### Educational Resource: Diffusion in Iron-Carbon Alloys**Overview:**An iron-carbon alloy initially containing 0.255 wt% C is exposed to an oxygen-rich and virtually carbon-free atmosphere at 1130°C. Under these circumstances, carbon diffuses from the alloy and reacts at the surface with the oxygen in the atmosphere; thus, the carbon concentration at the surface position is maintained essentially at 0.00 wt% C. The objective is to determine the position at which the carbon concentration will be 0.191 wt% after a 6-hour treatment. The diffusion coefficient (D) at 1130°C is given as 6.0 × 10⁻¹⁰ m²/s.**Table Explanation:**The table below provides values of the error function (erf(z)) for various values of the variable z.| z | erf(z) | z | erf(z) | z | erf(z) ||-------|---------|-------|---------|-----|---------|| 0.00 | 0.0000 | 0.55 | 0.5633 | 1.3 | 0.9340 || 0.025 | 0.0282 | 0.60 | 0.6039 | 1.4 | 0.9523 || 0.05 | 0.0564 | 0.65 | 0.6420 | 1.5 | 0.9661 || 0.10 | 0.1125 | 0.70 | 0.6778 | 1.6 | 0.9763 || 0.15 | 0.1680 | 0.75 | 0.7112 | 1.7 | 0.9838 || 0.20 | 0.2227 | 0.80 | 0.7421 | 1.8 | 0.9891 || 0.25 | 0.2763 | 0.85 | 0.7707 | 1.9 | 0.9929 || 0.30 | 0.3286 | 0

Given data: Initial concentration of C = 0.255 wt % Surface concentration = 0.0 wt % Temperature =…

Answered: An iron-carbon alloy initially…

Introduction to Chemical Engineering Thermodynamics

8th Edition

ISBN:9781259696527

Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Chapter1: Introduction

Section: Chapter Questions

Problem 1.1P

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Carbon tetra chloride is to be removed from a polymer solution by bubbling dry air through it at 297 K. The resulting mixture has % relative humidity of 70. It is required to remove 90% of carbon tetra chloride present by compressing to a suitable pressure and cooling to 283 K. What this pressure should be? Data: Vapor pressure of CCl4 at 297 K = 12.2 kN/m^2 and at 283 K = 6 kN/m^2
Ammonia gas is absorbed from an air stream by liquid water as a solvent. Conditions in theabsorber are 100°F and 50 psia. The liquid solution entering the top of the tower is 10.0 mol%ammonia. What is the composition of the gas stream leaving the top of the absorber if it can beassumed to be in equilibrium with the liquid phase? If the inlet gas contained 30.0 mol% ammonia,the equilibrium partial pressures of the two liquid components over a 10.0 mol% solution wouldbe 1.98 psia for ammonia and 0.45 psia for water. Determine the fraction of ammonia that wasremoved in the absorber.
EXAMPLE: Use the heat of solution data in Appendix I to determine the heat transferred per mole of entering solution into or out of (state which) a process in which 2 g mol of a 50 mol % solution of sulfuric acid at 25°C is mixed with water at 25°C to produce a solution at 25°C containing a mole ratio of 10 H₂O to 1 H₂SO4.
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2) A crystallizer is charged with 100 kg of a solution containing 25% Ba(NO;), in water. On cooling 10% of the original water present evaporates. Calculate the yield of crystals when the solution is cooled to 283 K. The solubility of mother liquor at 283 K is 7.0 kg Ba(NO3)2 100 kg total water W1 Evaporated vapor F W3 Solid crystals Ba(NO:)2 Crystallizer 100 kg solution 25% Ba(NO;)2 W2 Mother liquor
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Q1: An open tank is filled with 2ft of the top with pure methanol. The tank is tapered, as shown below. The air within the tank is stationary but cireulation of air immediately above the tank is adequate to assure a negligible concentration of methanol at this point. The tank and air space are at 77°F and 1 atm and the diffusivity of methanol in air in such conditions is 0.62 ft²/hr.Calculate the rate of loss of methanol from the tank at steady – state. Given that the partial pressure of methanol at 77°F is 135 mm Hg. d2 L-z 0 = 26.56° d1 = 4 ft Note: 1 atm = 760 mm Hg atm ft Use R = 0.7302- lbmol* R
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Cyclohexane vapor at 180°C and 5 atm is cooled and condensed to liquid then solidified to solid at 6.5°C and 1 atm write the hypothetical process path to calculate the specific enthalpy change of this process step by step the following blanks nothing that: 1-the condensation happens at 80.7°C and 1 atm 2- freezing happens at 6.5°C and 1 atm O 1. changing pressure at a constant temperature O 2. Changing the temperature at constant pressure O 3. Phase change at constant temperature and pressure O 4. the changing temperature at constant pressure O 5. phase change at constant temperature and pressure

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