Reactor.docx

Reactor The reaction in reactor is as follow

CH3OH + C4H8 → (CH)3COCH3 Material Balance across reactor Total MTBE production = 1000 tons MTBE production = 1000 ton × 1000 kg/1 ton = 1000000kg = 1000000/88 = 11363.63636 kg mole From reaction C4H8 required = 11363.63636 kg mole As conversion is 92.5 % C4H8 required = 11363.63636/0.925 = 12285.012285 kg mole C4H8 react = 11363.63636 kg mole C4H8 unreacted = 12285.012285 – 11363.63636 = 921.37589 kg mole Since, 𝑚𝑒𝑡ℎ𝑎𝑛𝑜𝑙 1.25 𝑏𝑢𝑡𝑦𝑙𝑒𝑛𝑒

=

1

Methanol in 1.25 % excess = 12285.012285 × 1.125 = 13820.63882 kg mole Methanol react = 11363.63636 kg mole Methanol unreacted = 13820.63882 - 11363.63636 = 2457.002 kg mole

Balance across Distillation column In Feed Methanol flow rate = F1 = 2457.00244 kg mole Butylene flow rate = F2 = 921.3759 kg mole MTBE = F3 = 11363.63636 kg mole Total feed flow rate = F = 14742.0147 kg mole

In Bottom As MTBE is heavy key-component so all of it goes to bottom and is 99 % pure. Wx3 = 11363.63636 kg mole W (0.99) = 11363.63636 W = 11478.42057 kg mole Component 1 is light key-component so it all goes to top. Wx1 = 0 Small amount of Component 2 goes to bottom product i.e. 1 % of W, Wx2 = 0.01 × 11478.42057 = 114.7842 kg mole

In Top F-D=W 14742.0147 - D = 11478.42057 D = 3263.5941 kg mole As component 1 is light key so it all goes to top. Dx1 = 2457.0024 kg mole Dx2 = 921.3759 – 114.7842 = 806.5917 kg mole Dx3 = 0

Balance across absorber At inlet Methanol flow rate = 2457.002 kg mole Butylene flow rate = 806.5918 kg mole Total flow rate = 3263.5938 kg mole Since the water used to extract the methanol is fed in a 4:1 molar ratio of water to all entering methanol.so, Water in = 2457.002 × 4 = 9828.008 kg mole

At outlet top Three percent of the water is absorbed into the organic stream. Water flow rate = 0.03 × 9828.008 = 294.8402952 kg mole Butylene goes 5 % of the entering butylene in bottom stream so at top Butylene flow rate = 0.95 × 806.5918 = 766.2622 kg mole Methanol at top is 4 % of the methanol fed in the absorber Methanol flow rate = 0.04 × 2457.002 = 98.2800984 kg mole

At outlet bottom Methanol flow rate = 2457.002 – 98.2800984 = 2358.7224 kg mole Butylene flow rate = 806.5918 - 766.2622 = 40.3296 kg mole Water flow rate = 9828.008 - 294.8402952 = 9533.1695 kg mole

Balance across Flash drum At Inlet Methanol flow rate = 2358.722 kg mole Butylene flow rate = 40.32959 kg mole Water flow rate = 9533.17 kg mole

At outlet Top The vapor leaving the flash drum contains 2.5% of the entering water and 3.5% of the entering methanol. Water flow rate = 0.025 × 9533.17 = 238.3292 kg mole Methanol flow rate = 0.035 × 2358.722 = 82.55528 kg mole Butylene all goes to top Butylene flow rate = 40.32959 kg mole

At outlet bottom Methanol flow rate = 2358.722 - 82.55528 = 2276.167 kg mole Water flow rate = 9533.17 - 238.3292 kg mole = 9294.84 kg mole

Balance across methanol recovery column At Inlet Methanol flow rate = 2276.167 kg mole Water flow rate = 9294.84 kg mole

At outlet top The overhead stream contains 99% of the methanol fed to the column

Methanol flow rate = 0.99 × 2276.167 = 2253.405 kg mole Water flow rate = 0.01 × 9294.84 = 92.9484 kg mole

At outlet bottom Methanol flow rate = 2276.167-2253.405 = 22.76167 kg mole Water flow rate = 9294.84 - 92.9484 = 9201.892 kg mole