Chapter 9 : Pressure Measurements

Example 9.1 Page No : 383

In [1]:
# Variables
T = 273+20;
P = 101.3*10**3;

# Calculations and Results
mfp = 22.7*10**-6*T/P;
print "length of mean free path when pressure is one atmospheric pressure(m)", mfp
P = 133;
mfp = 22.7*10**-6*T/P;
print "length of mean free path when pressure is one torr(m)", mfp
P = 133*10**-3;
mfp = 22.7*10**-6*T/P;
print "length of mean free path when pressure is one micrometer of Hg(m)", mfp
P = 249.1;
mfp = 22.7*10**-6*T/P;
print "length of mean free path when pressure is one inch of water(m)", mfp
P = 133*10**-6;
mfp = 22.7*10**-6*T/P;
print "length of mean free path when pressure is 10**-3 micrometer of Hg(m)", mfp

length of mean free path when pressure is one atmospheric pressure(m) 6.56574531096e-08
length of mean free path when pressure is one torr(m) 5.00082706767e-05
length of mean free path when pressure is one micrometer of Hg(m) 0.0500082706767
length of mean free path when pressure is one inch of water(m) 2.67005218788e-05
length of mean free path when pressure is 10**-3 micrometer of Hg(m) 50.0082706767

Example 9.2 Page No : 385

In [2]:
# Variables
T = 273.+25;
P = 99.22*10**3;
R = 288.;
df = P/(R*T);
dm = 0.82*996;
g = 9.81;
h = 200*10**-6;

# Calculations
P1 = g*h*(dm-df)*10**3;
Pa = P+P1;

# Results
print "Pressure of air source(N/m2)", Pa

Pressure of air source(N/m2) 100820.136398

Example 9.3 Page No : 387

In [3]:
# Variables
df = 1.*10**3;
dm = 13.56*10**3;
g = 9.81;
h = 130.*10**-3;

# Calculations
P = g*h*(dm-df);
Ph = P/9.81;

# Results
print "Pressure head(mm of water)", Ph

Pressure head(mm of water) 1632.8

Example 9.4 Page No : 389

In [4]:
# Variables
hn  = 250.;
d = 5.;
D = 25.;

# Calculations
h = hn*(1+(d/D)**2);

# Results
print "height", h

height 260.0

Example 9.6 Page No : 391

In [5]:
# Variables
P = 8.*133;
h = P/(800*9.81);
d = 2.;
D = 50.;

# Calculations
hn = h/(1+(d/D)**2);
e = (hn-h)/h*100;
eP = 0.8*1000*9.81*(hn-h);

# Results
print "error interms of pressure(N/m2)", eP

error interms of pressure(N/m2) -1.69968051118

Example 9.7 Page No : 393

In [7]:
import math

# Variables
P = 133.;
g = 9.81;
dm = 13.56*10**3;
R = 10.**-3;
d = 4.;
D = 20.;

# Calculations
th = math.asin(P/(g*dm*R*(1+(d/D)**2)));
thV = 90-th;

# Results
print "angle to which tube is incliend to vertical(degree)", thV

angle to which tube is incliend to vertical(degree) 88.708070889

Example 9.8 Page No : 395

In [8]:
import math


# Variables
P = 9.81;
g = 9.81;
dm = 0.864*10**3;
R = 4.*10**-3;
d = 2.;
D = 20.;

# Calculations
th = math.asin(P/(g*dm*R*(1+(d/D)**2)));

# Results
print "angle to which tube is incliend to horizontal(degree)", th

angle to which tube is incliend to horizontal(degree) 0.290558105825

Example 9.9 Page No : 397

In [9]:
import math 

# Variables
P = 500.*9.81;
g = 9.81;
d = 8.;
a =  (math.pi/4)*d**2;
A = 1200.;

# Calculations and Results
dm = 0.8*10**3;
hn = P/(g*dm*(1+(a/A)));
print "Length of scale(m)", hn
R = 0.6;
P1 = 50*9.81;
th = math.asin(P1/(g*dm*R*(1+(a/A))));
print "angle to which tube is incliend to horizontal(degree)", th

Length of scale(m) 0.599872595479
angle to which tube is incliend to horizontal(degree) 0.100146080124

Example 9.10 Page No : 399

In [10]:
import math 

# Variables
P = 100.*10**3;
g = 9.81;
di = 10.*10**-3;
D = 40.*10**-3;

# Calculations
A =  (math.pi/4)*D**2;
dm = 13.6*10**3;
a = A/(P/(dm*g*di)-1);
d = (4*a/math.pi)**0.5*10**3;

# Results
print "diameter of the tube(mm)", d

diameter of the tube(mm) 4.65136628261

Example 9.11 Page No : 401

In [11]:
import math 

# Calculations and Results
AR = 1./(0.83-0.8);
print "Amplification ratio", AR
D = 50.*10**-3;
A =  (math.pi/4)*D**2;
d = 6.*10**-3;
a =  (math.pi/4)*d**2;
PR = (a/A)*100;
print "percentage error", PR

Amplification ratio 33.3333333333
percentage error 1.44

Example 9.12 Page No : 403

In [12]:
import math 

# Variables
P = 981;
g = 9.81;
d = 500.*10**-3;

# Calculations
A =  (math.pi/4)*(10*10**-3)**2;
R = 275*10**-3;
th = 30;
W = A*d*P/(2*g*R*math.sin(math.radians(th)));

# Results
print "value of counter weight required(kg)", W

value of counter weight required(kg) 0.0142799666072

Example 9.13 Page No : 405

In [13]:
import math 

# Variables
Mp1 = 20./40;
Mp2 = 10./40;
Mp3 = 5./40;

# Calculations and Results
Eta = 0.225;
print "damping factor", Eta
Td = 1.2;
wd = 2*math.pi/Td;
wn = wd/((1-Eta**2)**0.5);
tc = 1/wn;
print "time constant(s)", tc
ess = 2*Eta/wn;
ess5 = 5*ess;
print "error for 5mm/s ramp(mm)", ess5
Tlag = 2*Eta*tc;
print "time lag(s)", Tlag
Eta1 = Eta*(0.5)**0.5;
print "New damping factor", Eta1
Td = 1.2;
wn1  =  wn*(0.5)**0.5;
print "New natural frequency(rad/s)", wn1
tc1 = 1/wn;
print "New time constant(s)", tc1
ess51 = ess5;
print "new error for 5mm/s ramp(mm)", ess51
Tlag1 = Tlag;
print "new time lag(s)", Tlag1

damping factor 0.225
time constant(s) 0.186088816266
error for 5mm/s ramp(mm) 0.418699836599
time lag(s) 0.0837399673198
New damping factor 0.159099025767
New natural frequency(rad/s) 3.79983491418
New time constant(s) 0.186088816266
new error for 5mm/s ramp(mm) 0.418699836599
new time lag(s) 0.0837399673198

Example 9.14 Page No : 407

In [14]:
import math 

# Variables
P = 7.*10**6;
R = 6.25*10**-3;
v = 0.28;
E = 200.*10**9;

# Calculations and Results
t = ((9*P*R**4*(1-v**2)/(16*E))**0.25)*10**3;
print "thickness of diaphram(mm)", t
ds = 7800;
fn = (2.5*t/(math.pi*R**2))*(E/(3*ds*(1-v**2)))**0.5;
print "natural frequency(Hz)", fn

thickness of diaphram(mm) 0.407908989726
natural frequency(Hz) 25306332.7381

Example 9.15 Page No : 409

In [15]:
# Variables
P = 100.*10**3;
A = 1500.*10**-6;
F = P*A;
Cs = F/3;

# Calculations and Results
Ls = Cs+40;
print "natural length of spring(mm)", Ls
P1 = 10*10**3;
F1 = P1*A;
Ss = 3+2*.5;
D = F1/Ss;
print "lacement (mm)",D

natural length of spring(mm) 90.0
lacement (mm) 3.75

Example 9.16 Page No : 411

In [16]:
# Variables
P = 200*10**3;
R = 70*10**-3;
v = 0.25;
t = 1*10**-3;
r = 60*10**-3;
E = 200*10**9;

# Calculations
Sr = (3*P*R**2*v/(8*t**2))*((1/v+1)-(3/v+1)*(r/R)**2);
St = (3*P*R**2*v/(8*t**2))*((1/v+1)-(1/v+3)*(r/R)**2);
Sta2 = (Sr-v*St)/E;
Sta3 = (Sr-v*St)/E;
r0 = 10*10**-3;
Sr1 = (3*P*R**2*v/(8*t**2))*((1/v+1)-(3/v+1)*(r0/R)**2);
St1 = (3*P*R**2*v/(8*t**2))*((1/v+1)-(1/v+3)*(r0/R)**2);
Sta1 = (Sr1-v*St1)/E;
Sta4 = (Sr1-v*St1)/E;
Gf = 1.8;
ei = 12;
eo = (Sta1+Sta4-Sta2-Sta3)*Gf*ei/4;

# Results
print "output voltage (V)", eo

output voltage (V) 0.0398671875

Example 9.17 Page No : 413

In [17]:
# Variables
Aou = 700.*25*1/100;
Aol = 100.*25*1/100;
AouPtP =  2*Aou;
AolPtP =  2*Aol;
Se1 = 1.;

# Calculations and Results
D1 = AouPtP/Se1;
print "deflection of screen corresponding to maximum pressure for sensitivity of 1mV/mm (mm)", D1
print "since the length of the screen is 100mm so waveform is out of range and hence sensitivity setting of 1mV/mm should not be used",
Se2 = 5;
D2 = AouPtP/Se2;
print "deflection of screen corresponding to maximum pressure for sensitivity of 5mV/mm (mm)", D2
print "delection is within the range",
Se3 = 20;
D3 = AouPtP/Se3;
print "deflection of screen corresponding to maximum pressure for sensitivity of 20mV/mm (mm)", D3
print "delection is within the range",
Se4 = 100;
D4 = AouPtP/Se4;
print "deflection of screen corresponding to maximum pressure for sensitivity of 100mV/mm (mm)", D4
print "delection is within the range",
Se5 = 500;
D5 = AouPtP/Se5;
print "deflection of screen corresponding to maximum pressure for sensitivity of 500mV/mm (mm)", D5
print "delection is within the range",
print "since the sensitivity of 5mV/mm gives higher deflection so it is the optimum sensitivity",

deflection of screen corresponding to maximum pressure for sensitivity of 1mV/mm (mm) 350.0
since the length of the screen is 100mm so waveform is out of range and hence sensitivity setting of 1mV/mm should not be used deflection of screen corresponding to maximum pressure for sensitivity of 5mV/mm (mm) 70.0
delection is within the range deflection of screen corresponding to maximum pressure for sensitivity of 20mV/mm (mm) 17.5
delection is within the range deflection of screen corresponding to maximum pressure for sensitivity of 100mV/mm (mm) 3.5
delection is within the range deflection of screen corresponding to maximum pressure for sensitivity of 500mV/mm (mm) 0.7
delection is within the range since the sensitivity of 5mV/mm gives higher deflection so it is the optimum sensitivity

Example 9.18 Page No : 415

In [18]:
# Variables
dP = (7000*10**3)-(100*10**3);
b = 25*10**-12;
R1 = 100;
dR = R1*b*dP;
ei = 5.;

# Calculations
deo = dR*ei/(4*R1)

# Results
print "output voltage of bridge(V)", deo

output voltage of bridge(V) 0.000215625

Example 9.19 Page No : 417

In [19]:
import math 

# Variables
T = 273+20;
P = 101.3*10**3;
R = 287;
de = P/(R*T);
C = 20.04*T**0.5;
r = 6.25*10**-3;
L = 0.6;

# Calculations
V = math.pi*((12.5*10**-3)**2)*(12.5*10**-3);
wn = C*r*(math.pi/(V*(L+0.5*math.pi*r)))**0.5;
fn = wn/(2*math.pi);
f = 1000;
u = f/fn;
mu = 19.1*10**-6;
eta = (2*mu/(de*C*r**3))*(3*L*V/math.pi)**0.5;
M = 1/(((1-u**2)**2)+((2*eta*u)**2))**0.5;
M = M*100;

# Results
print "attenuation = ", M

attenuation =  10.834485951

Example 9.20 Page No : 419

In [20]:
# Variables
d = 1;
At = (math.pi*d**2)*10**-6/4;
V = 100*10**-6;
h = 30*10**-3;

# Calculations
P1 = (At*h**2)/V;
P2 = (At*h**2)/(V-At*h);
e = P2-P1;

# Results
print "error =  ", e

error =   1.66588825836e-09