Fluid originally flows through a tube at a rate of 100 cm^3/s. To illustrate the sensitivity of the Poiseuille flow rate to various factors, calculate the new flow rate for the following changes with all other factors remaining the same as in the original conditions:

a. Another tube is used with a radius 0.100 times the original.
b. Pressure difference increases by a factor of 1.60.
c. A new fluid with 3.00 times greater viscosity is substituted.
d. The tube is replaced by one having 4.00 times the length.

a) 0.01

b) 150 cm^3/s

c) 300 cm^3/s

d) 25 cm^3/s

Explanation:

a) We know that :  

     Q=ΔP/R

     R=8ηl/π*r^4

Givens:

     r^2 = 0.1 r_1  

Plugging known information to get :  

      Q=ΔP/R

         =ΔP*π*r^4/8*η*l

Q_2/r_2^4 =Q_1/r_1^4

           Q_2=Q_1/r_1^4*r_2^4

                  =Q_1/r_1^4*r*0.0001*r_1^4

          Q_2 = 0.01

b) From the rate flow of the fluid we know that :  

            Q=ΔP/R                                   (1)

              F=η*Av/l                                 (2)

             R=8*ηl/π*r^4                           (3)

Where:

ΔP is the change in the pressure .  

r is the raduis of the tube .

l is the length of the tube .

η is the coefficient of the vescosity of the fluid .

R is the resistance of the fluid .

Givens: Q1 = 100 cm^3/s , ΔP= 1.5

Plugging known information into EQ.1 :  

                 Q=ΔP/R

     Q_2/ΔP2=Q_1/ΔP

              Q_2=150 cm^3/s

c) we know that :

      F = η*Av/l  

can be written as :  

     ΔP = F/A = η*v/l  

Givens: η_2 = 3η_1  

           Q=ΔP/R  

           Q=η*v/l*R

Q_2/η_2=Q_1/η_1

       Q_2=300 cm^3/s

d) We know that :  

           Q=ΔP/R

           R=8*ηl/π*r^4  

Givens: l_2 = 4*l_1

Plugging known information to get :  

              Q=ΔP/R  

              Q=ΔP*π*r^4/8*ηl

   Q_2/l_2=Q_1/l_1

          Q_2 = 25 cm^3/s

answer:

True

Explanation: