FE/EIT sample problems from EIT/FE exam for dynamics
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Dryer Design
this shows solutions to various problems on hyrdropower design
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- 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
