from __future__ import division
print "Example 32.5 page no 512\n\n"
# a fluid is transported 4 miles under turbulent flow conditions
#we have two choices in designing the system
OC_a=20000#per year pressure drop costs for the 2 inch ID pipe,$
CRF=0.1#capital recovery factor for both pipe
OC_b=OC_a/16#operating cost associated with the pressure drop cost per year for 4 inch pipe
d=4*5280#distance,feet
c_a=1# 2 inch ID pipe cost per feet,$
c_b=6# 4 inch ID pipe cost per feet,$
CC_a=d*c_a*CRF#capital cost for 2 inch ID pipe,$
CC_b=d*c_b*CRF#capital cost for 4 inch ID pipe,$
TC_a= OC_a +CC_a#total cost associated with 2 inch pipe
print "\n total cost with 2 inch pipe TC_a=%0.2f $"%(TC_a)#
TC_b=OC_b + CC_b#total cost associated with 4 inch pipe
print "\n total cost with 4 inch pipe TC_b=%0.2f $"%(TC_b)#
#from result we can conclude that 4 inch pipe is more economical
from scipy.optimize import fsolve
from __future__ import division
print " Example 32.6 page no 512\n\n"
#a process emits gas of containg dust,a particulate device is employed for particle capture
q=50000#vol. flow rate of dust,ft**3/min
c=2/7000#inlet loading of dust
DV=0.03#value of dust
#recovered value RV can be expressed in terms of pressure drop
#RV=q*c*DV*P1/(P1+15)
C_e=0.18#cost of electricity
E_f=0.55#fractional efficiency
def f(P1):
E=P1/(P1+15)#collection efficiency
RV=q*c*DV*E#recovered value in terms of E$/min
C_p=q*(C_e/44200)*P1/(E_f*60)
# x=q*c*DV*P1/(P1+15)-q*C_e*P1/E_f
x=RV-C_p
return x
P1=fsolve(f,100)
print "\n P1=%0.2f"%(P1)#
#calculation mistake in book
from __future__ import division
print "Example 32.8 page no 514\n\n"
#a filter press is in operation
#we have to determine the appraisal value of the press
i=0.03375#intrest on fund
n=9#time,year
SFDF=i/((1+i)**n -1)#sinking fund depreciation factor
P=60000#cost of filter press,$
L=500#salvage value,$
UAP= (P-L)*SFDF#uniform annual payment,$
print "\n uniform annual payment UAP=%0.2f $"%(UAP)#
#in determing the appraisel value where the straight line method of depreciation is used
# B = P -(P-L/n)x
#where x refers to any time the present before the end of usable
x=5#let for 5 year
B5=P-((P-L)/n)*x#appraissl value for 5 year
print "\n apprasial value B=%0.2f $"%(B5)#
from __future__ import division
print "Example 32.9 page no 516\n\n"
#we have to determine the annulized cost of a new processing plant of enviromental control
#input data
CC=150000#capital cost,$
I=.07#interst
n=5#time,year
CRF=(I*(1+I)**n)/((1+I)**5-1)#capital recovery factor CRF
IC=CRF*CC#installation cost,$
OC=15000#operation cost,$
AC=IC + OC#annulized cost
print "\n annulized cost AC=%0.2f $"%(AC)#