What is the pressure change in methyl alcohol (SG =0.791) flowing from 4.0 cm diameter pipe to a 1.5 cm diameter pipe if the velocity in the larger pipe is 0.40 m/s?

Answer 1

rho = density in kg/m^2 = 0.791* 1000 kg/m^3 = 791 kg/m^3

P1 + (1/2) rho V1^2 = P2 + (1/2) rho V2^2

V * area = Q the volume flow rate = constant

so
V1 * pi (2)^2 = V2 * pi * .75^2 (note using cm because only ratio important)
V2 = V1 * 4/.5625 = 0.40 m/s * 7.11 meters/second = 2.84 m/s
so
P1 + (1/2)(791) (.4^2) = P2 + (1/2)(791)(2.84)^2
P2 - P1 = (1/2)(791)(.4^2-2.84^2) Pascals or Newtons/m^2
note negative because it is a pressure drop

Answer 2

Thank you so much!