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jrodriguez2315

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.

  • 2 years ago
  • 2 years ago

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  1. jrodriguez2315
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    |dw:1327543065572:dw|

    • 2 years ago
  2. jrodriguez2315
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    What is the decrease of pressure in cm of water? What is the decrease of pressure in kPa?

    • 2 years ago
  3. eashmore
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    The device on the bottom is a mercury manometer. We know the different in height in proportional to the pressure. \[h_2 - h_1 = {P_A - P_B \over \rho_{Hg} \cdot g}\]

    • 2 years ago
  4. jrodriguez2315
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    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.

    • 2 years ago
  5. jrodriguez2315
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    I mean even after the due date when we have turned it in.

    • 2 years ago
  6. jrodriguez2315
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    Also, how do I take into account the pressure of the water?

    • 2 years ago
  7. eashmore
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    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. \[P_L - P_R = (\rho_{water} \cdot g \cdot h_1 + \rho_{Hg} \cdot g \cdot \Delta h) - (\rho_{water} \cdot g \cdot h_2 + \rho_{Hg} \cdot g \cdot \Delta h)\]\[P_L - P_R = (\rho_{water} \cdot g) \cdot (h_1 - h_2) + (\rho_{Hg} \cdot g) \]

    • 2 years ago
  8. jrodriguez2315
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    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?

    • 2 years ago
  9. jrodriguez2315
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    Maybe there is something wrong with the problem, I will ask the professor for help tomorrow. Thanks for the help eashmore.

    • 2 years ago
  10. eashmore
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    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.

    • 2 years ago
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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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