Question

Confused with a question: At a processing plant, olive oil of density 875 kg/m3 flows in a horizontal section of hose that constricts from a diameter of 2.80 cm to a diameter of 1.50 cm. Assume steady, ideal flow. change in pressure between the two sections of hose is 4.85 kPa.

Answer 1

Its continuous flow so A1V1=A2V2
and P2-P2=(1/2)p(V^2-V^2)

Answer 2

I know the answer, I just can't figure out how to work the solution

Answer 3

and the question is how to find the flow rate

Answer 4

Wow you're attractive, but anyway...

Flow rate is just the change in volume over time right?

Answer 5

yes in m^3/s

Answer 6

You have the two equations, so you can find the change in volume

Answer 7

i try solving for velocity and multiplying it by its corresponding area, but i dont get the right answer. The answer is 0.000614 m^3/s

Answer 8

I dont wanna write it out lol...but i will :p

Answer 9

Gimme a bit

Answer 10

What did you get for your velocity?

Answer 11

hold on, i erased it

Answer 12

i got 0.03 for V1

Answer 13

and .109 for V2

Answer 14

Ew my laptop is legit running out of battery...but basically with the P2-P1=(1/2)p(V2^2-V1^2) you can solve for distance, as this is the same as P2-P1=(1/2)p(A2^2-1^2)*d^2

Answer 15

Once you have distance i think this question is literally the same as this one:

https://answers.yahoo.com/question/index?qid=20120422090134AA5B9nS

but with different numbers

Answer 16

i miss typed...its (1/2)p(A2^2-A1^2)*d^2

Answer 17

If you still need help just search "water speed in a constricted pipe" most of the questions are the same anyway lol

Answer 18

oh okay thanks a lot!