Humidity from Vapor Pressure. The air in a room is at 37.8°C and a total pressure of 101.3 kPa abs containing water vapor with a partial pressure p_A = 3.59 kPa. Calculate:

Percentage and Relative Humidity. The air in a room has a humidity H of 0.021 kg H₂O/kg dry air at 32.2°C and 101.3 kPa abs pressure. Calculate:

Use of the Humidity Chart. The air entering a dryer has a temperature of 65.6°C (150°F) and dew point of 15.6°C (60°F). Using the humidity chart, determine the actual humidity and percentage humidity. Calculate the humid volume of this mixture and also calculate c_S using SI and English units.

Properties of Air to a Dryer. An air–water vapor mixture going to a drying process has a dry bulb temperature of 57.2°C and a humidity of 0.030 kg H₂O/kg dry air. Using the humidity chart and appropriate equations, determine the percentage humidity, saturation humidity at 57.2°C, dew point, humid heat, and humid volume.

Adiabatic Saturation Temperature. Air at 82.2°C having a humidity H = 0.0655 kg H₂O/kg dry air is contacted in an adiabatic saturator with water. It leaves at 80% saturation.

Adiabatic Saturation of Air. Air enters an adiabatic saturator having a temperature of 76.7°C and a dew-point temperature of 40.6°C. It leaves the saturator 90% saturated. What are the final values of H and T°C?

Humidity from Wet and Dry Bulb Temperatures. An air–water vapor mixture has a dry bulb temperature of 65.6°C and a wet bulb temperature of 32.2°C. What is the humidity of the mixture?

Humidity and Wet Bulb Temperature. The humidity of an air–water vapor mixture is H = 0.030 kg H₂O/kg dry air. The dry bulb temperature of the mixture is 60°C. What is the wet bulb temperature?

Dehumidification of Air. Air having a dry bulb temperature of 37.8°C and a wet bulb temperature of 26.7°C is to be dried by first cooling to 15.6°C to condense water vapor and then heating to 23.9°C.

Cooling and Dehumidifying Air. Air entering an adiabatic cooling chamber has a temperature of 32.2°C and a percentage humidity of 65%. It is cooled by a cold water spray and saturated with water vapor in the chamber. After leaving, it is heated to 23.9°C. The final air has a percentage humidity of 40%.

Time for Drying in Constant-Rate Period. A batch of wet solid was dried on a tray dryer using constant drying conditions and a thickness of material on the tray of 25.4 mm. Only the top surface was exposed. The drying rate during the constant-rate period was R = 2.05 kg H₂O/h · m² (0.42 lb_m H₂O/h · ft²). The ratio L_S/A used was 24.4 kg dry solid/m² exposed surface (5.0 lb_m dry solid/ft²). The initial free moisture was X₁ = 0.55 and the critical moisture content X_C = 0.22 kg free moisture/kg dry solid.

Prediction of Effect of Process Variables on Drying Rate. Using the conditions in Example 9.6-3 for the constant-rate drying period, do as follows:

Prediction in Constant-Rate Drying Region. A granular insoluble solid material wet with water is being dried in the constant-rate period in a pan 0.61 m × 0.61 m and the depth of material is 25.4 mm. The sides and bottom are insulated. Air flows parallel to the top drying surface at a velocity of 3.05 m/s and has a dry bulb temperature of 60°C and wet bulb temperature of 29.4°C. The pan contains 11.34 kg of dry solid having a free moisture content of 0.35 kg H₂O/kg dry solid, and the material is to be dried in the constant-rate period to 0.22 kg H₂O/kg dry solid.

Drying a Filter Cake in the Constant-Rate Region. A wet filter cake in a pan 1 ft × 1 ft square and 1 in. thick is dried on the top surface with air at a wet bulb temperature of 80°F and a dry bulb temperature of 120°F flowing parallel to the surface at a velocity of 2.5 ft/s. The dry density of the cake is 120 lb_m/ft³ and the critical free moisture content is 0.09 lb H₂O/lb dry solid. How long will it take to dry the material from a free moisture content of 0.20 lb H₂O/lb dry material to the critical moisture content?

Numerical Integration for Drying in Falling-Rate Region. A wet solid is to be dried in a tray dryer under steady-state conditions from a free moisture content of X₁ = 0.40 kg H₂O/kg dry solid to X₂ = 0.02 kg H₂O/kg dry solid. The dry solid weight is 99.8 kg dry solid and the top surface area for drying is 4.645 m². The drying-rate curve can be represented by Fig. 9.5-1b.

Drying Tests with a Foodstuff. In order to test the feasibility of drying a certain foodstuff, drying data were obtained in a tray dryer with air flow over the top exposed surface having an area of 0.186 m². The bone-dry sample weight was 3.765 kg dry solid. At equilibrium after a long period, the wet sample weight was 3.955 kg H₂O + solid. Hence, 3.955 − 3.765, or 0.190, kg of equilibrium moisture was present. The following sample weights versus time were obtained in the drying test:

Prediction of Drying Time. A material was dried in a tray-type batch dryer using constant-drying conditions. When the initial free moisture content was 0.28 kg free moisture/kg dry solid, 6.0 h was required to dry the material to a free moisture content of 0.08 kg free moisture/kg dry solid. The critical free moisture content is 0.14. Assuming a drying rate in the falling-rate region, where the rate is a straight line from the critical point to the origin, predict the time to dry a sample from a free moisture content of 0.33 to 0.04 kg free moisture/kg dry solid. (Hint: First use the analytical equations for the constant-rate and the linear falling-rate periods with the known total time of 6.0 h. Then use the same equations for the new conditions.)

Drying of Biological Material in Tray Dryer. A granular biological material wet with water is being dried in a pan 0.305 × 0.305 m and 38.1 mm deep. The material is 38.1 mm deep in the pan, which is insulated on the sides and bottom. Heat transfer is by convection from an air stream flowing parallel to the top surface at a velocity of 3.05 m/s, having a temperature of 65.6°C and humidity H = 0.010 kg H₂O/kg dry air. The top surface receives radiation from steam-heated pipes whose surface temperature T_R = 93.3°C. The emissivity of the solid is є = 0.95. It is desired to keep the surface temperature of the solid below 32.2°C so that decomposition will be kept low. Calculate the surface temperature and the rate of drying for the constant-rate period.

Drying When Radiation, Conduction, and Convection Are Present. A material is granular and wet with water and is being dried in a layer 25.4 mm deep in a batch-tray dryer pan. The pan has a metal bottom having a thermal conductivity of k_M = 43.3 W/m · K and a thickness of 1.59 mm. The thermal conductivity of the solid is k_S = 1.125 W/m · K. The air flows parallel to the top exposed surface and the bottom metal at a velocity of 3.05 m/s and a temperature of 60°C and humidity H = 0.010 kg H₂O/kg dry solid. Direct radiation heat from steam pipes having a surface temperature of 104.4°C falls on the exposed top surface, whose emissivity is 0.94. Estimate the surface temperature and the drying rate for the constant-rate period.

Diffusion Drying in Wood. Repeat Example 9.9-1 using the physical properties given but with the following changes:

Diffusivity in Drying Tapioca Root. Using the data given in Example 9.9-2, determine the average diffusivity of the moisture up to a value of X/X_C = 0.50.

Diffusion Coefficient. Experimental drying data for a typical nonporous biological material obtained under constant drying conditions in the falling-rate region are tabulated below.

Drying a Bed of Solids by Through Circulation. Repeat Example 9.10-1 for drying of a packed bed of wet cylinders by through circulation of the drying air. Use the same conditions except that the air velocity is 0.381 m/s.

Derivation of Equation for Through-Circulation Drying. Different forms of Eqs. (9.10-11) and (9.10-12) can be derived using humidity and mass-transfer equations rather than temperature and heat-transfer equations. This can be done by writing a mass-balance equation similar to Eq. (9.10-2) for a heat balance and a mass-transfer equation similar to Eq. (9.10-3).

Through-Circulation Drying in the Constant-Rate Period. Spherical wet catalyst pellets having a diameter of 12.7 mm are being dried in a through-circulation dryer. The pellets are in a bed 63.5 mm thick on a screen. The solids are being dried by air entering with a superficial velocity of 0.914 m/s at 82.2°C and having a humidity H = 0.01 kg H₂O/kg dry air. The dry solid density is determined as 1522 kg/m³, and the void fraction in the bed is 0.35. The initial free moisture content is 0.90 kg H₂O/kg solid and the solids are to be dried to a free moisture content of 0.45, which is above the critical free moisture content. Calculate the time for drying in this constant-rate period.

Material and Heat Balances on a Continuous Dryer. Repeat Example 9.10-2, making heat and material balances, but with the following changes. The solid enters at 15.6°C and leaves at 60°C. The gas enters at 87.8°C and leaves at 32.2°C. Heat losses from the dryer are estimated as 2931 W.

Drying in a Continuous Tunnel Dryer. A rate of feed of 700 lb_m dry solid/h containing a free moisture content of X₁ = 0.4133 lb H₂O/lb dry solid is to be dried to X₂ = 0.0374 lb H₂O/lb dry solid in a continuous-counterflow tunnel dryer. A flow of 13 280 lb_m dry air/h enters at 203°F with an H₂ = 0.0562 lb H₂O/lb dry air. The stock enters at the wet bulb temperature of 119°F and remains essentially constant in temperature in the dryer. The saturation humidity at 119°F from the humidity chart is H_W = 0.0786 lb H₂O/lb dry air. The surface area available for drying is (A/L_S) = 0.30 ft²/lb_m dry solid.

For the continuous dryer, calculate the time in the dryer in the constant-rate zone and the falling-rate zone.

Air Recirculation in a Continuous Dryer. The wet feed material to a continuous dryer contains 50 wt % water on a wet basis and is dried to 27 wt % by countercurrent air flow. The dried product leaves at the rate of 907.2 kg/h. Fresh air to the system is at 25.6°C and has a humidity of H = 0.007 kg H₂O/kg dry air. The moist air leaves the dryer at 37.8°C and H = 0.020 and part of it is recirculated and mixed with the fresh air before entering a heater. The heated mixed air enters the dryer at 65.6°C and H = 0.010. The solid enters at 26.7°C and leaves at 26.7°C. Calculate the fresh-air flow, the percent air leaving the dryer that is recycled, the heat added in the heater, and the heat loss from the dryer.

Sterilizing Canned Foods. In a sterilizing retort, cans of a given food were heated; the average temperature in the center of a can is approximately 98.9°C for the first 30 min. The average temperature for the next period is 110°C. If F₀ for the spore organism is 2.50 min and z = 10°C, calculate the time of heating at 110°C to make the process safe.

Temperature Effect on Decimal Reduction Time. Prove by combining Eqs. (9.12-6) and (9.12-8) that a plot of log₁₀ D_T versus 1/T (T in degrees absolute) is a straight line.

Thermal Process Time for Pea Purée. For cans of pea purée, F₀ = 2.45 min and z = 9.94°C (C2). Neglecting heatup time, determine the process time for adequate sterilization at 112.8°C at the center of the can.

Process Time for Adequate Sterilization. The F₀ value for a given canned food is 2.80 min and z is 18°F (10°C). The center temperatures of a can of this food when heated in a retort were as follows for the time periods given: t₁ (0–10 min), T₁ = 140°F; t₂ (10–30 min), T₂ = 185°F; t₃ (30–50 min), T₃ = 220°F; t₄ (50–80 min), T₄ = 230°F; t₅ (80–100 min), T₅ = 190°F. Determine if adequate sterilization is obtained.

Process Time and Numerical Integration. The following time–temperature data were obtained for the heating, holding, and cooling of a canned food product in a retort, the temperature being measured in the center of the can:

Sterility Level of Fermentation Medium. The aqueous medium in a fermentor is being sterilized and the time–temperature data obtained are as follows:

where T = K. The contamination level N₀ = 1 × 10¹² spores. Calculate the sterility level N at the end and ∇.

Time for Pasteurization of Milk. Calculate the time in min at 62.8°C for pasteurization of milk. The F₀ value to be used at 65.6°C is 9.0 min. The z value is 5°C.

Reduction in Number of Viable Cells in Pasteurization. In a given pasteurization process the reduction in the number of viable cells used is 10¹⁵ and the F₀ value is 9.0 min. If the reduction is to be increased to 10¹⁶ because of increased contamination, what would be the new F₀ value?