Question

A simplified version of the process for the manufacture of formaldehyde (HCHO) is shown in the Figure below. Methanol (CH3OH) is oxidized to formaldehyde using air as the source of oxygen as per the following reaction: CH2OH + 2202 HCHO + H2O The reactor is a fixed bed containing an iron-molybdenum oxide catalyst. The reactor feed, stream 4, contains 8 mol% methanol and 10 mol% oxygen. The methanol is completely converted in the reactor. A purge stream (primarily for nitrogen) is taken off the overhead product from the separator. The purge does not contain any water. The bottom product from the separator flows at 3,000 kg/min and is a 37 wt% solution of formaldehyde in water. The additional water required for this product is supplied to the separator in stream 5. Calculate the flow rates of the streams that have labels (i.e. 1-7). 1 Air feed Purge 5 Makeup water Methanol feed Reactor Formaldehyde solution : product Separator- distillation

Answer 1

Puge Air feed (N) * make up Water e = 0.08 Chooh Reactor Separator Distillation 3000 key/min flow - Floo - 30 as Hero CH OH + H2O₂ H CHO + H₂O (ACHO + H2O) Bottom Product from Separator = 3000 kg/min lot of H CHO = 30000 37 = 1110 kg/min lot of Ho = 1890 kg/min moles of Heho in Bottom Product a 3000 legfonin = 30 kg/fernal = 110= 37 Iesnakin As we know from equations that Holo much flow rate of CH₂OH required for this production is same maber flow rate of Chizohlo 37 kmalomin (Streson 2) mal fraction of Cha oh in stream is 0.08 Hence molar flow rate of Stream O= 37 = 462.5 kmal the produced in Reactor is same as Heho = 37 kmahorin toted molor flow rate of Ho in Bottom is = 18926 = 105 komolmin Them filow rate of make up water 105-37 = 68 komethin Stream ② min 0.08

0. 21 O reacted in the reactor is = 37 = 18.5 kemal and no entoer exit of O, is there so mates of malar flow rate of air should be I Ni O₂ in air Streason 1 = 18,5 = 88.09 krahin | 079:21 Na in feed= 69.59 konnthorin & No in pusg & 69.59 lemalios maler flow rate of streson ♡ = stream 6 - Stream o = 462.5 - 88.09 = 344.4) molar flow rate of stream ③ = stresom ® + streson ® = 374.41+ 69.59 = 444.0 bemalonin Note. - It is easy to maly size and Calculate flow rates in terms of knoles/min, you can change Them in legromin if necessory Thank you