Question

The figure shows a horizontal pipe with a circular cross section whose diameter varies. The cross-sectional area at X is 3.0×10−4m2 and at Y is 0.60×10−4m2. Water of density 1000kg/m3 fills the pipe and flows through it at a constant rate of 2.4×10−4m3/s. The difference in pressure between X and Y is most nearly 750 Pa 750 Pa A 1600 Pa 1600 Pa B 7700 Pa 7700 Pa C 8320 Pa 8320 Pa D

Answer 1

By Bernoulli's Principle :

$P_x+\dfrac{\rho v_x^2}{2}+\rho gz_x=P_y+\dfrac{\rho v_y^2}{2}+\rho gz_y$

Since , pipe is horizontal so every point is at same height .

So , $z_x=z_y$ .

The equation will reduced to :

$P_x+\dfrac{\rho v_x^2}{2}=P_y+\dfrac{\rho v_y^2}{2}$    ..... 1 )

Also flow rate will be constant :

$Q=A_xv_x=A_yv_y$

$v_x=\dfrac{Q}{A_x}\\\\v_x=\dfrac{2.4\times 10^{-4}}{3\times 10^{-4}}\ m/s\\\\v_x=0.8\ m/s$

$v_y=\dfrac{Q}{A_y}\\\\v_y=\dfrac{2.4\times 10^{-4}}{0.6\times 10^{-4}}\ m/s\\\\v_x=4\ m/s$

Now ,

$P_x-P_y=\dfrac{\rho v_y^2}{2}-\dfrac{\rho v_x^2}{2}\\\\P_x-P_y=\rho[\dfrac{ v_y^2}{2}-\dfrac{v_x^2}{2}]\\\\P_x-P_y=1000\times [\dfrac{ 4^2}{2}-\dfrac{0.8^2}{2}]\\\\P_x-P_y=7680\ Pa$

Difference in pressure between X and Y is most near to 7700 Pa.

Hence, this is the required solution.