In [1]:

```
import math
# Variables
V1 = 100.; # Initial velocity in m/s
g = 9.81; # Acceleration due to gravity in m/s2
z1 = 100.; # Initial elevation in m
# Calculation
V = math.sqrt(((2*g*z1)+(V1)**2)); # Final velocity in m/s2
# Results
print "The velocity of the object just before ir hits the ground is %.1f m/s"%V
# note : incorrect answer in the textbook
```

In [1]:

```
# Variables
dV = 0.5; # Change in volume in m3
P = 101.325e03; # Atmospheric pressure in N/m2
# Calculation
Wd = P*dV; # Work done in J
# Results
print "The amount of work done upon the atmosphere by the ballon is %.2f kJ"%(Wd/1000)
```

In [4]:

```
# Variables
dV = 0.6; # Change in volume in m3
P = 101.325e03; # Atmospheric pressure in N/m2
# Calculation
Wd = P*dV; # Work done in J
# Results
print "The print lacement work done by the air is %.1f KJ"%(Wd/1000)
```

In [5]:

```
import math
# Variables
T = 1.275e-03; # Torque acting against the fluid in N
N = 10000.; # Number of revolutions
W1 = 2*math.pi*T*N; # Work done by stirring device upon the system
P = 101.325e03; # Atmospheric pressure in N/m2
d = 0.6; # Piston diameter in m
# Calculation
A = (math.pi/4.)*(d)**2; # Piston area in m
L = 0.80; # Displacement of diameter in m
W2 = (P*A*L)/1000.; # Work done by the system on the surroundings i KJ
W = -W1+W2; # Net work tranfer for the system
# Results
print "The Net work tranfer for the system is %.1f KJ"%W
```

In [2]:

```
import math
# Variables
ad = 5.5e-04; # Area of indicator diagram
ld = 0.06; # Length of diagram
k = 147e06; # Spring consmath.tant in MPa/m
w = 150.; # Speed of engine
L = 1.2 ; # Stroke of piston
d = 0.8; # Diameter of the cylinder in m
# Calculation
A = (math.pi/4)*(0.8**2); # Area of cylinder
Pm = (ad/ld)*k; # Effective pressure
W1 = Pm*L*A*w; # Work done in 1 minute
W = (12*W1)/60; # The rate of work transfer gas to the piston in MJ/s
# Results
print "The rate of work transfer gas to the piston is %.0f kw"%(round(W/1000,-2))
```

In [3]:

```
import math
# Variables
Tm = 1535.; # Melting point of iron on degree
Ti = 15.; # Initial temperature
Tf = 1650.; # Final temperature
Lh = 270.e03; # Latent heat of iron in J/Kg
ml = 29.93; # Atomic weight of iron in liquid state
m = 56.; # Atomoc weight of iron
sh = 0.502e03; # Specific heat of iron in J/Kg
d = 6900.; # Density of molten metal in kg/m3
# Calculation and Results
H = (Tm-Ti)*sh + Lh + (ml/m)*(Tf-Tm)*1000; # Heat required
Mr = 5e03; # Melting rate in Kg/h
Hr = H*Mr ; # Rate of heat suppy
HrA = Hr/(0.7*3600) # Actual rate of heat supply
print "Rating of furnace would be %.2e"%HrA,"W"
V = (3*Mr)/d; # Volume required in m3
d = ((V/2.)*(4/math.pi))**(1./3); # Diameter of cylinder of furnace in m
l = 2*d; # Length of cylinder of furnace in m
print " Length of cylinder of furnace is %.2f m"%l
# rounding off error.
```

In [18]:

```
# Variables
SH = 0.9; # Specific heat of alluminium in solid state
L = 390.; # Latent heat
aw = 27.; # Atomic weight
D = 2400.; # Density in molten state
Tf = 700.+273; # Final temperature
Tm = 660.+273; # Melting point of aluminium
Ti = 15.+273; # Intial temperature
# Calculation
HR = round(SH*(Tm-Ti)+L+(29.93/27)*(Tf-Tm),1); # Heat requires
HS = round(HR/0.7,1) ; # Heat supplied
RM = 217*1000*3600/HS ; # From the data of problem 3.7
V = 2.18; # Volume
M = V*D;
# Results
print "Mass of alluminium that can be melted is %.2f"%(M/1000),"tonnes"
print "Rate at which alluminium can be melted is %.2f"%(RM/100000),"tonnes/h"
```

In [9]:

```
import math
# Variables
dd = 60e-06;
mw = 1.;
st = 0.07;
dw = 1000.;
dp = 15e-03;
# Calculation
N = (mw*6)/(math.pi*dd**3*dw);
Af = math.pi*dd**2*N;
S_L = 4/(dp*dw);
W = st*(100-S_L);
# Results
print "Work done during automization is %.2f J"%W
```

In [20]:

```
import math
# Variables
dc = 40e-02;
L = 30e-02;
P = 1e05; # Pressure in Pascal
I = 0.5;
V = 24.;
t = 15.*60; # in seconds
# Calculation
Wm = V*I*t;
Ws = 0.9*Wm;
W = P*(math.pi/4)*dc**2*L;
# Results
print "Work input to the motor is %.1f kJ"%(Wm/1000)
print "Work input to the stirrer is %.2f kJ"%(Ws/1000)
print "Work done by the fluid on the atmosphere is %.2f kJ"%(W/1000)
```

In [10]:

```
import math
# Variables
P1 = 100.
P2 = 37.9
P3 = 14.4;
V1 = 0.1
V2 = 0.2
V3 = 0.4;
# Calculation
n1 = (math.log(P1/P2))/(math.log(V2/V1));
n2 = (math.log(P2/P3))/(math.log(V3/V2));
# n1 = n2
W = ((P1*V1)-(P3*V3))/(n1-1);
# Results
print "Work done by the system is %.1f kJ"%W
```

In [25]:

```
# Variables
P1 = 20*1.01325e05;
V1 = 0.04
V2 = 2*V1;
n = 1.45;
# Calculation
P2 = round((V1/V2)**n*P1,-2)
W12 = ((P1*V1)-(P2*V2))/(n-1);
W23 = P2*(V2-V1);
Wc = W12-W23;
# Results
print "Work done in the cycle is %.2f kJ"%(Wc/1000)
# rounding off error. please check using calculator.
```