CHAPTER 4:The Material Balance for Chemical Reactors Copyright © 2020 by Nob Hill Publishing, LLC Rj V Q0 Q1 cj0 cj1 Conservation of mass rate of rate of inflow rate of outflow accumulation = of component j − of component j of component j rate of generation + of component j by (4.1) chemical reactions 1 / 152 General Mole Balance Rj V Q0 Q1 cj0 cj1 Conservation of mass d cj dV = Q0cj0 Q1cj1 + Rj dV (4.2) dt − ZV ZV 2 / 152 General Mole Balance d cj dV = Q0cj0 Q1cj1 + Rj dV dt − ZV ZV Equation 4.2 applies to every chemical component in the system, j = 1, 2,..., ns , including inerts, which do not take place in any reactions. Assuming component j enters and leaves the volume element only by convection with the inflow and outflow streams, i.e. neglecting diffusional flux through the boundary of the volume element due to a concentration gradient. The diffusional flux will be considered during the development of the material balance for the packed-bed reactor. 3 / 152 Rate expressions To solve the reactor material balance, we require an expression for the production rates, Rj Rj = νij ri i X Therefore we require ri as a function of cj This is the subject of chemical kinetics, Chapter 5 Here we use common reaction-rate expressions without derivation 4 / 152 The Batch Reactor Rj The batch reactor is assumed well stirred Let the entire reactor contents be the reactor volume element 5 / 152 Batch Reactor d cj dV = Q0cj0 Q1cj1 + Rj dV dt − ZV ZV Because the reactor is well stirred, the integrals in Equation 4.2 are simple to evaluate, cj dV = cj VR Rj dV = Rj VR ZVR ZVR The inflow and outflow stream flowrates are zero, Q0 = Q1 = 0. d (c V ) j R = R V (4.5) dt j R 6 / 152 Reactor Volume Equation 4.5 applies whether the reactor volume is constant or changes during the reaction. If the reactor volume is constant (liquid-phase reactions) dc j = R (4.6) dt j Use Equation 4.5 rather than Equation 4.6 if the reactor volume changes significantly during the course of the reaction. 7 / 152 Analytical Solutions for Simple Rate Laws In general the material balance must be solved numerically. If the reactor is isothermal, we have few components, the rate expressions are simple, then analytical solutions of the material balance are possible. We next examine derive analytical solutions for some classic cases. 8 / 152 First-order, irreversible Consider the first-order, irreversible reaction k A B, r = kcA −→ The material balance for a constant-volume reactor gives dcA = kcA (4.8) dt − Watch the sign! 9 / 152 First-order, irreversible We denote the initial concentration of A as cA0, cA(t) = cA0, t = 0 The solution to the differential equation with this boundary condition is kt cA = cA0e− (4.9) 10 / 152 First-order, irreversible 1 0.8 0.6 cA cA0 0.4 k = 0.5 k = 1 0.2 k = 2 k = 5 0 0 1 2 3 4 5 t 11 / 152 First-order, irreversible The A concentration decreases exponentially from its initial value to zero with increasing time. The rate constant determines the shape of this exponential decrease. Rearranging Equation 4.9 gives ln(cA/cA0) = kt − 12 / 152 First-order, irreversible 1 k = 0.5 k = 1 0.1 k = 2 cA cA0 0.01 k = 5 0.001 0 1 2 3 4 5 t One can get an approximate value of the rate constant from the slope of the straight line. This procedure is a poor way to determine a rate constant and should be viewed only as a rough approximation (Chapter 9). 13 / 152 First-order, irreversible The B concentration is determined from the A concentration. 1 Solve the material balance for component B, dc B = R = kc (4.10) dt B A with the initial condition for B, cB (0) = cB0 2 Note that the sum of cA and cB is constant. d(c + c ) A B = R + R = 0 dt A B Therefore, cA + cB is a constant. 14 / 152 First-order, reversible The value is known at t = 0, cA + cB = cA0 + cB0 So we have an expression for cB cB = cA0 + cB0 cA − kt cB = cB0 + cA0(1 e− ) (4.11) − 15 / 152 First-order, reversible Consider now the same first-order reaction, but assume it is reversible k1 A )* B −k −1 − The reaction rate is r = k1cA k 1cB . − − The material balances for A and B are now dcA = r = k1cA + k 1cB cA(0) = cA0 dt − − − dcB = r = k1cA k 1cB cB (0) = cB0 dt − − Notice that cA + cB = cA0 + cB0 remains constante 16 / 152 First-order, reversible Eliminate cB in the material balance for A gives dcA = k1cA + k 1(cA0 + cB0 cA) (4.13) dt − − − How do we want to solve this one? Particular solution and homogeneous solution (see text) Laplace transforms (control course) Separation! 17 / 152 First-order, reversible dc A = ac + b dt A cA dc t A = dt acA + b ZcA0 Z0 1 cA ln(ac + b) = t a A cA0 at b at cA = cA0e (1 e ) − a − Substitute a = (k1 + k 1), b = k 1(cA0 + cB0) − − − 18 / 152 First-order, reversible (k1+k 1)t k 1 (k1+k 1)t cA = cA0e− − + − (cA0 + cB0) 1 e− − (4.15) k1 + k 1 − − h i The B concentration can be determined by switching the roles of A and B and k1 and k 1 in Reaction 4.12, yielding − (k1+k 1)t k1 (k1+k 1)t cB = cB0e− − + (cA0 + cB0) 1 e− − (4.16) k1 + k 1 − − h i 19 / 152 First-order, reversible 1 0.8 cA(t) 0.6 cBs c 0.4 cAs 0.2 cB (t) 0 0 1 2 3 4 5 t Figure 4.5: First-order, reversible kinetics in a batch reactor, k1 = 1, k 1 = 0.5, cA0 = 1, cB0 = 0. − 20 / 152 Nonzero steady state For the reversible reaction, the concentration of A does not go to zero. Taking the limit t in Equation 4.15 gives −→ ∞ k 1 cAs = − (cA0 + cB0) k1 + k 1 − in which cAs is the steady-state concentration of A. 21 / 152 Nonzero steady state Defining K1 = k1/k 1 allows us to rewrite this as − 1 cAs = (cA0 + cB0) 1 + K1 Performing the same calculation for cB gives K1 cBs = (cA0 + cB0) 1 + K1 22 / 152 Second-order, irreversible Consider the irreversible reaction A k B −→ 2 in which the rate expression is second order, r = kcA. The material balance and initial condition are dcA 2 = kc , cA(0) = cA0 (4.18) dt − A Our first nonlinear differential equation. 23 / 152 Second-order, irreversible Separation works here dcA 2 = kdt cA − cA dc t A = k dt c2 − ZcA0 A Z0 1 1 = kt cA0 − cA − Solving for cA gives 1 1 − c = + kt (4.19) A c A0 Check that this solution satisfies the differential equation and initial condition 24 / 152 Second-order, irreversible 1 0.9 0.8 0.7 0.6 cA 0.5 cA0 0.4 second order 0.3 0.2 first order 0.1 0 0 1 2 3 4 5 The second-order reaction decays more slowly to zero than the first-order reaction. 25 / 152 Another second-order, irreversible k A + B C r = kcAcB −→ The material balance for components A and B are dcA = r = kcAcB dt − − dcB = r = kcAcB dt − − Subtract B’s material balance from A’s to obtain d(cA cB ) − = 0 dt 26 / 152 Another second-order, irreversible Therefore, cA cB is constant, and − cB = cA cA0 + cB0 (4.23) − Substituting this expression into the material balance for A yields dcA = kcA(cA cA0 + cB0) dt − − This equation also is separable and can be integrated to give (you should work through these steps), 1 − cB0 (cB0 cA0)kt cA = (cA0 cB0) 1 e − , cA0 = cB0 (4.24) − − c 6 A0 27 / 152 Another second-order, irreversible Component B can be computed from Equation 4.23, or by switching the roles of A and B in Reaction 4.20, giving 1 − cA0 (cA0 cB0)kt cB = (cB0 cA0) 1 e − − − c B0 What about component C? C’s material balance is dc C = kc c dt A B and therefore, d(cA + cC )/dt = 0. The concentration of C is given by cC = cA0 cA + cC0 − 28 / 152 Another second-order, irreversible Notice that if cA0 > cB0 (Excess A), the steady state cAs = cA0 cB0 − cBs = 0 cCs = cB0 + cC0 For cB0 > cA0 (Excess B), the steady state is cAs = 0 cBs = cB0 cA0 − cCs = cA0 + cC0 29 / 152 nth-order, irreversible n The nth-order rate expression r = kcA 5 4.5 4 n = 3 3.5 2 3 r 2.5 1 2 1/2 1.5 0 1 0.5 0 0 0.5 1 1.5 2 2.5 3 cA 30 / 152 nth-order, irreversible A k B r = kcn −→ A dc A = r = kcn dt − − A This equation also is separable and can be rearranged to dcA n = kdt cA − Performing the integration and solving for cA gives 1 n+1 n+1 cA = c− + (n 1)kt − , n = 1 A0 − 6 h i 31 / 152 nth-order, irreversible We can divide both sides by cA0 to obtain cA 1 n+1 = [1 + (n 1)k0t] − , n = 1 (4.25) cA0 − 6 in which n 1 k0 = kcA−0 has units of inverse time.