An ideal fluid undergoes streamline flow through a horizontal pipe. The pipe's cross-sectional area

Question

An ideal fluid undergoes streamline flow through a horizontal pipe. The pipe's cross-sectional area decreases from $A_1 = 40 \mathrm{~cm}^2$ to $A_2 = 20 \mathrm{~cm}^2$. If the pressure and velocity at $A_1$ are $P_1$ and $v_1$, and at $A_2$ are $P_2$ and $v_2$, which of the following relations is correct?

Professor PC Thomas Classes · Answer

(a) : From the equation of continuity, $A_1 v_1 = A_2 v_2 \implies 40 v_1 = 20 v_2 \implies v_2 = 2v_1$. The fluid speeds up as it enters the narrower section. According to Bernoulli's theorem for a horizontal pipe ($P + \frac{1}{2}ho v^2 = 	ext{constant}$), an increase in kinetic energy results in a decrease in pressure. Since $v_2 > v_1$, it must follow that $P_2 < P_1$.