# CHAPTER 1:Fundmentals¶

## EXAMPLE 1.1,PAGE NUMBER:3¶

In [1]:
# Variable Declaration
T_0=-5+273;# K
T_1=35+273;# K

# Calculation
COP=(T_0)/(T_1-T_0);# Coefficient of performance
print "Carnot COP=",round(COP,2),"(error)"

Carnot COP= 6.0 (error)


## EXAMPLE 1.2,PAGE NUMBER:4¶

In [2]:
# Variable Declaration
T_f=80;# Final Temperature in °C
T_i=0;# Initial Temperature in °C
h_f=334.91;# The specific enthalpy of water in kJ/kg

# Calculation
C=h_f/(T_f-T_i);# The average specific heat capacity in kJ/(kg K)
print "The average specific heat capacity is",round(C,3),"kJ/(kg K)"

The average specific heat capacity is 4.186 kJ/(kg K)


## EXAMPLE 1.3,PAGE NUMBER:4¶

In [3]:
# Variable Declaration
P=1.013;# Pressure in bar
h_fg=2257;# The latent heat of boiling water in kJ/kg
T_b=100; # The boiling point temperature of water in °C
m=1; # The mass of water in kg
T_i=30; # The initial temperature of water in °C
C_p=4.19;# The specific heat of water in kJ/kg°C

# Calculation
Q=m*((C_p*(T_b-T_i))+h_fg);# The quantity of heat added in kJ
print"The quantity of heat added is",round(Q,1),"kJ"

The quantity of heat added is 2550.3 kJ


## EXAMPLE 1.4,PAGE NUMBER:6¶

In [4]:
# Variable Declaration
V_1byV_2=2;# Volumetric ratio (given)
p_1=1.01325;# The atmospheric pressure in bar(101325 kPa)

# Calculation
p_2=V_1byV_2*p_1;# The new pressure in bar
print"The new pressure,p_2=",round(p_2,4),"bar(abs.)"

The new pressure,p_2= 2.0265 bar(abs.)


## EXAMPLE 1.5,PAGE NUMBER:7¶

In [5]:
# Variable Declaration
V_1=0.75;# The initial volume in m**3
T_1=273+20; # The initial temperature of water in K
T_2=273+90; # The final temperature of water in K

# Calculation
V_2=V_1*(T_2/T_1);# The final volume in m**3
print"The final volume,V_2=",round(V_2,2),"m**3"

The final volume,V_2= 0.75 m**3


## EXAMPLE 1.6,PAGE NUMBER:7¶

In [6]:
# Variable Declaration
R=287;# The specific gas constant in J/(kg K)
m=5; # The mass of ideal gas in kg
p=101.325;# The atmospheric pressure in kPa
T=273+25;# The temperature of an ideal gas in K

# Calculation
V=(m*R*T)/(p*1000);# The volume of an ideal gas in m**3
print"The volume of an ideal gas is",round(V,2),"m**3"

The volume of an ideal gas is 4.22 m**3


## EXAMPLE 1.7,PAGE NUMBER:7,8¶

In [7]:
# Variable Declaration
m_N=0.906;# The mass of nitrogen in a cubic metre of air in kg
R_N=297;# The specific gas constant of nitrogen in J/kg K
m_O=0.278;# The mass of oxygen in a cubic metre of air in kg
R_O=260;# The specific gas constant of oxygen in J/kg K
m_A=0.015;# The mass of argon in a cubic metre of air in kg
R_A=208;# The specific gas constant of argon in J/kg K
T=273.15+20;# The temperature of air in K

# Calculation
p_N=m_N*R_N*T;# The pressure of nitrogen in Pa
p_O=m_O*R_O*T;# The pressure of oxygen in Pa
p_A=m_A*R_A*T;# The pressure of argon in Pa
p_t=p_N+p_O+p_A;# The total pressure in Pa
print"The total pressure is",round(p_t,0),"Pa","(",round(p_t/10**5,5),"bar)"

The total pressure is 100985.0 Pa ( 1.00985 bar)


## EXAMPLE 1.8,PAGE NUMBER:8¶

In [8]:
# Variable declartion
t=225;# The wall thickness in mm
k=0.60;# Thermal conductivity in W/(m K)
L=10;# Length in m
h=3;# Height in m
delT=25;# The temperature difference between the inside and outside faces in K

# Calculation
Q_t=(L*h*k*delT*1000)/(t);# The rate of heat conduction in W
print"The rate of heat conduction,Q_t=",round(Q_t,0),"W""(or)",round(Q_t/1000,0),"kW)"

The rate of heat conduction,Q_t= 2000.0 W(or) 2.0 kW)


## EXAMPLE 1.9,PAGE NUMBER:10¶

In [9]:
# Variable declartion
R_i=0.3;# The inside surface resistance in (m**2 K)/W
R_c=1/2.8;# The thermal conductance of plastered surface in (m**2 K)/W
R_o=0.05;# The outside surface resistance in (m**2 K)/W

# Calculation
R_t=R_i+R_c+R_o;# The total thermal resistance in (m**2 K)/W
U=1/R_t;# The overall transmittance in W/(m**2 K)
print"The overall transmittance,U=",round(U,3)," W/(m**2 K)"

The overall transmittance,U= 1.414  W/(m**2 K)


## EXAMPLE 1.10,PAGE NUMBER:12¶

In [10]:
import math
# Variable declartion
T_f=3;# The temperature of fluid in °C
T_wi=11.5;# The temperature of water at inlet in °C
T_wo=6.4;# The temperature of water at outlet in °C
A=420;# The surface area in m**2
U=110;# The thermal transmittance in W/(m**2 K)

# Calculation
delT_max=T_wi-T_f;# The maximum temperature difference in K
delT_min=T_wo-T_f;# The minimum temperature difference in K
LMTD=(delT_max-delT_min)/math.log(delT_max/delT_min);
Q_f=U*A*LMTD;# The amount of heat transfer in W
print"The logarithmic mean temperature difference is",round(LMTD,3),"K"
print"The amount of heat transfer is",round(Q_f,0),"W (round off error)","or",round(Q_f/1000,0),"kW"

The logarithmic mean temperature difference is 5.566 K
The amount of heat transfer is 257145.0 W (round off error) or 257.0 kW