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Cooling tower and Dryer Principles

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- 1. COOLING TOWER By. Engr. Yuri G. Melliza
- 2. B Air Out hB,WB, ma Hot water 1 m1 t1 h1 Fan A Air in hA,WA, ma t 3 , h3 , m3 Make-up water 3 Cold water 2 t 2 , h 2 , m2 Catch Basin
- 3. Cooling Tower A Cooling tower is a wind braced enclosure or shell usually made of wood, concrete or metal with fillings on the inside to aid water exposure. The water to be cooled is pumped into a distributing header at the top of the tower from which it drops in sprays to the filling. The water spreads out in the filling thus exposing new water surfaces to the air circulating through the tower. The cooled water drops to the bottom of the tower called the catch basin. The air circulating through the tower becomes partially saturated with moisture by evaporating some amount of water. This evaporation is mostly what cools the water.
- 4. Fundamental Equations: 1. Actual Cooling Range (ACR) ACR = t1 - t2 2. Cooling Tower Approach (A) A = t2 - tWA 3. Theoretical Cooling Range (TCR) TCR = t1 – tWA 4. Cooling Tower Efficiency t1 t 2 e t 1 t WA x 100%
- 5. 5. Vapor Pressure PV = PW - PA(td - tW) where: A = 6.66 x 10-4 (For tW of equal or greater than 0C. A = 5.94 x 10-4 (For tW of less than 0C) 6. Specific Humidity or Humidity Ratio 0.622 Pv W P Pv kgm kgda 7. Relative Humidity Pv Φ x 100 % Pd
- 6. 8. Enthalpy h = 1.0045td + W(2501.3 + 1.86td) KJ/kgda 9. Specific Volume 0.287(t d 273) m 3 υ (P Pv) kgda 10. Degree of Saturation P Pd μ Φ P Pv
- 7. 11. By moisture balance in the tower: a) With make up water, m1 = m2 m3 = ma(WB - WA) kg/sec b) Without make up water available, m1 m2: m1 - m2 = ma (WB - WA) kg/sec 12. By Energy Balance in the tower a. Considering make up water m1 (h1 h2 ) ma kg/sec (hB h A ) (WB WA )h3 m1h1 ma (hB h A ) (WB WA )h3 h2 m1 KJ/kg
- 8. b. Without considering make up water m1 (h1 h 2 ) ma kg/sec (h B h A ) (WB WA )h 2 m1h1 m a (h B h A ) h2 m1 m a (WB WA ) KJ/kg 13. Driving Pressure gHρ o - ρ i ΔPd KPa 1000
- 9. 14. Mass Flow rate of air and vapor mixture m = ma(1+W) kg/sec m = ma + mv 15. Cooling water flow rate related to Brake Power of an Engine Brake Power m w 904.3 L/hr t1 - t 2
- 10. where: m1 - mass flow rate of water entering tower in t1 - temperature of hot water, C kg/sec t2 - temperature of cooled water, C m2 - mass flow rate of cooled water in kg/sec t3 - temperature of make up water, C m3 - make up water in kg/sec H - tower height, meters h1 - enthalpy of hot water in KJ/kg o - density of outside air and vapor mixture, h2 - enthalpy of cooled water in KJ/kg kg/m3 h3 - enthalpy of make up water in KJ/kg i - density of inside air and vapor mixture, hA - enthalpy of air entering tower in KJ/kgda taken at exit of the fill, kg/m3 hB - enthalpy of air leaving tower in KJ/kgda WA - humidity ratio of air entering tower in kgm/kgda WB - humidity ratio of air leaving tower in kgm/kgda ma - mass flow rate of dry air in kg/sec td - dry bulb temperature in C tw - wet bulb temperature in C
- 11. 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.
- 12. 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.
- 13. 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
- 14. HEAT REQUIREMENT BY THE PRODUCT Q = Q1 + Q2 + Q3 + Q4 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 Cp - specific heat of the product, KJ/kg-C or KJkg-K Cpw - specific heat of water, KJ/kg-C or KJ/kg-K
- 15. Condition A GWA MA Condition B GWB MB BDW MR(moisture removed) BDW (weight of product without moisture)
- 16. 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 27C with 40% RH will be preheated to 93C 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 = 1.05 m3/kgda At 93 C and 60% RH W = 0.54 kgm/kgda h = 1538.94 KJ/kgda
- 17. Q 0 Fresh air 1 heated air m 2 exhaust air Dryer m Air Preheater A GWA GW = BDW + M GW = BDW + Xm(GW) BDW GW (1 X m ) BDW GW(1 X m ) M X m (GW) B GWB Given: GWB = 680 kg/hr XmB = 0.035 ; XmA = 0.42 W0 = 0.0089 ; h0 = 49.8 W1 = 0.0089 ; h1 = 117.22 ; v1 = 1.05 W2 = 0.54 ;h2 = 1538.94
- 18. h2 h1 2 W2 h0 0 1 By moisture balance on dryer mW1 + MA = mW2 + MB M A MB m W2 W1 m = 850 kg/hr Qa1 = 850(1.05) = 892.43 m3/hr Qa1 = 0.25 m3/sec W0 = W 1 MB = 23.8 kg/hr BDW = 656.2 kg/hr GWA = 1131.4 kg/hr MA = 475.2 kg/hr By energy balance in the preheater: Q = m(h1 - h0) Q = 16 KW
- 19. Raw cotton has been stored in a warehouse at 29C 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 31C 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 GWB = 202.8 kg
- 20. 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
- 21. 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 30C and temperature of dried product is 115C. 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 30C = 125.79 KJ/kg hg at 101.325 KPa and 115C = 2706.12 KJ/kg Wet Feed (sand) Dried sand
- 22. Given: GWB = 20,000kg/hr; XmB = 0.005; XmA = 0.07 HHV =41,870 KJ/kg; Cp = 0.879 KJ/kg-C; tA = 30C ; tB = 115C 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 = Q1+ Q2+ Q3 +Q4 = 1425 KW

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