FS 231: Final Exam (5-5-03) Part A (Closed Book): 60 points

Transcription

1 Name: Start time: End time: FS 231: Final Exam (5-5-03) Part A (Closed Book): 60 points 1. Which of the following statements are true for a vapor compression refrigeration cycle operating under saturated conditions? EXPLAIN your answer. (3 points) a. The temperature of the refrigerant at the inlet and exit of the condenser are the same b. The temperature of the refrigerant at the inlet of the condenser is higher than that at the exit c. The temperature of the refrigerant at the inlet of the condenser is lower than that at the exit 2. Is the energy given out in the condenser greater than, equal to, or less than the energy absorbed in the evaporator? EXPLAIN. (2 points) 3. What happens to the Pressure, Temperature and Enthalpy of the refrigerant as the refrigerant goes from the inlet to the exit of each of the components in a refrigeration system operating under ideal conditions? (6 points) Compressor Condenser Expansion Valve Evaporator Pressure Temperature Enthalpy 4. Are the following statements true or false? (4 points) a. The freezing point of a solution (containing solids) is always less than the freezing point of the pure solvent. b. Rapid freezing results in extensive structural damage to the food product. c. Plank s equation is valid only when freezing is taking place from all three directions when dealing with a brick-shaped product. d. Slow freezing prevents the sugars in a popsicle from moving towards the center

2 5. a. List one advantage of using Freon as the refrigerant instead of Ammonia. (2 points) b. List one advantage of using Ammonia as the refrigerant instead of Freon. (2 points) 6. Under atmospheric conditions why does dry ice sublimate while regular ice melts? What conditions need to be modified in order to have dry ice melt? (3 points) 7. For most heating purposes, steam at high pressure (say 500 kpa) is desired. However, for evaporation, relatively low pressure (150 kpa) steam is often used. Why? (3 points) 8. In a multiple-effect evaporator, is the pressure in an evaporation chamber downstream lower or higher than the one upstream. Why? (3 points) 9. As the process of evaporation proceeds in an evaporator, and the solution gets more and more concentrated, will the rate of heat transfer increase or decrease? Why? (4 points) 10. In a psychrometric chart, constant dew-point temperature lines are the same as constant lines. (2 points) 11. In a psychrometric chart, constant enthalpy lines are the same as constant lines. (2 points)

3 12. In an adiabatic process, what thermodynamic properties remain constant? (3 points) 13. Are the following combinations possible? (2 points) a) Wet bulb temperature = 20 C & Dry bulb temperature = 15 C b) Dew-point temperature = Wet bulb temperature = Dry bulb temperature 14. If air is heated using a heater (similar process to that in the spray drying lab), what happens to: a. Specific volume b. Dew-point temperature (2 points) 15. Explain the following terms: (6 points) a. Blanching b. Pasteurization c. Sterilization 16. Explain the following terms: (5 points) a. D-value b. z value c. F value d. F value e. C value 0

4 17. a. What is an aseptic process? (3 points) b. What are the components of an aseptic processing system? (3 points) c. List 3 advantages and 3 disadvantages of aseptic processing. (3 points) d. What is the critical point in a multi-particle food product? (2 points)

5 Name: Start time: End time: FS 231: Final Exam (5-5-03) Part B (Open Book): 40 points Answer ANY 4 out of the following 5 questions 18. An ammonia refrigeration system is operating under ideal conditions. The refrigerant is 15% in vapor phase and at -24 C as it exits the expansion valve. If the cooling load is 2 tons, determine: a. Mass flow rate of refrigerant b. C.O.P. of the system (12 points) 19. A certain food product containing 75% water is placed in a metallic cylindrical container of height 65 cm and radius 3 cm. The thermal conductivity and specific heat of the unfrozen product are 0.4 W/m-K and 3500 J/kg-K and those of the frozen product are 1.5 W/m-K and J/kg-K respectively. The density of the product is 1000 kg/m. The initial temperature of the product is 5 C and its initial freezing point is -2 C. If the product is exposed to cold air at -35 C, how long will it take to freeze the product? The Biot number in this case is 75. (11 points) 20. A single effect evaporator is used to evaporate 90 kg of vapor every hour from a food product that contains 25% solids. The food product enters the evaporator at 10 C and the evaporator operates at 70 C. Steam (90% quality) enters the evaporator at the rate of 125 kg/hr and at kpa and exits the evaporator at 90 C. Determine the solids content of the final product if the specific heat of the feed and final product are 4000 J/kg-K. (12 points) 21. A spray dryer is used to dry milk containing (90% water). The mass flow rate of milk is 2 kg/hr and the mass flow rate of the hot air (dry bulb temperature of 90 C) is 85 kg/hr. If the hot air is obtained by using a heater (similar to the one used in the spray drying lab) to heat ambient air (dry bulb temperature of 25 C and a relative humidity of 50%), determine the relative humidity of the exit air. Make any necessary assumptions. (11 points) a. There are 10 spores (D 250 = 0.2 min, z = 10 C) in each of the unprocessed canned foods in a plant. If the spores receive a heat treatment of 268 F for 12 s, what is the F 0 value of the process? (5 points) b. What is the probability of spoilage of that lot of processed cans? (10 points)

6 FS 231 Final Exam (Solutions) Part A 1. B The compressor raises the temperature and pressure of the refrigerant so that it promotes heat transfer between the refrigerant and the surroundings in the condenser. During the process of compressing the refrigerant, the refrigerant is superheated. In the condenser, the refrigerant initially loses its degree of superheat and then condenses. Thus, the refrigerant is at a lower temperature at the end of the condenser than it was at the beginning. 2. The energy released by the refrigerant in the condenser is greater than that absorbed in the evaporator since the energy released in the condenser is the sum of the energy absorbed in the evaporator and the compressor. 3. Pressure Temperature Enthalpy Compressor Condenser Expansion Valve Evaporator 4. a. True b. False c. False d. False 5. a. Freon systems operate at a lower pressure difference than an Ammonia system and are hence easier to maintain. Freon is non-toxic and not flammable. b. Ammonia systems have the capability to extract more heat from a product due to the high latent heat of vaporization of Ammonia. 6. The triple point of regular ice is 0.01 C and the corresponding pressure is atm. The triple point of dry ice is C and the corresponding pressure is 5.1 atm. Since the pressure corresponding to the triple point of dry ice is higher than atmospheric pressure, it goes directly from solid to vapor phase, unlike regular ice. In order to have dry ice melt, we have to increase the pressure to above 5.1 atm. 7. In evaporation, the product is subjected to a decrease in pressure and hence it boils at a lower temperature than at atmospheric pressure. Thus, steam at a relatively lower pressure is sufficient to cause evaporation. In addition, it is necessary to minimize the heat treatment to the product, since most of the products undergoing evaporation are sensitive to high heat.

7 8. Lower. Since the boiling point of the concentrated product that enters the second effect is higher than that in the first effect, the pressure should be lower so as to facilitate boiling of the solution. Also, if the vapor produced in the first effect is used as the heating medium for the second effect, the pressure has to be lowered in order that the product in the second effect boil at temperatures lower than that of the vapor produced in the first effect. 9. Decrease. As the solution gets more and more concentrated, its boiling point increases. Thus, the temperature difference between the steam and product decreases and hence the rate of heat transfer decreases. In addition, with time, the walls of the heat exchanger may be fouled -- product burning at the surface. 10. Humidity ratio (or moisture content) 11. Wet bulb temperature 12. Enthalpy, wet bulb temperature 13. a. No b. Yes 14. a. Increases b. Remains constant 15. a. Blanching is the mild heat treatment of a food product to inactivate enzymes, reduce microbial load, wilt vegetables (to facilitate packaging), and to extend shelf life. b. Pasteurization refers to the heat treatment of a product to kill majority of the pathogenic organisms present in the food. Several spoilage microorganisms survive this process and hence the food should be refrigerated. c. Sterilization refers to the heat treatment of a product to (theoretically) destroy all organisms (vegetative cells and spores) present in the food. However, no food product is completely sterilized; instead they are rendered commercially sterile wherein all microorganisms of public health significance and spoilage microorganisms capable of growing at room temperature are destroyed. 16. a. It is the time required at a specific temperature for an order of magnitude change in the number of microorganisms. b. It is the temperature change required for an order of magnitude change in D. c. It is the time of processing at a reference temperature and reference z value that is equivalent to the process under consideration. d. It is the F value when the reference temperature is C and the z value is 10 C. e. It is the same concept as the F value, but applied to texture or nutrient retention attributes.

8 17. a. An aseptic process is one in which the product and package are sterilized separately and brought together in a sterile environment, with the resulting product being shelf-stable. b. The components of an aseptic processing system are pump, deaerator, heat exchanger, holding tube, cooling unit, back pressure valve, packaging unit, and aan aseptic surge tank. c. Advantages: Easy automation, unlimited package size, cheaper packaging, better product quality, less energy consumption, fewer operators, less space requirements, no need for refrigeration Disadvantages: Slower filler speeds, higher overall cost, need for better trained personnel, need to control ingredient quality better, more stringent process validation procedures d. The critical point in a multi-particle food product is the center of the slowest heating particle; it may not necessarily be the center of the fastest moving particle.

9 a. Cooling load = Q e = 2 tons = 2 (3516.8) W = W = ṁ(h 2 H 1 ) = ṁ ( ) x 10 3 Thus, ṁ kg/s t f ρλ fusion T F T " P D h FS 231: Final Exam (Solutions) Part B 18. We begin by locating the enthalpy of the refrigerant at the exit of the condenser, H (same as 1 at inlet of expansion valve). H = 0.15 (h ) (h ) = 0.15 ( ) (90.537) = kj/kg 1 g at -24 C f at -24 C We note that this value of enthalpy corresponds to a condenser pressure of ~850 kpa (by looking up Table A.6.2 on page 621 of textbook and finding the pressure at which h f = kj/kg) H = h = kj/kg 2 g at -24 C We can identify the condition of the refrigerant at the exit of the evaporator (point A ) by locating the point corresponding to -24 C on the saturated vapor line (page 623 of text). In order to determine the conditions of the refrigerant at the exit of the compressor, we follow a constant entropy line from point A till we intersect a horizontal line corresponding to 850 kpa. We note that the enthalpy at this point is 1675 kj/kg (= H ). 3 b. C.O.P. = (H - H )/(H - H ) = ( )/( ) Thus, C.O.P. = We use the following equation (Plank s equation) to determine the freezing time: R D 2 k 3 Here, ρ = 1000 kg/m, k = 1.5 W/m-K, λ fusion = 0.75 (333.5) = kj/kg The product can be considered to be an infinite cylinder of radius 3 cm (= D) since the height (= 65 cm) is large (greater than ten times) in comparison with the radius. Thus, P' = 1/4, R' = 1/16. Since the Biot number is 75, we can assume that the resistance to heat transfer at the surface is negligible (i.e., h = "). Also, T = 271 K, T = -35 C = 238 K F " Substituting these values, we get: t = 1,137 s f

10 20. The system diagram is as follows: a. Overall mass balance equation: m f = 90 + mp Total solids balance equation: 0.25 (m f) = x (m p) Energy balance equation: m f (c p(f) ) ( ) + m s (H s at kpa - H c at 90 C) = m v (H v at 70 C) + m p (c p(p) ) ( ) H = 0.9 (H ) (H ) = 0.9 (2660.1) (376.92) = kj/kg s v at kpa c at kpa Substituting the values, we get: 3 3 m (4000)( ) ( ) x 10 = 90 ( x 10 ) + m (4000)( ) f Solving the above three equations for the three unknowns, we get: m f = kg/hr, m p = kg/hr, and x = (= 47.5%) 21. We begin by identifying the conditions of ambient air by intersecting the dry bulb temperature line of 25 C with a relative humidity line of 50%. This is point A on the psychrometric chart. From this point, we draw a horizontal line and intersect it with the vertical line corresponding to a dry bulb temperature of 90 C (point B on chart). The humidity ratio at point A and B is 0.01 kg water / kg dry air (= W ). B In order to identify the condition of the exit air, we perform a water balance as follows (with the assumption that no moisture leaves with the dried milk powder): 0.9 (2) + 85 (W B) = 85 (W C) Substituting W B= 0.01, we get W C= kg water / kg dry air From point B, we follow a constant enthalpy line till we intersect a horizontal line corresponding to a humidity ratio of This represents the conditions of the exit air (point C on the chart). We note that the relative humidity at point C is 65 %. p

11 22. a. We start off with the following equation: F 0 10 (TT ref )/z t Here, T = 268 F, T ref = 250 F, z = 10 C (= 18 F), t = 12 s Thus, F 0 = 120 s (1) b. In order to determine the probability of spoilage, we need to determine the final microbial count. We do so as follows: N N 0 10 t/d T (2) We know the D value at 250 F and need to find the D value at the process temperature of 268 F. We use the following equation for that: D 268 D Thus, D = 0.02 min = 1.2 s 268 (3) Thus, from equation (2), we get: N /1.2 = 10-6 This implies that we have 10 spores per can OR 1 spore in every 10 cans. Thus, the -6 6 probability of spoilage is 1 in a million.