HW #3. Concentration of nitrogen within the surface (at the location in question), C B = 2 kg/m3

Transcription

1 HW #3 Problem 5.7 a. To Find: The distance from the high-pressure side at which the concentration of nitrogen in steel equals 2.0 kg/m 3 under conditions of steady-state diffusion. b. Given: D = m 2 /s at 1200 o J = kg/m 2 -s oncentration of nitrogen at high-pressure side, = 4 kg/m3. oncentration of nitrogen within the surface (at the location in question), = 2 kg/m3 c. ssumptions: (i) Linear concentration profile, i.e., steady state diffusion condition. (ii) One-dimensional diffusion, i.e. diffusion occurs only along the axis parallel to the thickness of the plate. (iii) Steel is a generic term. The diffusivity of nitrogen in steel may vary from grade to grade, depending on the concentrations of alloying elements. It is assumed that the diffusivity values mentioned correspond to the particular steel considered in the question. d. Solution: (It would be great but not necessary to have a figure here, illustrating the given situation.) Let the axis parallel to the thickness of the plate be x axis. Let the x co-ordinate of the high-pressure side be x = 0. Let the x co-ordinate where concentration of nitrogen is be x. Then, the distance from the high-pressure side at which the concentration of nitrogen in steel equals 2.0 kg/m 3 under conditions of steady-state diffusion = x -x.

2 From equation 5.3: J = D x x or J = D x x x x - = D J x kg / m 0 = ( 6 10 m /s kg / m kg / m - s - ) 7 2 x = m = 1 mm Hence, distance from high-pressure side = x - x = x - 0 = x equals m = 1 mm

3 Problem 5.8 a. To Find: The diffusion co-efficient of in Fe at 725 o b. Given: x x = 1 mm J = kg/m 2 -s oncentration of at x = x, = wt% oncentration of at x = x, = wt%. c. ssumptions: (i) (ii) (iii) Linear concentration profile, i.e., steady state diffusion condition. concentration at the two surfaces before quenching (suddenly cooling to room temperature = concentration at the two surfaces after quenching. One-dimensional diffusion, i.e. diffusion occurs only along the axis parallel to the thickness of the plate. d. Solution: Step 1: For the units to be consistent, the concentrations must be expressed in kg/m 3. This may be done using one of the following two methods: Method 1: Equation 4.9 (a) " (kg/m^3) = 10 Fe + ρ ρ Fe 3

4 Method 2: Let weight of =, weight of Fe = Fe, density of = ρ and density of Fe = ρ Fe. Then, g of is present in + Fe g of carburized Fe. g of is present in [( /ρ ) + ( Fe +ρ Fe )] cc of carburized Fe. g is present in 1 cc of carburized Fe. Fe ρ + ρfe * 10 3 kg of Fe is present in 1 m 3 of carburized Fe. Fe ρ + ρ Fe oth methods are equivalent. For = wt% " = g/cm 7.87 g/cm 3 = kg /m 3 For = wt% " = g/cm g/cm = kg /m 3

5 From equation 5.3 x D = J x or D = J x x => D = ( kg/m 2 - s) 10 3 m kg/m kg/m 3 D = m 2 /s D = m 2 /s

6 Problem 5.22 a. To Find: (i) Preexponential, D 0 (ii) ctivation energy, Q d (iii) Diffusion co-efficient, D at 875 o for the diffusion of g in u. b. Given: D = D 1 = at T = T 1 = 650 o D = D 2 = at T = T 2 = 900 o c. ssumptions: (i) (ii) The diffusion co-efficient values are accurate. Q d is independent of temperature. d. Solution: (a) Step 1: Formulate two simultaneous equations with Q d and D 0 as the two unknown variables using equation 5.9a lnd 1 = lnd 0 Qd R 1 T1 (1) lnd 2 = lnd 0 Qd R 1 T2 (2) Step 2: Express temperatures in Kelvin. T 1 = 923 K ; T 2 = 1173 K

7 Step 3: Solve for Q d Subtracting (2) from (1): lnd lnd = 1 2 Q d 1 1 * R T1 T2 Q d d = R lnd1 lnd 1 1 T T [ ] K K Q = (8.31 J/mol- K) ln ( ) ln ( ) Q d = 196,700 J/mol Step 4: Solve for D 0 Substituting the value of lnd 0 = lnd D 0 1 Qd + R 1 T1 Qd in (1) and re-arranging: ,700 J/mol = ( m /s) exp (8.31 J/mol - K)(923 K) D 0 = m 2 /s (b) Substituting the above values of D 0 and Q d in D = D 0 exp (-Q d /RT) D = ( m 2 196, 700 J/mol /s)exp (8.31 J/mol- K)(1148 K) D = m 2 /s

8 (a) Q d = 196,700 J/mol D 0 = m 2 /s (b) D = m 2 /s

9 Problem 5.23 a. To Find: ctivation energy, Q d and prponeeexntial, D 0 for the diffusion of Fe in r. b. Given: Plot of log 10 D versus 1000/T. c. ssumptions: (i) (ii) Q d is independent of temperature. The plot is accurate. d. Solution: Step 1: ln D = ln D 0 (Q d /RT) (1) Step 2: log D = log D 0 (Q d /2.303RT) The slope of the log D vs 1/T curve equals Q d /2.303 R. We could read the plot of log 10 D versus 1000/T as log 10 D versus 1/T if we multiplied the values on the 1000/T axis with There are a certain points to keep in mind while reading the plot, as mentioned in Example Problem 5.5: (i) (ii) The actual D values and not the log D values are noted on the vertical axis. The y-axis is on a log scale, so, a point midway between and is and not 5*10-16

10 Measured slope of log10d versus 1/T plot = = [ log (10-15 ) log ( (log D ) log D1 log D2 = T1 T2 T -3 ) ] / [ (0.614) (0.642) * 10 ] =[ ] / [ -2.8 * 10-5 ] = [ Note - If this line is extrapolated, it strikes the 1000/T axis at If this value is used to -15 calculate the slope, slope = [ log (10 ) log ( ) ] / [ (0.614) (0.671) * 10 ] = Depending on which points were chosen and how the values were read off the plot, it s possible for the calculated slope values to vary from person to person if the procedure is ok, it s ok. ]

11 Step 3: From the above steps, Q d /2.303 R = => Q d = * * = J/mol or 3.something * 10 5 J/mol Step 4: Using D 0 = D exp Q d D 0 = exp (341913*0.614*10-3 /8.314) D 0 = 9.25 * 10-5 m 2 /s [Note If a different value of Q d is used, the value of D 0 would be significantly different. If the calculated value of slope is taken to be , Q d = J/mol and D 0 = 5.93 * If the answer is of the same order and if the procedure is correct, it s ok. ] Q d 3 * 10 5 J/mol D 0 of the order of 10-5 m 2 /s

12 Problem 5.D2 a. To Find: The temperature/ temperature range and thickness/thickness range for which the purification of gas, as described in the problem, can be achieved. b. Given: 25.0 kj/mol = 200 p 2 exp = kj/mol p 2 exp Where the partial pressure values are in MPa. D (m 2 /s) = kj/mol exp D (m 2 /s) = exp 24.0 kj/mol RT p 2 on the high-pressure side = MPa (0.5 atm) p 2 in the enriched mixture = MPa (0.2 atm) p 2 on the high-pressure side = MPa (0.5 atm) p 2 in the enriched mixture = MPa (0.1 atm) c. ssumptions: (i) (ii) Linear concentration profile, i.e., steady state diffusion condition. One-dimensional diffusion, i.e. diffusion occurs only along the axis parallel to the thickness of the plate. (iii) J = 2.0 J

13 d. Solution: If it is assumed that J = 2.0 J, Only then, J 1 = x (200)( MPa kj MPa 10 7 m kj /s)exp 25.0 ) exp (4.0 = 2.0 J 2.0 = x ( kj )( MPa MPa)exp ( m kj /s)exp T = 395 K The x term always cancels out and hence, the purification occurs for all values of the thickness of the sheet. T = 395 K Purification occurs independent of the thickness of the sheet If no relation is assumed between Ja and Jb, it is possible to say that: ased on the partial pressure values given, it is always possible to calculate Ja and Jb for all temperatures and all thicknesses>0. s long as the 'concentrations' are reaching the desired values, we don't really need to compare Ja and Jb. Thus, given that diffusion is steady state, it appears that there are no restrictions on the values of temperature or thickness for the purification to take place.