# Chapter 3 : Work and Heat Transfer¶

## Example 3.1 Page No : 57¶

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

The velocity of the object just before ir hits the ground is 109.4 m/s


## Example 3.2 Page No : 57¶

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)

The amount of work done upon the atmosphere by the ballon is 50.66 kJ


## Example 3.3 Page No : 58¶

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)

The print lacement work done by the air is 60.8 KJ


## Example 3.4 Page No : 59¶

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

The Net work tranfer for the system is -57.2 KJ


## Example 3.5 Page No : 59¶

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))

The rate of work transfer gas to the piston is 24400 kw


## Example 3.6 Page No : 60¶

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.

Rating of furnace would be 2.17e+06 W
Length of cylinder of furnace is 2.23 m


## Example 3.7 Page No : 61¶

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"

Mass of alluminium that can be melted is 5.23 tonnes
Rate at which alluminium can be melted is 5.39 tonnes/h


## Example 3.8 Page No : 61¶

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

Work done during automization is 6.98 J


## Example 3.9 Page No : 62¶

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)

Work input to the motor is 10.8 kJ
Work input to the stirrer is 9.72 kJ
Work done by the fluid on the atmosphere is 3.77 kJ


## Example 3.10 Page No : 63¶

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

Work done by the system is 10.6 kJ


## Example 3.11 Page No : 63¶

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.

Work done in the cycle is 18.61 kJ