Determine
the specific agevity of a fluid
huiving viscosity 0.07 poise and kinematic
Wiscosity 0.042 stokes.
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Answer:
Specific gravity
S(for\ liquids)=\dfrac{\rho_L}{\rho_{H_2O}}S(for liquids)=
ρ
H
2
O
ρ
L
Density of water \rho_{H_2O}=1000kg/m^3ρ
H
2
O
=1000kg/m
3
Given
\mu_L=0.07\text{ poise}=7\times10^{-3}Ns/m^2,μ
L
=0.07 poise=7×10
−3
Ns/m
2
,
\nu_L=0.042St=4.2\times10^{-6}m^2/sν
L
=0.042St=4.2×10
−6
m
2
/s
Use the relation
\nu_L=\dfrac{\mu_L}{\rho_L}ν
L
=
ρ
L
μ
L
Then
S=\dfrac{\rho_L}{\rho_{H_2O}}=\dfrac{\mu_L}{\nu_L\cdot\rho_{H_2O}}S=
ρ
H
2
O
ρ
L
=
ν
L
⋅ρ
H
2
O
μ
L
S=\dfrac{7\times10^{-3}Ns/m^2}{4.2\times10^{-6}m^2/s\cdot1000kg/m^3}=\dfrac{5}{3}\approx1.667S=
4.2×10
−6
m
2
/s⋅1000kg/m
3
7×10
−3
Ns/m
2
=
3
5
≈1.667 answer
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