Normal octane at (68{ }^{circ} mathrm{F}left(u=8.31 times 10^{-6} mathrm{ft}^{2} / mathrm{s} ight)) is to be delivered at

Normal octane at \(68{ }^{\circ} \mathrm{F}\left(u=8.31 \times 10^{-6} \mathrm{ft}^{2} / \mathrm{s}\right)\) is to be delivered at a flow rate of \(3.0 \mathrm{gal} / \mathrm{min}\) through a \(2.0-\) in. schedule 40 commercial steel pipe (with an actual inside diameter of  2.067 in.). Energy losses are important, so consideration is being given to expanding the pipe to a 3.0-in. schedule 40 commercial steel pipe (with an actual inside diameter of 3.068 in.), using a \(10^{\circ}\) conical expansion. After a length \(\ell\) of the 3.0-in. pipe, it is decreased to a 2.0-in. schedule 40 commercial steel pipe with a gradual contraction \((\mathbf{K}=0.10\), based on the smaller diameter pipe flow). Find the minimum length \(\ell\) so that the energy loss will be the same as in a 2.0-in. pipe of length \(\ell+2\) (5.7 in.) See Fig. P8.114.

Figure P8.114 

Transcribed Image Text:

5.7 in. 2-in. pipe 5.7 in. Expansion 3-in. pipe Contraction