assuming that the critical velocity of viscous liquid is flowing through a capillary tube depends on radius r of tube , density p,and coefficient of viscosity of the liquid ,find expression for critical velocity.
Answers
Answer ⇒ V = kη/r²ρ.
Explanation ⇒
According to the question, Critical Velocity of viscous liquid is flowing through a capillary tube depends on radius r of tube , density p,and coefficient of viscosity of the liquid.
Let the critical velocity is proportional to rᵃ, ρᵇ , ηⁿ.
Then,
V = krᵃ, ρᵇ , ηⁿ, k is any dimension less constant.
Thus,
Dimension of V = LT⁻¹
Dimension of r = L
Dimension of ρ = ML⁻³
Dimension of η = ML⁻¹T⁻¹
Now, Putting dimension on both sides of equation,
[LT⁻¹] = [L]ᵃ[ML⁻³]ᵇ[ML⁻¹T⁻¹]ⁿ
[M°LT⁻¹] = [Mᵇ⁺ⁿ][Lᵃ⁻ⁿ⁻³ᵇ][T⁻ⁿ]
On comparing,
n = 1, a - n - 3b = 1 and b + n = 0
b = -n = -1
Now, in a - n - 3b =1,
a - 1 - 3(-1) = 1
a + 2 = 1
a = -1
Thus, V = kη/r²ρ.
This will be the relation. Now, k is constant. Its value was determined by Physicist 'Reynolds' and hence called as Reynolds number.
Hope it helps.
Answer:
Let the critical velocity is proportional to rᵃ, ρᵇ , ηⁿ.
V = krᵃ, ρᵇ , ηⁿ, k is any dimension less constant
Dimension of V = LT⁻¹
Dimension of r = L
Dimension of ρ = ML⁻³
Dimension of η = ML⁻¹T⁻¹
Putting dimension on both sides of equation,
[LT⁻¹] = [L]ᵃ[ML⁻³]ᵇ[ML⁻¹T⁻¹]ⁿ
[M°LT⁻¹] = [Mᵇ⁺ⁿ][Lᵃ⁻ⁿ⁻³ᵇ][T⁻ⁿ]
n = 1, a - n - 3b = 1 and b + n = 0
b = -n = -1
Now, in a - n - 3b =1,
a - 1 - 3(-1) = 1
a + 2 = 1
a = -1
V = kη/r²ρ.