# Chapter 32 : Economics and Finance¶

## Example 32.5 Page no 512¶

In :
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 $"%(TC_b)# #from result we can conclude that 4 inch pipe is more economical  Example 32.5 page no 512 total cost with 2 inch pipe TC_a=22112.00$

total cost with 4 inch pipe TC_b=13922.00 $ ## Example 32.6 Page no 512¶ In : 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

 Example 32.6 page no 512

P1=54.46


## Example 32.8 Page no 514¶

In :
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 $"%(B5)#  Example 32.8 page no 514 uniform annual payment UAP=5767.91$

apprasial value B=26944.44 $ ## Example 32.9 Page no 516¶ In : 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 $"%(AC)#  Example 32.9 page no 516 annulized cost AC=51583.60$