Ex Problem 10 - Countercurrent Packed Tower for Absorption of Ammonia Gas (Pilat) Given Gas (air flow) upward of 4,000 acfm at 68F or 528R and 1 atm pressure Inlet gaseous conc. of 6,000 ppm NH 3 (dry basis) Y in = 0.006 Inlet water conc of 0.0006 mole fraction NH 3 in water x in = 0.0006 Liquid flow rate = 500 lb liquid/ ft 2 hr Mol Wt H 2 O = M L =18.016 lb/ Gaseous flow rate of 500 lb gas (air) / ft 2 hr Mol Wt Gas (air) = 29 lb/ atomic wt N = 14.0067 atomic wt H = 1.0079 Assume Govt Emission Std is 1,000ppm NH 3 (dry basis) so y out = 0.001 Find a) H = Henry's law constant from solubility data of NH 3 in water at 20 o C b) H OG from ammonia-water data of Fellinger for 1.5" Raschig ring packing c) Liquid molar flow rate L m (lb moles/ft 2 hr) d) Gaseous molar flow rate (lb moles/ft 2 hr) e) out via Mass Balance around the absorption tower f) Number of gas phase mass transfer units N OG g) Height of packed section in tower Z = (H OG ) ( N OG ) h) Tower Diameter (ft) i) Gas phase overall mass transfer coefficient K G a (lb moles/ft 3 hr atm) j) Gas velocity in tower (ft/sec) k) Gas residence time in packed section of tower (seconds) L) L/G = Liquid/Gas flow rate ratio in gallons water / ft 3 1000 gas m) Draw Y- graph with equilibrium curve and tower operating line. n) Ammonia collection efficiency a) Calc Henry's law constant H using equilibrium solubility data for NH 3 in water Six NH 3 vapor pressures at temperatures in 20 o C range from Table B-4 page 682 Cooper & Alley (3rd Ed) i 0.. 5 (lb NH 3 per NH 3 Partial 100 lb H2O Pressure (mm Hg) 2 12.0 3 18.2 4 24.9 5 31.7 7.5 50.0 10 69.6 2 12.0 12 3 18.2 18.2 4 WtNH3 24.9 lb PNH3 24.9 mmhg PNH3 = torr 5 31.7 31.7 7.5 50.0 50 10 69.6 69.6 molesnh3inwater Mole fraction NH 3 in liquid phase = NH3 xnh3 TotalMolesinWater Mole fraction NH 3 in gas phase = Y NH3 Henry's Law Constant = H PNH3 i y NH3i Ptotal NH3i 0.016 0.021 0.7621 0.024 0.031 0.7786 y NH3i 0.033 H i y 0.041 NH3 = 0.8071 0.042 NH3 = H = 0.05 0.8303 NH3i 0.066 0.074 0.895 0.092 0.096 0.9573 Need to define unit ( mole) ( 453.6) M L M NH3 18.016lb 17.031lb lb M G 29 Y out 0.001 in 0.0006 Y in 0.006 out = Unknown Q Gas 4000ft 3 min 1 Ptotal 1atm P 1atm T 528R atm mmhg 760 WtNH3 i 17.031lb WtNH3 i 100lb + 17.031lb 18.016lb Assume H = 0.7614 = y NH3 / NH3 because the inlet liquid phase NH 3 concentration in is low at 0.0006 Henrys law constant H is in units of (mole fraction NH 3 in gas phase) / (mole fraction NH 3 in liquid phase) (a) H 0.7614 Ex 10 Absorption of Gaseous Ammonia.mcd 1 11/9/2009
b) Obtain height of gas phase mass transfer unit H OG = 3.0 ft from graph (shown below) for 1.5" Raschig rings, L = 500 lb liquid / ft 2 hr, G = 500 lb air / ft 2 hr. pp 456 Noll (b) H OG 3.0ft c) Liquid flow rate in units = L m Liquid flow rate in lb units = L d) Gaseous molar flow rate = Gm Gaseous mass flow rate = G M G = 29gmmole 1 G L 500lb 500lb ft hr 1 ft hr 1 L m L M L (c) M L = 18.016gmmole 1 L m = 27.753 ft 2 hr Y in = 0.006 Y out = 0.001 e) The mass balance equation NH 3 into tower = NH 3 out of tower Y in Gm + in Lm = Y out Gm + out Lm in was given as 0.0006 to can solve for the unknown out 17.241 ft 2 hr out2 27.753 (.006.001) +.0006 ft 2 hr Note that liquid conc out = 0.003706 is in mole fraction units, not ppm (in liquids, ppm means parts per million by weight whereas in gases, ppm means ppm by mole or by gaseous volume) To graph the operating line using the liquid phase concentration as the axis variable we will need a linear equation relating the tower gas phase NH 3 conc Y in terms of the liquid phase NH 3 conc. Starting at in = 0.0006 at the top of the tower and increasing to out one can obtain the linear operating equation for Y. Note that this equation represents the actual gas phase and liquid phase conc in the absorption tower (one could take a sample of the tower gases & liquids & measure the NH 3 conc). Operating line linear equation in = 0.0006 in = 0.0006 G M G out Y L in Y out m + in L m Y in (d) (e) + Y out = 17.241 ft 2 hr H = 0.7614 out = 0.003706 out2 = 0.003706 Ex 10 Absorption of Gaseous Ammonia.mcd 2 11/9/2009
f.) Number of Gas Phase Mass Transfer Units or N OG Ptotal = 1atm Because it is difficult to put an asterisk to denote the equilibrium conc for Y or Y* in Mathcad, let Y starin be the gas phase mole fraction conc of NH 3 in equilibrium with the liquid water at the gaseous inlet (bottom of this tower) and let Y starout be the gas phase conc. of NH 3 in equilibrium with the liquid water at the gaseous outlet (top of tower). Using nomenclature of Y in, Y out, in, out, Ystar in, and Ystar out should help to reduce confusion as to where these concentrations occur and make it easier to label diagrams and tower illustrations. Ystar in ( H) out Ystar in = 0.00282 Ystar out H in Y in Y out N OG Y in Ystar in ( Y out Ystar out ) ln Y in Ystar in Y out Ystar out Ystar out = 0.00046 (f) N OG = 3.352 there are 3.352 gas phase mass transfer units Note that the above equation for N OG assumes a straight equilibrium line and a straight operating line (which is an OK assumption for the absorption of dilute gas concentrations and air pollutants are usually in the dilute concentration range). g.) Height of Packed Tower Section = Z = HOG NOG Z H OG N OG (g) Z = 10.057 ft Z 2 ( 3.0ft) ( 3.352 h) Tower Diameter ) Z 2 = 10.056ft liter atm RG 0.082054 29gm The height of the 1.5 inch Raschig ring packed molek Mg mole section in the absorption tower = 10.057 ft G = lb 500 ft 2 ρ g PMg hr RGT T = 528 R P 1atm Q Gas ρ g Area = tower Area G cross-sectional area gas density = ρ g ρ g = 3 0.075217 lbft 4000ft 3 min 1 Area 0.07522 lb lb 500 ft 3 ft 2 hr Area = 36.106ft 2 0.5 Area4 Diameter π Diameter = 6.78 ft (h) Tower Diameter = 6.78 ft i) Gas Phase Overall Mass Transfer Coefficient K G a = 17.241ft hr 1 Graphs of the measured K G a versus liquid or gas flow rates are provided by equipment manufacturers for various absorption packing types and sizes. KGA 17.241 fthr ( 3ft) ( 1atm) KGA j) Gas Velocity (superficial) in packed tower Velocity Q Gas ( 36.106 ft 2 ) H OG P (i) Area 1 Vel 2 4000ft 3 min 1 1 Vel 2 = 1.846ftsec 1 (j) H OG = 3ft P = 1atm KGA = 1.752m ft 3 hratm Velocity = 1.846ftsec 1 k) Gas Residence time in packed absorption tower section ResidenceTime 2 ( 10.056ft) 1.846ftsec 1 1 (k) ResidenceTime Z Velocity 1 ResidenceTime = 5.447 sec ResidenceTime 2 = 5.447sec Ex 10 Absorption of Gaseous Ammonia.mcd 3 11/9/2009
L ) Liquid to gas flow rate ratio L/G in gallons/1000 cf gas L/G = 9.019 gallons water per 1000 ft 3 gas LG = 9.019 gal 1000 ft 3 LG L 8.34 lb 1 gal 1 Gρ g (l) m) Graph of Y- equilibrium curve and Mass Balance Operating Line j 0.. 7 Ystar = Gas phase equilibrium NH 3 conc in mole fraction units The 8 values shown in the matrix to the right are for the horizontal axis Y out = 0.001 in = 0.0006 Y in = 0.006 out = 0.003706 Now to put the operating line equation on an Y- graph YY j OpSlope j OpSlope = 1.6096803 in the graph and span the range from in to out L m OpSlope H = 0.761 Ystar j H( j ).0005.0006.001.002.0025.0030.0035.003706 YY j ( OpSlope) j in + Y out YY 7 = 0.006 7 = 0.003706 The operating line equation YY = 1.60968 is upper dash-dot line shown in below graph. The difference between the operating line & the lower solid equilibrium equation Ystar = H line shows the NH 3 concentration gradient driving force; the concentration gradient (y-y*) causes the NH 3 to move from the gas into the liquid. NH3 Equilibrium & Operating Line Y (mol fraction NH3 in gas phase) (m) 0.006 Ystar 0.004 YY 0.002 Yout =.001 Operating Line YY = 1.60968 Equilibrium Line Ystar = H Yin =.006 out =.003706 in =.0006 0 0 4 5. 10 Y in = 0.006 out = 0.003706 Y out = 0.001 in = 0.0006 0.001 0.0015 0.002 0.0025 0.003 0.0035 (mol fraction NH3 in liquid) Govt Emission Std is 1,000ppm NH 3 (dry basis) so Yout = 0.001 Remember, the operating line represents the actual concentrations of NH 3 in the gas and liquid inside the tower and thus these are measurable parameters. Y out and in are at the top of the tower and Y in and out are at the bottom of the tower for countercurrent flow (see illustration on next page) Ex 10 Absorption of Gaseous Ammonia.mcd 4 11/9/2009
n) Ammonia collection efficiency PercentCollectionEff ( Y in Y out ) 100 Y in PercentCollectionEff = 83.333 (n) Illustration below shows the ammonia absorption inlet and outlet concentrations for a countercurrent absorption tower. Y out = 0.001 in = 0.0006 Y in = 0.006 out = 0.003706 Y in = 0.006 out = 0.003706 Y = gas phase pollutant concentrations (mole fraction) = liquid phase pollutant concentrations (mole fraction) Pilat Comment: I prefer to use Y in, Y out, in, and out rather than Y 1, Y 2, 1, and 2 because I think it is too easy to forget whether the subscripts 2 and 1 indicate the top or bottom of an absorption tower. Ex 10 Absorption of Gaseous Ammonia.mcd 5 11/9/2009