# ch13 - Problem 13.1 CHAPTER THIRTEEN Methanol Production...

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Problem 13.1 CHAPTER THIRTEEN Methanol Production Rate 430,000 metric tons/year 51,114 kg/h 1,597 kmol/h Process stoichiometry: CH4 + H20 ---&amp;gt; CH30H + H2 So that the required feed rates (with given assumptions) are CH4 Feed Rate = 1,597 kmol/h 596 standard cubic meters/min Steam Feed Rate = 1,597 kmol/h 28,75lkg/h Problem 13.2 P = 1 atm r.h. = (pH20/p*H20 @ 30C) x 100% = 70% p*H20 = 0.0424 bar pH20 = 0.02968 bar yH20 0.029292 Basis: 1 mole of dry air (79 mole% N2, 21 mole% 02) component moles Mw mass mole frac N2 0.79 28 22.12 02 0.21 32 6.72 Water 0.030176 18 0.54 0.0293 Total 1.03 29.38 1.0000 component moles Mw mass N2 0.79 28 22.12 02 0.21 32 6.72 Total 1.00 . 28.84 Difference in avg molecular weight is due to presence of water; the difference is slight. Basis: 1 km01 of CH4 burned flow rate of air/km01 nat gas burned Problem 13.3 Composition of effluent gas from burners Component km01 mole frac kg mass frac 02 0.100 0.0088 3.2 0.0103 N2 7.900 0.6990 221.2 0.7139 co2 1.000 0.0885 44.0 0.1420 H20 2.302 0.2037 41.4 0.1337 total 11.302 1.0000 309.8 1.0000 p*H20 @ 150C = 4.74 bar &amp;lt; pH20 Therefore, there is no condensation in cooling the exhaust gases to 15OC, which means the effluent gas and stack gas have the same composition. Volumetric flow rate effluent gas 1,143 m3/kmol CH4 burned stack gas 392 m3/kmol CH4 burned density of air = 1.1471 kg/m3 density of stack gas = 0.7899 kg/m3 specific gravity = 1 0.6886 relative to ambient air I 13-1 Problem 13.4 Reformer Temperature 855 C Reformer Temperature 1128 K Reformer Pressure 15.8 atm Equilibrium Temperature 1128 K 1.6 Mpa CH4 + Hz0 ---&amp;gt; CO + 3Hz Production rates CH4 feed rate CH4 x (1 - fractional conversion) H20 feed rate H20 - fractional conversion x feed rate CH4 CO fractional conversion x feed rate CH4 CO2 feed rate CO2 H2 feed rate + 3 x feed rate CH4 x fractional conversion (a) methane:steam of 3:l Stoichiometric Table: Feed I Product (kmol/h)( (kmol/h) MolFrac (kg/h) MassFrac CH4 1600 170 0.0183 2,716 0.0242 H20 4800 3,370 0.3639 60,655 0.5416 co 1,430 0.1544 40,047 0.3576 co2 0.0000 0.0000 H2 4,291 0.4633 8,582 0.0766 Total 6400 9,261 1.0000 112,000 1.0000 KP~ 574.3668 = lO&amp;quot;(-(11,769/T(K))+l3.1927) Ratio1 574.366846 6.8939 ( (Yco x YH2A3) / (YCH4 x YH20) ) p2 fractional conversion of CH4* =-3.57E-07 Converge* = ((Kpl/Ratiol)-1)lOO * the Goal Seek tool is used to adjust the fractional conversion until Converge is close to zero Methane Conversion = 1 (b) Product Flow Rate = - I methane:steam of 1:l Stoichiometric Table: Feed I Product (kmol/h)l (kmol/h) MolFrac (kg/h) MassFrac CH4 1600 471 0.0863 7,538 0.1386 H20 1600 471 0.0863 8,480 0.1559 co 1,129 0.2068 31,608 0.5810 co2 0.0000 0.0000 H2 3,387 0.6205 6,773 0.1245 Total 3200 5,458 1.0000 54.400 1.0000 Kpl 574.3668 = lO&amp;quot;(-(11,769/T(K))+l3.1927) Ratio1 1653.61852 = ( k c 0 x YH2A3) / (YCH4 x Y H 2 0 ) Jp 2 fractional conversion of CH4 = 0.7055-65.266061Converge = ((Kpl/Ratiol)-1)lOO Methane Conversion -- I - [ Product Flow Rate =' 5,458 kmol/h 54,400 kg/h 13-2 Problem 13.4 Problem 13....
View Full Document

## This note was uploaded on 06/07/2008 for the course CHE 2171 taught by Professor Wetzel during the Spring '08 term at LSU.

### Page1 / 50

ch13 - Problem 13.1 CHAPTER THIRTEEN Methanol Production...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online