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+(\rho v_x^2)/(2)+\rho gz_x=P_y+(\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+(\rho v_x^2)/(2)=P_y+(\rho v_y^2)/(2)` ..... 1 )

Also flow rate will be constant :

`Q=A_xv_x=A_yv_y`

`v_x=(Q)/(A_x)\\\\v_x=(2.4* 10^(-4))/(3* 10^(-4))\ m/s\\\\v_x=0.8\ m/s`

`v_y=(Q)/(A_y)\\\\v_y=(2.4* 10^(-4))/(0.6* 10^(-4))\ m/s\\\\v_x=4\ m/s`

Now ,

`P_x-P_y=(\rho v_y^2)/(2)-(\rho v_x^2)/(2)\\\\P_x-P_y=\rho[( v_y^2)/(2)-(v_x^2)/(2)]\\\\P_x-P_y=1000* [( 4^2)/(2)-(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.