Chapter 2
2.1
a) Overall mass balance:
d (ρV )
= w1 + w2 − w3 (1)
dt
Energy balance:
d ρV (T3 − Tref )
C = w1C (T1 − Tref ) + w2C (T2 − Tref )
dt (2)
− w3C (T3 − Tref )
Because ρ = constant and V = V = constant, Eq. 1 becomes:
w3 = w1 + w2 (3)
b) From Eq. 2, substituting Eq. 3
d (T3 − Tref ) dT
ρCV = ρCV 3= w1C (T1 − Tref ) + w2C (T2 − Tref )
dt dt (4)
− ( w1 + w2 ) C (T3 − Tref )
Constants C and Tref can be cancelled:
dT3
ρV = w1T1 + w2T2 − ( w1 + w2 )T3 (5)
dt
The simplified model now consists only of Eq. 5.
Solution Manual for Process Dynamics and Control, 4th edition
Copyright © 2016 by Dale E. Seborg, Thomas F. Edgar, Duncan A. Mellichamp,
and Francis J. Doyle III
2-1
, Degrees of freedom for the simplified model:
Parameters : ρ, V
Variables : w1, w2, T1, T2, T3
NE = 1
NV = 5
Thus, NF = 5 – 1 = 4
Because w1, w2, T1 and T2 are determined by upstream units, we assume
they are known functions of time:
w1 = w1(t)
w2 = w2 (t)
T1 = T1(t)
T2 = T2(t)
Thus, NF is reduced to 0.
2.2
Energy balance:
d ρV (T − Tref )
Cp = wC p (Ti − Tref ) − wC p (T − Tref ) − UAs (T − Ta ) + Q
dt
Simplifying
dT
ρVC p = wC p Ti − wC p T − UAs (T − Ta ) + Q
dt
dT
ρVC p = wC p (Ti − T ) − UAs (T − Ta ) + Q
dt
b) T increases if Ti increases and vice versa.
T decreases if w increases and vice versa if (Ti – T) < 0. In other words, if
Q > UAs(T-Ta), the contents are heated, and T >Ti.
2-2
,2.3
a) Mass Balances:
dh1
ρA1 = w1 − w2 − w3 (1)
dt
dh2
ρA2 = w2 (2)
dt
Flow relations:
Let P1 be the pressure at the bottom of tank 1.
Let P2 be the pressure at the bottom of tank 2.
Let Pa be the ambient pressure.
P1 − P2 ρg
Then w2 = = (h1 − h2 ) (3)
R2 g c R2
P1 − Pa ρg
w3 = = h1 (4)
R3 g c R3
b) Seven parameters: ρ, A1, A2, g, gc, R2, R3
Five variables : h1, h2, w1, w2, w3
Four equations
Thus NF = 5 – 4 = 1
1 input = w1 (specified function of time)
4 outputs = h1, h2, w2, w3
2-3
, 2.4
Assume constant liquid density, ρ . The mass balance for the tank is
d (ρAh + m g )
= ρ(qi − q )
dt
Because ρ, A, and mg are constant, this equation becomes
dh
A = qi − q (1)
dt
The square-root relationship for flow through the control valve is
1/ 2
ρgh
q = C v Pg + − Pa (2)
gc
From the ideal gas law,
(m g / M ) RT
Pg = (3)
A( H − h)
where T is the absolute temperature of the gas.
Equation 1 gives the unsteady-state model upon substitution of q from Eq. 2 and
of Pg from Eq. 3:
1/ 2
dh (mg / M ) RT ρ gh
A = qi − Cv + − Pa (4)
dt A( H − h) gc
Because the model contains Pa, operation of the system is not independent of Pa.
For an open system Pg = Pa and Eq. 2 shows that the system is independent of Pa.
2-4
2.1
a) Overall mass balance:
d (ρV )
= w1 + w2 − w3 (1)
dt
Energy balance:
d ρV (T3 − Tref )
C = w1C (T1 − Tref ) + w2C (T2 − Tref )
dt (2)
− w3C (T3 − Tref )
Because ρ = constant and V = V = constant, Eq. 1 becomes:
w3 = w1 + w2 (3)
b) From Eq. 2, substituting Eq. 3
d (T3 − Tref ) dT
ρCV = ρCV 3= w1C (T1 − Tref ) + w2C (T2 − Tref )
dt dt (4)
− ( w1 + w2 ) C (T3 − Tref )
Constants C and Tref can be cancelled:
dT3
ρV = w1T1 + w2T2 − ( w1 + w2 )T3 (5)
dt
The simplified model now consists only of Eq. 5.
Solution Manual for Process Dynamics and Control, 4th edition
Copyright © 2016 by Dale E. Seborg, Thomas F. Edgar, Duncan A. Mellichamp,
and Francis J. Doyle III
2-1
, Degrees of freedom for the simplified model:
Parameters : ρ, V
Variables : w1, w2, T1, T2, T3
NE = 1
NV = 5
Thus, NF = 5 – 1 = 4
Because w1, w2, T1 and T2 are determined by upstream units, we assume
they are known functions of time:
w1 = w1(t)
w2 = w2 (t)
T1 = T1(t)
T2 = T2(t)
Thus, NF is reduced to 0.
2.2
Energy balance:
d ρV (T − Tref )
Cp = wC p (Ti − Tref ) − wC p (T − Tref ) − UAs (T − Ta ) + Q
dt
Simplifying
dT
ρVC p = wC p Ti − wC p T − UAs (T − Ta ) + Q
dt
dT
ρVC p = wC p (Ti − T ) − UAs (T − Ta ) + Q
dt
b) T increases if Ti increases and vice versa.
T decreases if w increases and vice versa if (Ti – T) < 0. In other words, if
Q > UAs(T-Ta), the contents are heated, and T >Ti.
2-2
,2.3
a) Mass Balances:
dh1
ρA1 = w1 − w2 − w3 (1)
dt
dh2
ρA2 = w2 (2)
dt
Flow relations:
Let P1 be the pressure at the bottom of tank 1.
Let P2 be the pressure at the bottom of tank 2.
Let Pa be the ambient pressure.
P1 − P2 ρg
Then w2 = = (h1 − h2 ) (3)
R2 g c R2
P1 − Pa ρg
w3 = = h1 (4)
R3 g c R3
b) Seven parameters: ρ, A1, A2, g, gc, R2, R3
Five variables : h1, h2, w1, w2, w3
Four equations
Thus NF = 5 – 4 = 1
1 input = w1 (specified function of time)
4 outputs = h1, h2, w2, w3
2-3
, 2.4
Assume constant liquid density, ρ . The mass balance for the tank is
d (ρAh + m g )
= ρ(qi − q )
dt
Because ρ, A, and mg are constant, this equation becomes
dh
A = qi − q (1)
dt
The square-root relationship for flow through the control valve is
1/ 2
ρgh
q = C v Pg + − Pa (2)
gc
From the ideal gas law,
(m g / M ) RT
Pg = (3)
A( H − h)
where T is the absolute temperature of the gas.
Equation 1 gives the unsteady-state model upon substitution of q from Eq. 2 and
of Pg from Eq. 3:
1/ 2
dh (mg / M ) RT ρ gh
A = qi − Cv + − Pa (4)
dt A( H − h) gc
Because the model contains Pa, operation of the system is not independent of Pa.
For an open system Pg = Pa and Eq. 2 shows that the system is independent of Pa.
2-4
