Question

Water, with a density of ?=1185 kg/m3 , flows in a horizontal pipe. In one segment of the pipe, the flow speed is ?1=7.13 m/s . In a second segment, the flow speed is ?2=1.57 m/s . What is the difference between the pressure in the second segment ( ?2 ) and the pressure in the first segment ( ?1 )?

P2-P1 =

A liquid of density 1110 kg/m3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the flow speed is 9.11 m/s and the pipe diameter ?1 is 10.3 cm. At Location 2, the pipe diameter ?2 is 14.9 cm. At Location 1, the pipe is Δ?=9.99 m higher than it is at Location 2. Ignoring viscosity, calculate the difference Δ? between the fluid pressure at Location 2 and the fluid pressure at Location 1.

ΔP=

Answer 1

1.) According to Bernoulli's theorem

2.) Bernoulli's equation says,

v²/2 + p/ρ + gy = constant,

This equation relates the pressure, velocity and height in the steady motion of an ideal fluid.

Where, v is the velocity at a point, p the pressure, ρ the density, g the acceleration of gravity, and y the height above an arbitrary reference level.

Also vA = constant, so vr^2 = constant.
Hence, at location 2, the speed is ,

(9.11 m/s)*(0.103^2/0.149^2),
that is, v2 = 4.353 m/s.

ΔP= P2 - P1

= ρ* { g(Δy) + (1/2)[(v1)^2 - (v2)^2] }
=(1110 kg/m^3) * [9.8 m/s^2 (9.99 m) + (1/2)(9.11^2 - 4.353^2)]

ΔP= 144215.35 N/m^2

ΔP = 1.442 ×10^5 Pa _Ans.