Chapter 01 - Introduction To Heat Transfer

Example No : 1.1 - Page : 8

In [15]:
from __future__ import division
# given data
t1=38   # in degree C
t2=21   # in degree C
k=0.19   # unit less
x=4   #in cm
x=x*10**-2   # in meter
# Formula q=k*A*(t1-t2)/x
q_by_A=k*(t1-t2)/x  
print "The rate of heat transfer is :",round(q_by_A,3),"W/m**2"

The rate of heat transfer is : 80.75 W/m**2

Example No : 1.2 - Page : 8

In [16]:
# given data
t_i=120   # in degree C
t_o=40   # in degree C
K=0.04   # unit less
x=0.06   #in m
Q=50   # in W
print "Assuming steady state heat transfer in the wall"
# Rate of heat transfer across the wall = Rate of electrical energy dissipation in the furnance
# Formula Q= K*A*(t_i-t_o)/x 
A=Q*x/(K*(t_i-t_o)) 
print "Area of wall = %0.4f square meter" %A

Assuming steady state heat transfer in the wall
Area of wall = 0.9375 square meter

Example No : 1.3 - Page : 10

In [17]:
from numpy import pi
# given data
t_f=30   # in degree C
t_s=400   # in degree C
d=0.04   #in m
h=20   # in W/m**2K
l=1   #in meter
A=pi*d*l#
q=h*A*(t_s-t_f)   # in W
print "Rate of heat loss = %0.3f watt" %q

Rate of heat loss = 929.911 watt

Example No : 1.4 - Page : 10

In [18]:
from numpy import pi
# given data
t_s=100   # in degree C
t_w=80   # in degree C
d=2*10**-3   #in m
h=3000   # in W/m**2 degree C
L=100   #in mm
L=L*10**-3   # in meter
A=pi*d*L#
# Heat loss by convection = Electric power supplied
# Formula h*A*(t_s-t_w) = Q
Q= h*A*(t_s-t_w)#
print "Electric power supplied = %0.1f watt" %Q

Electric power supplied = 37.7 watt

Example No : 1.5 - Page : 13

In [19]:
# given data
A=0.6*0.9   # in square meter
x=.025   # in meter
t_s=310   # in degree C
t_f=15   # in degree C
h=22   # in W/m**2 degree C
K=45   # in W/m degree C
Q_rad=250   # in W
# Heat transfer through the plate = Convection heat loss + radiation heat loss
# Formula Q_cond = Q_conv + Q_rad
# -K*A*dt/dx = h*A*(t_s-t_f)+ Fg12*sigmaA(Ts**4-Ta64)
t_i=x*(h*A*(t_s-t_f)+Q_rad)/(K*A)+t_s 
print "The inside plate temperature = %0.2f degree C" %t_i

The inside plate temperature = 313.86 degree C

Example No : 1.6 - Page : 13

In [20]:
from numpy import pi
# given data
T1=50   # in degree C
T1=T1+273   # in K
T2=20   # in degree C
T2=T2+273   # in K
d=5*10**-2   #in m
h=6.5   # in W/m**2K
l=1   #in meter
epsilon=0.8#
sigma=5.67*10**-8#
A=pi*d*l   # in Square meter
q_conv = h*A*(T1-T2)   # in W/m
print "The heat loss by convection = %0.1f W/m" %q_conv
# formula q= sigma*A*F_g12*(T1**4-T2**4) = sigma*A*epsilon*(T1**4-T2**4)    (since A1<<A2, so F_g12=epsilon)
q_rad = sigma*A*epsilon*(T1**4-T2**4)   # in W/m
print "Heat loss by radiation = %0.0f W/m" %q_rad
q_total= q_conv+q_rad#
print "Total heat loss = %0.3f W/m" %q_total

The heat loss by convection = 30.6 W/m
Heat loss by radiation = 25 W/m
Total heat loss = 55.672 W/m

Example No : 1.7 - Page : 19

In [21]:
from scipy.integrate import quad
# given data
T1=1350   # in degree C
T2=50   # in degree C
L=25*10**-2   #in meter
# Formula q= -k*A*dT/dx
# or q/A= -k*dT/dx
# let q/A = q_by_A
def integrand(T):
    return -0.838*(1+0.0007*T)
ans, err = quad(integrand, T1, T2)
def integrand(x):
    return 1
ans2, err2 = quad(integrand, 0, L)
q_by_A=ans/ans2#
print "Heat transfer rate per square meter through  the cylinder = %0.0f watt" %q_by_A

# Note : Answer in the book is incorrect

Heat transfer rate per square meter through  the cylinder = 6493 watt

Example No : 1.9 - Page : 21

In [22]:
# given data
K_A=0.5   # in W/m degree C
K_B=0.8   # in W/m degree C
Ti_A=600   # inside temp. of slab A in degree C
To_B=100   # outside temp. of slab B in degree C
t_A=4*10**-2   # thickness of slab A
t_B=6*10**-2   # thickness of slab B
# Heat transfer rate per square meter through the slab A
# q/A = +K_A * ( Ti_A - T) / t_A         (1)
# Heat transfer rate through slab B
# q/A = +K_B * ( T - To_B) / t_B        (2)
# Equating Eqns (1) and (2)
# K_A*(Ti_A - T)/t_A  =  K_B*(T - To_B)/t_B
T=t_A*t_B/(K_A*t_B+K_B*t_A)*(K_A*Ti_A/t_A + K_B*To_B/t_B)
print "T, intermediate temperature of slab A and B is :",round(T,3),"degree C"
#Putting the value of T in Eq(1), we get
q_by_A= K_A*( Ti_A - T) / t_A#
print "Steady state heat transfer rate per square meter is :",round(q_by_A,3),"W/m**2"
#Note : Answer in the book is wrong

T, intermediate temperature of slab A and B is : 341.935 degree C
Steady state heat transfer rate per square meter is : 3225.806 W/m**2

Example No : 1.10 - Page : 22

In [23]:
# given data
La=3*10**-2   # in meter
Aa=1   # in m**2
ka=150   # in W/m-K

Lb=8*10**-2   # in meter
Ab=0.5   # in m**2
kb=30   # in W/m-K

Lc=8*10**-2   # in meter
Ac=0.5   # in m**2
kc=65   # in W/m-K

Ld=5*10**-2   # in meter
Ad=1   # in m**2
kd=50   # in W/m-K

T1=400   # in degree C
T2=60   # in degree C

Ra=La/(ka*Aa)#
Rb=Lb/(kb*Ab)#
Rc=Lc/(kc*Ac)#
Rd=Ld/(kd*Ad)#
#The equivalent resistance for Rb and Rc
Re=Rb*Rc/(Rb+Rc)#
#Total Resistance
sigmaR=Ra+Re+Rd#
# heat transfer rate per square meter
q=(T1-T2)/sigmaR#
print "Heat transfer rate per square meter = %0.2e Watt" %q

Heat transfer rate per square meter = 1.18e+05 Watt

Example No : 1.11 - Page : 23

In [24]:
# given data
k_Al=202   # in W/mK
x_Al=0.005   # in m
del_T=80   # in degree C
R_contact=0.88*10**-4   # in m**2K/W
sigmaR=x_Al/k_Al+R_contact+x_Al/k_Al   # in m**2K/W
q=del_T/sigmaR   # in W/m**2
#Temperature drop across the rough surface
del_T=q*R_contact   #in degree C
print "Temperature drop across the rough surface = %0.3f degree C" %del_T

Temperature drop across the rough surface = 51.198 degree C

Example No : 1.12 - Page : 24

In [25]:
# given data
T1=100   # in degree C
T2=10   # in degree C
A=3*5   #in square meter
x=40*10**-2   # thickness in m**2
k=1.6   # in W/mk
h=10   # in W/m**2k
# Total resistance in heat flow path
sigmaR=x/(k*A)+1/(h*A)#
# so heat transfer rate
q=(T1-T2)/sigmaR   # in Watt
q=q*10**-3   #in kW
print "Heat transfer rate = %0.3f kW" %q

# Note: Answer in the book is wrong

Heat transfer rate = 3.857 kW

Example No : 1.13 - Page : 25

In [26]:
from scipy.integrate import quad
# given data
k='2.0+0.0005*T'   # in W/m-k
A=3*5   #in square meter
T1=150   # in degree C
T2=50   # in degree C
L=20*10**-2   # thickness in m**2
# Formula q= -k*A*dt/dx
def integrand(T):
    return 2.0+0.0005*T
ans, err = quad(integrand, T1, T2)
def integrand(x):
    return 1
ans2, err2 = quad(integrand, 0, L)

q=-A*ans/ans2   # in Watt
q=q*10**-3   #in kW
print "Rate of heat transfer = %0.3f kW" %q

Rate of heat transfer = 15.375 kW

Example No : 1.14 - Page : 26

In [27]:
# given data
T1=300   #in degree C
T2=50   #in degree C
x2=2*10**-2   # thickness of boiler wall in m
tc2=58   # thermal conductivity of wall in W/mk
x3=0.5*10**-2   # thickness of outer surface of the  wall in m
tc3=116*10**-3   # thermal conductivity of outer surface of the wall in W/mk
R1=2.3*10**-3   # in k/W
R2=x2/tc2#
R3=x3/tc3#
sigmaR=R1+R2+R3   # Total Resistance
q=(T1-T2)/sigmaR#
print "Heat transfer rate per unit area = %0.3f W/m**2" %q
# Note: Answer in the book is wrong

Heat transfer rate per unit area = 5464.687 W/m**2

Example No : 1.15 - Page : 26

In [28]:
from numpy import pi
# given data
Tf=80   # in degree C
I=200   # in amp
h=4000   # in W/m**2degree C
rho=70*10**-6#
L=100   # in cm
R=0.1   # in ohm
d=3   # in mm
d=d*10**-3#
As= pi*d#
#Formula I**2*R= h*As*(Tw-Tf)
Tw= I**2*R/(h*As)+Tf#
print "Central temperature of the wire = %0.f °C" %Tw

Central temperature of the wire = 186 °C

Example No : 1.16 - Page : 27

In [29]:
from __future__ import division
# given data
E=500   #Absorb solar energy in W/m**2
epsilon= 0.9#
T_s= 280   # in K
T_infinite=300   # in K
h_c=20   # in W/m**2degree C
T_sky=280   # in K
sigma=5.67*10**-8#
# Formula E= h_c*(T_p-T_infinite)+epsilon*sigma*(T_P**4-T_s**4)
# On simplication T_P= 340.6-0.255*T-p**4
T_p= 315.5   # in K
print "Equilibrium Temperature of the plate = %0.1f K" %T_p

Equilibrium Temperature of the plate = 315.5 K