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## GR0877 Problem 32

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (A) Who says Bernoulli’s principle is unlikely on the PGRE? According to this principle, one has $P_0 + \frac{\rho v_0^2}{2} = P + \frac{\rho v^2}{2}$ (there is no $\rho g z$ term here because of horizontality of the pipe). Conservation of mass gives: $\rho v_0 S = \rho v S$ $\implies$ $v=4v_0$, since $S = \pi r^2 = (\pi r_0^2)/4 = S_0/4$. Plugging $v = 4 v_0$ into the first equation one finally obtain $P = P_0 + \frac{\rho v_0^2}{2} - 16 \frac{\rho v_0^2}{2} = P_0 - \frac{15}{2} \rho v_0^2$.

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