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Ls ×2 A Example 2. A 2940 kg of solid is to be dried from 1 kg water/kg dry solids to 0.01 kg water/kg dry solids in a tray dryer consisting of 24.5 m² and 0.02 m deep which completely filled with T wet material. The mean air temperature is 350 K and the relative humidity across the trays may RH и be taken as constant at 10%. The mean air velocity is 2.0 m/s, air density 0.95 kg/m³ and the convective coefficient of heat transfer is given by: (h) h = 14.3G0.8; kW/m²K → 143 × (0.095 x 2 ) 8 where G is the mass velocity of the air in kg/m²s. The critical moisture contents of the solid are < 0.3 kg water/kg dry solids. what is the drying time? Take λ=2395 kJ/kg. Itotal Chapter Five Rc R₁ = h (T-Tw) Solution: S G = Pair XU = 0.95 x 2 = 1.9 kg m².sec kW m². K h = 14.3G0.8 = 14.3(1.9) 0.8 = 23.9 From Figure 1; at T=350 K and RH%-10% Rc (T-Tw) λ Tw = 317 K 23.9(350-317) = 0.328 2395 kgwater kgsolid.m².sec tTotal = ARC 2940 tTotal 1x [(×₁ = Xc) + Xc In XX |(1 − 0.3) + 0.3 In 24.5 x 0.328 0.3 0.01 tTotal = 630 sec Page 51

Ls ×2 A Example 2. A 2940 kg of solid is to be dried from 1 kg water/kg dry solids to 0.01 kg water/kg dry solids in a tray dryer consisting of 24.5 m² and 0.02 m deep which completely filled with T wet material. The mean air temperature is 350 K and the relative humidity across the trays may RH и be taken as constant at 10%. The mean air velocity is 2.0 m/s, air density 0.95 kg/m³ and the convective coefficient of heat transfer is given by: (h) h = 14.3G0.8; kW/m²K → 143 × (0.095 x 2 ) 8 where G is the mass velocity of the air in kg/m²s. The critical moisture contents of the solid are < 0.3 kg water/kg dry solids. what is the drying time? Take λ=2395 kJ/kg. Itotal Chapter Five Rc R₁ = h (T-Tw) Solution: S G = Pair XU = 0.95 x 2 = 1.9 kg m².sec kW m². K h = 14.3G0.8 = 14.3(1.9) 0.8 = 23.9 From Figure 1; at T=350 K and RH%-10% Rc (T-Tw) λ Tw = 317 K 23.9(350-317) = 0.328 2395 kgwater kgsolid.m².sec tTotal = ARC 2940 tTotal 1x [(×₁ = Xc) + Xc In XX |(1 − 0.3) + 0.3 In 24.5 x 0.328 0.3 0.01 tTotal = 630 sec Page 51

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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Ls ×2 A Example 2. A 2940 kg of solid is to be dried from 1 kg water/kg dry solids to 0.01 kg water/kg dry solids in a tray dryer consisting of 24.5 m² and 0.02 m deep which completely filled with T wet material. The mean air temperature is 350 K and the relative humidity across the trays may RH и be taken as constant at 10%. The mean air velocity is 2.0 m/s, air density 0.95 kg/m³ and the convective coefficient of heat transfer is given by: (h) h = 14.3G0.8; kW/m²K → 143 × (0.095 x 2 ) 8 where G is the mass velocity of the air in kg/m²s. The critical moisture contents of the solid are < 0.3 kg water/kg dry solids. what is the drying time? Take λ=2395 kJ/kg. Itotal Chapter Five Rc R₁ = h (T-Tw) Solution: S G = Pair XU = 0.95 x 2 = 1.9 kg m².sec kW m². K h = 14.3G0.8 = 14.3(1.9) 0.8 = 23.9 From Figure 1; at T=350 K and RH%-10% Rc (T-Tw) λ Tw = 317 K 23.9(350-317) = 0.328 2395 kgwater kgsolid.m².sec tTotal = ARC 2940 tTotal 1x [(×₁ = Xc) + Xc In XX |(1 − 0.3) + 0.3 In 24.5 x 0.328 0.3 0.01 tTotal = 630 sec Page 51
Transcribed Image Text:Ls ×2 A Example 2. A 2940 kg of solid is to be dried from 1 kg water/kg dry solids to 0.01 kg water/kg dry solids in a tray dryer consisting of 24.5 m² and 0.02 m deep which completely filled with T wet material. The mean air temperature is 350 K and the relative humidity across the trays may RH и be taken as constant at 10%. The mean air velocity is 2.0 m/s, air density 0.95 kg/m³ and the convective coefficient of heat transfer is given by: (h) h = 14.3G0.8; kW/m²K → 143 × (0.095 x 2 ) 8 where G is the mass velocity of the air in kg/m²s. The critical moisture contents of the solid are < 0.3 kg water/kg dry solids. what is the drying time? Take λ=2395 kJ/kg. Itotal Chapter Five Rc R₁ = h (T-Tw) Solution: S G = Pair XU = 0.95 x 2 = 1.9 kg m².sec kW m². K h = 14.3G0.8 = 14.3(1.9) 0.8 = 23.9 From Figure 1; at T=350 K and RH%-10% Rc (T-Tw) λ Tw = 317 K 23.9(350-317) = 0.328 2395 kgwater kgsolid.m².sec tTotal = ARC 2940 tTotal 1x [(×₁ = Xc) + Xc In XX |(1 − 0.3) + 0.3 In 24.5 x 0.328 0.3 0.01 tTotal = 630 sec Page 51
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