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Category > Engineering Posted 15 Jun 2017 My Price 4.00

One thousand cubic feet per hour of a 10 mol% NH3 in air mixture is required to produce nitrogen oxides

One thousand cubic feet per hour of a 10 mol% NH3 in air mixture is required to produce nitrogen oxides. This mixture is to be obtained by desorbing an aqueous 20 wt% NH3 solution with air at 20°C. The spent solution should not contain more than 1 wt% NH3. Calculate the volume of packing required for the desorption column. Vapor-liquid equilibrium data for Exercise 6.32 can be used and KGa = 4 lbmovh-ft3-atm partial pressure.

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Alpha Geek
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Status NEW Posted 15 Jun 2017 03:06 PM My Price 4.00

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file 1497540995-Answer.docx preview (128 words )
A-----------nsw-----------er:----------- v-----------olu-----------met-----------ric----------- fl-----------ow -----------rat-----------e o-----------f 1-----------0 m-----------ol -----------% i-----------n a-----------ir -----------mix-----------tur-----------e ------------NH3----------- ga-----------s =-----------Q =-----------100-----------0 f-----------t3 -----------/hr----------- aa-----------ver-----------age----------- mo-----------l.w-----------t a-----------ir ------------NH-----------3 m-----------ixt-----------ure----------- = -----------10 -----------/10-----------0 *-----------17+----------- 90----------- /1-----------00*-----------29 -----------= 1-----------.7 -----------+ 2-----------6.1----------- = -----------27.-----------8 l-----------b /-----------mol----------- sp-----------ent----------- li-----------quo-----------r s-----------hou-----------ld -----------not----------- co-----------nta-----------in -----------= 1----------- /1-----------7 =----------- 0.-----------059----------- mo-----------les----------- of----------- NH-----------3 L-----------iqu-----------or -----------mol-----------es -----------=1------------ 0.-----------059----------- = -----------0.9-----------41 -----------mol-----------es -----------T =----------- 20----------- oc-----------= 2-----------0 *-----------9 /-----------5 +-----------32 -----------= 6-----------8 o-----------F D-----------ens-----------ity----------- = -----------qy -----------= 2-----------7.8----------- *(-----------460----------- +3-----------2)/-----------359----------- * -----------( 4-----------60 -----------+68----------- ) -----------= 2-----------7.8----------- *4-----------92 -----------/35-----------9 *----------- 52-----------8 =-----------0.0-----------72 -----------lb -----------/ft-----------3 q-----------x =----------- 62-----------.3 -----------lb -----------/ft-----------3 a-----------ssu-----------min-----------g t-----------hat----------- , -----------for----------- 1 -----------in
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