DRYER & Sample Problems

- is an equipment used in removing moisture or solvents from a wet material or product.   - a substance that can contained bound moisture and is variable in moisture content which they posses at different times.  - amount of moisture present in the product at the start or at the end of the drying operation. - 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.  - it is the sum of the bone dry weight of the product and the weight of moisture. - it is the amount of moisture expressed as a percentage of the gross weight or the bone dry weight of the product. a)  - is the moisture content of the product in percent of the gross weight. b)  - it is the moisture content of the product in percent of tthe he bone dry weight. - is that type of drying operation in which the material to be dried is fed to and discharge from the dryer continuously.  - is that type of drying operation in which the material to be dried is done in batches at definite interval of time. 1. 2 3.

 - conduction heat transfer  - convection heat transfer  - radiation heat transfer

1. 2. 3.

(wet basis) (dry basis or regain)

where:  - gross weight  - bone dry weight  - weight of moisture  - moisture content Q = Q1 + Q2 + Q3 + Q4 Q1 = (BDW)Cp(tB - t A) kg/hr Q2 = MBCpw(tB - t A) kg/hr Q2 = MB(hf B - hf  A) kg/hr Q3 = (M A - MB)(hv B - hf  A) kg/hr Q4 = heat loss where: 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 or at the end of drying operation operation  t - temperature in C hf - enthalpy of water at saturated s aturated 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

1. 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 m 3/sec ( 0.25) b) the heat supplied to the preheater in KW (16) Q

 Air out 2

Outside air 

0

 Air in

1

Dryer  B

 A

Pre-Heater 

GWB MB XmB

GW A M A XmA

Psychrometric Chart

60% RH 2

W2

40% RH

W0 = W1

0 1

27°C

93°C

2.  A 400 kg/hr of ceramic powder is to be produced. The powder has a specific heat of 0.921 KJ/kg- C and a density of 1568 kg/m3. Initial moisture content is 19% and final moisture content is 5%. A continuous belt dryer is used. the drying time is 45 minutes at 71 C DB and 38 C WB. Product temperature is 38 C. Compute: a) the weight of moisture to be removed and weight of material entering the dryer b) the weight of air required in kg/hr if make up air is at 27 C DB and W = 0.0168 kgm/kgda. c) heat required by the dryer  Air out at 71 CDB and 38 C WB 2 W = 0.029148 kgm/kgda  Air in h = 148.1 KJ/kgda 1 v = 1.0211 m3/kgda Dryer  B

 A

GWB MB XmB

GW A M A XmA

38°C

2

W1 = 0.0168

1 1

27°C

Processes 1 to 2 – Humidification (Drying)

W2

38°C

71°C

GWB = 400 kg/hr XmB = 0.05 MB = 0.05(400) MB = 20 kg/hr BDW = 400(1 - 0.05) BDW = 380 kg/hr XmA  = 0.19 GW A = 380/(1 - 0.19) GW A = 469.14 kg/hr M A = 0.19(469.14) M A = 89.14 kg/hr Mr = moisture removed from the ceramic Mr = 89.14 - 20 Mr = 69.14 kg/hr at 71 C DB and 38 C WB W = 0.29148 kgm/kgda h = 148.1 KJ/kg da at 27 C DB and W = 0.0168 kgm/kgda h = 69.995 KJ/kgda m = Mr/(W2 - W1) m = 69.14(0.029148 - 0.0168) m = 0.854 kg/hr

Q1 = BDW(Cp)(71 - 38) Q1 = 380(0.921)(71 - 38) Q1 = 11,549.34 KJ/hr Q1 = 3.2 KW Q2 = MB(Cpw)(71 - 38) Q2 = 20(4.187)(71 - 38) Q2 = 2763.42 KJ/kr Q2 = 0.8 KW Q3 = Mr(hg71 - hf 38) Q3 = 69.14(2628.5 - 159.21) Q3 = 170,726.71 KJ/hr Q3 = 47.42 KW Qp = heat required by the product Qp = Q1 + Q2 + Q3 Qp = 51.42 KW Qa = heat required by air Qa = m(h2 - h1) Qa = 0.854(148.1 - 69.995) Qa = 66.7 KJ/hr Qa = 0.02 KW Q = Qp + Qa Q = 51.44 KW