math
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Fabric drying question
Hello,
I'd like to know if a wet (H2O) fabric would dry faster in a humid (95% RH) but slightly windy (6 km/h) environment or in a warmer (+10°C
than previous environment) and less humid (70% RH) but static (no significant airflow) environment.
I thought water vapor pressure plays a role, but I don't know if it's that big to make the second environment the faster one.
I apologize for the odd question.
Thank you
[Edited on 19-2-2013 by math]
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hissingnoise
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Most likely six-of-one, half a dozen the other . . . ?
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MrHomeScientist
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Sounds like a good science fair project!
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Magpie
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The less the relative humidity the faster the fabric would dry in a static environment. Normal convective currents and molecular diffusion regenerate
the low humidity near the surface.
Any kind of wind will speed up such a process.
I would guess that the low humidity situation would give faster drying. I might be hard pressed to come up with any calculations to
support that, however. I might give it a try.
The single most important condition for a successful synthesis is good mixing - Nicodem
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Magpie
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I was totally wrong in that guess. The wind greatly increased the drying rate, even at 95% RH.
In my analysis I modeled the wet fabric as a pan of water. I used two correlations for the wind case and one of the correlations for the no wind
case. The Langmuir evaporation equation gave complete crap so it's not used here.
1. For the 95% RH case with wind speed = 6 km/h (5.5ft/s)
Assume the air temperature = 86°F (30°C), and that the liquid temperature is the same.
Use the following constant rate drying equation
Rc = hV/λ (Tv – Ti) (ref 1)
Where Rc = the constant drying rate, lb/(h –ft2)
hV = gas phase heat transfer coefficient, BTUl/(h-ft2 - °F)
λ = latent heat of evaporation for water = 1044BTU/lb
Tv = bulk gas phase temperature = 86°F
Ti = liquid-gas interface temperature, assumed to equal the wet bulb temperature
= 84.6°F (from a psychometric chart for air-water)
For air drying, the extensive experimental results have been correlated by the relation
hV = 0.0128Gv^0.8 (ref 1)
where Gv = mass flow of the vapor phase, lb/(h- ft2)
at Tv = 86°F the density of air at a pressure of 1 atmosphere and 95% RH = 0.072lb/ft3
then Gv = (0.072lb/ft3)(5.5ft/s)(3600s/h) = 1426 lb/(h-ft2)
and hV = (0.0128)(1426)^0.8 = 4.27 BTU/(h-ft2 -°F )
and finally, Rc = [4.27 BTU/(h-ft2 -°F)]/(1044BTU/lb)[86°F – 84.6°F] =
0.0057 lb/(h – ft2) = 27.8 g/(h – m2)
--------------------------------------------------------------------------------
Using a second correlation from a different reference (ref 2):
Rc = (0.002198 + 0.0398Va^0.5756)(pw - paw)
for 0.0 ≤Va ≤ 5.36 m/s
Where Va = the velocity of the ambient air in m/s
pw = partial pressure of water vapor at saturation (= vapor pressure of water at the air temperature), mmHg
paw = partial pressure of the water vapor in the air, mmHg
Va = 6km/h = 1.67 m/s
pw = 32.1 mmHg
paw = 30.2 mmHg
Then Rc = [0.002198 + 0.0398(1.67)^0.5756 ](32.1 – 30.2) = 0.106 kg/(h-m2) = 106 g/(h-m2)
2. For the 70% RH case where the wind speed is zero.
Assume the ambient air temperature is 10°C cooler than in the previous situation, ie, 20°C (68°F). Assume that the water temperature is the same.
Here the ref 1 correlation cannot be used as there is no air flow over the water. The ref 2 correlation yields the following:
Va = 0
pw = 17.6 mmHg
paw = 12.1 mmHg
Then Rc = [0.002198](17.6 – 12.1) = 0.012 kg/(h-m2)
= 12 g/(h-m2)
References
1. Principles of Unit Operations, Foust et al, pp 324-325 (1962)
2. “A Unified Empirical Correlation for Evaporation of Water at Low Air Velocities,” Banzal and Xie, International Communications in Heat and
Mass Transfer, Volume 25, Issue 2, February 1998, Pages 183–190.
[Edited on 25-2-2013 by Magpie]
[Edited on 25-2-2013 by Magpie]
[Edited on 26-2-2013 by Magpie]
The single most important condition for a successful synthesis is good mixing - Nicodem
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