# Variables
deltax = 9./12; # thickness of wall in ft
k = 0.18; # thermal conductivity of wall in B/hr-ft-degF
t1 = 1500; # inside temperature of oven wall in degF
t2 = 400; # outside temperature of oven wall in degF
# Calculations
q = k*(t1-t2)/deltax; # heat loss in Btu/hr
# Results
print " The heat loss for each square foot of wall surface is %d Btu/hr-ft**2"%(q);
# Variables
x1 = 9./12; # thickness of firebrick in ft
k1 = 0.72; # thermal conductivity of firebrick in Btu/hr-ft-degF
x2 = 5./12; # thickness of insulating brick in ft
k2 = 0.08; # thermal conductivity of insulating brick in Btu/hr-ft-degF
x3 = 7.5/12; # thickness of redbrick in ft
k3 = 0.5; # thermal conductivity of firebrick in Btu/hr-ft-degF
t1 = 1500.; # inner temperature of wall in degF
t2 = 150.; # outer temperature of wall in degF
# Calculations and Results
# resistances of mortar joints are neglected
q = (t1-t2)/(x1/k1+x2/k2+x3/k3); # heat flow per square ft in Btu/hr
t2 = t1-(q*x1/k1); # first contact temperature in degF
print " The temperature at the contact of firebrick and insulating brick is %d degF"%(t2);
t3 = t2-(q*x2/k2); # second contact temperature in degF
print " The temperature at the contact of insulating brick and red brick is %d degF"%(t3);
import math
# Variables
d1 = 2.375/12; # internal diameter of pipe in ft
t = 1./12; # thickness of insulating material in ft
d2 = d1+2*t; # external (insulation)diameter of pipe in ft
k = 0.0375; # thermal conductivity of insulating material in Btu/hr-ft-F
l = 30.; # length of pipe in ft
t1 = 380.; # inner surface temperature of insulation
t2 = 80.; # outer surface temperature of insulation
# Calculations and Results
q = 2*math.pi*k*(t1-t2)/math.log(d2/d1); # heat loss per unit length
print " Heat loss per linear foot is %.f Btu/hr"%(q)
qtot = round(q)*l; # heat loss for 30 ft pipe
print " Total heat loss through 30 ft of pipe is %d Btu/hr"%(qtot)
import math
# Variables
d1 = 10.75/12; # outer diameter of pipe in ft
x1 = 1.5/12; # thickness of insulation 1 in ft
x2 = 2./12; # thickness of insulation 2 in ft
d2 = d1+2*x1; # diameter of insulation 1 in ft
d3 = d2+2*x2; # diameter of insulation 1 in ft
t1 = 700.; # inner surface temperature of composite insulation in degF
t2 = 110.; # outer surface temperature of composite insulation in degF
k1 = 0.05; #thermal conductivity of material 1 in Btu/hr-ft-degF
k2 = 0.039; # thermal conductivity of material 2 in Btu/hr-ft-degF
# Calculations
q = 2*math.pi*(t1-t2)/(math.log(d2/d1)/k1+math.log(d3/d2)/k2); # heat loss per linear foot in Btu/hr
# Results
print " The heat loss is found to be %d Btu/hr-ft"%( q);
# Variables
km = 0.0377; # Mean thermal conductivity at 220degF
t1 = 260.; # Inner surface temperature of slab in degF
t2 = 180.; # Outer surface temperature of slab in degF
A = 1.; # Area of slab in ft
x = 2./12; # Thickness of insulation in ft
# Calculations
q = km*A*(t1-t2)/x; # Heat loss through slab in Btu/hr
# Results
print " Heat loss through flat slab is %.1f Btu/hr"%(q);
# Variables
k = 0.8 # Avg. thermal conductivity in Btu/hr-ft-degF
T1 = 400. # Inner surface temperature of furnace in degF
T2 = 100. # Outer surface temperature of furnace in degF
a = 3. # Length of furnace in ft
b = 4. # Breadth of furnace in ft
c = 2.5 # Height of furnace in ft
Aa = 2*a*b # Area of surface A in ft**2
Ab = 2*b*c # Area of surface A in ft**2
Ac = 2*a*c # Area of surface A in ft**2
x = 4.5/12 # Thickness of insulation in ft
t = 24. # Time elapsed in hr
M = 4. # Number of edges
N = 8. # Number of corners
# Calculations
S = Aa/x+Ab/x+Ac/x+0.54*(a+b+c)*M+0.15*x*N # Shape factor
qo = round(S*k*(T1-T2),-2) # Heat flow per hour
q = qo*t # Heat loss in 24 hr
# Results
print "The heat loss in 24 hr is %d Btu"%(q)
# note : book answer is wrong. Kindly check.
# Variables
M = 8*9; # number of flow channels for the entire section
N = 8.37; # number of equal channel intervals
# the fractional part arises due to the fractional part of temperature close to border EG
# Calculations
k = M/N; # Ratio of shape factor to wall length
# Results
print " Shape factor for the special section where the ratio of radius of\
circle to half side length is 0.5,S is %.2fL"%( k );
from numpy import zeros
# Variables
t1 = 800.; # inner surface temperature of wall in degF
t4 = 200.; # outer surface temperature of wall in degF
#Grids are square in shape so delx = dely where delx,y sre dimensions of square grid
t2 = [700, 550 ,550, 587.5, 587.5, 596.9, 596.9, 599.3, 599.3, 599.8]; # Assumed temperature of grid point 1
t3 = [300, 300 ,375 ,375, 393.8, 393.8, 398.5, 398.5, 399.6, 399.6]; # Assumed temperature of grid point 2
th2 = zeros(9)
th3 = zeros(9)
# Calculations and Results
for i in range(9):
th2[i] = t1+t3[i]-2*t2[i]; # th1 = q/kz at grid pt1
th3[i] = t2[i]+t4-2*t3[i];# th2 = q/kz at grid pt2
print " Assuming t2 = %.1f degF and t2 = %.1f degF th1[%d] = %.1f degF and th2[%d] = %.1f degF "%(t2[i],t3[i],i,th2[i],i,th3[i]);
print " Since th2[%d] is not equal to th3[%d], hence other values of t2 and t3 are to be assumed"%(i,i);
print "Assuming t2 = 600 degF and t3 = 400 degF, th2 = th3."
print "Hence Steady state condition is satisfied at grid temperatures of 400 degF and 600 degF";
import math
# Variables
k = 0.9; # thermal conductivity of section material in Btu/hr-ft-degF
# Heat is considered to flow through fictitious rods and only half of the heat flows through symmetry axes
Ta = 300.;
Tb = 441.;
Tc = 600.;
Td = 300.;
Te = 432.;
Tf = 600.;
Tg = 600.;
Th = 600.;
Ti = 300.;
Tj = 384.;
Tk = 461.;
Tl = 485.;
Tm = 490.;
Tn = 300.;
To = 340.;
Tp = 372.;
Tq = 387.;
Tr = 391.;
Ts = 300.;
Tt = 300.;
Tu = 300.;
Tv = 300.;
Tw = 300.;
# Calculations and Results
# Above grid point temperatures are given in the question for the quarter section considered in degF(a,b,c...w are grid points)
q1 = 4*k*((Tc-Tb)/2+(Tf-Te)+(Tf-Tk)+(Tg-Tl)+(Th-Tm)/2); # Amount of heat coming from inside in Btu/hr
q2 = 4*k*((Tb-Ta)/2+(Te-Td)+(Tj-Ti)+(To-Tn)+(To-Tt)+(Tp-Tu)+(Tq-Tu)+(Tr-Tw)/2); # Amount of heat going outside in Btu/hr
q = (q1+q2)/2; # average of heat going in and heat coming out
print " Total heat flow per unit depth is %.1fBtu/hr"%(q);
S = q/(k*(Tc-Ta)); # shape factor in ft
print " Shape factor is %.2fft"%(S)