d_1^2v_1=d_2^2v_2$$

$$p_1+0.5\rho v_1^2+\rho gh=p_2+0.5\rho v_2^2\\ p_1+0.5\rho v_1^2+\rho gh=p_2+0.5\rho v_1^2\frac{d_1^4}{d_2^4}\\\\ 5\cdot10^4+0.5(1000) 3.5^2+(1000)(9.8)(13)\\=p_2+0.5(1000)3.5^2\frac{1}{2^4}\\p_2=1.83\cdot10^5\ Pa$$