Question

3. To match the Reynold’s number in an air flow and water flow using the same size model, which flow will require the higher flow speed? How much higher must it be? vair= 3.51x10-5 m2/s; vwater= 1.00x10-6 m2/s, where v is the kinematic viscosity

Answer 1

To develop this problem we will apply the principle of dimensional similarity between water and air. Later we will leave the speed units defined in terms of the kinematic viscosity to finally contrast the magnitude of difference between them with the given values. Reynolds number can be defined as,

`R_e = (\rho VL)/(\mu)`

Here,

`\rho`= Density

V = Velocity of flow

L = length

`\mu`= Dynamic viscosity

For dimensional similarity we have,

`\bigg ((\rho VL)/(\mu) \bigg)_a = \bigg ((\rho VL)/(\mu) \bigg)_w`

`\bigg ((\rho_a V_aL_a)/(\mu_a) \bigg) = \bigg ((\rho_w V_wL_w)/(\mu_w) \bigg)`

`(V_a)/(V_w) = (\rho_w)/(\rho_a) * (\mu_a)/(\mu_w)`

`(V_a)/(V_w) = (\\u_a)/(\\u_w)`

Here
`\\u` means the kinematic viscosity, then replacing with our values

`(V_a)/(V_w) = (3.51*10^(-5))/(1*10^(-6))`

`(V_a)/(V_w) = 35.1`

`V_a = 35.1V_w`

Therefore the velocity of air is higher than velocity of water by 35.1 times