Question

Consider two large parallel plates separated by a distance of 0.00914 m and filled with soybean oil of viscosity 40 mPa.s. The lower plate is being pulled at a relative velocity of 0.366 m/s greater than the top plate. (a) Calculate the shear stress and shear rate (b) If glycerol having a viscosity of 1.069 Pa.s is used instead of soybean oil, what relative velocity is needed using the same distance between plates so that the same shear stress is obtained as in part (a)? Also, what is new shear rate?

Answer 1

Distance between plates y = 0.00914 m

viscosity = 40 mPa.s x (1Pa.s/1000 mPa.s)

= 0.04 Pa.s

relative velocity = 0.366 m/s

Part a

Shear rate = relative velocity / distance

= (0.366 m/s) / (0.00914 m)

= 40.04 s-1

Newton's law of viscosity

Shear stress = viscosity x shear rate

= 0.04 Pa.s x 40.04 s-1

= 1.602 Pa

Part b

viscosity = 1.069 Pa.s

relative velocity =?

shear stress = 1.602 Pa

Distance between plates y = 0.00914 m

Newton's law of viscosity

Shear stress = viscosity x relative velocity / distance

1.602 Pa = 1.069 Pa.s x relative velocity / 0.00914 m

relative velocity = 0.0137 m/s

Shear rate = relative velocity / 0.00914 m

= (0.0137 m/s) / 0.00914 m

= 1.498 s-1