Answers_Heat & Mass Transfer - I

CHEMICAL ENGINEERING REVIEW Unit Operations: Heat and Mass Transfer - I 1. In Grosvenor pyschrometric chart, the point which was read from the x-axis that corresponds to the point at which constant humidity line touches the 100% humidity curve. ANSWER: DEW POINT

2. For the air-water system under ambient conditions, the adiabatic saturation temperature and the wet bulb temperature are nearly equal, because a. Water has higher latent heat of evaporation b. Lewis number is close to unity c. They are always equal under all circumstances d. Solubiity of the components of air in water is very small

3. The relative humidity of air decreases in spite of an increase in the absolute humidity when the a. Temperature rises b. Pressure rises c. Temperature falls d. Pressure falls

4. In this method, humidity is measured by determining the wet bulb and dry bulb temperatues simultaneously. Answer: PSYCHROMETRIC METHOD

5. During dehumidification of unsaturated air, wet bulb temperature & partial pressure of vapor are not constant. However, during evaporative cooling process with re-circulated water supply, the __________ remains constant. ANSWER; WET BULB TEMPERATURE

6. What is the difference in measuring wet bulb and dry bulb temperatures? ANSWER: in wet bulb, thermometer is covered with wet wick

7. For air-water system at ordinary conditions, the humid heat is almost equal to the specific heat, thus,

Where: hy = heat transfer coefficient between gas and surface of the liquid; M B = weight of the gas; ky = mass transfer coefficient; Cs = humid heat. This equation is also known as_______. Answer: LEWIS RELATION

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CHEMICAL ENGINEERING REVIEW Unit Operations: Heat and Mass Transfer - I 8. The plot below gives the saturated humidity (He) versus temperature (T). Line joining (H1,T1) and (H2,T2) is the constant enthalpy line. Identify T1, T2 and T3

ANSWERS: T1 = dry bulb temperature, T2 = wet bulb temperature, T3 = dew point

9. At 25°C and 90% relative humidity, water evaporates from the surface of the lake at the rate of 2 2 1.0 kg/m /h. What is the relative humidity that will lead to an evaporation rate of 3.0 kg/m /h, with other conditions remaining the same. Condition A

Condition B

Evaporation Rate, kg/m /h

1.0

3.0

Relative Humidity

90%

X

2

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70%

10. The equilibrium water absorbed by a certain silica gel in contact with moist air varies linearly with the humidity of air, , where X = kg water absorbed/kgof dry gel, Y = humidity of air in kg moisture/kg dry air. 0.5 kg of silica gel containing 5% (dry basis) absorbed water is 3 placed in a collapsible vessel in which there are 10 m of moist air particle with pressure of water

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CHEMICAL ENGINEERING REVIEW Unit Operations: Heat and Mass Transfer - I being 15 mm Hg. The total pressure and temperature are kept at 1 standard atmosphere and 25°C, respecetively. How much water is picked up from the moist air in the vessel by the silica gel? What is the final pressure of water vapor in the vessel?

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Initial weight of moisture in the vessel (

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Approximately air becomes dry, therefore, pressre of moisture is zero

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CHEMICAL ENGINEERING REVIEW Unit Operations: Heat and Mass Transfer - I 11. Match column A with column B A Desolventizing Spray Drying Lyophilization Dehumidification

I II III IV

B Banana Chips Cooling Spent Meal Milk Powder Fruits Daing Sugar

1 2 3 4 5 6 7

ANSWER: I-3; II-4; III-5; IV-2

12. Which is a part of a spray drying system? a. Plenum b. Atomizer c. Knife d. Steam heater

13. Drying of food is carried out in an insulated tray. The drying air has a partial pressure of water equals to 2,360 Pa and a wet bulb temperature 30°C. The product has has a drying surface of 2 0.05 m /kg dry solid. The material has critical rate in the falling rate period which is proportional -4 2 to the moisture content and the mass transfer coefficient is 5.34 x 10 kg/m ·h·Pa. Calculate: (a) 2 the drying rate in the constant rate period in kg/m ·h; (b) the time required to dry the material from a moisture content of 0.22 to 0.06 (both on dry basis). Vapor pressure of water at 30°C = 4,232 Pa. Critical moisture content is assumed to be 0.12. Equation 24.9 (McCabe) ( ) ̇ ̅̅̅̅̅̅̅̅̅̅ ( ) ( ̇

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14. It takes 6 hours to dry a wet solid from 50% moisture content to the critical moisture content of 15%. How much longer will it take to dry the solid to 10% moisture content, under the same drying conditions (the quilibrium moisture content of the solid is 5%)? [(

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15. A 50 cm x 50 cm x 1 cm flat wet sheet weighing 2 kg initiall was dried from both the sides under constant drying rate period. It took 1,000 seconds for the weight of the sheet to reduce to 1.75 kg. Another 1 m x 1 m x 1 cm flat sheet is to be dried from one side only. Under the same drying conditions, what is the required time (in seconds) to dry the sheet from initial weight of 4 kg to 3 kg? (

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16. Sheet material 0.5 cm thick containing 800 kg of dry stock/m of original wet stock is to be dried 2 at constant drying condition. The initial drying rate is 4 kg/h·m at the initial moisture content of 2 33%. The final drying rate is 1 kg/h·m at 6% final moisture content. The equilibrium moisture content is negligible. If drying is from two large surfaces only, and if the drying rate in the falling rate period is proportional to the free moisture content, calculate the total drying time. All moisture contents are on the dry basis.

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17. The equilibrium moisture curve for a solid is shown below:

The total moisture content of the solid is X and it is exposed to air of relative humidity H. Identify 1, 2, 3 and 4. Answer: 1- equilibrium noisture; 2- free moisture; 3 – bound moisture; 4 – unbound moisture

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CHEMICAL ENGINEERING REVIEW Unit Operations: Heat and Mass Transfer - I 18. To maintain rice grain quality, this treatment is included to allow for redistribution of internal moisture in the grain. In this treatment, drying is temporarily stopped and the moisture within the For air-water system at ordinary conditions, the humid heat is almost equal to the specific heat, thus, grain equalizes due to diffusion and would result to higher drying rate when re-started. ANSWER: tempering

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