Just need someone to talk to pls dont just use me for points
1 answer:
You might be interested in
Answer:
S = 5.7209 M
Explanation:
Given data:
B = 20.1 m
conductivity ( K ) = 14.9 m/day
Storativity ( s ) = 0.0051
1 gpm = 5.451 m^3/day
calculate the Transmissibility ( T ) = K * B
= 14.9 * 20.1 = 299.5 m^2/day
Note :
t = 1
U = ( r^2* S ) / (4*T*<em> t </em>)
= ( 7^2 * 0.0051 ) / ( 4 * 299.5 * 1 ) = 2.0859 * 10^-4
Applying the thesis method
W(u) = -0.5772 - In(U)
= 7.9
next we calculate the pumping rate from well ( Q ) in m^3/day
= 500 * 5.451 m^3 /day
= 2725.5 m^3 /day
Finally calculate the drawdown at a distance of 7.0 m form the well after 1 day of pumping
S =
where : Q = 2725.5
T = 299.5
W(u) = 7.9
substitute the given values into equation above
S = 5.7209 M
Answer:
a = 1.68m/
Explanation:
Please kindly find the attached file for explanations
Answer:
a) 280MPa
b) -100MPa
c) -0.35
d) 380 MPa
Explanation:
GIVEN DATA:
mean stress
stress amplitude
a)
--------------1
-----------2
solving 1 and 2 equation we get
b)
c)
stress ratio
d)magnitude of stress range
= 280 -(-100) = 380 MPa