Determine the range of average velocity of flow for which the flow would be in the critical region if SAE 10 oil at 60°F is flowing in a 2-in Schedule 40 steel pipe. The oil has a specific gravity of 0.89.

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The flow would be in the critical region if [latex] 2000 \lt N_R \lt 4000. [/latex] First, we use the Reynolds number and solve for velocity:

[latex] N_R = \frac{vD_ ho}{\eta} [/latex]

[latex] v= \frac{N_R \eta }{D ho} [/latex]

Then we find the values for [latex] \eta [/latex] , D, and [latex] ho [/latex] :

D = 0.1723 ft (from Appendix F)

[latex] \eta = 2.10 imes10^{-3} lb–s/ft^2 [/latex] (from Appendix D)

[latex] ho = \left(sg ight) \left(1.94 slugs/ft^3 ight)= \left(0.89 ight) \left(1.94 slugs/ft^3 ight) = 1.73 slugs/ft^3[/latex]

Substituting these values into Eq. (8–2), we get

[latex] v= \frac{N_R\left(2.10 imes 10^{-3} ight) }{\left(0.1723 ight) \left(1.73 ight) } =\left(7.05 imes 10^{-3} ight) N_R[/latex]

For [latex] N_R = 2000 [/latex],we have

[latex] v= \left(7.05 imes 10^{-3} ight) \left(2 imes 10^{3} ight) = 14.1 ft/s [/latex]

For [latex] N_R = 4000 [/latex],we have

[latex] v= \left(7.05 imes 10^{-3} ight) \left(4 imes 10^{3} ight) = 28.2 ft/s [/latex]

Therefore, if [latex] 14.1 \lt v\lt 28.2 ft/s [/latex], the flow will be in the critical region.

Question 8.3: Determine the range of average velocity of flow for which th...

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