Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Like this presentation? Why not share!

- Refrigeration system 2 by Yuri Melliza 876 views
- Methods of handling Supply air in H... by Yuri Melliza 1147 views
- Cooling Tower & Dryer Fundamentals by Yuri Melliza 1381 views
- ROTARY DRYER by Claire Canoy 2973 views
- Dryers by physics101 10363 views
- The Basics of a Rotary Dryer by Stelter & Brinck 853 views

No Downloads

Total views

1,024

On SlideShare

0

From Embeds

0

Number of Embeds

9

Shares

0

Downloads

38

Comments

0

Likes

1

No embeds

No notes for slide

- 1. 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.
- 2. 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
- 3. 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
- 4. Condition A Condition B GWA GWB MB MA BDWMR(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 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
- 6. 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
- 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 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
- 8. 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
- 9. 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
- 10. 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
- 11. 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

No public clipboards found for this slide

×
### Save the most important slides with Clipping

Clipping is a handy way to collect and organize the most important slides from a presentation. You can keep your great finds in clipboards organized around topics.

Be the first to comment