Question about density used in the calculations?

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lewhnh
Posts: 3
Joined: Wed Jan 26, 2011 2:04 am

Question about density used in the calculations?

Post by lewhnh »

Hello-
After using the "Air pressure drop calculator" tool a few times, I am unable to account for a discrepancy - wondering if you will help me out?

I'm passing air at -20°C through an 18" diameter pipe that is 100 ft long with an entrance pressure (p1) of 15.784 psia and a dP along the pipe of 0.00723 psi. The calculator determines that 17,114 lb/hr of gas is passing through the pipe, which it also says is 3,533.8 ft³/min (volumetric flow at normal conditions).

I've always considered normal conditions to be 14.7 psia = 0 psig = 1 atm and 293K = 20°C = 68°F, which would put the air density at approximately 0.075 lb/ft³ according to the ideal gas law. However, I know others use STP as 14.5 psia = -0.2 psig = 0.99 atm and 273K = 0°C = 32°F. These conditions result in a calculated density of 0.079 lb/ft³.

In order for the calculated flow values above to be "equal", the density used to convert would need to be (17,114 lb/hr × 1 hr/60 minutes) ÷ 3,534 ft³/min = 0.081 lb/ft³. So that's my first question - what is your definition of "normal conditions"?

Using the mass rate going forward, since that is a more concrete number, I wonder how the calculator determines the inlet velocity. Again, it appears that there is a discrepancy with the density calculation since at the entrance conditions, the ideal gas law predicts the density wil be
(15.784 psia × 28.8 lb/mol) ÷ (10.73 lbf/in²-ft³/mol³°R × (-20°C × 1.8°R/°C + 459 + 32)) = 0.093 lb/ft³
At that density, the average velocity through a 18" diameter pipe (1.77 ft² area) should be:
(3,534 lb/hr × 1 hr/60 min ÷ 0.093 lb/ft³) ÷ 1.77 ft² = 357 ft/min
However, the calculator returns a v1 value of 1,590 ft/min - over 4x what I would expect.

And finally, I used the calculated Reynolds number (351,859) to back out the viscosity of the air,
Re1 = 4 × w ÷ (pi × D × mu)
=> mu = 4 × w ÷ (Re1 × D × pi) = 4 (17,115 lb/hr) ÷ (351,859 × 1.5 ft × 3.14159)
= 68,460 lb/hr ÷ 1,658,100 ft = 0.0413 lb/ft-hr
When I repeat the calculation at 45°C, the back-calculation looks like this:
=> mu = 4 (15,013 lb/hr) ÷ (308,650 × 1.5 ft × 3.14159)
= 60,052 lb/hr ÷ 1,454,500 ft = 0.0413 lb/ft-hr
So it appears that you use the same viscosity for air at T = -20°C as you do for T = 45°C. References I've found indicate that air's viscosity decreases with temperature, from about 0.039 lb/ft-hr at -20°C to 0.046 lb/ft-hr at 45°C.
Is that accurate? I'm interested in the viscous effect on the total mass flow as well as the density effect, so I'd like to know that both are accounted for in your model.

Thanks!
admin
Site Admin
Posts: 375
Joined: Mon Feb 08, 2010 7:47 pm

Re: Question about density used in the calculations?

Post by admin »

Calculator uses density rho=1.293 kg/m3 and nu=13.2x10-6 m2/s, which is in Imperial units rho=0.0807358 lb/ft3 and nu=0.020460041 in2/s.

Norma conditions used here are for t=0C=273 K, p=101325 Pa.

Also you can find model used on:
http://www.pipeflowcalculations.com/pip ... l-flow.php
first equation on the page.
Pipe flow calculations - since 2000
lewhnh
Posts: 3
Joined: Wed Jan 26, 2011 2:04 am

Re: Question about density used in the calculations?

Post by lewhnh »

OK r.e. the normal conditions.

In the equation/ theory description, you specify v1 as specific volume (and later specific weight), but the units displayed in the table on the bottom of the calculator are in distance / time (i.e. ft/s), which indicates velocity and not volume. So the question is: "How is the v1 that is displayed as a result actually calculated - and is it linear velocity, specific volume, or what?"

Thanks again-
admin
Site Admin
Posts: 375
Joined: Mon Feb 08, 2010 7:47 pm

Re: Question about density used in the calculations?

Post by admin »

In theory page v1 - is spec weight - m3/kg
In calculator v1 and v2 are velocities and in calculator it is not required to input spec. weight or spec. volume as calculator is for air only and spec. weight can be calculated for known pressure and temperature.
Also velocities can be calculated once flow rate is calculated.
Pipe flow calculations - since 2000
lewhnh
Posts: 3
Joined: Wed Jan 26, 2011 2:04 am

Re: Question about density used in the calculations?

Post by lewhnh »

OK. That makes sense

So the reported v1 & v2 are ARE velocities, and as you said you can determine these by multiplying the mass flow rate by the specific weight (I would divide by density, but that is the same thing).
In either case, as you said the specific weight can be calculated using the ideal gas law.

That law predicts the specific weight to be
(10.73 lbf/in²-ft³/mol³°R × (-20°C × 1.8°R/°C + 459 + 32)) ÷ (15.784 psia × 28.8 lb/mol) = 10.75 ft³/lb
At that specific weight, the average velocity through a 18" diameter pipe (1.77 ft² area) should be:
(3,534 lb/hr × 1 hr/60 min × 10.75 ft³/lb) ÷ 1.77 ft² = 357 ft/min

However, the calculator returns a v1 value of 1,590 ft/min - over 4x what I would expect.
What am I doing wrong?
admin
Site Admin
Posts: 375
Joined: Mon Feb 08, 2010 7:47 pm

Re: Question about density used in the calculations?

Post by admin »

Here is what have with pipe diameter calculator:

2. mass flow rate (w): w = 3534 lb/h
3. diameter (D): D = 18 in
4. velocity (v): v = 328.5156 ft/min
5. cross section area (A): A = 1.7671458 ft2
6. pressure of gas (p): p = 15.784 psi
7. density (ρ): ρ = 0.09358722 lb/ft3
8. temperature (T): T = -20.0 C
9. gas constant (R): R = 287 J/kgK



And with air pressure drop calculator

2. mass flow rate (w): w = 3534 lb/h
4. diameter (D): D = 18 in
7. velocity (v1): v1 = 328.29324 ft/min
9. cross section area (A): A = 1.7671458 ft2
13. pressure on the pipe start (p1): p1 = 15.784 psi
16. temperature (T): T = -20.0 C


with other values excluded as not important now.

Now, I do not how you get 1,590 ft/min in calculator.
Pipe flow calculations - since 2000
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