# CHM ENG 142 Study Guide - Midterm Guide: Volumetric Flow Rate, Mass Flow Rate, Rate Equation

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School
Berkeley
Department
Chemical and Biomolecular Engineering
Course
CHM ENG 142
Professor CBE 142 Homework #6
Assigned: Wednesday, March 4, 2020
Due: Wednesday, March 11, 2020
1. Leaking reactor
An elementary first order gas-phase reaction AB with rate constant k is carried out in an
isothermal, isobaric PFR with a continuous leak along the entire length of the reactor. As a
result of this leak, the volumetric flow rate is v(z) = v0[1 - sqrt(z/L)] at any position z along
the reactor from 0 to length L. Assume that all gases behave ideally.
a. Write an overall mass balance for a differential slice of the reactor using volumetric
flow rate v, fluid density ρ, and the symbol M to indicate a mass flow rate.
M|z - M|z+Δz - Mleak = 0
Given M = ρv and constant density, v|z - v|z+Δz = vleak
Divide both sides by Δz to get (v|z - v|z+Δz)/Δz = dv/dz = -vleak/Δz
Plug in expression for v(z) to get vleak/Δz = v0 / [2 sqrt(zL)]
b. Write a mole balance of species A for a differential slice of the reactor.
F|z - F|z+Δz - Fleak + rAΔV = 0
F|z - F|z+Δz - CAvleak = kCAΔV
c. Allow Δz → 0 and obtain a differential equation for the conversion XA in terms of z.
Solve for XA as a function of z. What is XA when z = L?
FA|z - FA|z+Δz = CA(vleak + kΔV)
Divide both sides by Δz to get -dFA/dz = CA (vleak/Δz) + kCAAc
FA0 (dXA/dz) = (FA0/v) (1 - XA) (kAC + v0 / [2 sqrt(zL)])
dXA/dz = (1 - XA) (kAC/v0 + 1 / [2 sqrt(zL)]) / [1 - sqrt(z/L)]
Integrate by u-substitution with u = sqrt(z/L), du = dz / [2 sqrt(zL)] = dz / (2L*u)
Right side of equation: [a + 1 / (2L*u)] / (1 - u) (2L*u) du = (2aL * u + 1) / (1 - u) du
Integrate to get -2aL u - (2aL+1) ln (1 - u) = -2aL sqrt(z/L) - (2aL+1) ln [1 - sqrt(z/L)]
Simplify to get -2L (kAc/v0) (sqrt(z/L) + ln [1 - sqrt(z/L)]) - ln [1 - sqrt(z/L)]
Therefore XA = 1 - exp(2kLAc/v0 (sqrt(z/L) + ln [1 - sqrt(z/L)]) + ln [1 - sqrt(z/L)])
This equation is undefined at z = L, but if 2L kAc/v0 << 1, XA = 1 - [1 - sqrt(z/L)] = 1
Note: L*Ac = V, so kLAc/v0 = kV/v0 = kτ = Da. When Da << 1, reaction >> residence
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Unlock all 5 pages and 3 million more documents. 2. Reactive distillation
Elementary, reversible reaction A(l) + B(l) ↔ C(l) + D(l,g) is carried out in a semi-batch reactor.
Since product D is volatile, as the reaction proceeds, pressure PD builds up in the headspace of
the reactor as dictated by Henry’s law: CD = HDPD for HD = 5e-5 mol/(L atm). The reactor is
maintained at constant temperature and the headspace is maintained at PD = 5 atm.
a. Solve for dV/dt as a function of the flow rate of D out of the liquid phase FD,out using an
overall mass balance.
Since the only thing leaving the system is product D, dm/dt = - mass of D per time
Therefore, dV/dt = - FD,out MWD / ρD.
b. Solve for dXA/dt as a function of volume and conversion using a mole balance on A.
For a batch reactor, dNA/dt = rAV = -k1CACBV + k-1CCCDV
Based on stoichiometry, CA = NA0/V (1 - XA), CB = NA0/V (ΘB - XA), CC = NA0/V (XA)
Given a constant pressure of D, CD = HDPD = constant
Plug in values: dXA/dt = [(NA0/V) k1 (1 - XA)(ΘB - XA) - k-1(HDPD) XA]
c. Solve for the flow rate of D out of the liquid phase FD,out as a function of volume and
conversion using a mole balance on D.
For product D, dND/dt = rDV - FD,out
FD,out = d(CDV)/dt + k1CACBV - k-1CCCDV
FD,out = HDPD dV/dt + [(NA0/V) k1 (1 - XA)(ΘB - XA) - k-1(HDPD) XA]
Plug in dV/dt: FD,out (1 - HDPD MWD / ρD) = [(NA0/V) k1 (1 - XA)(ΘB - XA) - k-1(HDPD) XA]
Therefore FD,out = [(NA0/V) k1 (1 - XA)(ΘB - XA) - k-1(HDPD) XA] / (1 - HDPD MWD / ρD)
d. Given initial conditions NA0 = NB0 = 50 moles and V0,liquid = 300 L, liquid = 1000 g/L,
MWD = 60 g/mol, k1 = 1 mol/(L hr), and KC = 0.05, numerically solve the system of
equations from parts a-c and plot XA from t = 0 to 10 seconds. On the same graph, plot
XA from t = 0 to 10 seconds in the case in which D remains liquid.
Mass balance: dV/dt = - FD,out MWD / ρD = -0.06 FD,out
Mole balance on A: dXA/dt = - [50/V (1 - XA)2 - 0.005 XA]
Mole balance on D: FD,out = [50/V (1 - XA)2 - 0.005 XA] / 0.999985
Plug FD,out into equation for volume: dV/dt = -0.06 [50/V (1 - XA)2 - 0.005 XA]
Now there are two variables, V and XA, which depend on time, and two equations
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## Document Summary

Due: wednesday, march 11, 2020: leaking reactor. An elementary first order gas-phase reaction a b with rate constant k is carried out in an isothermal, isobaric pfr with a continuous leak along the entire length of the reactor. Given m = v and constant density, v| z - v| z+ z = v leak. Divide both sides by z to get (v| z - v| z+ z )/ z = dv/dz = -v leak / z. Plug in expression for v(z) to get v leak / z = v 0 / [2 sqrt(zl): write a mole balance of species a for a differential slice of the reactor. F| z - f| z+ z - f leak + r a v = 0. F| z - f| z+ z - c a v leak = kc a v: allow z 0 and obtain a differential equation for the conversion x a in terms of z.