PSE-823: Advanced Process Dynamics and Control¶

Lecture 2 Companion -- Full Derivations and Worked Answers¶

(Coughanowr & LeBlanc, Ch. 2-3)¶

Setup (run once)¶

  • Imports used below
In [1]:
import sympy as sp
t, s = sp.symbols("t s", positive=True)

Part A¶

Mixing Scenario: Separation of Variables and Prob. 2.9¶

(Coughanowr & LeBlanc, Ch. 2, Sec. 2.1)

Eq. 2.2 by separation¶

$5\,dC_a/dt = 2 - C_a$. Separate: $\frac{dC_a}{2 - C_a} = \frac{dt}{5}$. Integrate from $(0, 3)$: $-\ln(2 - C_a)\big|_3^{C_a} = t/5$, so $\ln\frac{2 - C_a}{2 - 3} = -t/5$, i.e. $C_a - 2 = e^{-t/5}$.

Prob. 2.9 (stream 1 raised to 20 L/min)¶

$v_3 = 40$ L/min, $C_{a3} = (20 + 80)/40 = 2.5$ g/L, $\tau = 150/40 = 3.75$ min. With $C_a(0) = 3$: $3.75\,dC_a/dt + C_a = 2.5$, so $C_a = 2.5 + 0.5e^{-t/3.75}$.

Check -- Prob. 2.9 by dsolve¶

  • The same ODE with the new $\tau$ and inlet
In [2]:
C = sp.Function("C")
ode = sp.Eq(sp.Rational(15, 4) * C(t).diff(t) + C(t),
            sp.Rational(5, 2))
print(sp.dsolve(ode, C(t), ics={C(0): 3}).rhs)

5/2 + exp(-4*t/15)/2

Code walkthrough. Rational numbers keep the result exact: 15/4 = 3.75 and 5/2 = 2.5. dsolve returns 5/2 + exp(-4 t/15)/2, i.e. 2.5 + 0.5 e^(-t/3.75).

Part B¶

Energy Balance: Where the Reference Temperature Goes¶

(Coughanowr & LeBlanc, Ch. 2, Sec. 2.1)

Tee¶

$\rho v_1C_p(T_1 - T_{ref}) + \rho v_2C_p(T_2 - T_{ref}) = \rho v_3C_p(T_3 - T_{ref})$. Since $v_1 + v_2 = v_3$, the $T_{ref}$ terms cancel: $v_1T_1 + v_2T_2 = v_3T_3$.

Vessel¶

$\rho v_3C_p(T_3 - T_{ref}) - \rho v_3C_p(T - T_{ref}) + Q = \frac{d}{dt}[\rho VC_p(T - T_{ref})]$. Constant $\rho, V, C_p$: divide by $\rho v_3C_p$ to get Eq. 2.3, $\tau\,dT/dt + T = T_3 + Q/(\rho v_3C_p)$, $\tau = V/v_3$.

Prob. 2.9, energy part¶

$T_3 = (500 + 1100)/40 = 40$ C, $\tau = 3.75$ min, $Q/(\rho v_3C_p) = 1.05\times10^6/40000 = 26.25$ C, final $T = 66.25$ C, $T(t) = 66.25 + 13.75\,e^{-t/3.75}$.

Check -- the reference temperature drops out¶

  • Keep $T_{ref}$ symbolic and simplify
In [3]:
v1, v2, T1, T2, T3, Tr = sp.symbols("v1 v2 T1 T2 T3 T_ref")
tee = sp.Eq(v1*(T1 - Tr) + v2*(T2 - Tr), (v1 + v2)*(T3 - Tr))
print(sp.solve(tee, T3)[0])

(T1*v1 + T2*v2)/(v1 + v2)

Code walkthrough. The tee balance is typed with T_ref everywhere and v3 = v1 + v2. Solving for T3 gives (T1 v1 + T2 v2)/(v1 + v2): T_ref has vanished, as the book states.

Part C¶

The Derivative Rule, Proved¶

(Coughanowr & LeBlanc, Ch. 2, Sec. 2.2)

Eq. 2.6 by parts¶

$\int_0^\infty f'(t)e^{-st}dt$ with $u = e^{-st}$, $dv = f'dt$, $du = -se^{-st}dt$, $v = f$: $= [fe^{-st}]_0^\infty + s\int_0^\infty fe^{-st}dt = -f(0) + sF(s)$. Applied twice: $\mathcal{L}\{f''\} = s^2F - sf(0) - f'(0)$.

Ex. 2.4¶

$sx - 2 + 3x = 0 \Rightarrow x(s) = 2/(s+3) \Rightarrow x(t) = 2e^{-3t}$.

Check -- second-derivative rule¶

  • Same test as in the lecture, one order higher
In [4]:
f = sp.cos(3 * t) + t**2
F = sp.laplace_transform(f, t, s, noconds=True)
lhs = sp.laplace_transform(f.diff(t, 2), t, s, noconds=True)
rhs = s**2*F - s*f.subs(t, 0) - f.diff(t).subs(t, 0)
print(sp.simplify(lhs - rhs))

0

Code walkthrough. f(0) = 1 and f'(0) = 0 for this test function; the difference between the direct transform of f'' and s^2 F - s f(0) - f'(0) simplifies to 0.

Part D¶

Partial Fractions in Full¶

(Coughanowr & LeBlanc, Ch. 3, Sec. 3.1)

Ex. 3.3 cover-up table¶

$x(s) = \frac{s^4 - 6s^2 + 9s - 8}{s(s-2)(s+1)(s+2)(s-1)}$. Cover each factor and set $s$ to its root: $A = \frac{-8}{(-2)(1)(2)(-1)} = -2$, $B = \frac{16 - 24 + 18 - 8}{(2)(3)(4)(1)} = \frac{1}{12}$, $C = \frac{1 - 6 - 9 - 8}{(-1)(-3)(1)(-2)} = \frac{11}{3}$, $D = \frac{16 - 24 - 18 - 8}{(-2)(-4)(-1)(-3)} = -\frac{17}{12}$, $E = \frac{1 - 6 + 9 - 8}{(1)(-1)(2)(3)} = \frac{2}{3}$.

Ex. 3.5 coefficient matching¶

With $A = 1$, $B = -1$: $1 = (1 + D)s^3 + (3 + C + 2D)s^2 + (2 + C + D)s + 1$, so $D = -1$, $C = -1$.

Check -- cover-up as a limit¶

  • Cover-up = multiply by the factor, then let $s$ go to the root
In [5]:
X3 = (s**4 - 6*s**2 + 9*s - 8) / (s*(s-2)*(s+1)*(s+2)*(s-1))
for root in [0, 2, -1, -2, 1]:
    print(root, sp.limit(X3 * (s - root), s, root))

0 -2
2 1/12
-1 11/3
-2 -17/12
1 2/3

Code walkthrough. Multiplying by (s - root) cancels that factor; the limit evaluates the rest at the root -- the cover-up rule written as calculus. Prints -2, 1/12, 11/3, -17/12, 2/3, the five coefficients in the book.

Prob. 2.9 inverted (Lecture Your Turn D)¶

$3.75s + 1 = 3.75(s + 0.2667)$. First term: $\frac{2.5/3.75}{s(s + 0.2667)} = \frac{2.5}{s} - \frac{2.5}{s + 0.2667}$. Second term: $\frac{11.25/3.75}{s + 0.2667} = \frac{3}{s + 0.2667}$. Sum $\frac{2.5}{s} + \frac{0.5}{s + 0.2667}$, so $C_a(t) = 2.5 + 0.5e^{-t/3.75}$, matching Part A.