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"
# 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
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
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
# 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
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
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
# 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
# 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
# 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
# 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
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
# 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
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
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