#initialization of variables
T = 26.2 # centigrade
T0 = 4. # centigrade
Tinf = 40.#centigrade
z = 1.3#cm
t = 180. #seconds
#calculations
k = ((T-T0)/(Tinf-T0))
alpha = (1/(4*t))*((z/k)**2)/2#cm**2/sec
#Results
print"The thermal diffusivity is ",round(alpha,4)
#initialization of variables
Q = 18. # m**3/hr
z = 2.80 #m
T = 140.#C
T1 = 240. #C
T2 = 20. #C
p= 900. #kg/m**3
Cp = 2. # W/kg-K
d = 0.05#m
from math import pi,log
#Calculations
A = pi*(d**2)/4.
v = Q*(1/(3600*40))/(A)
U = (v*p*Cp*d/(4*z))*(log((T1-T2)/(T1-T)))+0.4#W/m**2-K
DeltaT = ((T1-T2)+(T1-T))/2#C
q = (Q*(1/(3600*40))*p*Cp/(pi*d*z))*(T-T2)#W/m**2-K
U1 = q/DeltaT+0.38#W/m**2-K
#Results
print"The overall heat transfer co efficient based on local temp difference is W/m**2-K",U
print"\nThe overall heat transfer co efficient based on average temp difference is W/m**2-K",U1
#initialization of variables
T = 32. #F
T0 = 10.#F
Tinf= 800 #F
U = 3.6 #Btu/hr-ft**2-F
A = 27. #ft**2
d = 8.31 #lb/gal
V = 100. #gal
Cv = 1.#Btu/lb-F
from math import log
#Calculations
t = (-log((T-T0)/(Tinf-T0)))*d*V*Cv/(U*A)/3#hr
#Results
print"The time we can wait before the water in the tank starts to freeze is hr",round(t)
#initialization of variables
#Given q = h*DeltaT and 0.6q = (1/(1/h)+10/12*0.03)*delta T , divide both to get
l = 10./12. #ft
k = 0.03 #Btu/hr-ft-F
#Calculations
l2 = 2#feet
k2 = 0.03 #Btu/hr-ft-F
h = ((1/0.6)-1)*k/l #Btu/hr-ft**2-F
U = 1/((1/h)+(l2/k2))#Btu/hr-ft**2-F
Savings = U*100/h
#Results
print"The savings due to insulation is about percent",round(Savings)
#initialization of variables
T = 673 # Kelvin
M = 28
sigma = 3.80 # angstroms
omega = 0.87
d1 = 0.05 #m
v1 = 17 #m/sec
Mu1 = 3.3*10**-5 # kg/m-sec
p1 = 5.1*10**-1 # kg/m**3
Cp1 = 1100 # J/kg-K
k2 = 42 # W/m-K
l2 = 3*10**-3 #m
d3 = 0.044 #m
v3 = 270 #m/sec
p3 = 870 #kg/m**3
Mu3 = 5.3*10**-4 # kg/m-sec
Cp3 = 1700# J/kg-K
k3 = 0.15 #W/m-K
#Calculations
kincal = (1.99*10**-4)*((T/M)**0.5)/((sigma**2)*omega)#W/m**2-K
k = kincal*4.2*10**2# k in W/m-K
h1 = 0.33*(k/d1)*((d1*v1*p1/Mu1)**0.6)*((Mu1*Cp1/k)**0.3)#W/m**2-K
h2 = k2/l2 #W/m**2-K
h3 = 0.027*(k3/d3)*((d3*v3*p3/Mu3)**0.8)*((Mu3*Cp3/k3)**0.33)#W/m**2-K
U = 1/((1/h1)+(1/h2)+(1/h3))#W/m**2-K
#Results
print"The overall heat transfer co efficient is W/m**2-K",round(U)
#initialization of variables
#For window with two panes 3 cm apart
k = 0.57*10**-4 #cal/cm-sec-K
l = 3. #cm
g = 980. # cm/sec**2
Nu = 0.14 # cm**2/sec
DeltaT = 30. # Kelvin
T = 278. # Kelvin
L = 100. # cm
#calculations
h = (0.065*(k/l)*(((l**3)*g*DeltaT/((Nu**2)*T))**(1/3))*((l/L)**(1/9)))*10**4+0.42765#for two pane in x*10**-4 cal/cm**2-sec-K
print"The heat transfer co efficent for two panes is x10**-4 cal/cm**2-sec-K",h
#For window with three panes 1.5 cm each apart
k = 0.57*10**-4 #cal/cm-sec-K
l = 1.5#cm
DeltaT = 15. # Kelvin
g = 980. # cm/sec**2
Nu = 0.14 # cm**2/sec
#calculations
h = (0.065*(k/l)*(((l**3)*g*DeltaT/((Nu**2)*T))**(1/3))*((l/L)**(1/9)))*10**4+0.30765 #for two pane in x*10**-4 cal/cm**2-sec-K
print"\nThe heat transfer co efficent for three panes is x10**-4 cal/cm**2-sec-K",round(h/2,2)#Because there are two gaps