Unformatted text preview: ated, the humidity of the air would be:
10.57/280 = 0.0378 lb water/lb dry air, which, from Figure 18.17 corresponds to
87% relative humidity. This is an upper limit. Iteration 1:
Assume a relative humidity, HR, of 60%. From Fig. 18.17, H = 0.025 lb/lb dry air.
Therefore, moisture in the air = 0.025(280) = 7 lb moisture.
Moisture remaining in the cotton cloth = 10.57 – 7 = 3.57 lb moisture
This gives a moisture content of the solid = 3.57/41.67 = 0.0857 lb H2O/lb dry air
From Figure 18.25 at 100oF, the equilibrium HR = 59% Exercise 18.26 (continued)
Assume a relative humidity, HR, of 59%. From Fig. 18.17, H = 0.0245 lb/lb dry air.
Therefore, moisture in the air = 0.0245(280) = 6.86 lb moisture.
Moisture remaining in the cotton cloth = 10.57 – 6.86 = 3.71 lb moisture
This gives a moisture content of the solid = 3.71/41.67 = 0.089 lb H2O/lb dry air
From Figure 18.25 at 100oF, the equilibrium HR = 61%
By interpolation, the final equilibrium moisture content of the solid is 8.8 wt% (dry basis)
and the relative humidity of the air is 59.5%.
Now calculate the final pressure in the room. Initially, have:
2.24 lb water = 2.24/18.02 = 0.124 lbmol
280 lb air = 280/28.97 = 9.665 lbmol
Total lbmol of initial gas = 0.124 + 9.665 = 9.789 lbmol
At equilibrium, have:
10.57 – 0.088(41.67) = 6.903 lb water in gas or 6.903/18.02 = 0.383 lbmol
Total lbmol of final gas = 0.383 + 9.665 = 10.048 lbmol
Assuming an ideal gas, final pressure = 1(10.048/9.789) = 1.026 atm Exercise 18.27
Subject: Batch drying of raw cotton in a cross-circulation tray dryer.
Given: Raw cotton with an initial total moisture content of 95 wt% (dry basis) with a dry
density of 43.7 lb/ft3, to be dried to 10 wt% (dry basis). Dryer trays are 3 cm high with an area
of 1.5 m2 and are filled with cotton. Tray bottoms are insulated. Heating medium is air at 160oF
and 1 atm, with a relative humidity of 30%. The air flows at mass velocity across the tray of 500
lb/h-ft2. Under these conditions, the critical moisture content is 0.4 lb water/lb bone-dry cotton,
with a falling-rate drying period like Figure 18.31a, based on free-moisture content.
Assumptions: Applicability of Figure 18.25 for equilibrium moisture content.
Find: (a) Amount of raw cotton in lb (wet basis) that can be dried in one batch for 16 trays.
(b) Drying time for the constant-rate period.
(c) Drying time for the falling-rate period.
(d) Total drying time if preheat period is 1 h.
Analysis: (a) Compute the amount of cotton in 16 full trays.
Assume cotton volume does not change when it is wetted.
Volume of one tray = 3(1.5)(104) = 45,000 cm3
Amount of dry cotton per tray = 45,000(43.7/62.4) = 31,500 g or 69.4 lb
Total mass of wet cotton = 69.4(1.95) = 135 lb of wet cotton
For 16 trays, can dry 16(135) = 2,160 lb of wet cotton
(b) Assume during the constant-rate drying period that the relative humidity and
temperature of the air remains constant as it passes over the trays.
For 30% relative humidity at 160oF, from Figure 18.25...
View Full Document
- Spring '11
- The Land