- #1
RGNEM
- 5
- 0
I am working on a fluid mechanics problem modeling something I've come into contact with at work.
I have an open to atmosphere reservoir of a fluid with properties as follows
SG=1.48
Viscosity= 47000 Cp
Height of fluid column, 3 inches or so. i.e. Neglible
I then have a pressure vessel with air inside of it at
190 Torr = 3.67 psi
If I immerse a plastic tube in the fluid, connect it to the pressure vessel, and open the clamp, I will establish some flow.
Inner Diameter of Tube is .19 in
I have been trying to determine the flow speed and pressure in the flow.
I have been using the equation
[tex] \frac{P_1}{\gamma} + \frac{V_1^2}{2*g} + Z_1= \frac{P_2}{\gamma} + \frac{V_2^2}{2*g} + Z_2 [/tex]
Considering change in height to be negligible (only a few inches), my equation reduces to
[tex] \frac{P_1}{\gamma} + \frac{V_1^2}{2*g} = \frac{P_2}{\gamma} + \frac{V_2^2}{2*g} [/tex]
Taking initial fluid velocity to be zero, as it is coming from the reservoir of fluid, my equation further simplifies to
[tex] \frac{P_1}{\gamma}= \frac{P_2}{\gamma} + \frac{V_2^2}{2*g} [/tex]
Plugging in values, I find my V_2 to be about 399 in/s
This is too high, but it must be due to the incorrect assumption of inviscid flow.
My next step is to add head loss to my equation:
[tex] \frac{P_1}{\gamma} = \frac{P_2}{\gamma} + \frac{V_2^2}{2*g} + H_L [/tex]
To determine this, I first calculate my Reynolds Number using the equation
[tex]R_e=\frac{ρVD}{\mu}[/tex]
then Friction Factor using the equation
[tex]f=\frac{64}{R_e}[/tex]
Using this value I find my Head Loss using the equation
[tex] HL=f\frac{l}{D}\frac{v^2}{2g}[/tex]
My problem is that, after using an assumed initial velocity to begin the iterative process,
my Reynold's Number comes out very low, my Friction Factor very high and my Head Loss astronomical.
For example, for an assumed V1 of 20 in/s
Re= .077
f=828.99
HL=4043058523 in
Plugging into the next iteration of my flow equation, the inside of the eventual square root required for solving for V_2 comes out to be negative, which is just a small problem...
I hope I'm missing something obvious, and after spending many hours just to get to this point, I'd appreciate any ideas/comments/help that any of you could provide.
Thank you in advance.
I have an open to atmosphere reservoir of a fluid with properties as follows
SG=1.48
Viscosity= 47000 Cp
Height of fluid column, 3 inches or so. i.e. Neglible
I then have a pressure vessel with air inside of it at
190 Torr = 3.67 psi
If I immerse a plastic tube in the fluid, connect it to the pressure vessel, and open the clamp, I will establish some flow.
Inner Diameter of Tube is .19 in
I have been trying to determine the flow speed and pressure in the flow.
I have been using the equation
[tex] \frac{P_1}{\gamma} + \frac{V_1^2}{2*g} + Z_1= \frac{P_2}{\gamma} + \frac{V_2^2}{2*g} + Z_2 [/tex]
Considering change in height to be negligible (only a few inches), my equation reduces to
[tex] \frac{P_1}{\gamma} + \frac{V_1^2}{2*g} = \frac{P_2}{\gamma} + \frac{V_2^2}{2*g} [/tex]
Taking initial fluid velocity to be zero, as it is coming from the reservoir of fluid, my equation further simplifies to
[tex] \frac{P_1}{\gamma}= \frac{P_2}{\gamma} + \frac{V_2^2}{2*g} [/tex]
Plugging in values, I find my V_2 to be about 399 in/s
This is too high, but it must be due to the incorrect assumption of inviscid flow.
My next step is to add head loss to my equation:
[tex] \frac{P_1}{\gamma} = \frac{P_2}{\gamma} + \frac{V_2^2}{2*g} + H_L [/tex]
To determine this, I first calculate my Reynolds Number using the equation
[tex]R_e=\frac{ρVD}{\mu}[/tex]
then Friction Factor using the equation
[tex]f=\frac{64}{R_e}[/tex]
Using this value I find my Head Loss using the equation
[tex] HL=f\frac{l}{D}\frac{v^2}{2g}[/tex]
My problem is that, after using an assumed initial velocity to begin the iterative process,
my Reynold's Number comes out very low, my Friction Factor very high and my Head Loss astronomical.
For example, for an assumed V1 of 20 in/s
Re= .077
f=828.99
HL=4043058523 in
Plugging into the next iteration of my flow equation, the inside of the eventual square root required for solving for V_2 comes out to be negative, which is just a small problem...
I hope I'm missing something obvious, and after spending many hours just to get to this point, I'd appreciate any ideas/comments/help that any of you could provide.
Thank you in advance.