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2 . 3 Sampling losses in tubes ( after Nazaroff 2 0 0 8 and Friedlander 1 9 7 7 ) Outside air is delivered

2.3 Sampling losses in tubes (after Nazaroff 2008 and Friedlander 1977)
Outside air is delivered to the instruments of an air monitoring station through a D =2 cm (insidediameter) copper tube at volumetric flow rate of Q =20 L/min. The tube is horizontally oriented and is L =5 m long. Answer the following questions to evaluate the transmission efficiency of particles through this tube. Consider unit density particles with a size range of 0.01m dp 10m. Deposition occurs by means of Brownian diffusion and gravitational settling.
(a) If the Reynolds number (Re) based on tube diameter is less than 2100, then the flow in the boundary layer along the pipe walls will remain laminar. What is the Reynolds number forthis system?
(b) For laminar flow, the air velocity profile evolves from uniform at the inlet to parabolic over a distance of approximately 0.04 D Re. Over what fraction of the length of this tube is the flow undeveloped (i.e., not yet fully parabolic)?
(c) Use the approach developed in lecture, based on magnitude analysis, to predict the particlepenetration efficiency as a function of diameter. (Equations are reproduced below.)
(d) Repeat the analysis of part (c), using the results from the papers by D. B. Ingham(Journal ofAerosol Science, 6,125,1975) and J. Pich (Aerosol Science, 3,351,1972) to evaluate the penetration efficiency. (Equations are reproduced below.)
(e) Plot your results from (c) and (d). How well do the results of the magnitude Equations:
a. Results of magnitude analysis for particle penetration through a tube:
Pd=(1-2LDbQ2)2,(only valid if 2LDbQ1; otherwise set Pd=0)
L= tube length; D= tube diameter; vg= gravitational settling velocity; Q= volume flow rate
through tube; Db= Brownian diffusivity of particles
b. Results for gravitational penetration, from Pich (1972)
Ps=1-2{21-232-131-232+arcsin(13)} valid if1; otherwise Ps=0
=3LDvg16Q
c. Results for Brownian diffusion, from Ingham (1975)
Pd=0.819exp(-14.63)+0.0976exp(-89.22)+0.0325exp(-228)+0.0509exp(-125.923)
=DbL4Qanalysis (c)compare with the more detailed analysis results (d)?
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