EXAMPLES ON FLUID, THERMAL AND MIXED SYSTEMS
MEEN 364 Lecture 10, 11, 7
Parasuram August 19, 2001
HANDOOUT E.9 - EXAMPLES ON FLUID, THERMAL AND MIXED SYSTEMS Example 1: A thermal system The following figure shows a simple model of an industrial furnace. A packing of temperature T1 is being heated in the furnace by an electric heater supplying heat at the rate Qi(t). The temperature inside the furnace is T2, the walls are at temperature T3 and the ambient temperature is Ta. The thermal capacitances of the packing, the air inside the furnace and the furnace walls are Ch1, Ch2 and Ch3 respectively. Derive the state-variable equations for this system assuming that the heat is transferred by convection only, with the convective heat transfer coefficients hc1 (air-packing), hc2 (air-inside walls) and hc3 (outside walls-ambient air).
T3, Ch3
Ta
T2, Ch2
T1, Ch1 Qi(t)
The rate of heat transfer, Q between a solid wall and a fluid flowing over it is given by Q = hc A(Tw − T f ),
(1)
where hc is the convective heat transfer coefficient, A is the area of heat transfer and Tw and Tf represent the wall and fluid temperatures respectively. Using the above relations for the packing, we have Q1 = m1c1
dT1 dT = C h1 1 = hc1 A1 (T2 − T1 ). dt dt
(2)
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MEEN 364 Lecture 10, 11, 7
Parasuram August 19, 2001
Similarly applying the relation for the furnace, we have Q2 = m 2 c 2
dT2 dT = C h 2 2 = Qi (t ) − hc1 A1 (T2 − T1 ) − hc 2 A2 (T2 − T3 ). dt dt
(3)
Applying the relation given by equation (1) to the walls, we get Q3 = m3 c3
dT3 dT = C h 3 3 = hc 2 A2 (T2 − T3 ) − hc 3 A3 (T3 − Ta ). dt dt
(4)
Equations (2), (3) and (4) represent the governing differential equations of motion for the above-defined system. State-space representation Let the states of the system be defined as T1 = x1 , T2 = x 2 ,
(5)
T3 = x3 . Substituting the above relation in equation (2), we have dT1 = hc1 A1 (T2 − T1 ), dt . h A h A ⇒ x1 = − c1 1 x1 + c1 1 x 2 . C h1 C h1
C h1
(6)
Similarly substituting the relations given by equation (5) in equation (3), we have dT2 = Qi (t ) − hc1 A1 (T2 − T1 ) − hc 2 A2 (T2 − T3 ), dt . h A h A h A h A 1 ⇒ x 2 = c1 1 x1 − c1 1 + c 2 2 x 2 + c 2 2 x3 + Qi (t ). Ch2 Ch2 Ch2 Ch2 Ch2 Ch2
(7)
Substituting the relations given by equation (5) in equation (4), we get dT3 = hc 2 A2 (T2 − T3 ) − hc 3 A3 (T3 − Ta ), dt . h A h A h A h A ⇒ x 3 = c 2 2 x 2 − c 2 2 + c 3 3 x 3 + c 3 3 Ta . C h3 C h3 C h3 C h3 C h3
(8)
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MEEN 364 Lecture 10, 11, 7
Parasuram August 19, 2001
Rewriting equations (6), (7) and (8) in matrix format, we have hc1 A1 − . C h1 x. 1 h A x 2 = c1 1 . Ch2 x3 0
hc1 A1 0 C h1 x 0 1 1 h A h A hc 2 A2 − c1 1 + c 2 2 x2 + Ch2 Ch2 Ch2 x Ch2 h A hc 2 A2 h A 3 0 − c2 2 + c3 3 C h3 C h 3 C h3
0 Q (t ) 0 i . Ta hc 3 A3 C h 3
Example 2: A mixed system A simplified sketch of a computer tape drive is shown below. Write the equations of motion in terms of the parameters listed below.
Free body diagram of the take-up capstan T
J1, B1
ω1
Tm r1
3
MEEN 364 Lecture 10, 11, 7
Parasuram August 19, 2001
where J1 is the inertial of the motor, B1 is the motor damping constant, Tm is the torque developed by the motor, T is the tension in the string. Writing the torque balance equation, we have J1
dω 1 + B1ω dt
1
− Tr1 = Tm .
(9)
But from the figure it can be concluded that .
.
T = B( x 2 − x 1 ) + k ( x 2 − x1 ).
(10)
We also know that the torque developed by the motor is proportional to the armature current. Hence Tm = k t i a .
(11)
Substituting the equations (10) and (11) in equation (9), we have dω 1 J1 + B1ω 1 − Tr1 − Tm = 0, dt . . dω 1 ⇒ J1 + B1ω 1 − Br1 ( x 2 − x1 ) − kr1 ( x 2 − x1 ) − k t ia = 0. dt
(12)
Free body diagram of the idler wheel T2
ω2
J2, B2
r2 F
Writing the Torque balance equation, we get J2
dω 2 + B 2ω dt
2
+ T2 r2 − Fr2 = 0.
(13)
But again, .
.
T2 = B( x 2 − x1 ) + k ( x 2 − x1 ).
(14)
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MEEN 364 Lecture 10, 11, 7
Parasuram August 19, 2001
Substituting equation (14) in equation (13), we get dω 2 + B2ω 2 + T2 r2 − Fr2 = 0, dt . . dω 2 ⇒ J2 + B2ω 2 + Br2 ( x 2 − x1 ) + kr2 ( x 2 − x1 ) − Fr2 = 0. dt J2
(15)
From the figure, the following relation can be concluded, .
x1 = r1ω 1 ,
(16)
.
x 2 = r2ω 2 . Equations (12), (15) and (16) represent the governing differential equation of motion. State-space representation Let the states of the system be defined as x1 = X 1 ,
ω 1 = X2, x2 = X 3 , ω
(17)
= X 4.
2
Substituting the above relation in equation (16), we have .
x1 = r1ω 1 , .
⇒ X 1 = r1 X 2 .
(18)
.
x 2 = r2ω 2 , .
⇒ X 3 = r2 X 4 .
(19)
Substituting the relation given by equation (17) in equation (12), we have J1
dω 1 + B1ω dt
.
1
.
− Br1 ( x 2 − x1 ) − kr1 ( x 2 − x1 ) − k t ia = 0,
.
⇒ J 1 X 2 + B1 X 2 − Br1 (r2 X 4 − r1 X 2 ) − kr1 ( X 3 − X 1 ) − k t ia = 0, .
⇒ X2 =−
(
)
k kr1 B + Br12 kr Br r X1 − 1 X 2 + 1 X 3 + 1 2 X 4 + t ia . J1 J1 J1 J1 J1
(20)
Substituting the relation given by equation (17) in equation (15), we have 5
MEEN 364 Lecture 10, 11, 7
J2
dω 2 + B 2ω dt
Parasuram August 19, 2001 .
2
.
+ Br2 ( x 2 − x1 ) + kr2 ( x 2 − x1 ) − Fr2 = 0,
.
⇒ J 2 X 4 + B2 X 4 + Br2 (r2 X 4 − r1 X 2 ) + kr2 ( X 3 − X 1 ) − Fr2 = 0,
(
)
kr2 Br1 r2 kr2 B2 + Br22 r ⇒ X4 = X1 + X2 − X3 − X 4 + 2 F. J2 J2 J2 J2 J2 .
(21)
Rewriting equations (18), (19), (20) and (21) in matrix format, we have 0 . X. 1 − kr1 X 2 J1 . = 0 X. 3 X 4 kr2 J 2
r1 B1 + Br12 − J1 0 Br1 r2 J2
(
)
0 kr1 J1 0 kr − 2 J2
0 Br1 r2 J1 r2 B2 + Br22 − J2
(
0 X 1 k t X 2 + J1 X 3 0 X 4 0
)
0 0 i a . 0 F r2 J 2
(22)
Example 3: A fluid system Introduction Fluid capacitor A fluid capacitor is shown in the following figure
P1
Qc
Pr Cf
The pressure in a fluid capacitor must be referred to a reference pressure Pr. The volume flow rate Qc is given by Qc = C f
dP1r , where Cf is the fluid capacitance. dt
Fluid inertor The symbolic diagram of a fluid inertor is shown in the following figure. QI
P1
P2
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MEEN 364 Lecture 10, 11, 7
Parasuram August 19, 2001 I
The elemental equation for the inertor is dQI , where I is the fluid inertance. For frictionless incompressible flow in a dt ρ L uniform passage having cross sectional area A and length L, the inertance I = , A where ρ is the mass density of the fluid. P12 = I
Fluid resistor The symbolic diagram of a fluid resistor is shown below. QR
P1
P2
Rf The elemental equation of an ideal resistor is P12 = R f QR . Problem: Develop the input-output differential equation relating the output pressure to the input pressure for the fluid system shown below.
For the fluid resistor, we have P12 = R f QR .
(23)
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MEEN 364 Lecture 10, 11, 7
Parasuram August 19, 2001
For the inertor, we get P23 = I
dQR . dt
(24)
For the fluid capacitor, QR = C f
dP3r . dt
(25)
Writing the pressure balance equation, we have Ps = P1r = P12 + P23 + P3 r dQR + P3r dt dP d 2 P3 r ⇒ Ps = R f C f 3r + C f I + P3r , dt dt 2 d 2 P3r dP ⇒ Cf I + R f C f 3 r + P3r = Ps . 2 dt dt ⇒ Ps = R f QR + I
(26)
Equation (26) represents the governing differential equation of motion for the fluid system shown. State-space representation Let the states of the system be defined as P3r = x1 , dP3r = x2 . dt
(27)
From the above relation, we get .
x1 = x 2 .
(28)
Substituting the relation given by equation (27) in equation (26), we have d 2 P3r dP Cf I + R f C f 3r + P3r = Ps , 2 dt dt .
⇒ C f I x 2 + R f C f x 2 + x1 = Ps ,
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MEEN 364 Lecture 10, 11, 7 .
⇒ x2 = −
Parasuram August 19, 2001
Rf 1 1 x1 − x2 + Ps . Cf I I Cf I
(29)
Rewriting equations (28) and (29) in matrix format, we get . 0 x. 1 = − 1 x 2 C I f
1 0 R f x1 + 1 P . s − I x 2 C f I
(30)
Assignment 1) The sewage system leading to a treatment plant is shown. The variables qA and qB are input flow rates into tanks 1 and 2 respectively. Pipes 1, 2 and 3 have resistances as shown. Derive the state equations. qA
qB
Tank 1
A1 h1
A2 h2
R1
q1
R2
Pipe 1
Tank 2
Pipe 2 q2
q3
R3 Pipe 3
2) The temperatures of the side surfaces of the composite slab shown below areT1 and T2. The other surfaces are perfectly insulated. The cross sectional areas of the two parts of the slab are A1 and A2 and their conductivities are k1 and k2 respectively. The length of the slab is L. a) Find the equivalent thermal resistance of the slab and express it in terms of the thermal resistances of the two parts.
A1 T1
k1
T2
k2
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MEEN 364 Lecture 10, 11, 7
Parasuram August 19, 2001 A2
L 3) Write the equations of motion for the hanging crane shown below. Assume that the driving force on the hanging crane is provided by the motor mounted on the cab with one of the support wheels connected directly to the armature shaft. The motor constants are Ke, Kt and the circuit driving the motor has a resistance Ra and no inductance. The wheel has a radius ‘r’.
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Parasuram August 19, 2001
HANDOOUT E.9 - EXAMPLES ON FLUID, THERMAL AND MIXED SYSTEMS Example 1: A thermal system The following figure shows a simple model of an industrial furnace. A packing of temperature T1 is being heated in the furnace by an electric heater supplying heat at the rate Qi(t). The temperature inside the furnace is T2, the walls are at temperature T3 and the ambient temperature is Ta. The thermal capacitances of the packing, the air inside the furnace and the furnace walls are Ch1, Ch2 and Ch3 respectively. Derive the state-variable equations for this system assuming that the heat is transferred by convection only, with the convective heat transfer coefficients hc1 (air-packing), hc2 (air-inside walls) and hc3 (outside walls-ambient air).
T3, Ch3
Ta
T2, Ch2
T1, Ch1 Qi(t)
The rate of heat transfer, Q between a solid wall and a fluid flowing over it is given by Q = hc A(Tw − T f ),
(1)
where hc is the convective heat transfer coefficient, A is the area of heat transfer and Tw and Tf represent the wall and fluid temperatures respectively. Using the above relations for the packing, we have Q1 = m1c1
dT1 dT = C h1 1 = hc1 A1 (T2 − T1 ). dt dt
(2)
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MEEN 364 Lecture 10, 11, 7
Parasuram August 19, 2001
Similarly applying the relation for the furnace, we have Q2 = m 2 c 2
dT2 dT = C h 2 2 = Qi (t ) − hc1 A1 (T2 − T1 ) − hc 2 A2 (T2 − T3 ). dt dt
(3)
Applying the relation given by equation (1) to the walls, we get Q3 = m3 c3
dT3 dT = C h 3 3 = hc 2 A2 (T2 − T3 ) − hc 3 A3 (T3 − Ta ). dt dt
(4)
Equations (2), (3) and (4) represent the governing differential equations of motion for the above-defined system. State-space representation Let the states of the system be defined as T1 = x1 , T2 = x 2 ,
(5)
T3 = x3 . Substituting the above relation in equation (2), we have dT1 = hc1 A1 (T2 − T1 ), dt . h A h A ⇒ x1 = − c1 1 x1 + c1 1 x 2 . C h1 C h1
C h1
(6)
Similarly substituting the relations given by equation (5) in equation (3), we have dT2 = Qi (t ) − hc1 A1 (T2 − T1 ) − hc 2 A2 (T2 − T3 ), dt . h A h A h A h A 1 ⇒ x 2 = c1 1 x1 − c1 1 + c 2 2 x 2 + c 2 2 x3 + Qi (t ). Ch2 Ch2 Ch2 Ch2 Ch2 Ch2
(7)
Substituting the relations given by equation (5) in equation (4), we get dT3 = hc 2 A2 (T2 − T3 ) − hc 3 A3 (T3 − Ta ), dt . h A h A h A h A ⇒ x 3 = c 2 2 x 2 − c 2 2 + c 3 3 x 3 + c 3 3 Ta . C h3 C h3 C h3 C h3 C h3
(8)
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MEEN 364 Lecture 10, 11, 7
Parasuram August 19, 2001
Rewriting equations (6), (7) and (8) in matrix format, we have hc1 A1 − . C h1 x. 1 h A x 2 = c1 1 . Ch2 x3 0
hc1 A1 0 C h1 x 0 1 1 h A h A hc 2 A2 − c1 1 + c 2 2 x2 + Ch2 Ch2 Ch2 x Ch2 h A hc 2 A2 h A 3 0 − c2 2 + c3 3 C h3 C h 3 C h3
0 Q (t ) 0 i . Ta hc 3 A3 C h 3
Example 2: A mixed system A simplified sketch of a computer tape drive is shown below. Write the equations of motion in terms of the parameters listed below.
Free body diagram of the take-up capstan T
J1, B1
ω1
Tm r1
3
MEEN 364 Lecture 10, 11, 7
Parasuram August 19, 2001
where J1 is the inertial of the motor, B1 is the motor damping constant, Tm is the torque developed by the motor, T is the tension in the string. Writing the torque balance equation, we have J1
dω 1 + B1ω dt
1
− Tr1 = Tm .
(9)
But from the figure it can be concluded that .
.
T = B( x 2 − x 1 ) + k ( x 2 − x1 ).
(10)
We also know that the torque developed by the motor is proportional to the armature current. Hence Tm = k t i a .
(11)
Substituting the equations (10) and (11) in equation (9), we have dω 1 J1 + B1ω 1 − Tr1 − Tm = 0, dt . . dω 1 ⇒ J1 + B1ω 1 − Br1 ( x 2 − x1 ) − kr1 ( x 2 − x1 ) − k t ia = 0. dt
(12)
Free body diagram of the idler wheel T2
ω2
J2, B2
r2 F
Writing the Torque balance equation, we get J2
dω 2 + B 2ω dt
2
+ T2 r2 − Fr2 = 0.
(13)
But again, .
.
T2 = B( x 2 − x1 ) + k ( x 2 − x1 ).
(14)
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MEEN 364 Lecture 10, 11, 7
Parasuram August 19, 2001
Substituting equation (14) in equation (13), we get dω 2 + B2ω 2 + T2 r2 − Fr2 = 0, dt . . dω 2 ⇒ J2 + B2ω 2 + Br2 ( x 2 − x1 ) + kr2 ( x 2 − x1 ) − Fr2 = 0. dt J2
(15)
From the figure, the following relation can be concluded, .
x1 = r1ω 1 ,
(16)
.
x 2 = r2ω 2 . Equations (12), (15) and (16) represent the governing differential equation of motion. State-space representation Let the states of the system be defined as x1 = X 1 ,
ω 1 = X2, x2 = X 3 , ω
(17)
= X 4.
2
Substituting the above relation in equation (16), we have .
x1 = r1ω 1 , .
⇒ X 1 = r1 X 2 .
(18)
.
x 2 = r2ω 2 , .
⇒ X 3 = r2 X 4 .
(19)
Substituting the relation given by equation (17) in equation (12), we have J1
dω 1 + B1ω dt
.
1
.
− Br1 ( x 2 − x1 ) − kr1 ( x 2 − x1 ) − k t ia = 0,
.
⇒ J 1 X 2 + B1 X 2 − Br1 (r2 X 4 − r1 X 2 ) − kr1 ( X 3 − X 1 ) − k t ia = 0, .
⇒ X2 =−
(
)
k kr1 B + Br12 kr Br r X1 − 1 X 2 + 1 X 3 + 1 2 X 4 + t ia . J1 J1 J1 J1 J1
(20)
Substituting the relation given by equation (17) in equation (15), we have 5
MEEN 364 Lecture 10, 11, 7
J2
dω 2 + B 2ω dt
Parasuram August 19, 2001 .
2
.
+ Br2 ( x 2 − x1 ) + kr2 ( x 2 − x1 ) − Fr2 = 0,
.
⇒ J 2 X 4 + B2 X 4 + Br2 (r2 X 4 − r1 X 2 ) + kr2 ( X 3 − X 1 ) − Fr2 = 0,
(
)
kr2 Br1 r2 kr2 B2 + Br22 r ⇒ X4 = X1 + X2 − X3 − X 4 + 2 F. J2 J2 J2 J2 J2 .
(21)
Rewriting equations (18), (19), (20) and (21) in matrix format, we have 0 . X. 1 − kr1 X 2 J1 . = 0 X. 3 X 4 kr2 J 2
r1 B1 + Br12 − J1 0 Br1 r2 J2
(
)
0 kr1 J1 0 kr − 2 J2
0 Br1 r2 J1 r2 B2 + Br22 − J2
(
0 X 1 k t X 2 + J1 X 3 0 X 4 0
)
0 0 i a . 0 F r2 J 2
(22)
Example 3: A fluid system Introduction Fluid capacitor A fluid capacitor is shown in the following figure
P1
Qc
Pr Cf
The pressure in a fluid capacitor must be referred to a reference pressure Pr. The volume flow rate Qc is given by Qc = C f
dP1r , where Cf is the fluid capacitance. dt
Fluid inertor The symbolic diagram of a fluid inertor is shown in the following figure. QI
P1
P2
6
MEEN 364 Lecture 10, 11, 7
Parasuram August 19, 2001 I
The elemental equation for the inertor is dQI , where I is the fluid inertance. For frictionless incompressible flow in a dt ρ L uniform passage having cross sectional area A and length L, the inertance I = , A where ρ is the mass density of the fluid. P12 = I
Fluid resistor The symbolic diagram of a fluid resistor is shown below. QR
P1
P2
Rf The elemental equation of an ideal resistor is P12 = R f QR . Problem: Develop the input-output differential equation relating the output pressure to the input pressure for the fluid system shown below.
For the fluid resistor, we have P12 = R f QR .
(23)
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MEEN 364 Lecture 10, 11, 7
Parasuram August 19, 2001
For the inertor, we get P23 = I
dQR . dt
(24)
For the fluid capacitor, QR = C f
dP3r . dt
(25)
Writing the pressure balance equation, we have Ps = P1r = P12 + P23 + P3 r dQR + P3r dt dP d 2 P3 r ⇒ Ps = R f C f 3r + C f I + P3r , dt dt 2 d 2 P3r dP ⇒ Cf I + R f C f 3 r + P3r = Ps . 2 dt dt ⇒ Ps = R f QR + I
(26)
Equation (26) represents the governing differential equation of motion for the fluid system shown. State-space representation Let the states of the system be defined as P3r = x1 , dP3r = x2 . dt
(27)
From the above relation, we get .
x1 = x 2 .
(28)
Substituting the relation given by equation (27) in equation (26), we have d 2 P3r dP Cf I + R f C f 3r + P3r = Ps , 2 dt dt .
⇒ C f I x 2 + R f C f x 2 + x1 = Ps ,
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MEEN 364 Lecture 10, 11, 7 .
⇒ x2 = −
Parasuram August 19, 2001
Rf 1 1 x1 − x2 + Ps . Cf I I Cf I
(29)
Rewriting equations (28) and (29) in matrix format, we get . 0 x. 1 = − 1 x 2 C I f
1 0 R f x1 + 1 P . s − I x 2 C f I
(30)
Assignment 1) The sewage system leading to a treatment plant is shown. The variables qA and qB are input flow rates into tanks 1 and 2 respectively. Pipes 1, 2 and 3 have resistances as shown. Derive the state equations. qA
qB
Tank 1
A1 h1
A2 h2
R1
q1
R2
Pipe 1
Tank 2
Pipe 2 q2
q3
R3 Pipe 3
2) The temperatures of the side surfaces of the composite slab shown below areT1 and T2. The other surfaces are perfectly insulated. The cross sectional areas of the two parts of the slab are A1 and A2 and their conductivities are k1 and k2 respectively. The length of the slab is L. a) Find the equivalent thermal resistance of the slab and express it in terms of the thermal resistances of the two parts.
A1 T1
k1
T2
k2
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MEEN 364 Lecture 10, 11, 7
Parasuram August 19, 2001 A2
L 3) Write the equations of motion for the hanging crane shown below. Assume that the driving force on the hanging crane is provided by the motor mounted on the cab with one of the support wheels connected directly to the armature shaft. The motor constants are Ke, Kt and the circuit driving the motor has a resistance Ra and no inductance. The wheel has a radius ‘r’.
10
