In studying for my final, I'm working through the problems I don't understand from my old tests. The only one I can't go back and figure out is the second part of a 2 part question. I'll write it all so you can read it as I would, just ignore part c (it was discussed in another post.(adsbygoogle = window.adsbygoogle || []).push({});

A liquid of density equal to 1.85 g/cm^3 flows through two horizontal sections of tubing joined end to end. In the first section (large section) the cross-sectional area is 10 cm^2, the flow speed is 275 cm/s, and the pressure is 1.2E5 Pa. In the second section (small section), the diameter of the tube is 2.5 cm.

c) Find the flow speed, flow rate, and the pressure in the small tube section.

d) The system is modified adding a vertical section between the 2 horizontal tubes. The first section is now 1.5m below the 2nd section. In the first section, the flow and the pressure remains the same as in part c). Find the pressure and the flow rate in the small tube section.

Now, I'm also assuming that the cross-sectional areas are the same as in the other part (.1 m^2 for the big section and .049 m^2 in the small section). I then use bernoulli's equation and get:

(1.2E5) + (.5)(1850)(2.75^2) = P2 + (1850)(9.8)(1.5) + (.5)(1850)(v2^2)

This leaves me with 2 unknowns. Can I use A1v1 = A2v2 to get v2? That just make the flow rate the same in part d) as in part c) but the pressure would still be different. I just didn't think I could do that. If not, what should I do to get the second equation?

Thank you for your time!

-edge

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# Homework Help: Old test question.

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