An ideal fluid of density $800\ \text{kg m}^{-3}$ flows smoothly through a bent pipe (as shown in figure) that tapers in cross-sectional area from $a$ to $\dfrac a2$. The pressure difference between the wide and narrow sections of the pipe is $4100\ \text{Pa}$. At the wider section, the velocity of the fluid is $\dfrac{\sqrt x}{6}\ \text{m s}^{-1}$ for $x$ = ______. (Given $g = 10\ \text{m s}^{-2}$)
Numerical value type. Enter your answer.
Answer: 363
Continuity: $v_2 = 2v_1$. The wide end is $1\ \text{m}$ higher. Bernoulli:
$$P_1 + \frac12\rho v_1^2 + \rho g(1) = P_2 + \frac12\rho(2v_1)^2$$
$$P_1 - P_2 = \frac32\rho v_1^2 - \rho g = 1200v_1^2 - 8000$$
With $P_1 - P_2 = 4100\ \text{Pa}$: $v_1^2 = \dfrac{12100}{1200}$, so $v_1 = \dfrac{110}{\sqrt{1200}} = \dfrac{110}{20\sqrt3} = \dfrac{\sqrt{363}}{6}\ \text{m s}^{-1}$.
So $x = 363$.
Solution by Sreeraj P, M.Sc Physics