A 2. = and. Solution:

Transcription

1 The ollowg gas phase reaction is conducted a PFR, under isothermal conditions. k k B. However, another reaction also occurs parallel. The rate equations are r = ( k + k, rb = k and r = k, with k = 0.05 s - and k = m 3 /mol/s. The eed 5 mol/s, mol/s mol/s. The total volumetric low rate o the eed is 00 lit/s. The PFR diameter is 0 cm and length is 00 m. The temperature is K and the let pressure is atm (absolute. ssumg ideal gas law, determe the pressure and concentrations o, B and at the outlet. The viscosity o the gas remas constant the reactor and is. 0-5 Pa-s. The density o the gas at the let is 4 kg m -3. The riction actor 6.9 or a smooth pipe is given by the equation =.8log0 Re Solution: Note: The number o moles lowg depends on the extent o reaction. The pressure drop may or may not be signiicant. So we should take pressure drop to consideration. Sce there is more than one reaction, it is better to use concentration or molar low rate than conversion the design equation. We will approximate atm = 0 5 Pa. R = 8.34 mol/j/k. = 0 - m 3 s -. P = 0 5 Pa. F - = 5 mol s -, F B- = mol s - and F - = mol s -. k = 0.05 s - and k = m 3 mol - s -. Total molar low rate at the let is F T- = 8 mol s -. The volume o the reactor is.5708 m 3. The mean residence time is 5.7 seconds. t the let average velocity is.73 m/s. The design equation or all the three species is given by df r dv =, dfb rb dv = and df r dv = The rate equations are given terms o concentrations, which can be written terms o molar low rates F = and ( + + FT RT F F F P P B = = RT. Hence = FP F F F RT ( + + B Note that R and T are constants. The design equations become, then df FP FP = r = ( k + k = k + k dv ( F + FB + F RT ( F + FB + F RT

2 dfb FP = rb = k = k dv F F F RT ( + + B df FP = r = k = k dv ( F + FB + F RT and The Re is given by 4( ρ 5 dvav ρ 4ρ = = = µ π dµ π dµ The riction actor or a smooth pipe is given by the equation = log0 Re Usg this we get = This should be used the pressure drop equation. τ The riction actor or low a pipe is given by = and the orce balance gives ρ Vav π d Pd P = π dlτ. Thereore, τ = ρ Vav =, which leads to 4 4L dp ρ Vav 3ρ 3( ρ 3( ρ ( F + FB + F = = = = RT dz d π d π d π d P ( F + F + F dp 4 B = dz P Divide by tube cross sectional area to get dv on the LHS denomator dp 8( ρ ( F + FB + F = RT 3 7 dv π d P Instead o usg pressure directly, we can dee a new variable P-new = P/0 5. This ensures that all the dependent variables are o similar order o magnitude dp 8( ρ new = dv 0 π d = The three mass balance equations (i.e. design equations should be solved along with the pressure drop equation. The itial conditions are known. The results are shown the plots below.

3 V av- =.734 m/s, Re = dp, = , new dv = I we calculate dp/dv at the let and assume it is a constant, we get a pressure drop o atm, which is not correct (real value is approx 0.04 atm but is a reasonable approximation. t the exit, F =.30 mol/s, F B = mol/s, F =.6785 mol/s, P = Pa. F T = mol/s = F T R T / P = m 3 /s = 3.44 lit/s. Flow rate at the let was 00 lit/s. The crease is due to a signiicant crease molar low rate (8 to 0. mol/s and a slight decrease pressure (rom atm to.9 atm.

4 I we provide the same volume by slightly decreasg the diameter (rom 0 cm to 7.5 cm, and creasg the length (00 to 356 m, then the results are t the exit, F =.4775 mol/s, F B = 5.9 mol/s, F =.575 mol/s, P = Pa. I we assume dp/dz is a constant, we get P = 0.67 atm, which is not correct, the right value beg close to atm. F T = mol/s, = F T R T / P = m 3 /s = 33.3 lit/s. This is maly due to the decrease the pressure and to some extent due to the crease number o moles. What i we try to use a smaller tube? Diameter, d = 6 cm and length z = 556 m, so that the volume remas the same?

5 Pressure decreases rapidly and the program is not able to tegrate beyond a volume o 450 lit. The pressure drop is very sharp that region and is likely to go to zero and lower (which is meangless. ll that it means is We can t pump the gas at the required low rate with only 7 atm pressure at the let. Either we need to provide a higher pressure at the let OR we have to settle or a lower low rate. Gettg the value o dp/dz at the let and assumg it is a constant gives us the total pressure drop as.85 atm, which is not correct. However, this dicates that we should be usg the correct pressure drop equation and that we should tegrate it, rather than neglect the pressure drop or assume that it is a constant. Fally, i we use d = 0 cm, and z = 50 m, we get almost negligible pressure drop ( atm to.999 atm. lso, Re = 0 5, =0.0038, P = atm, F T = 0. mol/s, all o which is due to the change number o moles due to the reaction. (Plots are not given.