# Case 2

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```					Theoretical Analysis for Optimal SAW Flow Sensor Design
(Case 2: Quartz Crystal) - All calculations in MKS units.
An analog thermal circuit of Figure 2.1(a) SAW flow sensor is shown in Figure 2.2. Following
analysis presents thermal losses for various dimensions of the SAW substrate. Theoretical
calculation of thermal resistances (conduction, convection, and radiation) for various
dimensions of the SAW substrate are presented and analysis to optimize time response
are made here. (These calculation uses quartz as the substrate material.)

2                 2            2
LS  1.0 10  0.866 10  0.5 10
Side of the substrate in m
4               4              4
AS  1.0 10               0.25 10
 0.75 10
Area of the substrate in m2
Thermal conductivity of quartz substrate
k L  1.38
Thickness range of the substrate in m
4            4            4
tS  5.0 10         4.0 10      1.0 10
tS
        
RL AS  tS 
kL  AS
Thermal resistance (degC/W)
AS 
1·10 -4
7.5·10 -5
5·10 -5
2.5·10 -5


RL AS  0.0005      
3.623188
4.830918
7.246377
14.492754


RL AS  0.0004      
2.898551
3.864734
5.797101
11.594203


RL A S  0.0003     
2.173913
2.898551
4.347826
8.695652

RL AS  .0002 
1.449275
1.932367
2.898551
5.797101

Plots of Thermal Resistance versus various values of Thickness, and Area of the Crystal.
4

R  0.0001t
            2

0
0         2 10     4 10         6 10
t

15


R  A  0.0004         7.5

0
0               5 10           1 10
A
trace 1

tS  0.0005
tS
       
RL AS  tS 
kL  AS
For
Derivative of thermal resistance w.r.t area of the SAW substrate, for a fixed thickness, yields:
0

R A  t        5 10
d
dA

1 10
0                   5 10           1 10
A
d

RL AS  tS        
dAS
-3.62318 8·10 4
-6.44122 4·10 4
-1.44927 5·10 5
-5.79710 1·10 5

HEAT LOSS DUE TO CONDUCTION

Let's define Qcond as the heat loss due to conduction. Three cases should be considered:
1. Heat loss due to convection and radiation is zero, i.e., value of convection and radiation
thermal resistances is infinite.

2. Heat loss due to convection and radiation is infinite, i.e., value of convection and radiation
thermal equivalent resistances is zero

3. Heat loss due to convection and radiation is finite. For this case, the conduction loss
would be calculated by first calculating total heat loss from the thermal resistance analog circuit,
and then subtracting convection and radiation components of heat loss.

Case 1:
Assuming that heat loss is only due to conduction, then theoretically the heat loss, for a
given temperature range, for various dimensions of the SAW substrate is calculated as follows:

T L  50  80
Temperature range in degree C (50 to 80)

Heat loss in watts
TL

Qcond AS  tS   
RL AS tS
           
50
Qcond AS  tS 
 RL  AS  tS 
For temperature of 50 degC
AS 
1·10 -4
7.5·10 -5
5·10 -5
2.5·10 -5
15

10
Q       A          
 0.0005
5

0
0    5 10          1 10
A


Qcond A S  0.0005   
13.8
10.35
6.9
3.45

Analysis: For infinite value of convection and radiation thermal resistances (or zero
convection and radiation losses), it would take extremely high power (42 to 10.5 watts), for
a given range of SAW substrate area, to raise to only 50 degC . Therefore, these analysis are not acceptable.

Case 2: For zero convection and radiation thermal resistances (or infinite convection and radiation losses), the
thermal equivalent circuit would be short circuited.

Case 3: Finite Convection and Radiation Losses
8
  5.67  10
T amb 25  273
Temperature in degK
Assume
T L  50
Temperature in degC
e  0.04
emissivity of Al (assuming top of substrate has Al
SAW transducers and pads and also the top cover
of the flow cell is made of aluminum). Note that no
data for the quartz substrate emissivity
is available. Al is among the materials of higher
emissivity. Therefore, if we take "e" of Al, the
heat loss due to radiation is calculated as follows:
1
Fe 
 2     
   1
 e     
                             4
Qrad AS    AS Fe  T L  273  T amb 
                   
4

  T L  273  4          
Qrad AS  5.67  AS Fe               ( 2.98) 4
  100                   
4 10

Q        
 A 2 10

0
0     5 10    1 10
A

 
3.469565·10 -4
2.602174·10 -4
1.734783·10 -4
8.673913·10 -5

Analysis: Radiation losses are extremely small and can be ignored.
Convection Losses:
hTOP  4.08

 
1
RTOP AS 
 hTOP AS
(degC/W)
1 10

R        A  5000

0
0       5 10     1 10
A

 
RTOP AS 
2.45098·10 3
3.267974·10 3
4.901961·10 3
9.803922·10 3

(degC/W)
(sq.m)
hBOT 7.56

 
1
RBOT AS 
 hBOT AS
6000

4000
R    A 
2000

0
0                5 10           1 10
A

 
RBOT AS 
1.322751·10 3
1.763668·10 3
2.645503·10 3
5.291005·10 3

hVER  6.35
6           6             6
AV  5 10       4 10          1 10

 
1
RVER AV 
 hVER AV
2 10

R     A 1 10

0
0        2 10           4 10   6 10
A

 
RVER AV 
3.149606·10 4
3.937008·10 4
5.249344·10 4
7.874016·10 4
1.574803·10 5

Equivalent convection resistance of the top and bottom surfaces of the SAW substrate is given by:
RTOP AS  RBOT AS
RTBCONV AS 
RTOPAS  RBOTAS
RTBCONV AS  
859.10652 9
1.145475·10 3
1.718213·10 3
3.436426·10 3

        
RTOP AS  RBOT AS
 
RTBCONV AS 
RTOPAS  RBOTAS
Including convection from the vertical surfaces, results in the following equivalent convection resistance:
RTBCONV AS  RVER AV 
RCONV  AS  AV  
RTBCONVAS  RVERAV
4000

3000

R       A    5 10

 2000
1000

0
0        5 10   1 10
A


RCON V A S  5 10
6

836.295212
1.105278·10 3
1.629328·10 3
3.098373·10 3


RCON V A S  4 10
6

840.760047
1.11309·10 3
1.646362·10 3
3.160556·10 3


RCONV AS  10
6

854.44525 1
1.137204·10 3
1.699669·10 3
3.36304·10 3

            
50
QCONV AS  AV 
RCONV AS  AV     

QCONV AS  5 10          
6

0.059788
0.045238
0.030688
0.016138

Total Heat Loss:

From the equivalent circuit (Figure 2.4), the total thermal resistance (ignoring radiation losses) is given by:

                           
RTot AS  tS  AV  RL AS  tS  RCONV AS  AV                                 
4000

3000

R    A    0.0005 5 10


 2000
1000

0
0               5 10               1 10
A


RTot AS  0.0005  5 10
6

839.918401
1.110109·10 3
1.636574·10 3
3.112866·10 3

For simplification, assume fixed thickness of the substrate of 0.0005 m; then we can replace A S and AV by L2S
and 0.0005*LS, respectively
tS  0.0005
                           
RTot AS  tS  AV  RL AS  tS  RCONV AS  AV                                 
Derivative of the total thermal resistance w.r.t area of the SAW substrate, for a fixed thickness, is given below:

0

5 10
d
R    A    0.0005 5 10



dA
1 10

1.5 10
0           5 10           1 10
A
d

RTot AS  0.0005  5 10
6

dAS
-8.1771 28·10 6
-1.4284 29·10 7
-3.1045 75·10 7
-1.1232 27·10 8

Total heat loss for a given temperature rise is calculated as follows:

T  50
Temperature difference has a range of values, let's pick 50
degC for the following analysis
T
QTot AS  tS  AV  
 RTot AS  tS AV 
0.06

0.04

Q         A          

 0.0005 5 10    
0.02

0
0       5 10   1 10
A


QTot AS  0.0005  5 10             
6

0.05953
0.045041
0.030552
0.016062

Case 3: Conduction Loss Calculation

tS  0.0005
6
AV  5 10
                             
Qcond AS  tS  QCONV AS  AV  QTot AS  tS  AV                
for
T  50

Qcond AS  tS       
2.579076·10 -4
1.968624·10 -4
1.358772·10 -4
7.513231·10 -5

QCONV AS  AV              
0.059788
0.045238
0.030688
0.016138


QTot AS  tS  AV          
0.05953
0.045041
0.030552
0.016062

3 10

2 10
Q            A          
 0.0005
1 10

0
0   5 10   1 10
A


Qcond AS  0.0005              
2.579076·10   -4

1.968624·10 -4
1.358772·10 -4
7.513231·10 -5

Time Response:

3
cL  0.7536 10
3
  2.2  10
L C  0.01  0.008  0.002
Characteristic length of the substrate:
                   LC
2   
T L C   cL   
                                        

                         
2 
kL   
15

10

T L   
5

0
0   0.002   0.004       0.006   0.008   0.01
L

 
T LC 
12.172639
7.790489
4.38215
1.947622
0.486906

```
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