Freeze drying is a technique for dehydrating substances at low temperatures, thereby avoiding the degradation that may accompany heating. The material to be dried is cooled to a temperature at which all of the water present turns to ice. The frozen substance is then placed in a vacuum chamber and may also be subjected to radiant or microwave heating; the ice in the food sublimes, and the vapor is carried off by the vacuum pump.
Steaks are to be freeze-dried in a heated chamber at 1 torr (1 mm Hg). The steaks, which contain 72% water by mass, enter the chamber at —26°C at a rate of 50 kg/min. Of the water entering with the steaks, 96% leaves as a vapor at 60°C; the remainder leaves as a liquid with the steaks at 50°C.
- Use the heat capacity data given below and additional tabulated data for water to calculate the required heat input in kilowatts.
- When large temperature changes are not involved in a phase-change operation, a reasonable estimate of the required heat transfer rate may be obtained by neglecting contributions of temperature changes to die overall process enthalpy change (i.e., by taking only phase changes into account). Moreover, it is often reasonable to use any available values of latent heats, neglecting their dependence on temperature and pressure. In the case of the freeze-drying process, the approximation might be to calculate only the heat needed to melt all die water and vaporize 96% of it, using latent heats at the normal melting and boiling points (Table B.l) and neglecting the heat required to raise the temperature of the meat and water. What percentage error in the calculated value of Q would result from this approximation? Take the value determined in Part (a) to be exact. (c) Many substances, such as food and drugs, spoil if exposed too long to high temperatures (which accelerate rates of degradation) or to liquid water (which provides an environment for growth of microbial species that cause degradation). Also, rates of evaporation and sublimation increase as
temperature increases and pressure decreases. Use those observations to construct a one- paragraph explanation of how freeze-drying works and the reason for each step of the process. (For example, why is the sublimation done in a vacuum chamber?) Your explanation should be clear to someone with a nontechnical or nonscientific background.
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- □ 40 a' 12 o' a O d'arrow_forwardTL = 85 D. Problem 9.12 The top view of a line measures 60 mm. The line is parallel to the V.P. and inclined at 45° to the H.P. One end of the line is 25 mm in front of the V.P. and lies on the H.P. Draw its projections and determine the true length. Interpretation Let the line be PQ parallel to the V.P. The front view has true length and the top view is parallel to xy. Construction Refer to Fig. 9.12. 1. Draw a reference line xy. Mark point p' on xy and point p 25 mm below xy. 2. Draw a 60 mm long line pq parallel to xy. This repre- sents the top view. 3. Draw line from point p', inclined at 45° to xy to meet the projector from point q at point q'. Join p'a' to represent the front view. Measure length of p'a' as true length of line PQ. Here T.L. = 85 mm. Result True length of line PQ is p'q' = 85 mm. 5 Fig. 9.12 p 60 σarrow_forwardProblem 9.13 A 70 mm long line PQ does not have H.T. and V.T. One end of the line is 30 mm in front of the V.P. and 20 mm above the H.P. Draw its projections. Interpretation As the line PQ does not have H.T. and V.T., it is parallel to both H.P. and V.P. Construction Refer to Fig. 9.13. 1. Draw a reference line xy. Mark point p' 20 mm above xy and point p 30 mm below xy. 2. Draw a 70 mm long line p'a' parallel to xy to repre- sent the front view. X 20 p Fig. 9.13 3. Also, draw a 70 mm long line pq parallel to xy to represent the top view. 70 q yarrow_forwardProblem 10.19 A square lamina ABCD of side 40 mm is suspended from a point O such that its surface is inclined at 30° to the V.P. The point O lies on the side AB 12 mm away from A. Draw its projections. Construction Refer to Fig. 10.19. 1. First stage Draw a square a'b'c'd keeping a'd' parallel to xy. Mark a point o' on a'ď at a distance 12 mm from end a' as the point of suspension. Also, mark the centre of the square g' to represent the centre of gravity. 2. Second stage Reproduce the front view of first stage such that o'g' is perpendicular to xy. Project corners and obtain bd as the top view. 3. Third stage Reproduce the top view keeping bd inclined at 30° to xy. Obtain new points a', b', c' and ď' in the front view by joining the points of intersection of the vertical projectors drawn from points a, b, c and d of the third stage with the corresponding horizontal locus lines drawn from points a', b', c' and ď of the second stage. Join new a'b'c'd to represent the final front view.arrow_forwardProblem 10.15 A circular plane of diameter 50 mm is resting on a point of the circumference on the V.P. The plane is inclined at 30° to the V.P. and the centre is 35 mm above the H.P. Draw its projections.arrow_forwardYou are asked to manufacture 10 kg of polyester with a number-average molecular weight of 1000 by polymerizing butane-1,4-diol(HO(CH2)4OH) with adipic acid (HOOC-(CH2)4-COOH).a) What weight of diol and diacid do you need, respectively? To whatextent, p, should the reaction be carried out to? Assume a stoichiometricbalance.b) What are the number and weight fractions of dimer, trimer and tetramerat this point in the reaction?c) Because of the polymerization by dehydration to olefin, 3 mol% of thediol will be lost. What would be the number-average molecular weightwhen the reaction is carried out to the same extent? How could you offsetthis loss so that the desired molecular weightarrow_forward9.4. A PID temperature controller is at steady state with an output pressure of 9 psig. The set point and process temperature are initially the same. At time = 0, the set point is increased at the rate of 0.5°F/min. The motion of the set point is in the direction of lower temperatures. If the current settings are PART 3 LINEAR CLOSED-LOOP SYSTEMS Ke = 2 psig/°F Ti = 1.25 min TD = 0.4 min plot the output pressure versus time.arrow_forward9.6. A PI controller has the transfer function Determine the values of K, and T. 5s + 10 Ge Sarrow_forward9.5. The input & to a PI controller is shown in Fig. P9-5. Plot the output of the controller if Ke 2 and 0.50 min. - E 0.5 0 -0.5 0 FIGURE P9-5 2 4 t, minarrow_forward9.3. An ideal PD controller has the transfer function P Ke (TDs + 1) E An actual PD controller had the transfer function P = Ke E TDS +1 (TDIẞ)s +1 where ẞis a large constant in an industrial controller. If a unit-step change in error is introduced into a controller having the second transfer function, show that P(1) = Kc (1 + Ae˜¯BD) where A is a function of ẞwhich you are to determine. For ẞ=5 and K = 0.5, plot P(t) versus tl tp. As ẞ, show that the unit-step response approaches that for the ideal controller.arrow_forward9.1. A pneumatic PI temperature controller has an output pressure of 10 psig when the set point and process temperature coincide. The set point is suddenly increased by 10°F (i.e., a step change in error is introduced), and the following data are obtained: Time, s psig 0- 10 0+ 8 20 7 60 90 5 3.5 Determine the actual gain (psig per degree Fahrenheit) and the integral time.arrow_forward2. A unit-step change in error is introduced into a PID controller. If Ke TD = 0.5, plot the response of the controller P(t). = =10, 1, andarrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
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