Please help me FLOW RATE!!?

Question

Please help me FLOW RATE!!?
please answer these questions

show your work.. please thank you soooooo much!


a certain pipe is observed to have a flow rate of 600 cc/min. determine the new flow rate if the following changes are enacted one at a time, with all other quantities remaining at their original values.

a) the pressure is increased from 20 psi to 60 psi

b) the diameter of the pipe is changed from 6mm to 18mm

c) the viscosity in the pipe changes to a new value only 1/2 of its original value

d) the length of the pipe changes from 50 cm to 25 cm

it does not say if it is turbuland or laminar... sorry

PLEASE SHOW ALL YOUR WORK THANK YOU!

Answer 1

for the formula, Check this out: http://en.wikipedia.org/wiki/Hagen%E2%80%93Poiseuille_equation#Standard_fluid_dynamics_notation



a)1800 cc/min (gets tripled, since pressure is tripled and they are directly proportional. Others in the same way, just look at the equation to see the relation bet. Q and the sought quantity)

b)81 x 600 cc/min

c)1200 cc/min

d) 1200 cc/min

Answer 2

v = velocity

a = cross sectional area of pipe

h = head applied

d = density of fluid

g = gravitational constant





formula:

rate of flow = q = v x a

velocity of flow = v = ( 2 x g x h)^0.5

pressure = d x h x g

a) the pressure is increased from 20 psi to 60 psiSo new h will be 3h in pressure

New v = ( 3 x 2 x g x h )^0.5 = 3^0.5 x ( 2 x g x h)^0.5 = 1.732051 x old v

New flow rate q = 1.732051 x old flow rate = 1.732051 x 600 = 1039.23 cc/min ans.

b) the diameter of the pipe is changed from 6mm to 18mm

Old Area of cross section of pipe = ( pi /4) x D^2 where D = diameter

New dia is 18/6 = 3 times D

New area of cross section = (pi/4)x(3 D)^2 = 9 times the old area

New rate of flow = v x 9a = 9x600 = 5400 cc/min ans.



c) the viscosity in the pipe changes to a new value only 1/2 of its original value

velocity of flow is inversely proportional to viscosity.

So whenviscosity is ½, velocity will be 2 times original velocity

New flow will be 2 x a x v = 2 x600 = 1200 cc/min ans



d) the length of the pipe changes from 50 cm to 25 cm

h/l = is original gradient as l becomes l/2, new gradient will be 2 times old gradient.



As velocity varies with gradient^0.5 , new velocity will be proportional to ( 2 x old gradient)^0.5

i.e. 2^0.5 x old velocity = 1.414x old velocity.

New flow shall be 1.414x600 = 848.4 cc/min ans.

Answer 3

Q Vxa