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Mechanical Engineer

- 1. DRYER Dryer - is an equipment used in removing moisture or solvents from a wet material or product. Hygroscopic Substance - a substance that can contained bound moisture and is variable in moisture content which they posses at different times. Weight of Moisture - amount of moisture present in the product at the start or at the end of the drying operation. Bone Dry Weight - it is the final constant weight reached by a hygroscopic material 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 the weight 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.
- 2. Continuous Drying - is that type of drying operation in which the material to 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 DRYERS 1. Direct Dryers - conduction heat transfer 2. Indirect Dryers - convection heat transfer 3. Infra-red Dryers - radiation heat transfer PRODUCT SYMBOLS 1. GW = BDW + M 2. 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
- 3. HEAT REQUIREMENT BY THE PRODUCT Q = Q 1 + Q 2 + Q3 + Q 4 Q1 = (BDW)Cp(tB - tA) kg/hr Q2 = MBCpw(tB - tA) kg/hr Q2 = MB(hfB - hfA) kg/hr Q3 = (MA - MB)(hvB - hfA) kg/hr = MR(hvB - hfA) Q4 = heat loss Q1 - sensible heat of product, KJ/hr Q2 - sensible heat of moisture remaining in the product, KJ/hr Q3 - heat required to evaporate and superheat moisture removed from the product in KJ/hr Q4 - heat losses, KJ/hr A,B - conditions at the start or at the end of drying operation t - temperature in C hf - enthalpy of water at saturated liquid, KJ/kg hv - enthalpy of vapor, KJ/kg
- 4. Condition A Condition B GWA GWB MB MA BDW MR(moisture removed) BDW (weight of product without moisture)
- 5. It is desired to designed a drying plant to have a capacity of 680 kg/hr of product 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 entering the dryer and will leave the dryer with the same temperature but with a 60% 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
- 6. Q 0 Fresh air 1 heated air 2 exhaust air Dryer m m Air Preheater A GWA B GWB GW = BDW + M Given: GW = BDW + Xm(GW) GWB = 680 kg/hr BDW GW = XmB = 0.035 ; XmA = 0.42 (1 − X m ) W0 = 0.0089 ; h0 = 49.8 BDW = GW(1 − X m ) W1 = 0.0089 ; h1 = 117.22 ; v1 = 1.05 M = X m (GW) W2 = 0.54 ;h2 = 1538.94
- 7. 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 1 By moisture balance on dryer By energy balance in the mW1 + 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
- 8. Raw cotton has been stored in a warehouse at 29°C and 50% relative humidity, with a regain of 6.6%. (a) the cotton goes through a mill and passes through the weaving room kept at 31°C and 70% relative humidity with a regain of 8.1%. What is the moisture in 200 kg of cotton? (b) for 200 kg of cotton from the warehouse, how many kilograms should appear in the woven cloth, neglecting lintage and thread losses? ANSWER: a) 12.4 kg ; b) 202.8 kg GW = BDW + M M = Xm(BDW) BDW = GW/(1+Xm) Given: XmA = 0.066 ; XmB = 0.081 GWA = 200 kg BDW = 187.61 kg MA = 12.4 kg MB = 15.2 kg GW = 202.8 kg
- 9. A 10 kg sample from a batch of material under test is found to have a BDW of 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 - BDW MA = 1.5 kg MB = XmB(BDW) MB = 1.7 kg GW = BDW + M GWB = 10.2 kg GWB/GWA = 1.02
- 10. A rotary dryer is fired with bunker oil of 41 870 KJ/kg HHV is to produce 20 metric tons per hour of dried sand with 0.5% moisture from a wet feed containing 7% moisture, specific heat of sand is 0.879 KJ/kg-°C, temperature of wet feed is 30°C and temperature of dried product is 115°C. Calculate the L/hr of bunker oil consumed if the specific gravity of bunker oil is 0.90 and 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
- 11. Given: GWB = 20,000kg/hr; XmB = 0.005; XmA = 0.07 HHV =41,870 KJ/kg; Cp = 0.879 KJ/kg-C; tA = 30°C ; tB = 115°C S = 0.90; e = 60% GW = BDW + M ; GW = BDW/(1-Xm) M = Xm(GW) BDW = GW - M MB = 100 kg/hr ; e = Q/mf(HHV) BDW = 19,900 kg/hr mf = 204.2 kg/sec GWA = 21,398 kg/hr ; MA = 1498 kg/hr df = 900 kg/m3 MA - MB = MR ; MR = 1398 kg/hr Vf = 0.227 m3 Q1 = BDW(Cp)(tB - t=A) = 413 KW Vf = 227 Liters Q2 = MB(Cpw)(tB - tA) = 10 KW Q3 = MR(hg -hf) = 1002 KW Q4 = 0 Q = Q + Q + Q +Q = 1425 KW

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