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Problem-1 (Drying)

Problem-1 (Drying).

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Problem-1 (Drying)

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  1. Problem-1 (Drying) • An insoluble wet granular material is dried in a pan 0.457 x 0.457 m and 25.4 mm deep. The material is 25.4 mm deep in the pan, and the sides and bottom can be considered to be insulated. Heat transfer is by convection from an air stream flowing parallel to the surface at a velocity of 6.1 m/s. The air is at 65.6 oC and has a humidity of 0.01 kg of water/kg dry air. Estimate: • The rate of drying for the constant-rate period. Assume convective heat transfer coefficient for the parallel flow can be expressed by the following empirical equation: h=0.0204 G0.8, where mass velocity, G = air density x air flow velocity. • If the air stream were to flow normal to the drying surface (all other conditions remaining same) what would be the new drying rate? Assume convective heat transfer coefficient for air flow normal to the drying surfacecan be expressed by the following empirical equation: h=1.17 G0.37. (Assume the correlation holds for the G values used). • If the air velocity is decreased 50% and its temperature is increased 10 oC, what should be the new drying rate?

  2. Problem-1 (Drying) Given: Velocity of drying air, Vair = 6.1 m/s Temperature of drying air, Tair = 65.6 oC Absolute humidity of drying air,  = 0.01 kg water/kg dry air Estimate: Drying rate for the constant-rate period, Rc = ? If air were to flow normal to drying surface, Rc = ? If Vair = 0.5 x 6.1 m/s, Tair = (65.6+10) oC, Rc = ?

  3. Problem-1 (Drying) Solution: Part-1 At constant drying-rate period, moisture will evaporate from the product at wet bulb temperature, Twb. From humidity chart, following adiabatic saturation line from Tair = 65.6 oC and  = 0.01 kg water/kg dry air, wet bulb temperature, Twb can be determined as: Twb = 29 oC and corresponding absolute humidity wb = 0.026 kg water/kg dry air Humid volume, VH = (22.41/273) Tair (1/28.97 + /18.02) = Tair (2.83 x 10-3 + 4.56 x 10-3 ) = (65.6+273)(2.83 x 10-3 + 4.56 x 10-3 x 0.01) = 0.974 m3/kg dry air Density of moist air, moist air: 1 kg dry air  (1+0.01) kg moist air  1.01 kg moist air moist air = 1.01 / 0.974 = 1.037 kg/m3 Mass velocity, G = moist air x Vair = 1.037 x 6.1 x 3600 = 22772.5 kg/m2 hr

  4. Problem-1 (Drying) For parallel flow, convective heat transfer coefficient, h = 0.0204 G0.8 = 0.0204 x (22772.5)0.8 = 62.45 W/m2 K = 62.45 J/s m2 K From steam table, latent heat of vaporization at Twb = 29 oC is hfg = 2433 kJ/kg = 2433000 J/kg For constant rate period: Rc hfg = h(Tair – Twb) Rc x 2433000 = 62.45 (65.6 – 29) Rc = 9.394 x10-4 kg/m2 s = 3.38 kg/m2 hr For surface area A = 0.457 x 0.457 = 0.208849 m2 Drying rate = RcA = 3.38 x 0.208849 = 0.706 kg of water/hour Note: If thickness of layer is increased, no change of Rc in the constant rate period

  5. Problem-1 (Drying) Part-2 As G = 22772.5 kg/m2 hr > 19,500, a higher degree of uncertainty exist in estimation of Rc For flow normal to the drying surface: convective heat transfer coefficient, h = 1.17 G0.37 = 1.17 x (22772.5)0.37 = 47.91 W/m2 K = 47.91 J/s m2 K For constant rate period: Rc hfg = h(Tair – Twb) Rc x 2433000 = 47.91 (65.6 – 29) Rc = 7.207x10-4 kg/m2 s = 2.594 kg/m2 hr For surface area A = 0.457 x 0.457 = 0.208849 m2 Drying rate = RcA = 2.594 x 0.208849 = 0.54 kg of water/hour

  6. Problem-1 (Drying) Solution: Part-3 Velocity of drying air, Vair = 0.5 x 6.1 m/s Temperature of drying air, Tair = (65.6+10) oC = 75.6 oC Absolute humidity of drying air,  = 0.01 kg water/kg dry air Again from humidity chart, following adiabatic saturation line from Tair = 75.6 oC and  = 0.01 kg water/kg dry air, wet bulb temperature Twb can be determined as: Twb = 30.5 oC Humid volume, VH = (22.41/273) Tair (1/28.97 + /18.02) = Tair (2.83 x 10-3 + 4.56 x 10-3 ) = (75.6+273)(2.83 x 10-3 + 4.56 x 10-3 x 0.01) = 1.0024 m3/kg dry air Density of moist air, moist air: 1 kg dry air  (1+0.01) kg moist air  1.01 kg moist air moist air = 1.01 / 1.0024 = 1.0076 kg/m3 Mass velocity, G = moist air x Vair = 1.00076 x (0.5 x 6.1) x 3600 = 11063.45kg/m2 hr

  7. Problem-1 (Drying) For parallel flow, convective heat transfer coefficient, h = 0.0204 G0.8 = 0.0204 x (11063.45)0.8 = 35.05 W/m2 K = 35.05 J/s m2 K From steam table, latent heat of vaporization at Twb = 30.5 oC is hfg = 2429 kJ/kg = 2429000 J/kg For constant rate period: Rc hfg = h(Tair – Twb) Rc x 2429000 = 35.05 (75.6 – 30.5) Rc = 6.5 x10-4 kg/m2 s = 2.34 kg/m2 hr For surface area A = 0.457 x 0.457 = 0.208849 m2 Drying rate = RcA = 2.34 x 0.208849 = 0.489 kg of water/hour

  8. Problem-2 (Drying) • The experimental average diffusion coefficient of moisture in a given wood is 2.97 x 10-6 m2/hr. Large planks of this wood 25.4 mm thick are dried from both sides by air having a humidity such that the equilibrium moisture content in the wood is 0.04 kg water/kg dry wood. The wood is to be dried from a total moisture content of 0.29 to 0.09 kg water/kg dry wood. Calculate • Drying time. • If the drying takes place at a higher temperature when the effective diffusivity of water is increased tenfold, what would be the final moisture content of the plank if the drying time is still kept the same? (Note that the simple diffusion model assumes drying to occur very slowly under isothermal conditions) • How long will it take to dry a ball made of the same wood, 2.54 cm in diameter? All other conditions are the same as in (a)

  9. Problem-2 (Drying) Given: Diffusivity of water in wood DL = 2.97 x 10-6 m2/hr Air relative humidity is such that the equilibrium moisture content X*m = 0.04 kg water/kg dry wood Initial moisture content Xm,initial = 0.29 kg water/kg dry wood Final moisture content Xm,final = 0.09 kg water/kg dry wood Estimate: Drying time, t = ?

  10. Problem-2 (Drying) Part-a Only free moisture can participate in transport process One dimensional diffusion equation in Cartesian coordinates: Where Xm = free moisture content = Xm,total – X*m If Xm,total = X*m; free moisture Xm = 0 and no more drying for this drying condition

  11. Problem-2 (Drying) Keeping only the first term of the infinite series (see any book on conduction) and integrating across thickness to obtain average moisture content, (1) Xm,initial = 0.29 – 0.04 = 0.25 kg water/kg dry wood Xm,final = 0.09 – 0.04 = 0.05 kg water/kg dry wood

  12. Problem-2 (Drying) DL = 2.97 x 10-6 m2/hr H/2 = 25.4/2 = 12.7 mm = 0.0127 m Part-b DL = 2.97 x 10-6 x 10 = 2.97 x 10-5 m2/hr t = 30.8 hours Insert in equation (1) above to find (Other parameters held constant) Part-c Solution (approximate) to the one-dimensional diffusion equation in spherical geometry: R = Radius of sphere

  13. Problem-2 (Drying) Note: We can use equation (1) to determine DL for a given solid by properly designing a drying experiment to ensure one dimensional diffusion. Once DL is determined, it can be used to estimate the falling rate period for any other geometry of the same solid over the same moisture content range and at the same temperature. DL can vary by two to three orders-of-magnitude for a given solid during a single drying run. It is a sensitive function of both moisture content and temperature.

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