Separation Process Principles- 2n - Seader &amp; Henley - Solutions Manual

# Volume of one extrusion d2l4 31403752054 00552

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Unformatted text preview: 1,488 – 5,512(0.05) = 1,212 lb/h (c) Q to the feed for sensible and latent heat = 5,512(0.3)[(50 – 15)1.8] + 1,488(1)[(50 – 15)1.8] + 1,212[(1024.6) + (0.45)[(75 – 15)1.8] = 1,499,000 Btu/h (a) Q from the air = 1,499,000 = mair (0.25)[525 – 75)1.8] Solving, mair = 7,400 lb air/h Exercise 18.39 Subject: Drying of isophthalic acid crystals in an indirect-heat, steam-tube, rotary dryer. Given: 5,000 lb/h of wet, isophthalic acid crystals at 30oC and 1 atm with 30 wt% moisture (wet basis) to be dried to a moisture content of 2 wt% (wet basis). Evaporation takes place at 100oC, which is also the crystals exit temperature. Dryer uses condensing steam at 25 psig (barometer = 15 psia). Specific heat of isophthalic acid = 0.2 cal/g-oC. Assumptions: Adiabatic dryer. Heating steam enters as a saturated vapor and leaves as a saturated liquid. Find: (a) The rate of evaporation. (b) The rate of heat transfer. (c) The flow rate of steam. Analysis: (a) mass flow rate of isophthalic acid = 5,000(0.70) = 3,500 lb/h mass flow rate of entering moisture = 5,000 – 3,500 = 1,500 lb/h mass flow rate of moisture in exiting crystals = 3,500(0.02/0.98) = 71.4 lb/h rate of evaporation = 1,500 – 71.4 = 1,429 lb/h (b) Evaporation at 100oC = 212oF. From steam tables, heat of vaporization = 970.3 Btu/lb Rate of heat transfer = 3,500(0.2)[(100 – 30)1.8] + 1,500(1)[(100 – 30)1.8] + 1,429(970.3) = 1,664,000 Btu/h (c) From steam tables, heat of vaporization at 25 + 15 = 40 psia = 933.7 Btu/lb Flow rate of heating steam = 1,664,000/933.7 = 1,782 lb/h Exercise 18.40 Subject: Drying of extruded filter cake of calcium carbonate with a through-circulation belt dryer having 3 zones. Given: Calcium carbonate extruded into cylindrical pieces of 1/4-inch diameter and 1/2-inch length with an initial moisture content of 30% (dry basis) and a critical moisture content of 10% (dry basis). Belt dryer is 6-ft wide with three drying zones of 8-ft long each, running with a belt speed, S, of 1 ft/min (0.00508 m/s). Carbonate bed height on the belt is 2-inches with an external porosity of 50%. Air at 170oF (76.7oC) and 10% relative humidity flows through the bed at a superficial velocity, us,of 2 m/s, upward in the first and third zones and downward in the zone 2. Assumptions: Negligible preheat period. Constant-rate drying at the wet-bulb temperature of the entering air. Find: Moisture content distribution with height at the end of each zone and the final average moisture content. Analysis: This exercise is exactly the same as Example 18.18 except that three zones of 8-ft length each are used instead of two zones of 12-ft length each. Thus, the following preliminary calculations from Example 18.18 apply here: Entering air has a wet-bulb temperature of 110.2oF = 37.8oC = Tw Entering air has a humidity of 0.0265 lb H2O/lb dry air vap Heat of vaporization of water at 110.2oF = 2413 kJ/kg = ∆ H w Particle area /volume = a = 395 m2/m3 Convective heat-transfer coefficient = 0.188 kJ/s-m2-K = h Specific heat to the ai...
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## This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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