Respuesta :
Answer:
(a) dynamic viscosity = [tex]1.812\times 10^{-5}Pa-sec[/tex]
(b) kinematic viscosity = [tex]1.4732\times 10^{-5}m^2/sec[/tex]
Explanation:
We have given temperature T = 288.15 K
Density [tex]d=1.23kg/m^3[/tex]
According to Sutherland's Formula dynamic viscosity is given by
[tex]{\mu} = {\mu}_0 \frac {T_0+C} {T + C} \left (\frac {T} {T_0} \right )^{3/2}[/tex], here
μ = dynamic viscosity in (Pa·s) at input temperature T,
[tex]\mu _0[/tex]= reference viscosity in(Pa·s) at reference temperature T0,
T = input temperature in kelvin,
[tex]T_0[/tex] = reference temperature in kelvin,
C = Sutherland's constant for the gaseous material in question here C =120
[tex]\mu _0=4\pi \times 10^{-7}[/tex]
[tex]T_0[/tex] = 291.15
[tex]\mu =4\pi \times 10^{-7}\times \frac{291.15+120}{285.15+120}\times \left ( \frac{288.15}{291.15} \right )^{\frac{3}{2}}=1.812\times 10^{-5}Pa-s[/tex]when T = 288.15 K
For kinematic viscosity :
[tex]\nu = \frac {\mu} {\rho}[/tex]
[tex]kinemic\ viscosity=\frac{1.812\times 10^{-5}}{1.23}=1.4732\times 10^{-5}m^2/sec[/tex]
