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Old Feb 14th 2017, 06:23 AM   #1
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Please Help - Calculate Steam Flow Rate

Hi,

I'm sure I'm missing something here. I have a small steam vessel (internal volume 380mL) with an inlet pressure of 55psig (380 kPa) and existing to atmospheric with no change in elevation. I believe that Bernoulli states that velocity is the pressure change over density times volume, but this number is outlandish (and the units end up being 1/(m*second squared)). What am I missing here. Please let me know. Thank you.
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Old Feb 14th 2017, 02:09 PM   #2
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Bernoulli's Principle states:

$\displaystyle \frac {v^2} 2 + gz + \frac P {\rho} $ = constant

See if that works for you.
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Old Feb 15th 2017, 05:45 AM   #3
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Originally Posted by ChipB View Post
Bernoulli's Principle states:

$\displaystyle \frac {v^2} 2 + gz + \frac P {\rho} $ = constant

See if that works for you.
Hi,

Yes, that would work. However, if I take $\displaystyle \frac P {\rho} $, the units drop out to just inches. I'll assume the units drop out some other way (going back to BTU/hr or watts), but my only otherquestion is what the correct units would be. I should be able to keep everything in psi and pounds per cubic inch to give me inches per second, but not sure what the correct metric units would be. Thank for the help though.
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Old Feb 15th 2017, 08:36 AM   #4
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Originally Posted by josephmatchett View Post
Hi,

Yes, that would work. However, if I take $\displaystyle \frac P {\rho} $, the units drop out to just inches.
Let's do a little dimensional analysis:

In SI the units of P is N/m^2 (otherwise known as Pascals). Recall that 1 Newton = 1 Kg-m/s^2, so that P has fundamental units of Kg/(m-s^2). Rho (density) has units of Kg/m^3. So:

$\displaystyle \frac P {\rho} = \frac {Kg/(m-s^2)}{Kg/m^3} = (m/s)^2$

In imperial units you need to be careful to note the difference between pound-mass and pound-force. I suspect you made this mistake:

$\displaystyle \frac P {\rho} = \frac {lb/in^2}{lb/in^3} = in$

But pressure is units of pound-force/area, whereas density is in units of pound-mass/volume. Recall that 1 lb-force = 1 lb-mass x 32.2 ft/s^2:

$\displaystyle \frac P {\rho} = \frac {lb_f/in^2}{lb_m/in^3} = \frac {32.2 lb_m ft/(s^2 \ in^2)}{lb_m/in^3}$

which has units of distance^2 per second^2.
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Old Feb 15th 2017, 09:10 AM   #5
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Originally Posted by ChipB View Post
Let's do a little dimensional analysis:

In SI the units of P is N/m^2 (otherwise known as Pascals). Recall that 1 Newton = 1 Kg-m/s^2, so that P has fundamental units of Kg/(m-s^2). Rho (density) has units of Kg/m^3. So:

$\displaystyle \frac P {\rho} = \frac {Kg/(m-s^2)}{Kg/m^3} = (m/s)^2$

In imperial units you need to be careful to note the difference between pound-mass and pound-force. I suspect you made this mistake:

$\displaystyle \frac P {\rho} = \frac {lb/in^2}{lb/in^3} = in$

But pressure is units of pound-force/area, whereas density is in units of pound-mass/volume. Recall that 1 lb-force = 1 lb-mass x 32.2 ft/s^2:

$\displaystyle \frac P {\rho} = \frac {lb_f/in^2}{lb_m/in^3} = \frac {32.2 lb_m ft/(s^2 \ in^2)}{lb_m/in^3}$

which has units of distance^2 per second^2.
Wow...can't believe I made the force/mass mistake. Thank you.
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