Question about density used in the calculations?
Posted: Wed Jan 26, 2011 11:53 pm
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!
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!