Colorado School of Mines CHEN403
Developing Transfer Functions from Heat & Material Balances
Example Transfer Functions — Stirred Tank Heater
F0, T0, ρ0
h, Ac, T, ρ
F s, Ts
F1, T1, ρ1
We will develop the transfer functions for a stirred tank heater by setting up the heat &
material balance equations for the case where the volume might change (since flow out of
the tank is controlled by a valve). The overall mass balance on this system will be:
d ( ρ1V )
dt
= ρ0F0 − ρ1F1
where the volume in the tank, V , might be changing with time (since the liquid level might
be changing). If we simplify the physical properties by assuming a constant liquid density,
ρ1 = ρ0 ≡ ρ , and if the valve has linear flow characteristics, F1 = C v h , then:
d ( ρAc h)
dt
= ρF0 − ρF1 ⇒
Ac
dh
= F0 − C v h .
dt
Remember that the steady state version of this equation is:
0 = F0* − C v h* .
The energy balance will be:
(
d ρ1VHˆ 1
dt
) = ρ F Hˆ
0 0
0
− ρ1F1 Hˆ 1 + UA (Ts − T1 )
We can split apart the time derivative term:
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December 21, 2008
Colorado School of Mines CHEN403
ρ1V
( ) + Hˆ
d Hˆ 1
dt
d ( ρ1V )
1
dt
= ρ0F0 Hˆ 0 − ρ1 F1 Hˆ 1 + UA (Ts − T1 )
and then insert the mass balance ODE for the second time derivative:
ρ1V
( ) + Hˆ [F ρ
d Hˆ 1
dt
1
0 0
− F1ρ1 ] = ρ0F0 Hˆ 0 − ρ1F1 Hˆ 1 + UA (Ts − T1 ) .
Doing a little bit of math gives:
ρ1V
ρ1V
ρ1V
ρ1V
( ) + F ρ Hˆ
1
− F1ρ1 Hˆ 1 = ρ0F0 Hˆ 0 − ρ1F1 Hˆ 1 + UA (Ts − T1 )
( ) + F ρ Hˆ
1
= ρ0F0 Hˆ 0 + UA (Ts − T1 )
d Hˆ 1
dt
0 0
d Hˆ 1
dt
0 0
( ) = ρ F Hˆ
d Hˆ 1
dt
0 0
0
− F0ρ0 Hˆ 1 + UA (Ts − T1 )
dHˆ 1
= ρ0F0 Hˆ 0 − Hˆ 1 + UA (Ts − T1 ) .
dt
(
)
We would still like to convert the enthalpy terms into terms that directly relate to
temperature. Even with a temperature dependent heat capacity the time derivative can be
split into:
ρ1VCˆ p1
(
)
dT1
= ρ0F0 Hˆ 0 − Hˆ 1 + UA (Ts − T1 ) .
dt
If we assume that the heat capacities are constant and that there is no reaction, then:
Hˆ (T ) = Cˆ p (T − Tref ) + Hˆ ref and Hˆ ref ,1 = Hˆ ref ,0
and:
ρ1VCˆ p
dT1
= ρ0F0Cˆ p (T0 − T1 ) + UA (Ts − T1 ) .
dt
For final simplicity let’s assume the density is essentially constant, so ρ1 = ρ0 ≡ ρ and:
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December 21, 2008
Colorado School of Mines CHEN403
ρVCˆ p
dT1
= ρF0Cˆ p (T0 − T1 ) + UA (Ts − T1 ) .
dt
The steady state version of this equation is:
(
)
(
)
0 = ρF0*Cˆ p T0* − T1* + UA Ts* − T1* .
Changing temperatures, constant inlet flow rate
Let’s develop the transfer functions using these ODEs as our starting point. Let’s assume
that the only independent variables that can change are T0 and Ts . We will hold F0
constant, so V (and hence h ) will be constant, too. So we only need to deal with the energy
balance equation. When we linearize this we need to put it in terms of the deviation
variables T0′ and Ts′ (for the independent variables) and T1′ (for the dependent variable):
*
′
ρVCˆ p  dT1 = ∂ ρF0Cˆ p (T0 − T1 ) + UA (Ts − T1 )  ⋅ T0′

 dt ∂T 

0
+
*
∂ 
ρF0Cˆ p (T0 − T1 ) + UA (Ts − T1 )  ⋅ Ts′

∂Ts 
+
*
∂ 
ρF0Cˆ p (T0 − T1 ) + UA (Ts − T1 )  ⋅ T1′

∂T1 
*
*
′
ρVCˆ p  dT1 = ρF0Cˆ p  ⋅ T0′ + [UA]* ⋅ Ts′ +  −ρF0Cˆ p − UA ⋅ T1′

 dt 



′
ρVCˆ p  dT1 + ρF0Cˆ p + UA  ⋅ T1′ = ρF0Cˆ p  ⋅ T0′ + [UA] ⋅ Ts′ .

 dt 



Note that this equation shows how the stirred tank fluid temperature is affected by changes
in the other temperatures.
This ODE gets converted to transfer functions by taking the Laplace transform :
ρVCˆ p  s ⋅ T1′ + ρF0Cˆ p + UA  ⋅ T1′ = ρF0Cˆ p  ⋅ T0′ + [UA] ⋅ Ts′






( ρVCˆ  s + ρF Cˆ
p
0
p
)
+ UA  ⋅ T1′ = ρF0Cˆ p  ⋅ T0′ + [UA] ⋅ Ts′
ρF0Cˆ p 
[UA]


T1′ =
⋅ T0′ +
⋅ Ts′ .
ρVCˆ p  s + ρF0Cˆ p + UA 
ρVCˆ p  s + ρF0Cˆ p + UA 

 


 

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December 21, 2008
Colorado School of Mines CHEN403
This shows that we have two transfer functions:
T1′ = G0 ( s )T0′ + Gs ( s )Ts′
where:
ρF0Cˆ p 
[UA]


and Gs ( s ) ≡
.
G0 ( s ) ≡
ρVCˆ p  s + ρF0Cˆ p + UA
ρVCˆ p  s + ρF0Cˆ p + UA 

 


 

Notice that both of these transfer functions come from the same first order ODE, so both
are referred to as first order transfer functions. We’ll see in a later section that we would
like to put the equations into a standard form that arises when the multiplier on the T1′
term is unity (i.e., one). This can be done by dividing everything by the existing factor on
the T1′ term, ρF0Cˆ p + UA  . Starting with the deviation variable form of the ODE:


 ρVCˆ p  dT ′
 ρF0Cˆ p 


UA

 1 + T1′ = 
 ⋅ T0′ + 
 ⋅ Ts′ .
 ρF0Cˆ p + UA  dt
 ρF0Cˆ p + UA 
 ρF0Cˆ p + UA 
We can simplify the form of the equation by defining new symbols for the groupings:
ρVCˆ p
ρF0Cˆ p
UA
., K 0 ≡
, and K s ≡
τ≡
ˆ
ˆ
ρF0Cˆ p + UA
ρF0C p + UA
ρF0C p + UA
so:
τ⋅
dT1′
+ T1′ = K 0 ⋅ T0′ + K s ⋅ Ts′
dt
( τs + 1)T1′ = K 0 ⋅T0′ + K s ⋅ Ts′
T1′ =
K0
Ks
⋅ T0′ +
⋅ Ts′
τs + 1
τs + 1
and the transfer functions are now:
G0 ( s ) =
K0
Ks
and Gs ( s ) =
.
τs + 1
τs + 1
A block diagram for the stirred tank heater can be drawn as follows.
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December 21, 2008
Colorado School of Mines CHEN403
T0′
G0
+
Ts′
+
+
T1′
Gs
Allow Inlet Flow Rate to Change
The previous set of transfer functions were derived from the energy balance ODE
linearized only for changes in temperature. If the inlet flow rate also changes then we must
also take this into account in both ODEs. When we linearize these ODEs we need to put
them in terms of the deviation variables F0′ , T0′ , and Ts′ (for the independent variables) and
h′ & T1′ (for the dependent variables). The overall material balance becomes:
Ac
dh′ ∂
*
=
[F0 − Cv h] ⋅ F0′
dt ∂F0
+
∂
*
[ F0 − Cv h] ⋅ T0′
∂T0
+
∂
*
F0 − C v h] ⋅ Ts′
[
∂Ts
∂
*
F0 − Cv h] ⋅ h′
[
∂h
∂
*
+ [ F0 − C v h] ⋅ T ′
∂T
+
and:
Ac
dh′
= [1] ⋅ F0′ + [0] ⋅ T0′ + [0] ⋅ Ts′ + [ −C v ] ⋅ h′ + [0] ⋅ T ′
dt
Ac
dh′
= F0′ − Cv h′
dt
dh′
[ Ac ] dt + [Cv ] h′ = F0′ .
The energy balance becomes (remember to first divide by h to put all variables on the lefthand side of the ODE):
Stirred Tank Heater
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December 21, 2008
Colorado School of Mines CHEN403
*
 ρF Cˆ

ρAcCˆ p  dT1 = ∂  0 p (T0 − T1 ) + UA (Ts − T1 )  ⋅ F0′

 dt ∂F
h

0 
 h
*

∂  ρF0Cˆ p
UA
+

(T0 − T1 ) + (Ts − T1 )  ⋅ T0′
∂T0  h
h

∂
+
∂Ts
*
 ρF0Cˆ p

UA

(T0 − T1 ) + (Ts − T1 ) ⋅ Ts′
h
 h

*

∂  ρF Cˆ
UA
+  0 p (T0 − T1 ) +
(Ts − T1 ) ⋅ h′
∂h  h
h

*

∂  ρF0Cˆ p
UA
+

(T0 − T1 ) + (Ts − T1 )  ⋅ T1′
∂T1  h
h

*
*
*
 ρCˆ

 ρF Cˆ 
ρAcCˆ p  dT1 =  p (T0 − T1 )  ⋅ F0′ +  0 p  ⋅ T0′ +  UA  ⋅ Ts′
 h 

 dt
 
 h

 h 
*
 ρF0Cˆ p

UA
+  − 2 (T0 − T1 ) − 2 (Ts − T1 )  ⋅ h′
h
 h

*
 ρF0Cˆ p UA 
+ −
−  ⋅ T1′
h
h 

 ρCˆ

 ρF *Cˆ 

ρAcCˆ p  dT1 =  *p T0* − T1*  ⋅ F0′ +  0* p  ⋅ T0′ +  UA
⋅ Ts′
* 


 dt
h 
 h

 h 
 ρF *Cˆ

UA
+  − 0*2 p T0* − T1* − *2 Ts* − T1*  ⋅ h′
h
 h

 ρF0*Cˆ p UA 
+  − * − *  ⋅ T1′
h
h 

(
)
(
)
(
)
 ρF *Cˆ

 ρCˆ

 ρF *Cˆ 
ρAcCˆ p  dT1 +  0* p + UA*  ⋅ T1′ =  *p T0* − T1*  ⋅ F0′ +  0* p  ⋅ T0′ +  UA*  ⋅ Ts′
h 

 dt
h 
 
 h
 h

 h 
 ρF *Cˆ

UA
−  0*2 p T0* − T1* + *2 Ts* − T1*  ⋅ h′
h
 h

(
)
(
Stirred Tank Heater
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)
(
)
December 21, 2008
Colorado School of Mines CHEN403
We can multiply through by the steady state level, h* :
ρAc h*Cˆ p  dT1 + ρF0*Cˆ p + UA  ⋅ T1′ = ρCˆ p T0* − T1*  ⋅ F0′ + ρF0*Cˆ p  ⋅ T0′ + [UA] ⋅ Ts′

 dt 





(
)
(
)
(
 ρF0*Cˆ p T0* − T1* + UA Ts* − T1*
−
h*

)  ⋅ h′

Notice that the term multiplying h′ is the steady state energy balance and is zero, so:
ρAc h*Cˆ p  dT1 + ρF0*Cˆ p + UA  ⋅ T1′ = ρCˆ p (T0* − T1* )  ⋅ F0′ + ρF0*Cˆ p  ⋅ T0′ + [UA] ⋅ Ts′ .

 dt 





Now we can get the transfer functions by converting to Laplace variables & algebraically
manipulating. First from the mass balance:
[ Ac ] ⋅ s ⋅ h′ + [Cv ] ⋅ h′ = F0′
( Ac s + Cv ) ⋅ h′ = F0′
h′ =
1
F0′
Ac s + Cv
Next from the energy balance:
( ρA h Cˆ  s + ρF Cˆ
*
c
*
0
p
(
p
)
(
)
+ UA  ⋅ T1′ = ρCˆ p T0* − T1*  ⋅ F0′ + ρF0*Cˆ p  ⋅ T0′ + [UA] ⋅ Ts′
)
ρCˆ p T0* − T1* 


T1′ =
⋅ F0′
* ˆ
* ˆ
ρAc h C p  s + ρF0 C p + UA

 

* ˆ
ρF0 C p 
[UA]


+
⋅ T0′ +
⋅ Ts′
ρAc h*Cˆ p  s + ρF0*Cˆ p + UA 
ρAc h*Cˆ p  s + ρF0*Cˆ p + UA 

 


 

Now there are three transfer functions from the energy balance:
T1′ = GF ⋅ F0′ + G0 ⋅ T0′ + Gs ⋅ Ts′
where:
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December 21, 2008
Colorado School of Mines CHEN403
(
)
ρCˆ p T0* − T1* 


GF ≡
* ˆ
* ˆ
ρAc h C p  s + ρF0 C p + UA 

 

ρF0*Cˆ p 


G0 ( s ) =
* ˆ
ρAc h C p  s + ρF0*Cˆ p + UA

 

Gs ( s ) =
[UA]
ρAc h*Cˆ p  s + ρF0*Cˆ p + UA

 

.
There is only one transfer function from the mass balance:
h′ = Gh ⋅ F0′ where Gh ≡
1
.
Ac s + C v
The full block diagram relating the independent & dependent variables are shown in the
following figure.
h′
Gh
F0′
GF
Stirred Tank Heater
T0′
G0
Ts′
Gs
+
-8-
+
+
T1′
December 21, 2008