Unformatted text preview: f sublimation
ρc = density of the crystal layer
kc = thermal conductivity of the crystal layer
L = length of the tube
Tg = bulk temperature of the gas carrying the component that desublimes
Tc = bulk temperature of the coolant
ALM = log mean area for conduction through the crystal layer ALM = 2πrL − 2πro L
2πrL
ln
2πro L (3) Combining (2) and (3), simplifying, separating variables, and putting in integral form,
rs
ro r ln kc (Tg − Tc )
r
dr =
ro
ρc ∆H s Carrying out the integration and applying the limits gives (1). t 0 dt (4) Exercise 17.39
Subject: Evaporation of aqueous NaOH in a singleeffect evaporator
Given: Feed of 50,000 lb/h of a 20 wt% aqueous solution of NaOH at 120oF. Concentration to
40 wt% NaOH at a pressure of 3.7 psia. Heating medium is saturated steam at a temperature
40oF higher than the solution temperature in the evaporator.
Assumptions: Perfect mixing in the evaporator. No heat losses.
Find: (a) Boilingpoint elevation of the solution
(b) Saturated heating steam temperature and pressure
(c) Evaporation rate
(d) Heat transfer rate
(e) Heating steam flow rate
(f) Economy
(g) Heattransfer area if U = 300 Btu/hft2oF
Analysis:
(a) From the steam tables, at 3.7 psia, saturated steam temperature = 122.3oF. From the
During chart of Fig. 17.32, the boiling temperature of 40 wt% NaOH = 166oF. Therefore the
boilingpoint elevation = 166 – 122 = 44oF. This result is consistent with Fig. 17.33.
(b) Saturated heating steam temperature = 166 + 40 = 206oF, with a pressure of 13 psia.
Note: Best not to use steam under vacuum. Should change to atmospheric pressure.
(c) NaOH flow rate = 0.2(50,000) = 10,000 lb/h. Entering water rate = 50,000 – 10,000 =
40,000 lb/h. Water in concentrated NaOH = (60/40)10,000 = 15,000 lb/h. Therefore
evaporation rate = 40,000 – 15,000 = 25,000 lb/h.
(d) Calculate heat transfer rate by an enthalpy balance around the evaporator using Fig.
17.35 and the steam tables for enthalpies.
Q = mvHv + mpHp  mfHf
Note that the vapor produced is superheated steam at 3.7 psia and 166oF. Q = 25,000(1,133.7) + 25,000(145)  50,000(72) = 28,400,000 Btu/h
(e) Heating steam = ms = Q/∆Hs = 28,400,000/974.2 = 29,100 lb/h
(f) Economy = 25,000/29,100 = 0.859 = 85.9%
(g) From (1781), A= Q
28, 400, 000
=
= 2,370 ft 2
U (Ts − Tp ) 300 ( 206 − 166 ) (1) Exercise 17.40
Subject: Comparison of singleeffect to doubleeffect evaporation.
Given: Feed of 30,000 lb/h of 10 wt% NaOH at 100oF. Concentrated to 50 wt% using saturated
steam at 115 psia.
Assumptions: Perfect mixing in the evaporator. No heat losses. Equal heattransfer areas for
the doubleeffect system.
Find: (a) Heattransfer area and economy for a single effect, with U = 400 Btu/hft2oF and
vaporspace pressure of 4 in Hg.
(b) Heattransfer areas and overall economy for a doubleeffect system, with forward
feed and U = 450 for the first effect and 350 for the second effect. Vaporspace pressure of 4 in
Hg for the second effect
Analysis: (a) Singleeffect
First compute the mass balance.
NaOH in t...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

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