Question 1033473: Please help, I am very stuck, and it's a 3 part question:
Poiseuille's Law gives the rate of flow, R, of a gas through a cylindrical pipe in terms of the radius of the pipe, r,, for a fixed drop in pressure between the two ends of the pipe.
(a) Find a formula for Poiseuille's Law, given that the rate of flow is proportional to the fourth power of the radius.
Use k as the constant of proportionality.
R=
(b) If R= 550cm3/sec in a pipe of radius cm for a certain gas, find a formula for the rate of flow of that gas through a pipe of radius r cm.
R=
(c) What is the rate of flow of the gas in part (b) through a pipe with a 7 cm radius? Round your answer to two decimal places.
R= ___ cm3/sec
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! *************************************************************************
Poiseuille's Law is
:
delta R = (pi * Pressure difference * radius^4) / (8 * viscosity * length)
:
in our problem the constant of proportionality (k) is equal to (pi * Pressure difference) / (8 * viscosity * length)
:
we can find the value of k when given delta R and its associated radius
:
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The problem states that delta R is proportional to the fourth power of r
:
a) delta R = k * r^4, where r is the radius and k is the constant of proportionality
:
b) 550 = k * r^4
k = 550 / r^4
*** we need the associated r to find the value of k, let's call it ro, then
k = 550 / ro
delta R = (550 / ro) * r^4
:
c) delta R = k * 7^4
delta R = (550 / ro) * 2401
delta R = 1320550 / ro
*** this is as far as it goes without the value for ro
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