Question

A 4.70 cm radius pipe has ethanol flowing through it at 6.00 m/s. The pipe then rises by 7.00 m and increases in radius to 11.3 cm. The pipe then empties into the air.
(a) What is the pressure within the ethanol in the first 4.70 cm radius segment?
____ Pa
(b) How fast is the ethanol in the second 11.3 cm radius segment?
___ m/s

Answer 1

from equation of continuity

volume flow rate remains same in a fluid flow

A1*v1 = A2*v2

A1 = pi*r1^2

A2 = pi*r2^2

r1 = radius of smaller pipe = 4.7 cm = 0.047 m

r2 = radius of larger pipe = 11.3 cm = 0.113 m

v1 = 6 m/s

v2 = ?

r1^2*v1 = r2^2*v2

v2 = v1*(r1/r2)^2

v2 = 6*(0.047/0.113)^2

v2 = 1.04 m/s

==================

from Bernoullis principle

P1 + (1/2)*rho*v1^2 + rho*g*h1 = P2 + (1/2)*rho*v2^2 + rho*g*h2

P1 = P2 + (1/2)*rho*(v2^2 - v1^2) + rho*g*(h2 - h1)

h1 = 0

h2 = 7 m

rho = density of ethanol = 789 kg/m^3

P2 = 10^5 Pa

v1 = 6 m/s

v2 = 1.04 m/s

P1 = 10^5 + (1/2)*789*(1.04^2-6^2) + 789*9.8*7

P1 = 1.4*10^5 Pa <<<<<----------ANSWER

=======================

part(b)

v2 = 1.04 m/s