{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Separation Process Principles- 2n - Seader & Henley - Solutions Manual

37 subject bulk phase desublimation of benzoic acid

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: , simplifying, separating variables, and integrating both sides, − rs ri r ln k (T − T ) ri dr = c m c r ρc ∆H f t dt (4) r2 r2 r2 r 1 r − ln r ri + = s ln s + ( ri 2 − rs2 ) s 2 4r 2 ri 4 (5) 0 The left side of (4) becomes, r2 r ln ri − r ln r ]dr = ln ri [ rs 2 ri ri rs Combining (4) and (5) and integrating the right side gives (1). ri s Exercise 17.33 Subject: Zone melting with a single or partial pass. Given: A crystal layer undergoing zone melting to remove impurities. Assumptions: Melt zone of width, l, is perfectly mixed with impurity concentration, w. Diffusion of the impurity does not occur in the solid phase. Initial impurity concentration is uniform at wo. Impurity concentration in the melt zone is in equilibrium with that in the solid phase upstream of the melt zone. Find: Derive an expression for the average impurity concentration over a particular length of crystal layer, z2 – z1, after one pass or partial pass of zone melting. Also calculate for Example 17.14, wavg for z1 = 0 and z2/l = 9. Analysis: The given expression to be derived is: wavg = wo where, K= l (1 − K ) K ( z2 − z1 ) exp − z2 K zK − exp − 1 l l +1 (1) impurity concentration in the solid phase impurity concentration in the melt phase Using the definition of wavg with (17-73), z2 wavg = z1 ws dz z2 − z1 z2 = z1 wo 1 − (1 − K ) exp − Kz l dz (2) z2 − z1 Integrating (2) and applying the limits, wavg = wo l (1 − K ) l (1 − K ) z2 − z1 zK zK + exp − 2 − exp − 1 z2 − z1 K ( z2 − z1 ) l K ( z2 − z1 ) l which simplifies to (1). From Example 17.14, K = 0.36 and wo = 0.01. From above, (z2 – z1)/l = 9 Substitution into (1) gives, wavg = 0.01 (1 − 0.36 ) 0.36 ( 9 ) exp [ −9(0.36] − exp ( 0 ) + 1 = 0.0081 , Exercise 17.34 Subject: Zone melting with a single or partial pass. Given: A crystal layer undergoing the zone melting of Example 17.14, where the last 20% of the crystal layer is removed following the first pass to z/l = 9. Assumptions: Melt zone of width, l, is perfectly mixed with impurity concentration, w. Diffusion of the impurity does not occur in the solid phase. Initial impurity concentration is uniform at wo. Impurity concentration in the melt zone is in equilibrium with that in the solid phase upstream of the melt zone. Find: The average impurity concentration in the remaining crystal layer using the expression derived in Exercise 17.33. Analysis: The expression in Exercise 17.33 is: wavg = wo l (1 − K ) K ( z2 − z1 ) exp − z2 K zK − exp − 1 l l +1 where, from Example 17.14, K = 0.36 , wo = 0.01 , z1 = 0 , l/L = 0.1, and given z2 = 0.80 . Therefore, and, l l/L 0.1 0.1 = = = z2 − z1 z2 / L − z1 / L 0.8 − 0 0.8 z2 z2 / L 0.8 = = =8 l l / L 0.1 Substitution into (1) gives, wavg = 0.01 (1 − 0.36 ) (0.1) 0.36 ( 0.8 ) exp [ −8(0.36] − exp ( 0 ) + 1 = 0.0079 (1) Exercise 17.35 Subject: Zone melting with a single pass. Given: A bar of 98 wt% Al with 2 wt% Fe impurity subjected to one pass of zone refining. K = 0.29 for the impurity. The resulting bar is cut off at z2 = 0.75 z and z/l =10. Assumptions: Melt zone of width, l, is perfectly mixed with impurity concentration, w. Diffusion of...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern