Drying-Worked-Problems (2).doc

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  Worked Examples: 1) Air containing 0.005 kg of water vapour per kg of dry air is preheated to 52°C in adryer and passed to the lower shelves. It leaves these shelves at 60% relativehumidity and is reheated to 52°C and passed over another set of shelves, againleaving at 60% relative humidity. This is again repeated for the third and fourthsets of shelves, after which the air leaves the dryer. On the assumption that thematerial in each shelf has reached the wet bulb temperature and heat loss isnegligible, estimate: (i) the temperature of the material on each tray; (ii) theamount of water removed, in kg/hr, if 300m 3 /min of moist air leaves the dryer.(i) Air leaves the pre-heater of the dryer at 325°K Humidity of incoming air = 0.005 kg water/kg dry air It enters the first shelf. So, wet bulb temperature = 25°CMoisture is removed along constant wet bulb temperature line till 60% R.H. isreachedThis gives the exit condition of air from first shelf.From the chart,Humidity of air leaving first shelf = 0.016 kg water/kg dry air. Dry bulbtemperature of exit air is at 27°C and is at humidity of 0.016 kg water/kg dry air. Thisair is again heated to 52°C dry bulb temperature in second heater. So, air leavesheater at 52°C and humidity of 0.016 kg water/kg dry air. When it leaves the Secondshelf, the corresponding dry bulb temperature is 34°C and humidity is 0.023 kg water/kg dry air. This air enters the third shelf after preheating to 52 °C. Similarly for thirdshelf, exit air has a humidity of 0.028 kg water/ kg dry air and a dry bulb temperatureis 39°C. The air leaving the fourth shelf has a humidity of 0.032 kg water/ kg dry air and a dry bulb temperature of 42°C Thesolidtemperaturescorrespond toWBT and theyare 23 °C, 27°C,32 °C and 34 °Crespectively.111    Fig. : P 6.1 Humidity vs Temperature (ii)Final moist air conditions: (Y) = 0.032 kg water/ kg dry air Dry bulb temperature = 42°CV H  = 8315 [(1/M air  ) + (Y’/M water  )] [(t G  + 273)/P t ] V H  = 8315 [(1/28.84) + (0.032/18)] [(42 + 273)/1.013×10 5 ]V H  = 0.945 m 3 /kg dry air.Amount of dry air leaving/hr = (300 × 60)/0.945 = 1.905×10 4  kgWater removed/hr = 1.905 ×10 4 (0.032 – 0.005) = 514.35 kg/hr.2) A batch of the solid, for which the following table of data applies, is to be dried from 25 to6 percent moisture under conditions identical to those for which the data were tabulated. Theinitial weight of the wet solid is 350 kg, and the drying surface is 1 sq m/8 kg dry weight.Determine the time for drying.X × 100, kg moisture kg dry solid 352520181614121090 . 80. 76.4 N × 100, kg moisture evaporated /hr.m 2 30303026.623.920.818159.774.32.5 112  Fig. : P6.2 1/N vs X for falling rate period  X 1 = 0.25/ (1 – 0.25) = 0.333,X 2  = 0.06/ (1 – 0.06) = 0.0638,Initial weight of wet solid = 350 kgInitial moisture content = 0.333 kg moisture/kg dry solid So total moisture present in wet solid (initially) = 350 × 0.25 = 87.5 Kg moistureWeight of dry solid = 262.5 kg = L S A = 262.5/8 = 32.8125 m 2 ,or L S /A = 8 kg/m 2 X Cr   = 0. 20,N C  = 0.3 kg/m 2 hr So for constant rate period t I  = L S /AN C  [X 1  – X Cr  ] = [262.5/ (32.8125 × 0.3)] [0.333 – 0.2] = 3.55 hr.For falling rate period, we finding time graphically113  X0.20.1800.160.140.1200.1000.0900.0800.070.0641/N3.335.566.257.148.3210.0011.1112.514.2915.625Area = 1.116,  time = Area under the curve × L S /A = 1.116 × L S /A = 1.116 × 8 = 8.928 hr.  Total time = 8.928 + 3.55 = 12.478 hr.3) A wet slab of material weighing 5 kg srcinally contains 50 percent moisture onwet basis. The slab is 1 m x 0.6 m x 7.5 cm thick. The equilibrium moisture is 5 percent on wet basis. When in contact with air, the drying rate is given in the table below. Wet slab wt, kg5.04.03.63.53.43.062.85Drying rate, kg/(hr)(m 2 )5.05.04.54.03.52.001.00Drying takes place from one face only. (i)Plot the rate of drying curve and find the critical moisture content.(ii)How long will it take to dry the wet slab to 15 percent moisture on wet basis? Weight of wet solid = 5 kgMoisture = 0.50 moisture/kg wet solid = 0.5/ [(0.5 moisture) + (0.5 dry solid)] = 0.5/ (1 – 0.5) = moisture/dry solid  X 1  = 1For 5 kg wet solid, moisture = 5 × 0.5 = 2.5 kg.114
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