Chapter 7:Sensors And Transducers

Example 7.2,Page No:401

In [1]:
import math

#variable declaration
R   =10000;                           #resistance in Ω
R1  = 3850;                           #resistance of potentiometer Ω
R2  = 7560;                           #resistance of potentiometer Ω
l   = 50*10**-3;                      #length of uniform wound wire in m
x   = 10;

#calculations

R3  = (R/float(2));                  #resistance of potentiometer in .normal position in Ω
r   = (R/float(l));                  #resistance of potentiometer wire per unit length Ω/mm
dR1  = R3-R1;                        #change in resistance of potentiometer from its normal position Ω
D1  = (dR1/float(r));                #displacement in mm
dR2  = (R2-R3);                      #change in resistance of potentiometer from its normal position in Ω
D2  = (dR2/float(r));                #displacement in mm
RE  = (x/float(r));                  #resolution of potentiometer in mm

#result
print'displacement %3.2f'%(D1*10**3),'mm';
print'displacement %3.3f'%(D2*10**3),'mm';
print'resolution of potentiometer %3.3f'%(RE*10**3),'mm';
 
displacement 5.75 mm
displacement 12.800 mm
resolution of potentiometer 0.050 mm

Example:7.3,Page No:403

In [2]:
import math

#variable declaration
R25  = 100;                             #resistance of thermistor at 25°C
t2   = 35;                              #temperature in °C
t1   = 25;                              #temperature in  °C
alpha = 0.05;                           #temperature coefficient

#calculations
t   = t2-t1;                            #temperaturre difference in °C
x   = alpha*t;
R35 = (R25)*(1-x);                      #resistance in Ω

#result
print'resistance at 35°C is %d'%R35,'Ω';
resistance at 35°C is 50 Ω

Example:7.4,Page No:406

In [3]:
import math

#variable declaration
l    = 1.00;                       #length in mm
L    = 2;                          #inductance in mH
d    = 0.02;                       #displacement in mm

#calculations
la    = l-d;                     #length of air gap when d=0.02
dl    = (2*(1/float(la)))-L;     #change in inductance in mH
r     =  dl/L;                   #ratio of change in inductance to the original inductance
dd    =  r/l;                    #ratio of displacement to original gap length

#result
print'inductance = %3.2f'%dl,'mH';
print'ratio of change in inductance to the original inductance =%3.2f'%r;
print'ratio of change in inductance to the original inductance =%3.2f'%dd;
print'hence dl is directly proportional to displacement';
inductance = 0.04 mH
ratio of change in inductance to the original inductance =0.02
ratio of change in inductance to the original inductance =0.02
hence dl is directly proportional to displacement

Example:7.5,Page No:409

In [4]:
import math

#variable declaration
d    = 1.8;                       #output voltage at maximum displacement in V
de   = 0.0045;                    #deviation from straight line through the origin

#calculations 
a     = (de/float(d))*100;         #percentage linearity indicating in both -ve and +ve

#result
print'percentage linearity %3.2f'%a,'%';
percentage linearity 0.25 %

Example:7.6,Page No:409

In [5]:
import math

#variable declaration
Vo    = 1.8;                            #output voltage in mV
Vi    = 0.6;                            #input voltage in mV;
a     = 500;                            #amplification factor
r     = 1/float(4);                     #scale can read 
v     = 4;                              #output of voltmetr in V
D     = 100;                            #millivoltmeter readings

#calculation
s     = Vo/float(Vi);                        #sensitivity in mV/mm
sm    =  a*s;                                #sensitivity of measurement in mV/mm
s1    = (v/float(D))*10**3;                  # 1 scale division in mV
Vm    = r*s1;                                #minimum voltage  that can be read on voltmeter
R     = Vm/float(sm);                        #resolution in mm

#result 
print'sensitivty of LVDT %3.2f'%s,'mV/mm';
print'resolution %3.4f'%R,'mm';
sensitivty of LVDT 3.00 mV/mm
resolution 0.0067 mm

Example:7.7,Page No:413

In [6]:
import math

#variable declaration
A     = 300*10**-6;                     #area of plate in m**2
d     = 0.2*10**-3;                     #distance between plates in mm
e0    = 8.85*10**-12;                   #permittivity in F/m
er2    = 8;                             #dielectric constant of mica 
d1    = 0.18*10**-3;                    #distance between plates in mm
er1   = 1;                              #dielectric constant
D1    = 0.19;
D2    =  0.01;                          #thickness of mica sheet in mm
D3    = 0.17;                           #displacement in mm
D4    = 0.01;




#calculation
C    = ((e0*A)/float(d));                    #value of capacitance in pF
dD   = d-d1;                                 #change in displacement in mm
dC   = ((e0*A)/(float(d1)))-C;               #change in capacitance in capacitance
x1   = (dC/float(C));                        #per unit change in capacitance 
x2   = (dD/float(d));                        #per unit change of displacement
d3   =  d-d1;                                #length of air gap between plates in mm
x    =  x1/float(x2);                        #ratio of unit change of capacitance to unit change in displacement
D    =  (D1/(float(er1)))+((D2/float(er2)));
C1   =  (e0*A)/float(D*10**-3);              #initial capacitance of transducer in mm
d4   =  d1-d3;                               #length of air gap in mm
d1   = (D3/float(er1))+(D4/float(er2));
C2   =  (e0*A)/float(d1*10**-3);             # capacitance with displacement is applien in pF
dC2  = C2-C1;                                #change in capacitance in pF
y1   = (dC2/float(C1));                      #per unit change in capacitance 
y2   = (dD/float(d));                        #per unit change of displacement
Y    = y1/float(y2);                         #ratio of unit change of capacitance to unit change in displacement

#result
print'capacitance = %2.3f'%(C*10**12),'pF';
print'change in capacitance %3.3f'%(dC*10**12),'pF';
print'ratio ofper unit change of capacitance to per unit change in displacement = %f'%x;
print'capcitance when mica is inserted = %3.2f'%(C1*10**12),'pF';
print'change in capacitance when mica sheet is inserted = %2.2f'%(dC2*10**12),'pF';
print'ratio ofper unit change of capacitance to per unit change in displacement = %3.3f'%Y;
capacitance = 13.275 pF
change in capacitance 1.475 pF
ratio ofper unit change of capacitance to per unit change in displacement = 1.111111
capcitance when mica is inserted = 13.88 pF
change in capacitance when mica sheet is inserted = 1.62 pF
ratio ofper unit change of capacitance to per unit change in displacement = 1.168

Example:7.8,Page No:417

In [7]:
import math

#variable declaration
t    = 2.5*10**-3;                               #thickness in m
g    = 0.055;                                    #voltage intensity in Vm/N
p    = 1.4*10**6;                                #pressure in N/m**2
e    = 40.6*10**-12;                             #permittivity of quartz in F/m

#calculation
E     =  g*t*p;                                 #output voltage in V
q     =  e*g;                                   #charge sensitivity in pC/N

#result
print'output voltage = %3.2f'%E,'V';
print'charge sensitivity = %3.3f'%(q*10**12),'pC/N';
output voltage = 192.50 V
charge sensitivity = 2.233 pC/N

Example:7.9,Page No:417

In [8]:
import math

#variable declaration
r   = 6*10**-3;                      #radius in m
t   = 1.8*10**-3;                    #thickness in m
g   = 0.055;                         #voltage intensity in Vm/N
E   = 120;                           #voltage developed in V

#calculation
A    = r*r;                          #area in m**2
p    = E/(float(g*t));               #pressure in N/m**2
F    = p*A;                          #force in N


#result
print'force = %3.2f'%F,'N';
force = 43.64 N

Example:7.10,Page No:417

In [1]:
import math

#variable declaration
r  = 6*10**-3;                       #radius in m
t  = 1.5*10**-3;                     #thickness in m
e  = 12.5*10**-9;                    #permittivity in F/m
F  = 6;                              #force in N
d  = 150*10**-12;                    #charge density in pC/N
E  = 12*10**6;                       #modulus of elasticity in N/m**2
s  = 0.167*10**6;                    #stress 

#calculation
A  = r*r;
p  = F/float(A);                       #pressure in MN/m**2
p1 = p*10**-6;
e1  = s/float(E);                      #strain 
g  = d/float(e);                       #voltage sensitivity in V*m/N;
E1 = g*t*p;                            #voltage generated in V
Q  = d*F;                              #charge in C
C  = (Q)/float(E1);                    #capacitance in F

#result
print'strain = %3.5f'%e;
print'charge = %3.1f'%(Q*10**12),'pC';
print'capacitance = %3.3d'%(C*10**12),'pf';
strain = 0.00000
charge = 900.0 pC
capacitance = 300 pf

Example:7.11,Page No:421

In [2]:
import math

#variable declaration
p  =  0.00912;                        #resistivity in Ωm
B  =  0.48;                           #flux density in Wb/m**2
RH =  3.55*10**-4;                    #hall coefficient in m**3/C

#calculation
Ex  = p;                              #Ex in terms of Jx in °
Ey  = RH*B;                           #ey interms of Jx in °
x= Ex/float(Ey);
t  = math.atan(x);

print'hall angle %3.2f'%t,'°(Equal to 1 minute 4 seconds)';
hall angle 1.55 °(Equal to 1 minute 4 seconds)

Example:7.12,Page No:421

In [11]:
import math

#variable declaration
p  =  0.00912;                        #resistivity in Ωm
B  =  0.48;                           #flux density in Wb/m**2
RH =  3.55*10**-4;                    #hall coefficient in m**3/C
I  = 0.015;                           # current in A
l  = 15*10**-3;                       #length in m
b  = 10**-3;                          #breadth in m


#calculation
A   = l*b;                            #area in m**2
Jx  = I/float(A);                     #current density in A/m**2
Ey  = RH*B*Jx;                        #Ey in V/m
V   = Ey*I;                           #voltage between contacts in V

#result
print'voltage between contacts = %5.5f'%V,'V';
voltage between contacts = 0.00256 V

Example:7.13,Page No:432

In [12]:
import math

#variable declaration
Gf  =  4.2;                       #guage factor of resistance 

#calculation
u  =(Gf-1)/float(2);              #poisson's  ratio

#result
print'poissons ratio = %1.1f'%u;
poissons ratio = 1.6

Example:7.14,Page No:432

In [13]:
import math

#variable declaration
R      = 120;                                 #resistance in Ω
Gf     = 2;                                   #guage factor 
s      = 400*10**6;                           #elastic limit stress in N/m**2
E      = 200*10**9;                           #modulus of elasticity in N/m**2
alpha  = 20*10**-6;                           #resistance temperature coefficient in /°C
x      =  1/float(10);                        #cahnge in stress 
dt     = 20;                                  #change in temperature in °C

#calculations
sc     =  s*x;                              #change in stress in N/m**2
e      =  sc/float(E);                      #strain 
dR     =  Gf*e*R;                           #change in resistance in mΩ
dR1    =  R*alpha*dt;                       #change in resistance in mΩ

#result
print'change in resistance = %3.2f'%(dR*10**3),'mΩ';
print'Note:printing mistake in textbook';
print'change in resistance = %3.2f'%(dR1*10**3),'mΩ';
change in resistance = 48.00 mΩ
Note:printing mistake in textbook
change in resistance = 48.00 mΩ

Example:7.15,Page No:433

In [14]:
import math

#variable declaration
L      =  0.12;                        #length in m
A      = 3.8*10**-4;                   #area in m**2
R      = 220;                          #resistance in Ω
Gf     = 2.2;                          #guage factor
dR     = 0.015;                        #change in resistance in Ω
E      = 207*10**9;                    #elasticity in N/m**2

#calculations
dL     =  (dR*L)/float(R*Gf);    #change in length in m      
s      =  (E*dL)/float(L);                              
F      =  s*A;                   #force in kN            

#result
print'change in length = %2.2e'%dL,'m';
print'force = %3.3f'%(F*10**-3),'kN';
change in length = 3.72e-06 m
force = 2.438 kN

Example:7.16,Page No:444

In [15]:
import math

#variable declaration
Rg     =  100;                 #resistance in Ω
Rsh    =  80000;               #resistance in Ω
Gf     = 2.1;

#calculations
x    = (Rg/float(Rg+Rsh));    #equivalent strain
eeq  = x/(float(Gf));         #strain  in microstrain

#result
print'strain  = %3.1f'%(eeq*10**6),'microstrain';
strain  = 594.5 microstrain

Example:7.17,Page No:445

In [16]:
import math

#variable declaration
n     =  4;                                 #four arm bridge
Rg    =  200;                               #resistance in Ω
Rsh   = 100*10**3;                          #resistance in Ω
x     = 140;                                #number of divisions
Gf    = 2.0;                                #guage factor

#calculation
eeff  = Rg/float(n*Gf*(Rg+Rsh));            #effective strain
d     = eeff/float(x);                      #1 division scale
s     = float(d)*Rg;                        #strain when loaded

#result
print'strain = %3.2f'%(s*10**6),'microstrain';
print'Note:calculation mistake in text book,Rg value is taken wrong in calculating s';
strain = 356.43 microstrain
Note:calculation mistake in text book,Rg value is taken wrong in calculating s

Example:7.18,Page No:447

In [17]:
import math

#variable declaration
ex     =   0.00016;                    #strain values in axial 
ey     =   0.00064;                    #strain values in circumferential direction
E      =   200*10**9;                  #modulus of elasticity in N/,**2
u      =   0.26;                       #poisson's ratio 

#calculation
sigmax  =  (E*(ex+(u*ey)))/float(1-(u**2));         #longitudinal stress in N/m**2
sigmay  =  (E*(ey+(u*ex)))/float(1-(u**2));         #hoop stress in N/m**2

#result

print'longitudinal stress = %3.2f'%(sigmax/10**6),'MN/m**2';
print'longitudinal stress = %3.1f'%(sigmay/10**6),'MN/m**2';
longitudinal stress = 70.01 MN/m**2
longitudinal stress = 146.2 MN/m**2

Example:7.19,Page No:447

In [18]:
import math

#variable declaration
A     =  110*10**-6;                 #area in m**2
P     =  25;                         #load in kN
ex    =  1540;                       #strain values in axial direction
ey    =  -420;                       #strain values in transvers  direction

#calculation
sigmax    =  P/float(A);               #axial stress in N/M**2
E         =  sigmax/float(ex);         #modulus of elasticity in N/M**2
u         =  (-ey*E)/float(sigmax);    #poisson's ratio

#result
print'modulus of elasticity = %3.1f'%E,'N/M**2'
print'poissons ratio  = %3.4f'%u;
modulus of elasticity = 147.6 N/M**2
poissons ratio  = 0.2727

Example:7.21,Page No:450

In [9]:
import math

#variable declaration
e1    =  60*10**-6;                        #strain in microstrains
e2    =  48*10**-6;                        #strain in microstrain
e3    =  -12*10**-6;                       #strain in microstrain
E     =  200*10**9;                        #modulus of elsticity in N/m**2
u     =  0.3;

#calculation
x      = (e1+e3)/float(2);                 #average of strains
a      = math.sqrt(((e1-e2)**2)+((e2-e3)**2));
b      = 1/math.sqrt(2);
y      = a*b;
emax   = x+y;                             #principle strains
emin   = x-y;                             #principle strains
x1     = x/float(1-u);
y1     = y/float(1+u); 
sigmamax  = E*(x1+y1);                    #principle stress
sigmamin  = E*(x1-y1);                    #principle stress
tmax      = E*y1;                         #maximum shear stress in MN/m**2
k         = ((2*e2)-e1-e3)/float((e1-e3));
theta     = (math.atan(k));      #location of principle planes
theta1    =(theta*180)/float(math.pi);
theta2    =theta1+180;
theta11   = (theta1)/float(2);
theta22   = (theta2)/float(2);



print'emax  = %2.2e'%(emax);
print'emin  = %2.3e'%(emin);
print'sigmamax = %3.3f'%(sigmamax*10**-6),'MN/m**2';
print'sigmamin  = %3.3f'%(sigmamin*10**-6),'MN/m**2';
print'maximum shear stress = %3.3f'%(tmax*10**-6),'MN/m**2';
print'location of principle planes = %f'%theta11,'°';
print'location of principle planes = %f'%theta22,'°';
 
emax  = 6.73e-05
emin  = -1.927e-05
sigmamax = 13.514 MN/m**2
sigmamin  = 0.201 MN/m**2
maximum shear stress = 6.656 MN/m**2
location of principle planes = 16.845034 °
location of principle planes = 106.845034 °

Example:7.22,Page No:454

In [20]:
import math

#variable declaration
d      = 0.06;                    #diameter in m
Rg     = 120;                     #nominal resistance of each guage Ω
Gf     = 2.0;                     #guage factor 
v      = 6;                       #supply voltage in V
E      = 200*10**9;               #modulus of elasticity in N/m**2
u      = 0.3;                     #poisson's ratio
P      = 1000;                    #load in N

#calculation

A      = (math.pi/float(4))*d*d;
s      =  P/float(A);                      #stress in N/m**2
e      =  s/float(E);                      #strain 
x      = Gf*e;                             #fraction change in resistence i.e dR/R
a      = v/float(4);
y      = 2*(1+u)*(x)*a;                    #output volatge in uV
                  
#result
print'sensitivity of load = %3.2f'%(y*10**6),'uV/kN';
sensitivity of load = 13.79 uV/kN