In [2]:

```
#Variable Declaration:
Q1 = 8.03 #Years(part 1)
D = 365 #Days in a year
H = 24 #Hours in a day
M = 60 #Minutes in an hour
S = 60 #Seconds in a minute
Q2 = 150 #Miles per hour(part 2)
FM = 5280 #Feet in a mile
YF = 1.0/3.0 #Yard in a feet
Q3 = 100 #Meter per second square(part 3)
Cmm = 100 #Centimeter in a meter
FC = 1.0/30.48 #Feet in a centimeter
SsMs = 60**2 #Second square in a minute square
Q4 = 0.03 #Gram per centimeter cube (part 4)
PG = 1.0/454.0 #Pound in a gram
CF = (30.48)**3 #Centimeter in a feet
#Calculation:
A1 = Q1*D*H*M*S #Seconds (s)
A2 = Q2*FM*YF #Yards per hour (yd/hr)
A3 = Q3*Cmm*FC*SsMs #Feet per min square (ft/min^2)
A4 = Q4*PG*CF #Pound per feet cube (lb/ft^3)
#Results:
print "1. Seconds in",Q1,"year is:",round(A1/10**8,2)," x 10**8 s"
print "2. Yards per hour in",Q2,"miles per hour is:",round(A2/10**5,1)," x 10**5 yd/h"
print "3. Feets per minute square in",Q3,"meter per square is:",round(A3/10**6,3)," x 10**6 ft/min^2"
print "4. Pounds per feet cube in",Q4,"gram per centimeter cube is:",round(A4),"lb/ft^3"
```

In [3]:

```
#Variable Declaration:
Q1 = 32.2 #Gravitational acceleration (ft/s^2) (part 1)
CF = 32.2 #Conversion factor (lb.ft/lbf.s^2)
M = 100 #Mass (lb)
SA = 3 #Surface area (in^2)
FsIs = (1.0/12.0)**2 #Feet square in a inch square
Q2 = 14.7 #Atmospheric pressure (psi) (part 2)
GP = 35 #Gauge Pressure (psig)
#Caculations:
F = M*Q1/CF #Force (lbf)
P = F/SA/FsIs #Pressure at the base (lbf/ft^2)
Pa = GP+Q2 #Absolute pressure (psia)
#Results:
print "1. Pressure at the base is:",round(P),"lbf/ft^2"
print "2. Absolute pressure is:",round(Pa,1),"psia"
```

In [4]:

```
#Variable Declaration:
Q1 = 20.0 #Mass (lb) (part 1)
MH = 1.008 #Molecular weight of H (lb/lbmol)
MO = 15.999 #Molecular weight of O (lb/lbmol)
Q2 = 454 #Gram in pound (part 2)
Q3 = 6.023*10**23 #Avogadro nuber (part 3)
#Calculations:
Mol = 2*MH+MO #Molecular weight of water (lb/lbmol)
A1 = Q1/Mol #Pound.moles of water (lbmol)
A2 = Q1*Q2/Mol #Gram.moles of water (gmol)
A3 = A2*Q3 #Molecules of water (molecules)
#Results:
print "1. Pound.moles of water is:",round(A1,2),"lbmol water"
print "2. Gram.moles of water is:",round(A2),"gmol water"
print "3. Molecules of water is:",round(A3/10**26,3)," x 10**26 molecules"
```

In [7]:

```
#Variable declaration:
SG = 0.92 #Specific gavity of liquid, methanol
DW = 62.4 #Density of reference substance, water (lb/ft^3)
#Calculation:
DM = SG*DW #Density of methanol (lb/ft^3)
#Result:
print "Density of methanol is:",round(DM,1),"lb/ft^3"
```

In [8]:

```
#Variable declaration:
SG = 0.8 #Specific Gravity
AV = 0.02 #Absolute Viscosity (cP)
cP = 1 #Viscosity of centipoise (cP)
VcP = 6.72 * 10**-4 #Pound per feet.sec in a centipoise (lb/ft.s)
pR = 62.43 #Reference density (lb/ft^3)
#Calculations:
u = AV*VcP/cP #Viscosity of gas (lb/ft.s)
p = SG*pR #Density of gas (lb/ft^3)
v = u/p #Kinematic viscosity of gas (ft^2/s)
#Result:
print "Kinematic viscosity of gas is:",round(v/10**-7,3),"x 10**-7 ft^2/s"
```

In [9]:

```
#Variable declaration:
X = 7.0 #Coordinate X of H2SO4
Y = 24.8 #Coordinate Y of H2SO4
S = 45 #Slope
#Calculations:
#From the figure C.1 we found the intersection of curves mu = 12cP
mu = 12
#Results:
print "Absolute viscosity of a 98% sulfuric acid solution at 45° C is :",mu*10**-2," g/cm.s"
```

In [10]:

```
#Variable declaration:
CpM = 0.61 #Heat capacity of methanol (cal/g.°C)
G = 454 #Grams in a pound
B = 1.0/252.0 #Btu in a calorie
C = 1.0/1.8 #Degree celsius in a degree fahrenheit
#Calculation:
Cp = CpM*G*B*C #Heat capacity in English units (Btu/lb.°F)
#Result:
print "Heat capacity in English units is: ",round(Cp,2)," Btu/lb.°F"
```

In [11]:

```
#Variable declaration:
kM = 0.0512 #Thermal conductivity of methanol (cal/m.s°C)
B = 1.0/252.0 #Btu in a calorie
M = 0.3048 #Meters in a feet
S = 3600 #Seconds in an hour
C = 1.0/1.8 #Degree celsius in a degree fahrenheit
#Calculation:
k = kM*B*M*S*C #Thermal conductivity in English units (Btu/ft.h.°F)
#Result:
print "Thermal coductivity in English units is:",round(k,3),"Btu/ft.h.°F"
```

In [13]:

```
#Variable declaration:
D = 5 #Diameter of pipe (ft)
V = 10 #Fluid velocity (ft/s)
p = 50 #Fluid density (lb/ft^3)
u = 0.65 #Fluid viscosity (lb/ft.s)
F = 1.0/12.0 #Feet in an inch
VCp = 6.72*10**-4 #Viscosity of centipoise (lb/ft.s)
#Calculation:
A = D*V*p*F/u/VCp #Reynolds Number
#Result:
if(A>2100):
print "The Reynolds number is :",round(A,-2),"; therefore, the flow is turbulent."
elif(A<2100):
print "The Reynolds number is :",round(A,-2),"; therefore, the flow is not turbulent."
```

In [14]:

```
#Variable declaration:
#For the problem at hand, take as a basis 1 kilogram of water and assume the potential energy to be zero at ground level conditions.
z1 = 0 #Intial height from ground level (m)
z2 = 10 #Final height from ground level (m)
PE1 = 0 #Initial potential energy at z1 (J)
m = 1 #Mass of water (kg)
g = 9.8 #Gravitational acceleration (m/s^2)
gc = 1 #Conversion factor
#Calculations:
PE2 = m*(g/gc)*z2 #Final potential energy at z2 (J)
#Result:
print "The potential energy of water is :",PE2,"J "
```