#initialisation of variables
c = 100 #in
a = 10 #in
Q = 0.976 #ft
#CALCULATIONS
G = a*Q #ft
#RESULTS
print 'the graphical basic = %.2f ft'%(G)
import math
#initialisation of variables
a = 27.6 #sq ft
h = 1.37 #ft
d = 1.53*(27.9)**0.38*(1.36)**0.24 #ft
#CALCULATIONS
R = d/4 #ft
A = (math.pi*d**2)/4 #sq ft
#RESULTS
print 'The diameter hydraulics radius and area of the hydraulically equivalent circular conduit = %.1f sq ft'%(A)
#initialisation of variables
h1 = 13.5 #ft
h2 = 19.0 #ft
h3 = 27.5 #ft
c1 = 2.0*10**4 #ft
c2 = 2.1*10**4 #ft
c3 = 2.2*10**4 #ft
#CALCULATIONS
H = h1+h2+h3 #ft
C = c1+c2+c3 #ft
#RESULTS
print 'the most economical distributions of the available head = %.1e ft'%(C)
import math
#initialisation of variables
p = 60 #in
h = 20 #percent
a = 1000 #ft
h1 = 40 #percent
c = 0.5 #ft
p1 = 14.3 #ft
p2 = 6.1 #ft
d = 11.7*10**-2 #ft
#CALCULATIONS
deltaV = 26.7 - 18.3 # fps
eq = 3.9*10**-2 * math.sqrt(deltaV/c)*(0.426)**0.356
D = p*eq #ft
#RESULTS
print 'the air valve with a discharge the change in slop = %.2f in.'%(D)
# note : slightly different because of rounding off error.
import math
#initialisation of variables
p = 90. #deg
h = 48. #in
p1 = 100. #psig
P = (1/2*math.pi)*h**2*p1*0.7071 #lb
r = 3000./54-31 #ft
l = 170. #in
b = 6.5*10**-6 #ft
w = 46. #ft
w1 = 1000. #ft
#CALCULATIONS
s = b*w*30*10**6
D = (1./4*math.pi)*h**2*p1 #lb
P = (r)*h**2 #lb
T = math.pi*h*(1/4)*s #lb
T1 = (1./2)*l #tons
Del = b*w*w1 #ft per
#RESULTS
print 'The expansion and contraction of the steel line can be as great as = %.1f ft per 1000ft of length, if unrestrained'%(Del)