# 	**Problem 5.1**
# A community is installing a new well in a regionally confined aquifer 
# with a transmissivity of 1589 ft2/day and a storativity of 0.0005. The 
# planned pumping rate is 325 gal/min. There are several nearby wells 
# tapping the same aquifer, and the project manager needs to know if the 
# new well will cause significant interference with these wells. Compute 
# the theoretical drawdown caused by the new well after 30 days 
To = 1589 ft^2/day
So = 0.0005
Q = 325 gal/min
t = 30 day
r1 = 50 ft
r2 = 150 ft
r3 = 250 ft
r4 = 500 ft
r5 = 1000 ft
r6 = 3000 ft
r7 = 6000 ft
r8 = 10000 ft
u1 = r1^2 * So / (4 * To * t)
u2 = r2^2 * So / (4 * To * t)
u3 = r3^2 * So / (4 * To * t)
u4 = r4^2 * So / (4 * To * t)
u5 = r5^2 * So / (4 * To * t)
u6 = r6^2 * So / (4 * To * t)
u7 = r7^2 * So / (4 * To * t)
u8 = r8^2 * So / (4 * To * t)
Wu1 = ExpInt(u1)  # The function W(u) = ExpInt(u)
Wu2 = ExpInt(u2)
Wu3 = ExpInt(u3)
Wu4 = ExpInt(u4)
Wu5 = ExpInt(u5)
Wu6 = ExpInt(u6)
Wu7 = ExpInt(u7)
Wu8 = ExpInt(u8)
s1 = Q * Wu1 / (4 * pi() * To); ft
s2 = Q * Wu2 / (4 * pi() * To); ft
s3 = Q * Wu3 / (4 * pi() * To); ft
s4 = Q * Wu4 / (4 * pi() * To); ft
s5 = Q * Wu5 / (4 * pi() * To); ft
s6 = Q * Wu6 / (4 * pi() * To); ft
s7 = Q * Wu7 / (4 * pi() * To); ft
s8 = Q * Wu8 / (4 * pi() * To); ft

# 	**Problem 5.3**
# Plot the distance-drawdown data from Problem 1 on semilog paper.

# See answer on page 564 of text.

# 	**Problem 5.5**
# Plot the distance-drawdown data from Problem 2 on semilog paper

# See answer on page 565 of text.

# 	**Problem 5.7**
# If the aquifer in Problem 1 is not fully confined, but is overlain by a 
# 13.7-ft-thick confining layer with a vertical hydraulic conductivity of 
# 0.13 ft/day and no storativity, what would be the drawdown values after 
# 30 days of pumping at 325 gal/min at the indicated distances?
To = 1589 ft^2/day
So = 0.0005
Q = 325 gal/min
t = 30 day
b_prime = 13.7 ft
K_prime = 0.13 ft/day
r1 = 50 ft
r2 = 150 ft
r3 = 250 ft
r4 = 500 ft
r5 = 1000 ft
r6 = 3000 ft
r7 = 6000 ft
r8 = 10000 ft
u1 = r1^2 * So / (4 * To * t)
u2 = r2^2 * So / (4 * To * t)
u3 = r3^2 * So / (4 * To * t)
u4 = r4^2 * So / (4 * To * t)
u5 = r5^2 * So / (4 * To * t)
u6 = r6^2 * So / (4 * To * t)
u7 = r7^2 * So / (4 * To * t)
u8 = r8^2 * So / (4 * To * t)
B = sqrt(To*b_prime/K_prime)
c1 = r1/B
c2 = r2/B
c3 = r3/B
c4 = r4/B
c5 = r5/B
c6 = r6/B
c7 = r7/B
c8 = r8/B
Wu_rB1 = 4.56  # from Appendix 3
Wu_rB2 = 2.44
Wu_rB3 = 1.53
Wu_rB4 = 0.71
Wu_rB5 = 0.18
Wu_rB6 = 0.001
Wu_rB7 = 0
Wu_rB8 = 0
s1 = Q * Wu_rB1 / (4 * pi() * To); ft
s2 = Q * Wu_rB2 / (4 * pi() * To); ft
s3 = Q * Wu_rB3 / (4 * pi() * To); ft
s4 = Q * Wu_rB4 / (4 * pi() * To); ft
s5 = Q * Wu_rB5 / (4 * pi() * To); ft
s6 = Q * Wu_rB6 / (4 * pi() * To); ft
s7 = Q * Wu_rB7 / (4 * pi() * To); ft
s8 = Q * Wu_rB8 / (4 * pi() * To); ft

# 	**Problem 5.9**
# With reference to the well and aquifer system in Problem 1, compute the
# drawdown at a distance of 250 ft at the following times: 1, 2, 5, 10
#  15, 30, and 60 min; 2, 5, and 12 h; and 1, 5, 10, 20, and 30 days. 
To = 1589 ft^2/day
So = 0.0005
Q = 325 gal/min
r = 250 ft
t1 = 1 min
t2 = 2 min
t3 = 5 min
t4 = 10 min
t5 = 15 min
t6 = 30 min
t7 = 60 min
t8 = 2 hr
t9 = 5 hr
t10 = 12 hr
t11 = 1 day
t12 = 5 days
t13 = 10 days
t14 = 20 days
t15 = 30 days
u1 = r^2 * So / (4 * To * t1)
u2 = r^2 * So / (4 * To * t2)
u3 = r^2 * So / (4 * To * t3)
u4 = r^2 * So / (4 * To * t4)
u5 = r^2 * So / (4 * To * t5)
u6 = r^2 * So / (4 * To * t6)
u7 = r^2 * So / (4 * To * t7)
u8 = r^2 * So / (4 * To * t8)
u9 = r^2 * So / (4 * To * t9)
u10 = r^2 * So / (4 * To * t10)
u11 = r^2 * So / (4 * To * t11)
u12 = r^2 * So / (4 * To * t12)
u13 = r^2 * So / (4 * To * t13)
u14 = r^2 * So / (4 * To * t14)
u15 = r^2 * So / (4 * To * t15)
Wu1 = ExpInt(u1)  # The function W(u) = ExpInt(u)
Wu2 = ExpInt(u2)
Wu3 = ExpInt(u3)
Wu4 = ExpInt(u4)
Wu5 = ExpInt(u5)
Wu6 = ExpInt(u6)
Wu7 = ExpInt(u7)
Wu8 = ExpInt(u8)
Wu9 = ExpInt(u9)  
Wu10 = ExpInt(u10)  
Wu11 = ExpInt(u11)  
Wu12 = ExpInt(u12)  
Wu13 = ExpInt(u13)  
Wu14 = ExpInt(u14)  
Wu15 = ExpInt(u15)  
s1 = Q * Wu1 / (4 * pi() * To); ft
s2 = Q * Wu2 / (4 * pi() * To); ft
s3 = Q * Wu3 / (4 * pi() * To); ft
s4 = Q * Wu4 / (4 * pi() * To); ft
s5 = Q * Wu5 / (4 * pi() * To); ft
s6 = Q * Wu6 / (4 * pi() * To); ft
s7 = Q * Wu7 / (4 * pi() * To); ft
s8 = Q * Wu8 / (4 * pi() * To); ft
s9 = Q * Wu9 / (4 * pi() * To); ft
s10 = Q * Wu10 / (4 * pi() * To); ft
s11 = Q * Wu11 / (4 * pi() * To); ft
s12 = Q * Wu12 / (4 * pi() * To); ft
s13 = Q * Wu13 / (4 * pi() * To); ft
s14 = Q * Wu14 / (4 * pi() * To); ft
s15 = Q * Wu15 / (4 * pi() * To); ft

# 	**Problem 5.11**
# Plot the time-drawdown data from Problem 9 on semilog paper.

# See answer on page 566 of text.

# 	**Problem 5.13**
# A well that pumps at a constant rate of 78,000 ft3/day has achieved 
# equilibrium so that there is no change in the drawdown with time. (The 
# cone of depression has expanded to include a recharge zone equal to the
# amount of water being pumped.) The well taps a confined aquifer that is 
# 18 ft thick. An observation well 125 ft away has a head of 277 ft above 
# sea level; another observation well 385 ft away has a head of 291 ft. 
# Compute the value of aquifer transmissivity using the Thiem equation.
Q = 78000 ft^3/day
b = 18 ft
h1 = 277 ft
r1 = 125 ft
h2 = 291 ft
r2 = 385 ft
To = (Q/(2*pi()*(h2-h1)))*ln(r2/r1); ft^2/day