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

Volume of one extrusion d2l4 31403752054 00552

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

Ask a homework question - tutors are online