Glycerine flows steadily through a horizontal tube of length 1.5m and radius 1.0 cm . If theamount of glycerine collected per second at one is 4.0xx10^(-3)kgs^(-1) , what is the pressure difference between the two ends ofthe tube? (Density of glycerine =1.3xx10^(10^(3)kgm^(-3) and viscosity of glycerine =0.83 Pa s). [ you may also like to check if the assumption of laminar flow in the tube is correct]

Glycerine flows steadily through a horizontal tube of length 1.5m and

1.	Glycerine flows steadily through a horizontal tube of length 1.5m and radius 1.0 cm . If theamount of glycerine collected per second at one is 4.0xx10^(-3)kgs^(-1) , what is the pressure difference between the two ends ofthe tube? (Density of glycerine =1.3xx10^(10^(3)kgm^(-3) and viscosity of glycerine =0.83 Pa s). [ you may also like to check if the assumption of laminar flow in the tube is correct]
Answer» SOLUTION :VOLUME of flowing glycerine in one second, `V=(M)/(rho)` `=(4xx10^(-3))/(1.3xx10^(3))` `=3.1xx10^(-6)m^(3)s^(-1)` From Poiseille.s law, `V=(piPr^(4))/(8etal)` `P=(8etalV)/(pie^(4))` `=(8xx0.83xx1.5xx3.1xx10^(-6))/(3.14xx(10^(-2))^(4))` `=9.833xx10^(2)` `=9.8xx10^(2)P_(a)` Value of Raynolds number for laminar flow in tube , `R_(e)=(rhod)/(eta)xx(m)/(rhopir^(2))` `=(dm)/(etapir^(2))=(2rm)/(etapir^(2))` `=(2M)/(etapir)` `=(2xx4xx10^(-3))/(0.83xx3.14xx10^(-2))` `=3.069xx10^(-1)` `=0.31` Here , `R_(e)lt2000`, the flow is steady that means laminar.