Principles of Food and Bioprocess Engineering (FS 231) Solutions to Example Problems on Freezing

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Darlene Newton
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1 Principles o Food and Bioprocess Engineering (FS 21) Solutions to Example Problems on Freezing 1. In order to determine the a to be used in the above equation, e recall that Plank s equation is valid or 1-D reezing only. Thus, the sides o the container that are 80 cm and 90 cm long are considered to be ininitely long, and hence a = 5 cm = 0.05 m For a slab, P' = 1/2, R' = 1/8 2 = 1,100 kg/m, usion (product) = 0.8 (.5 kj/kg) = kj/kg, h = 140 W/m K k = 1.6 W/m K (make sure that the properties used are or the rozen product and not liquid) T F= -1.5 C, T a = -0 C Thus, t = 850 s (= 64 min) 2. We use the olloing equation (Plank s equation) to determine the reezing time: Here, = 1,000 kg/m, k = 1.5 W/m K, usion (product) = 0.75 (.5) = kj/kg The product can be considered to be an ininite cylinder o radius cm (a = 6 cm = 0.06 m) since the height (= 65 cm) is large (greater than ten times) in comparison ith the diameter. Thus, P' = 1/4, R' = 1/16. Using our knoledge on unsteady state heat transer, e kno that as long as N Bi > 40 (here, N Bi = 75), e can assume that the resistance to heat transer at the surace is negligible (i.e., h = ). Also, T F = 271 K, T = -5 C = 28 K Substituting these values, e get: t = 1,17 s. T = 27 K, T' = 271 K, usion (solvent) = 6,00 J/mol, R g = 8.14 J/mol K Thus, X = 0.98 = (500/18)/[(500/18) + (m s/40)] Thus, m = g s 4. We irst determine the initial reezing temperature o the solution as ollos: In the above equation, Here, m = 1,500 g, M = 18 g, m s = 60 g, M s = 45. Thus, e get: X = 0.984

2 Substituting X = 0.984, T = 27 K, R g = 8.14 J/mol K, usion (solvent) = 6,00 J/mol, yields: T' = K. This is the initial reezing point (= T F) We then use the olloing equation (Plank s equation) to determine the reezing time: We assume that the product is an ininite cylinder and hence, P' = 1/4, R' = 1/16 2 a = 0.05 m, h = 120 W/m K, k = 0.6 W/m K, = 1,000 kg/m, T a = 2 K usion (product) = (1500/1560) (.5 x 10 ) = 20.7 kj/kg This yields: t = 045 s (= 51 min) 5. We use the olloing equation: usion (solvent) = 6,00 J/mol, R g = 8.14 J/mol K, T = 27 K, T' = 271 K This yields, X = 0.98 Considering a 100 g solution, e have m = 90 g, m = 10 g. Also, M = 18 g. s Substituting these values in:, yields: M = 98 g s 6. The irst method involves the use o the Clausius-Clapeyron equation given belo: ith Here, m = 600 g, M = 18 g, m s = 400 g, M s = 0 g Thus, e get: X = 0.71 Substituting, usion (solvent) = 6,00 J/mol, R g = 8.14 J/mol K, T = 27 K, X = 0.71, e get: T = K Thus, depression in reezing point = = K = 1.2 C The second method involves the use o the olloing equation: Here, m = number o moles o solute per kg o solvent = (400/0)*(1000/600) = 22.2 Also, k = 1.86 C/mol Thus, T = (1.86)(22.2) = 41. C There is a dierence in the ansers obtained by the to methods because the second method is not valid in this case since e do not have a dilute solution. 7. We use the olloing equation (Plank s equation) to determine the reezing time:

3 Here, = 950 kg/m, k = 0.5 W/m K, usion = 0.8 (.5) = kj/kg The product can be considered to be an ininite slab o thickness 5 cm (= a) since the other to dimensions are large in comparison to this dimension. Thus, P' = 1/2, R' = 1/8. Since the Biot number is 100, e can assume that the resistance to heat transer at the surace is negligible (i.e., h = ). Also, T F= 271 K, T a = -40 C = 2 K Substituting these values, e get: t = 4169 s 8. a. In order to determine the reezing point o the solution, e need to use the Calusius- Clapeyron equation given belo: (1) ith, (2) In equation 2, m = 2 kg, M = 18 g, m s = 1 kg, M s = 110 g Thus, X = In equation 1, usion (solvent) = 6,00 J/mol, R = 8.14 J/mol K, T = 27 K Substituting, e get: T' = 265 K (= - 8 C) b. I NaCl (molecular eight = 58.5 g) as used instead o CaCl 2, X ould be loer than and hence T' ould be loer than -8 C. 9. We begin by using Plank s equation (given belo): In this problem, e do not truly have an ininite cylinder (since height 10*diameter). Hoever, it is close enough that e can use Plank s equation and e ill be able to obtain an estimate o the reezing time. Here, P' = 1/4, R' = 1/16 2 a = 0.1 m, h = 500 W/m K, k = 0.5 W/m K, = 950 kg/m usion (product) = (0.9) (.5 kj/kg) = kj/kg = 00,150 J/kg R g = 8.14 J/mol K T = - 5 C a Thus, e can solve or the reezing time i e kno the initial reezing point, T F. The initial reezing point can be determined using the Clausius-Clapeyron equation (given belo): (1)

4 = 6,00 J/mol, R = 8.14 J/mol K usion (solvent) g and (2) Since the product has 10% solids, the remaining 90% is ater. In computing, X, e only need the ratio o mass o ater to mass o solids and do not actually need the exact amount o solids and ater. I e consider the total product eight to be 100 g, then: m = 90 g, m s = 10 g No, M = 18 g, M s = 50 g Substituting these values in equation 2, e get: X = 0.96 Substituting this in equation 1, e get: T' = K = -4.1 C (= T F) Substituting this in Plank s equation, e get: t = 11,996 s (= ~200 min = ~. hrs) 10. Consider a 100 g product. Initially, the product has 90 g ater and 10 g solids. Ater 40 % o this ater has rozen, the product has 54 g ater (40 % o 90 g = 6 g; 6 g o ater is no rozen, thereby leaving 54 g o ater) and 10 g o solids The mole raction o ater in solution is given by: We then use the olloing equation (Clausius-Clapeyron equation): Here, usion (solvent) = 6,00 J/mol, R g = 8.14 J/mol K Solving, e get: T' = K = C 11. We begin by using Plank s equation (given belo): In this problem, e have an ininite cylinder since the height is at least 10 times the diameter. Hence, the value o a in this problem is 0.08 m. Thus, P' = 1/4, R' = 1/16 2 h = 250 W/m K, k = 0.6 W/m K, = 850 kg/m usion = (0.75) (.5 kj/kg) = kj/kg = 250,125 J/kg R g = 8.14 J/mol K T = - 0 C = 24 K a Thus, e can solve or the reezing time i e kno the initial reezing point, T F. The initial reezing point (T ) can be determined using the Clausius-Clapeyron equation (given F

5 belo): (1) = 6,00 J/mol, R = 8.14 J/mol K usion g ith (2) Since the product has 25% solids, the remaining 75% is ater. In equation 2, m = 0.75, M = 18 g, m s = 0.25, M s = 175 g (Note that e do not need the actual values o the mass o solids & ater, but only the ratios hence e use m = 0.75 and m s = 0.25) Thus, X = 0.97 In equation 1, usion = 600 J/mol, R = 8.14 J/mol K, T = 27 K Substituting, e get: T' = K = T F Substituting this in Plank s equation, e get: t = 5,901 s (= ~1.64 h) 12. We begin ith the Clausius-Clapeyron equation: Here, T = 27 K, T = 268 K, usion (solvent) = 600 J/mol, R g = 8.14 J/mol K Solving, e get: X = No, Here, m = 2000, M = 18, M s = 60 Solving, e get: m =.6 g s We then use the Plank s equation (belo) to determine the reezing time: Here, = 975 kg/m k = 0.45 W/m K usion (product) = m.c. o product ( usion (ater)) = [200/( )]{.5 x 10 }J/kg = x 10 J/kg T F = -5 C = 268 K T a = -40 C = 2 K h = 200 W/m 2 K

6 Since to o the dimensions o the tray are more than 10 times the third dimension, e can approximate the tray to be an ininite slab. For an ininite slab, a = thickness = 0.04 m, P' = ½, R' = 1/8 Substituting these values, e get: t = 41 s

Principles of Food and Bioprocess Engineering (FS 231) Exam 2 Part A -- Closed Book (50 points)

Principles of Food and Bioprocess Engineering (FS 231) Exam 2 Part A -- Closed Book (50 points) 1. Are the following statements true or false? (20 points) a. Thermal conductivity of a substance is a measure