Glycerine flows steadily through a horizontal tube of length 1.5m and radius 1.0 cm . If the amount of glycerine collected per second at one is `4.0xx10^(-3)kgs^(-1)` , what is the pressure difference between the two ends of the 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]

Explanation

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.