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Air containing 5 mol% NH3 at a total flow rate of 40 lbmol/hr enters a packed co

ID: 527473 • Letter: A

Question

Air containing 5 mol% NH3 at a total flow rate of 40 lbmol/hr enters a packed column operating at 25 degree C and 1 atm, where 90% of the ammonia is scrubbed by 3000 lb/hr of water flowing countercurrently. Use the generalized chart with packing factor FP to estimate the superficial gas flooding velocity (ft/s), the column inside diameter (in) for operation at 70% of flooding, and the pressure drop per foot of packing (in H_2 O/ft) for one-inch ceramic Raschig rings (F_p = 179 ft^2/ft^3). b) Calculate the number of transfer units and the length of the tower (ft) required. The following equilibrium data are available for the ammonia-water system at 25 degree C and 1 atm.

Explanation / Answer

Ans:- a) Given molar mass of ammonia NH3 = 17

molar mass of air = 32

Average molecular weight = 0.9x17+0.1x32

                                      = 18.5g

gas flow rate = 40 x 18.5/3000

                    = 0.2466kg

Density of the gas = total pressure x Average molecular weight / RT

                            = 101.3 x 18.5 /8.314x298

                             =0.756kg/m3

volumetric flow rate = gas flow rate / density

                              = 0.2466/0.756

                               = 0.3275m3/s

Now total NH3 absorbed = 40x0.1x0.9x17

                                     = 61.2kg/hr

liquid flow rate = 40x18+61.2/3000

                       = 0.2604kg/s

density of the liquid = 1000kg/m3

flow parameter = volumetric flow rate/ gas flow rate (density of gas/liquid flow rate)0.5

                      = 0.3275/0.2466(0.756/0.2604)0.5

                      = 2.24

Now n = exp(flow parameter)

                    =exp(2.24)

                    =9.39

viscocity of the liquid= 0.84x10-3 , here liquid = water

Packing factor Fp = 179ft2/ft3

superficial gas flooding velocity V= (n/Fp x viscocity)1/2

                                                = (9.94/ 179x0.84x10-3 )1/2

                                                 = 0.41m/s

From the generalised chart pressure drop P = (Fp)0.7

                                                                 = (179)0.7

                                                                  = 16054Pa

column diameter D = 4 x volumetric flow rate / 0.7xVxflow parameter)

                             = 4x 0.3275/ 0.7x0.41

                              = 4.67m

Ans b) Let A be the amount

A = liquid flow rate / flow parameter x gas flow rate

= 0.2604/2.24x0.2466

= 0.471

Total xNH3 = 0.02+0.04+0.06+0.07+0.082 =0.272

Total yNH3 =0.025+0.08+0.021+0.26+0.40

                = 0.786

Number of transit unit = 1/1-Aln[flowparameter(XNH3 ) -YNH3] / 1-1/A (1-(flow parameter XNH3) -YNH3]

                                  =1/1-0.471ln(2.24x0.272 -0.786)/1-1/0.47(1-2.24x0.272-0.786)

                                   = 5.236

length of the tower = height of the system/ number of transit unit

                            = 1/5.236

                            =19.09ft

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