Chemical Engineering: An Introduction (2012) – Solutions Manual & PPT Slides – Denn

Chapters 2-15 & Powerpoint Slides Covered




SOLUTIONS MANUAL

, Solutions Manual
for

Chemical Engineering: An Introduction
Morton M. Denn




Copyright © 2011, Morton M. Denn




1

,Chapter 2

2.1 The equation for conservation of mass with a constant density simplifies to
dV / dt = q f − q , where V is the volume of liquid at any time. V(t) = BLh2/2H. With a bit

of manipulation we then obtain
dh 2 2 H
= (q f −q ), which integrates for constant q f − q
dt BL
to
2H
h 2 − ho2 = (q f − q )(t − to ) , where ho = h(to). The height is then
BL

h(t ) = ho2 +
2H
(q f − q )(t − to ) .
BL

2.2 This is hokey problem, but it is a good exercise in writing a mass balance.
M < Mo, dM/dt = W1; M ≥ Mo, dM/dt = W1 – k(M – Mo)
(a) M = W1t for M < Mo . Thus to = Mo/W1.
(b) The simplest way to integrate the equation for t > to (i.e., M > Mo) is to define a new
variable u = M – Mo – W1/k, which turns the equation into du/dt = –ku, which is
⎡ −k ⎜ t − o ⎟ ⎤
⎛ M ⎞
W1 ⎢ ⎜ W ⎟
separable. u = –W1/k at t = Mo/W1. The solution is M (t ) = M o+ 1− e ⎝ 1 ⎠⎥ . A
k ⎢ ⎥
⎣ ⎦
W
steady state M = M o+ 1 is reached as t → ∞
k
d (h − h* ) K fb Q(t )
2.3 (a) The equation for the height becomes =− (h − h* ) + (1 − K ff ) ,
dt A A
−K t / A t
e fb K τ/A
with a solution h(t ) − h = ∫ e fb (1 − K ff )Q(τ )dτ in analogy to the development
*

A 0
leading to Eq. 2.9. If Kff = 1 the solution is h = h* for all time; i.e., perfect control. If Kff <
1 the steady-state offset is reduced to Q*(1 – Kff)/Kfb. There is no reason to use Kfb > 1,
which would change the sign of the steady-state offset.
−K t / A t
e fb K τ/A
(b) Now h(t ) − h = ∫ e fb [Qu (τ ) + (1 − K ff )Q m (τ )]dτ . Clearly the design
*

A 0
equation for Kff = 1 is identical the that in Section 2.6.3, with Q(t) replaced by Qu(t). A
conservative design for the feedback controller design for Kff < 1 would use the
maximum expected value of Qu + (1 – Kff)Qm in the development in Section 2.6.3.
(c) Let u = h – h*. The equation for the height then becomes
t
du
A = − K fbu − K I ∫ u (τ )dτ +Q(t ) . If we differentiate this once with respect to t and
dt 0
t
d d 2u du dQ
recall that
dt 0∫ u (τ ) d τ = u (t ) we obtain A
dt 2
+ K fb
dt
+ KIu =
dt
. This is identical to

the equation for a forced damped harmonic oscillator. If Q = constant the right-hand side
goes to zero, and we know that an unforced damped oscillator will decay to u = 0, or h =
h*.


2

, 2.4 (a) Following the development in Section 2.7.2, k/A = 0.0125s-1. From Eq. 2.14, h =
6.56cm, compared to an experimental value of 2.5.
(b) A plot of h1/2 vs t gives a best slope k/2A = 0.0338cm1/2s-1. With k = CAog1/2 and the
specified diameters of the tank and orifice we obtain a value C = 1.17, which is rather
large. The value of C computed from the data in Table 2.1 is 0.91

2.5 (a) The slope is 5.2x10-3m/s, so the tank is predicted to drain in 587 seconds.
(b) k/2A = 0.0016m1/2s-1, so the tank is predicted from Eq. 2.18 to reach h = 0 at 1,090s.
(c) C = 0.84.

2.6 dV/dt = q – kh1/2 . At steady state dV/dt = 0 and h = (q/k)2.
h
2.7 This requires a bit of calculus that should be familiar. V (t ) = ∫ A( y )dy , where y is a
0

dV dh
dummy variable of integration, so = A(h) = −kh < 0. We want dh/dt to be
dt dt
constant regardless of the liquid level, so A(h) must be proportional to h1/2. The area is a
circle of radius R(h) at each value of h, so R(h) is proportional to h1/4.

2D.1 There are four variables: m, g, s, and t, and there are three dimensions, so there will
be one independent dimensionless group. m is the only variable involving mass, so it
cannot appear in the group. The result is s/gt2 = C = constant. From elementary physics,
the rate of change of momentum (mass times acceleration) = the sum of imposed forces:
md2s/dt2 = mg. Dividing by m and integrating twice gives s = ½gt2.

2D.2 (a) The variables are p(M/Lθ2), u(L/θ), V(L3), n(moles), and Mw(M/moles). There
are two equivalent ways to do this problem. Either we consider five variables and four
dimensions, or we recognize that the product nMw must always appear as a single
variable with dimensions of mass, in which case we have four variables and three
dimensions. In either case there is one dimensionless group, which must be a constant.
That group is clearly pV/nMwu2, so pV =C nMwu2, where C is a constant.
(b) Mwu2 = KT, so pV = nRgT, where Rg = CK is a constant to be determined
experimentally.

2D.3 (a) There are five variables: Q(L3/θ), ρ(M/L3), Δp/L ((M/L2θ2), R (L), and η(M/Lθ),
and three dimensions. Hence, there will be two independent dimensionless groups. The
way to ensure independence is to have at least one variable, say Q, appear in one group
but not in the other. Two (non-unique) groups are then ρQ/Rη and ρR3Δp /η2L, so ρQ/Rη
= f(ρR3Δp/η2L), or Q = (Rη/ρ) f(ρR3Δp/η2L), where f is an unknown function.
(b) If Q is proportional to Δp/L then the function f must be proportional to its argument;
hence, Q = CR4Δp/ηL, where C is a constant. Q is independent of the density, which is a
consequence of the unimportance of inertia in this flow. In a fluid mechanics course it is
shown that the flow of fluid elements in the “laminar” regime is parallel to the pipe wall
at a constant speed, so there is no change in momentum, hence no inertial contribution.


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