Dryers
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Dryers

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Dryers Dryers Presentation Transcript

  • DRYERDryer - is an equipment used in removing moisture or solvents from a wetmaterial or product.Hygroscopic Substance - a substance that can contained bound moistureand is variable in moisture content which they posses at different times.Weight of Moisture - amount of moisture present in the product at the startor at the end of the drying operation.Bone Dry Weight - it is the final constant weight reached by a hygroscopicmaterial when it is completely dried out. It is the weight of the product without the presence of moisture.Gross Weight - it is the sum of the bone-dry weight of the product and theweight of moisture.Moisture Content - it is the amount of moisture expressed as a percentage of the gross weight or the bone dry weight of the product. A) Wet Basis - is the moisture content of the product in percent of the gross weight. B) Dry Basis 0r Regain - it is the moisture content of the product in percent of the bone dry weight.
  • Continuous Drying - is that type of drying operation in which the materialto be dried is fed to and discharge from the dryer continuously.Batch Drying - is that type of drying operation in which the material to be dried is done in batches at definite interval of time.CLASSIFICATION OF DRYERS1. Direct Dryers - conduction heat transfer2. Indirect Dryers - convection heat transfer3. Infra-red Dryers - radiation heat transferPRODUCT SYMBOLS1. GW = BDW + M2. Xm = [M/GW] x 100% (wet basis)3. Xm = [M/BDW] x 100% (dry basis or regain)where: GW - gross weight BDW - bone dry weight M - weight of moisture Xm - moisture content
  • HEAT REQUIREMENT BY THE PRODUCTQ = Q 1 + Q 2 + Q3 + Q 4Q1 = (BDW)Cp(tB - tA) kg/hrQ2 = MBCpw(tB - tA) kg/hrQ2 = MB(hfB - hfA) kg/hrQ3 = (MA - MB)(hvB - hfA) kg/hr = MR(hvB - hfA)Q4 = heat lossQ1 - sensible heat of product, KJ/hrQ2 - sensible heat of moisture remaining in the product, KJ/hrQ3 - heat required to evaporate and superheat moisture removed from the product in KJ/hrQ4 - heat losses, KJ/hrA,B - conditions at the start or at the end of drying operationt - temperature in Chf - enthalpy of water at saturated liquid, KJ/kghv - enthalpy of vapor, KJ/kg
  • Condition A Condition B GWA GWB MB MA BDWMR(moisture removed) BDW (weight of product without moisture)
  • It is desired to designed a drying plant to have a capacity of 680 kg/hr ofproduct 3.5% moisture content from a wet feed containing 42% moisture.Fresh air at 27°C with 40% RH will be preheated to 93°C before enteringthe dryer and will leave the dryer with the same temperature but with a60% RH. Find: a) the amount of air to dryer in m3/sec ( 0.25) b) the heat supplied to the preheater in KW (16) At 27 °C DB and 40% RH At 93° C and W = .0089 kgm/kgda W = .0089 kgm/kgda h = 117.22 KJ/kgda h = 49.8 KJ/kgda υ = 1.05 m3/kgda At 93 °C and 60% RH W = 0.54 kgm/kgda h = 1538.94 KJ/kgda
  • Q 0 Fresh air 1 heated air 2 exhaust air Dryer m m Air Preheater A GWA B GWBGW = BDW + M Given:GW = BDW + Xm(GW) GWB = 680 kg/hr BDWGW = XmB = 0.035 ; XmA = 0.42 (1 − X m ) W0 = 0.0089 ; h0 = 49.8BDW = GW(1 − X m ) W1 = 0.0089 ; h1 = 117.22 ; v1 = 1.05M = X m (GW) W2 = 0.54 ;h2 = 1538.94
  • h2 2 MB = 23.8 kg/hr h1 W2 h0 BDW = 656.2 kg/hr GWA = 1131.4 kg/hr W0 = W 1 MA = 475.2 kg/hr 0 1By moisture balance on dryer By energy balance in themW1 + MA = mW2 + MB preheater: M A − MB Q = m(h1 - h0) m= W2 − W1 Q = 16 KW m = 850 kg/hr Qa1 = 850(1.05) = 892.43 m3/hr Qa1 = 0.25 m3/sec
  • Raw cotton has been stored in a warehouse at 29°C and 50% relativehumidity, with a regain of 6.6%. (a) the cotton goes through a mill andpasses through the weaving room kept at 31°C and 70% relative humiditywith a regain of 8.1%. What is the moisture in 200 kg of cotton? (b) for200 kg of cotton from the warehouse, how many kilograms should appearin the woven cloth, neglecting lintage and threadlosses? ANSWER: a) 12.4 kg ; b) 202.8 kgGW = BDW + MM = Xm(BDW)BDW = GW/(1+Xm)Given:XmA = 0.066 ; XmB = 0.081GWA = 200 kgBDW = 187.61 kgMA = 12.4 kgMB = 15.2 kgGW = 202.8 kg
  • A 10 kg sample from a batch of material under test is found to have a BDWof 8.5 kg. This material is processed and is then found to have a regain(dry basis moisture content) of 20%. How much weight of product appears for each kilogram of original material. (1.02 kg/kg)Given:GWA = 10 kg ; BDW = 8.5 kg ; XmB = 0.20 (dry basis)M = GW - BDWMA = 1.5 kgMB = XmB(BDW)MB = 1.7 kgGW = BDW + MGWB = 10.2 kgGWB/GWA = 1.02
  • A rotary dryer is fired with bunker oil of 41 870 KJ/kg HHV is to produce20 metric tons per hour of dried sand with 0.5% moisture from a wet feedcontaining 7% moisture, specific heat of sand is 0.879 KJ/kg-°C, temperatureof wet feed is 30°C and temperature of dried product is 115°C. Calculatethe L/hr of bunker oil consumed if the specific gravity of bunker oil is 0.90and dryer efficiency of 60%. hf at 30°C = 125.79 KJ/kg hg at 101.325 KPa and 115°C = 2706.12 KJ/kg ut Wet Feed (sand) Gas o Flue Dried sand G as in Flue
  • Given:GWB = 20,000kg/hr; XmB = 0.005; XmA = 0.07HHV =41,870 KJ/kg; Cp = 0.879 KJ/kg-C; tA = 30°C ; tB = 115°CS = 0.90; e = 60%GW = BDW + M ; GW = BDW/(1-Xm)M = Xm(GW)BDW = GW - MMB = 100 kg/hr ; e = Q/mf(HHV)BDW = 19,900 kg/hr mf = 204.2 kg/secGWA = 21,398 kg/hr ; MA = 1498 kg/hr df = 900 kg/m3MA - MB = MR ; MR = 1398 kg/hr Vf = 0.227 m3Q1 = BDW(Cp)(tB - t=A) = 413 KW Vf = 227 LitersQ2 = MB(Cpw)(tB - tA) = 10 KWQ3 = MR(hg -hf) = 1002 KWQ4 = 0Q = Q + Q + Q +Q = 1425 KW