| The Breakroom > The Water Cooler |
| Anyone any good at physics? |
| << < (4/4) |
| Ross:
Though not having the original velocity at one end wouldn't enable me to be able to do any of the above right? |
| andyf:
That's right, Ross. Anyway, the question you quoted doesn't ask anything about velocity, only about "flow rate" and "mass flow rate". The latter being weight per second (or minute, or hour) I think you can assume that the former must refer to volume per sec, min or hour. All you can say is that both flow rates will, midway along the nozzle, be the same as the input rate, though it might impress if you conclude with a comment on the velocity with a recital of Lew's mathematics. |
| Lew_Merrick_PE:
Ross, Did you get this straightened out? Irrational minds want to know.... ;-) |
| Ross:
Getting there, I got the "flow rate". Which is 10 m/s this is actually the velocity of flow. though no info of where the point of measurement was. :palm: So I've worked out the surface area of the midpoint of the tube to be 1.227. And the velocity of the flow rate would be 8.1499. Flow rate/surface area. Then 8.1499x1.227=9.999 which is the flow rate through that point. I'm assuming that mu tutor mixed up the difference between flow rate and velocity of flow. This is probably why this has all been so hard to understand :palm: Make any sense? |
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