Question

If the kinematic viscosity of water at S.TP is 1:0038 x 10 m²/a and the mass density is 998.23 kg/m³, calculate its absolute viscosity

Answer 1

Absolute viscosity=(998.23$\frac{kg}{m^3}$) /(1:0038$\times$ 10m²/a)

Convert 998.23$\frac{kg}{m^3}$ to g/mm³= $\frac{998.23}{1000}$= 0.99823$\frac{g}{mm^3}$

1:0038$\times$ 10m²/a= $\frac{10m^2}{38a}$ = $\frac{5m^2}{19a}$

Absolute viscosity= (0.99823$\frac{g}{mm^3}$)/($\frac{5m^2}{19a}$ )

Absolute viscosity= 0.26269Pa.S