Question

A liquid of density 1.37 times 10^3 kg/m^3 flows steadily through a pipe of varying diameter and height. At location 1 along the pipe the flow speed is 9.89 m/s and the pipe diameter is 11.7 cm. At location 2 the pipe diameter is 15.5 cm. At location 1 the pipe is 9.23 m higher than it is at location 2. Ignoring viscosity, calculate the difference between the fluid pressure at location 2 and the fluid pressure at location 1.

Answer 1

Answer 2

SOLUTION :

Let pressure at point 1 and 2 be p1 and p2.

VH 1 = Velocity head at point 1 

= v1^2 / (2g) 

= 9.89^2 / (2 * 9.8)) 

= 4.99 m 

VH 2 = Velocity head at point 2

 = v2^2 / (2g) 

= (A1 * v1 / A2)^2 / (2g) 

= ((pi/4 * 0.00117^2 * 9.89)  / (pi/4 * 0.00155^2))/ (2 * 9.8)

= 0.29 m 

As per energy conservation :

VH1 + PH1 + Position head1 = VH2 + PH2 + Position head2

=> PH2 - PH1 = VH1 - VH2 + (Position head1 - Position head2)

=> (p2 - p1)/ w = 4.99 - 0.29 + 9.23

=> (p2 - p1) = 13.93 * 1.37 * 10^3 

=> pressure difference = 19084 Pa (ANSWER)