Thermodynamics and Fluid Mechanics Thermodynamics and Fluid Mechanics Physics Help Forum

 Feb 14th 2017, 06:23 AM #1 Junior Member   Join Date: Feb 2017 Posts: 3 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.
 Feb 14th 2017, 02:09 PM #2 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,347 Bernoulli's Principle states: $\displaystyle \frac {v^2} 2 + gz + \frac P {\rho}$ = constant See if that works for you. topsquark likes this.
Feb 15th 2017, 05:45 AM   #3
Junior Member

Join Date: Feb 2017
Posts: 3
 Originally Posted by ChipB 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.

Feb 15th 2017, 08:36 AM   #4
Physics Team

Join Date: Jun 2010
Location: Morristown, NJ USA
Posts: 2,347
 Originally Posted by josephmatchett 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.

Feb 15th 2017, 09:10 AM   #5
Junior Member

Join Date: Feb 2017
Posts: 3
 Originally Posted by ChipB 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.

 Tags calculate, flow, rate, steam

 Thread Tools Display Modes Linear Mode

 Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post ling233 Advanced Mechanics 0 Apr 27th 2016 06:28 PM ling233 Advanced Mechanics 3 Jan 17th 2016 05:18 AM robhare General Physics 1 Apr 25th 2011 10:20 AM alexito01 Thermodynamics and Fluid Mechanics 3 Mar 11th 2009 08:32 AM Devlan Thermodynamics and Fluid Mechanics 0 Nov 10th 2008 08:48 AM