import math
ho = 12.0 # Outside convective heat transfer coefficient in W/m**2K
x1 = 0.23# Thickness of brick in m
k1 = 0.98 # Thermal conductivity of brick in W/mK
x2 = 0.08 # Thickness of foam in m
k2 = 0.02# Thermal conductivity of foam in W/mK
x3 = 1.5# Thickness of wood in cm
k3 = 0.17# Thermal conductivity of wood in W/cmK
hi = 29.0# Inside convective heat transfer coefficient in W/m**2K
A = 90.0 # Total wall area in m**2
to = 22.0# outside air temperature in degree Celsius
ti = -2.0 # Inside air temperature in degree Celsius
print "\n Example 18.1\n"
U = (1/((1/ho)+(x1/k1)+(x2/k2)+(x3*1e-2/k3)+(1/hi)))# Overall heat transfer coefficient
Q = U*A*(to-ti) # Rate of heat transfer
R = (1/ho)+(x1/k1)
t2 = to-Q*R/A # Temperature at inside surface of brick
print "\n The rate of heat removal is ",Q ," W"
print "\n Temperature at inside surface of brick is ",t2 ," degree celcius"
#The answers vary due to round off error
import math
r1 = 5.0 # Inner radius of steel pipe in cm
r2 = 10.0 # Extreme radius of inner insulation in cm
r3 = 13.0# Extreme radius of outer insulation in cm
K1 = 0.23 # Thermal conductivity of inner insulation in W/mK
K2 = 0.37 # Thermal conductivity of outer insulation in W/mK
hi = 58.0 # Inner heat transfer coefficient in W/m**2K
h0 = 12.0 # Inner heat transfer coefficient in W/m**2K
ti = 60.0 # Inner temperature in degree Celsius
to = 25.0 # Outer temperature in degree Celsius
L = 50.0 # Length of pipe in m
print "\n Example 18.2\n"
Q =((2*math.pi*L*(ti-to))/((1/(hi*r1*1e-2))+(math.log(r2/r1)/(K1))+(math.log(r3/r2)/(K2))+(1/(h0*r3*1e-2))))
# Rate of heat transfer
print "\n Heat transfer rate is ",Q/1e3 ," kW"
#The answers vary due to round off error
import math
to = 20 # Environment temperature in degree Celsius
t = 100# Temperature of steam path in degree Celsius
ta1 = 26.76 # Temperature at other end in degree Celsius for rod A
d = 10 # diameter of rod in mm
L = 0.25 # length of rod in m
h = 23 # heat transfer coefficient in W/m**2 K
tb1 = 32.00 # Temperature at other end in degree Celsius for rod B
tc1 = 36.93 # Temperature at other end in degree Celsius for rod C
print "\n Example 18.3\n"
A = math.pi/4 * (d*1e-3)**2 #Area of rod
p = math.pi*d*1e-3 # perimeter of rod
# For rod A
a = (ta1-to)/(t-to)
ma = (math.acosh(1/a))/L
Ka = (h*p)/(ma**2*A) # Thermal conductivity of rod A
print "\n Thermal conductivity of rod A is ",Ka ," W/mK"
# For rod B
b = (tb1-to)/(t-to)
mb = (math.acosh(1/b))/L
Kb = (h*p)/(mb**2*A) # Thermal conductivity of rod B
print "\n Thermal conductivity of rod B is ",Kb ," W/mK"
c = (tc1-to)/(t-to)
mc = (math.acosh(1/c))/L
Kc = (h*p)/(mc**2*A) # Thermal conductivity of rod A
print "\n Thermal conductivity of rod C is ",math. ceil(Kc) ," W/mK"
#The answers vary due to round off error
import math
h = 17.4 # Convective heat transfer coefficient in W/m**2K
K = 52.2 # Thermal conductivity in W/mK
t = 120 # Heat reservoir wall temperature in degree celcius
t0 = 35 # Ambient temperature in degree celcius
L = 0.4 # Lenght of rod in m
b = .050 # width of rod in mm
H = .050 # Heigth of rod in mm
print "\n Example 18.4\n"
l= L/2
A = b*H
m = math.sqrt(4*h*b/(K*b*H))
t1 = (t-t0)/math.cosh(m*l) + t0 # Midway temperature of rod
Q1 = 2*5.12*K*A*(t-t0)*math.tanh(m*l) # Heat loss rate
print "\n Midway temperature of rod is ",t1 ," degree Celcius"
print "\n Heat loss rate is ",Q1 ,"W"
#The answers vary due to round off error
import math
d = 8.0 # Average diameter in mm
r = 750.0 # Density in Kg/m**3
t = 2.0 # Intermediate temperature in degree celcius
t_inf = 1.0 # Ambient temperature in degree celcius
t0 = 25.0 # Initial temperature in degree celcius
c = 3.35 # Specific heat in kJ/KgK
h = 5.8 # Heat transfer coeeficient in W/m**2K
T1 = 10.0 # time period in minutes
T2 = 30.0 # time period in minutes
t1 = 5.0 # Intermediate temperature in degree celcius
print "\n Example 18.5\n"
tau1 = c*1e3*math.log((t0-t_inf)/(t-t_inf))/(h*60) # Time to cool down to 2 degree celcius
tau2 = (t0-t_inf)*(math.exp(-(c*T1*60)/(c*1e3))) # Temperature of peas after 10 minutes
Y = math.exp(-1*(c*T2*60)/(c*1e3))
tau3 = (t0*Y-t1)/(Y-1)
print "\n Time to cool down to 2 degree celcius is ",tau1 ," min"
print "\n Temperature of peas after 10 minutes is ",tau2 ," degree celcius"
print "\n Temperature of peas after 30 minutes is ",tau3 ," degree celcius"
#The answers given in book are incorrect
import math
mh = 1000 # mass flow rate of hot fluid in Kg/h
mc = 1000 # mass flow rate of cold fluid in Kg/h
ch = 2.09 # Specific heat capacity of hot fluid in kJ/kgK
cc = 4.187 #Specific heat capacity of cold fluid in kJ/kgK
th1 = 80# Inlet temperature of hot fluid in degree celcius
th2 = 40 # Exit temperature of hot fluid in degree Celsius
tc1 = 30 # Inlet temperature of cold fluid in degree Celsius
U = 24 # heat transfer coefficient in W/m**2K
print "\n Example 18.6\n"
Q = mh*ch*(th1-th2)
tc2 = Q/(mc*cc) + tc1# outlet temperature of cold fluid
te = th2-tc1 # Exit end temperature difference in degree Celsius
ti = th1 - tc2 # Inlet end temperature difference in degree Celsius
t_lm = (ti-te)/(math.log(ti/te))
A = Q / (U*t_lm*3.6) # Surface are of heat exchanger
print "\n Surface area of heat exchanger is ",A ," m**2"
#The answers vary due to round off error
import math
Hfg = 2257.0 # Latent heat at 100 degree Celsius
ma = 500.0 # mass flow rate of air in Kg/h
ch = 1.005 # Specific heat capacity of hot air in kJ/kgK
ta1 = 260.0 # Inlet temperature of hot air in degree Celsius
ta2 = 150.0 # Inlet temperature of cold air in degree Celsius
tc1 = 100.0 # Inlet temperature of steam
tc2 = tc1 # Exit temperature of steam
U = 46.0 # heat transfer coefficient in W/m**2K
print "\n Example 18.7\n"
Q = ma*ch*(ta1-ta2)
m = Q/Hfg # mass flow rate of steam
te = ta2-tc1 # Exit end temperature difference in degree Celsius
ti = ta1 - tc2 # Inlet end temperature difference in degree Celsius
t_lm = (ti-te)/(math.log(ti/te))
A = Q / (U*t_lm*3.6) # Surface are of heat exchanger
print "\n Surface area of heat exchanger is ",A ," m**2"
#The answers vary due to round off error
import math
mh = 20.15 # mass flow rate of hot fluid in Kg/s
mc = 5.04 # mass flow rate of cold fluid in Kg/h
ch = 2.094 # Specific heat capacity of hot fluid in kJ/kgK
cc = 4.2 #Specific heat capacity of cold fluid in kJ/kgK
th1 = 121# Inlet temperature of hot fluid in degree Celsius
th2 = 40 # Exit temperature of hot fluid in degree Celsius
tc1 = 10 # Inlet temperature of cold fluid in degree Celsius
U = 0.34 # heat transfer coefficient in kW/m**2K
n = 200 # total number of tubes
l = 4.87 # length of tube in m
d = 1.97 # Outer diameter in cm
print "\n Example 18.8\n"
A = math.pi*n*d*1e-2*l # Total surface area
mc_oil = mh*ch
mc_water = mc*cc
c_min = mc_water
c_max =mc_oil
if (mc_oil<mc_water):
c_min = mc_oil
c_max =mc_water
R = c_min/c_max
NTU = U*A/c_min
e = (1-math.exp(-1*NTU*(1-R)))/(1-R*math.exp(-1*NTU*(1-R)))
t_larger = e*(th1-tc1)
t_water = t_larger
t_oil = t_water*mc_water/mc_oil
th2 = th1 - t_oil # Exit temperature of oil
Q = mh*ch*(th1-th2) # Rate of heat transfer
print "\n Exit temperature of oil is ",th2 ," degree celcius"
print "\n Rate of heat transfer is ",Q ," kW"
#The answers vary due to round off error
import math
u_m = 0.8 # mean velocity in m/s
D = 5 # Diameter in cm
v = 4.78e-7 # dynamic coefficient of viscosity
Pr = 2.98 # Prantl number
K = 0.66 # Thermal conductivity in W/mK
l = 3 # length of pipe in m
tw = 70 # Wall temperature
tf = 50 # mean water temperature
print "\n Example 18.9\n"
Re = u_m*D*1e-2/v # Reynold number
Nu = 0.023*(Re**0.8)*(Pr**0.4)
h = K*Nu/(D*1e-2) # Heat transfer coefficient
A = math.pi*D*1e-2*l # Surface area
Q = h*A*(tw-tf) # Rate of heat transfer
print "\n Heat transfer coefficient is ",h ," W/m**2K"
print "\n Rate of heat transfer is ",Q/1e3 ," kW"
#The answers vary due to round off error
import math
b = 10 # width of plate in cm
h = 15 # Height of plate in cm
hr = 8.72 # Radiative heat transfer coefficient in W/m**2K
tw = 140 # temperature of wall in degree Celsius
tf = 20 # Atmospheric temperature in degree Celsius
v = 2.109e-5 # Coefficient of dynamic viscosity in m**2/s
Pr = 0.692 # Prantl number
K = 0.0305 # Thermal conductivity in W/mK
L = 0.15 # characteristic length in m
g = 9.81 # Gravitational acceleration in m/s**2
print "\n Example 18.10\n"
A = 2*b*1e-2*h*1e-2 # total area of plate
t_mean = (tw+tf)/2 +273
B = 1/t_mean
del_t = tw-tf
Gr = g*B*del_t*L**3/v**2 # Grashoff number
x = Gr*Pr
Nu = 0.59*(Gr*Pr)**0.25
hc = Nu*K/L
Q = (hc+hr)*A*del_t # Rate of heat dissipation
print "\n Rate of heat dissipation is ",Q ," W"
#The answers vary due to round off error
import math
d1 = 2.0 # Diameter of steel rod in cm
d2 = 16.0 # Diameter of cylindrical furnace in cm
e1 = 0.6 # emissivity of inner surface
e2 = 0.85 # emissivity of rod surface
T = 1093.0 # Inner surface temperature of furncae in degree celcius
Tr1 = 427.0 # Initial temperature of rod in degree celcius
Tr2 = 538.0 # Initial temperature of rod in degree celcius
sigma = 5.67e-8 # Constant
rho = 7845.0 # density in kg/ m**3
c = 0.67 # Specific heat capacity in kJ/kgK
print "\n Example 18.11\n"
A_ratio = d1/d2 # Surface area ratio of cylindrical bodies
F12 = (1/((1/e1)+(A_ratio*(1/e2 -1))))
A1 = math.pi*d1*1e-2*1 # Surface area of rod
T1 = Tr1+273
T2 = T +273
T3 = Tr2 +273
Qi = sigma*A1*F12*(T1**4-T2**4)
Qe = sigma*A1*F12*(T3**4-T2**4)
Q_avg = abs((Qi+Qe)/2)
tau = rho*c*(1e-4)*math.pi*(Tr2-Tr1)/(Q_avg*(1e-3))
# Time required for heating operation
print "\n Time required for heating operation is ",tau ," s"
#The answers vary due to round off error
import math
d1 = 10.0 # Diameter of inner cylinder in cm
d2 = 20.0 # Diameter of outer cylinder in cm
e1 = 0.65 # emissivity of inner surface
e2 = 0.4 # emissivity of outer surface
T1 = 1000.0 # Inner surface temperature in K
T2 = 500.0 # outer suface temperature in K
sigma = 5.67e-8 # Constant
print "\n Example 18.12\n"
A1 = math.pi*d1*1e-2
A2 = math.pi*d2*1e-2
R =(((1-e1)/(e1*A1))+((1-e2)/(e2*A2))+(1/(A1*1)))
Eb1 = sigma*T1**4
Eb2 = sigma*T2**4
Q = (Eb1-Eb2)/R # Net heat transfer between two cylinders
print "\n Net heat transfer between two cylinders is ",Q ," W/m length"
#The answers vary due to round off error
d1 = 10.0 # Diameter of inner cylinder in cm
d2 = 20.0 # Diameter of outer cylinder in cm
e1 = 0.65 # emissivity of inner surface
e2 = 0.4 # emissivity of outer surface
T1 = 1000.0 # Inner surface temperature in K
T2 = 500.0 # outer surface temperature in K
sigma = 5.67e-8 # Constant
print "\n Example 18.12\n"
A1 = math.pi*d1*1e-2
A2 = math.pi*d2*1e-2
R =(((1-e1)/(e1*A1))+((1-e2)/(e2*A2))+(1/(A1*1)))
Eb1 = sigma*T1**4
Eb2 = sigma*T2**4
Q = (Eb1-Eb2)/R # Net heat transfer between two cylinders
print "\n Net heat transfer between two cylinders is ",Q ," W/m length"
#The answers vary due to round off error