Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Orifices are commonly placed in flow lines to measure the flow rate through the line. The flow rate is proportional to the square root of the pressure decrease across the orifice. The mercury manometer shown here is used to measure this pressure decrease between points A and B at the pipe wall. Answer the questions below for h1 = 19.2 cm and h2 = 3 cm.

Physics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
|dw:1327543065572:dw|
What is the decrease of pressure in cm of water? What is the decrease of pressure in kPa?
The device on the bottom is a mercury manometer. We know the different in height in proportional to the pressure. \[h_2 - h_1 = {P_A - P_B \over \rho_{Hg} \cdot g}\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

im sorry, im not familiar with that equation. Im guessing my professor has not gone over it? I have a wack job of a professor who will not tell us the answers to any homework problems if after the due date.
I mean even after the due date when we have turned it in.
Also, how do I take into account the pressure of the water?
Good point. Let's take a closer look at the problem. On the left, we have\[P_L = \rho_{water} \cdot g \cdot h_1 + \rho_{Hg} \cdot g \cdot \Delta h\]The pressure on the right\[P_R = \rho_{water} \cdot g \cdot h_2 + \rho_{Hg} \cdot g \cdot \Delta h\]Taking the difference of the left and right pressure. \[P_L - P_R = (\rho_{water} \cdot g \cdot h_1 + \rho_{Hg} \cdot g \cdot \Delta h) - (\rho_{water} \cdot g \cdot h_2 + \rho_{Hg} \cdot g \cdot \Delta h)\]\[P_L - P_R = (\rho_{water} \cdot g) \cdot (h_1 - h_2) + (\rho_{Hg} \cdot g) \]
This is exactly what I did! I was going crazy thinking I had actually done something wrong. Would you mind actually solving this for me?
Maybe there is something wrong with the problem, I will ask the professor for help tomorrow. Thanks for the help eashmore.
The earlier equation isn't right. Sorry. I'm a little off today. :-/ Let's do it like this. \[\left(\begin{matrix}P_A - \rho_W \cdot g \cdot h_1 = \rho_M \cdot g \cdot (H-h_2) \\ P_B - \rho_W \cdot g \cdot h_2 = \rho_M \cdot g \cdot (H - h_1)\end{matrix}\right)\]\[P_A - P_B = \left( \rho_W \cdot g \cdot (h_1 - h_2) \right) + \left( \rho_M \cdot g \cdot (h_1 - h_2) \right)\] The reason we use \(h_2\) when solving for \(P_A\) is because the pressure will equal the amount of mercury that was displaced, not the amount that is left. This should be right. Sorry for the confusion.

Not the answer you are looking for?

Search for more explanations.

Ask your own question