2 π R T M Area = R L. where R P R T

Transcription

1 Leak Rate Problem Using a specification from the Parenteral Society, the leak rate for a new, clean empty freeze ryer shoul be less than 0.02mBar-L/sec. a) If one ha a leak of that magnitue, an if it came from multiple sources, where each source was a roun hole with iameter 0.2 microns, then how many such holes woul be present? b) If the leak rate were change to X mbar-l/sec, then how many such holes woul be present? What follows will be the erivation of an equation that will allow us to solve the problem above as well as many similar problems. The equation that will be erive is 2 π R T Area = R L where R L is the "Leak Rate" in mbar-liters/sec or PV / time. is ecular weight of air, P is pressure outsie the LYO (or more technically, it is P between the insie an outsie of the Lyo), R is the olar Gas constant, joule an T is room temperature in egrees K. [DO ALL CALCULATIONS e K IN SI UNITS ALL THE TIE AND THEN CONVERT THE FINAL ANSWER TO WHATEVER UNITS YOU WANT.] I use athca. I can multiply ( 2ft 3mm 0.05furlong) = L an athca assures that my answer an units are correct. If you aren't using athca, Be Careful. Kinetic theory efines ν as the number of collisions m 2 that will occur between a gas some sec area of a wall. ν = EQ #1. n' is efine as the number of per unit volume. n' = m 3 an ν is the mean velocity of gas from kinetic theory m ν m = 8 R T

2 Thus we began with 1 equation an 2 efinitions of terms. ν = EQ #1 Let ν = the number of collisions per square meter in a sec. n' = m 3 Definition n' is a concentration 'like' term (conc. is es m 3 ) 8 R T J ν m = Defiition Note: R := := kg = air ecular K wt. an T := K = room temperature. Since we know that the collisions are one with air, we can estimate the mass, w, of the collisions. w = ν EQ#2 w = total mass of collisions per m 2 in a secon. = mass of each air ecule Na = Avogoro's number := e 1 as a check on what has been one so far, we can look at a units only calculation: collisions kg.029 w = e 1 = kg Substituting ν from EQ#1 into EQ#2 an solving for the number of /m 3 ν = EQ#1 w = ν EQ#2 w = <= This is the substitution w n' = EQ#3 <= Here it is solve for the concentration 'like' term. ν m n' m 3 Now verify the units for n', the concentration 'like' term: = = m 3 e note that n'/na is an actual concentration term an can be substitute into the gas law.

3 n Gas Law: P = V R T please see that n V concentration. n' is the same as. Both are Na P = n' w R T = R T EQ# Notice that for n' we substitute in ν m EQ#3 an cancelle out w P = R T EQ# ν m Now substitute in an expression for the mean velocity of, ν m w P R T = an simplify to get P = w 2 π R T EQ#5 8 R T 'w' is the mass of /sec striking some area an is easily obtaine. From the problem, we know that 0.02 mbar-liter /s of are striking an unknown area. An we can safely assume that they are air with = 29 g/. The mass of the was given by the problem, where the Leak Rate was efine as mbar-liter/sec. Writing a ifferential form of the gas law, t n V t P = we shoul note that R T V t P is in fact our leak rate. We shoul call it R L = V t P but notice that t n t P V m 2 = w = R T Area Not obvious? w has units of So substitute in R L an we get R L w = R T Area which can now be put back into EQ#5 to get kg R L 2 π P R T 2 π R T = which simplifies to Area = R R T Area L

4 Now let's put in some numbers so that we can compute. mbar :=.001bar <= efinition mbar liter R L := 0.02 s R := J K Leak Rate olar Gas Constant P := Pa Pa Pressure ifferential between atmospheric an Lyo T := K := 29 gm olecular weight of Air 2 π R T Area := R L Area = m 2 ia := 2 ia = µm <= iameter if it were a single hole The original question was how many 0.2µm holes must exist to obtain the observe leak rate? Area of a.2µm hole is Area BugArea Area 2.2µm BugArea := π BugArea = 0.031µm = That many pores of 0.2µm iameter can exist with the Society Leak Rate. π

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