Dynamic Energy Balance
Dynamic Energy Balance
• Last time:
T
ri
dT
  W
n
Cp



C
Cp
dT


H
V

Q
 i i dt
o  io 
i
Ri
s

i
To
well-mixed CSTR w/flow & reaction
• for multiple reactions:
Rxnterm  H Ri1
ri1
i
1
rxn #
V  H Ri2
ri2
i
2
V
Simplifications
• ACC Term:
For moderate T changes use average Cpi
 assume Cpi are constant
 n Cp
i
i
 C V Cpi
or
  V Cp m
so :
dT
dT
 niCpi dt  V Cpm dt
Simplifications
• Also:
T
o  Cio  Cpi dT  o  Cpm T  To 
To
or
 o
 C Cp T  T 
io
i
o
• Simplified Form:
ri
dT
V  Cpm
 o  Cpm T  To   H Ri V  Q  W s
dt
i
well-mixed CSTR w/flow, reaction & avg. Cp
In-Class Exercise
Liquid flows continuously into an initially empty tank, which contains a full-depth
heating coil. As the tank fills, an increasing proportion of the coil is covered
by liquid. Once the tank is full, the liquid starts to overflow into the discharge
pipe, but heating is maintained.
How long does it take the system to reach steady state, what is the final exit
temperature, and how long does it take before the tank “overflows”?
vo
To
Steam in
Ts
v
T

Condensate
In-Class Exercise
In-Class Exercise
In-Class Exercise
In-Class Exercise
In-Class Exercise
Energy Balance on Vessel Jacket
• If Constant Tj:
Q  UATj  T 
Assumptions:
•jacket well insulated
•liquid in jacket is well mixed
•Vj (Volume of jacket fluid) is constant
•j is constant
•no work nor reaction in jacket
• Jacket energy balance:
assume j1 = j2
Tj2 = constant
ri
dT
V  Cpm
 o  Cpm T  To   H Ri V  Q  W s
dt
i
 j V j Cpmj
dTj
dt
  j  j Cpmj T j1  T j   UAT j  T 
Energy Balance on Vessel Jacket
(especially if Tj varies significantly)
• If Tj not constant: (for small systems, low flow, or not well mixed)
Q  UATj  T 
Tj 
assume j1 = j2
i.e. constant 
T j1  T j 2
2
[1]
Energy Balance on Vessel Jacket
(Tj not constant)
• Energy balance on heat transferred:
Q  UATj  T    j  j Cpmj Tj1  Tj 2  [2]
Rate of energy
Rate of energy
transfer from jacket = loss by jacket
to the reactor
fluid
from [1] Tj 2  2Tj  Tj1
then into [2]
Tj 
T j1  T j 2
2
[1]
Energy Balance on Vessel Jacket
(Tj not constant)
Tj 
2 j  j Cpmj T j1  UAT
2 j  j Cpmj  UA
Kj
Q  UATj  T 
 2 j  j Cpmj 

T j1  T 
Q  UA
 2  Cp  UA 
 j j mj

Q  UAKj Tj1  T 
K j  f  j 
Effect of Flow on Kj
 2 j  j Cpmj 

Kj 
 2  Cp  UA 
 j j mj

so as j  Kj  ??
j
Energy Balance on Vessel Jacket
(Tj not constant)
• Energy balance on jacket:
Assumptions:
•jacket well insulated
•liquid in jacket is more like plug flow (i.e cooling coil)
•Vj (Volume of jacket fluid) is constant
•j is constant
•no work nor reaction in jacket
 j V j Cpmj
dT j
dt
  j  j Cpmj T j1  T j 2   UAT j  T 
and:
Tj 2  2Tj  Tj1
Small Vessel Jackets
•
For small vessels or high pressure
systems the thickness of the vessel
wall can be significant…thus one
needs to consider the thermal
capacity of the wall.
• Energy balance on wall:
dTw
 w Vw Cp w
 hi Ai T  Tw   ho Ao Tw  T j 
dt
• Energy balance for jacket:
 j V j Cpmj
dTj
dt
  j  j Cpmj T j1  T j   ho Ao Tw  T j 
Tj constant
In-Class Exercise
Liquid flows continuously into continuous stirred tank reactor, which is fully-jacketed and
well-mixed. At a certain time, reactant A is introduced into the feed liquid, such that the
volumetric flowrate remains constant.
1) Show that the steady state solution to the problem gives a reactor T of ~331 K.
2) With no control (Kc=0), use the dynamic model to find the SS solution.
3) Add proportional control to the dynamic model. Examine the effect of varying Tset from
300 - 375 K, and Kc from -1 to 5.
4) Add integral control to the controller equation. What effect does this have?
TC
o, To, Cao
j, Tj
TE
V, T
Ca, Cb

T

jo, Tjo
In-Class Exercise
• From before:
Q  UAKj Tjo  T 
• Let:
UAKj  U 
• So:
Q  U Tjo  T 
 2 j  j Cpmj 

Kj 
 2  Cp  UA 
 j j mj

In-Class Exercise
• P only controller: U   Uo  Kc T  Tset  with: Kc = controller gain
• PI controller:
U   U o  K c T  Tset  
Kc
I
t
 T  T dt
set
0
with: I = integral time
• Let:
t
errsum   T  Tset dt
or: d errsum  T  Tset dt
0
• Then:
d errsum 
 T  Tset 
dt
initial condition?
In-Class Exercise
In-Class Exercise
In-Class Exercise
In-Class Exercise
In-Class Exercise : SS Solution
In-Class Exercise: SS Solution
In-Class Exercise: dynamic - SS
In-Class Exercise : dynamic - SS
In-Class Exercise
In-Class Exercise