(10 Points) A contractor is trying to estimate the distance to be expected of a drawdown
of 4.81 m from a pumping well under the following conditions:
Pumping rate = 0.0280 cubic meter per seconds
Pumping time = 1066 d
Drawdown in observation well = 9.52 m
Observation well is located 10.00 m from the pumping well
Aquifer material = medium sand
Aquifer thickness = 14.05 m
Assume that the well is fully penetrating in an unconfined aquifer.
Q=PI()*K(H2^2-H1^2)/(ln(R2/R1))
27.7
To estimate the distance of the drawdown, we can use the Theis equation, which is commonly used in groundwater hydrology. The Theis equation relates the drawdown in an observation well to the pumping rate, pumping time, hydraulic conductivity, distance to the observation well, and aquifer parameters.
The Theis equation is given by:
S = (Q / (4 * π * T)) * W(u)
Where:
- S is the drawdown in the observation well
- Q is the pumping rate
- T is the transmissivity of the aquifer
- W(u) is the well function, which depends on the dimensionless variable "u"
To calculate the distance of the drawdown, we need to solve the Theis equation for the distance (r) when the drawdown (S) is equal to 4.81 m.
First, we need to calculate the transmissivity of the aquifer:
T = K * b
Where:
- K is the hydraulic conductivity of the aquifer
- b is the aquifer thickness
Given:
- Pumping rate (Q) = 0.0280 cubic meter per second
- Pumping time (t) = 1066 days
- Drawdown (S) = 4.81 m
- Distance to the observation well (x) = 10.00 m
- Aquifer material (Medium sand)
- Aquifer thickness (b) = 14.05 m
Let's calculate the transmissivity (T) first:
K (hydraulic conductivity):
The hydraulic conductivity of medium sand can vary, but let's assume a value of 10 meters/day for this calculation.
K = 10 meters/day
T (transmissivity):
T = K * b
T = 10 meters/day * 14.05 meters
T = 140.5 m²/day
Now, we need to calculate the dimensionless variable "u":
u = (S * r²) / (4 * T * t)
Substituting the given values:
4.81 = (0.0280 * r²) / (4 * 140.5 * 1066)
Now, rearrange the equation to solve for r:
r² = (4 * T * t * S) / Q
r = √((4 * T * t * S) / Q)
Substituting the values:
r = √((4 * 140.5 * 1066 * 4.81) / 0.0280)
Calculating r:
r = √(3476111.352858571)
r = 1864.375 m
Therefore, the estimated distance of the drawdown from the pumping well is approximately 1864.375 meters.
To estimate the distance of the drawdown, we can use Theis' equation, which is commonly used to estimate drawdowns in unconfined aquifers. The equation is as follows:
S = (Q/4πT) * W(u)
Where:
S is the drawdown at distance r from the well
Q is the pumping rate
T is the transmissivity of the aquifer
W(u) is a well function that depends on the dimensionless parameter u
u = (r^2 * S) / (4Tt)
To solve this equation, we need to determine the transmissivity of the aquifer and calculate the value of u. The transmissivity can be estimated using the equation:
T = K * H
Where:
T is transmissivity
K is the hydraulic conductivity of the aquifer material
H is the thickness of the aquifer
In this case, the pumping rate is 0.0280 cubic meters per second, the pumping time is 1066 days, the drawdown in the observation well is 9.52 meters, the distance between the pumping well and the observation well is 10.00 meters, the aquifer material is medium sand, and the aquifer thickness is 14.05 meters.
First, we need to determine the transmissivity:
K (hydraulic conductivity) can vary depending on the type of aquifer material. For medium sand, a typical range of K is 10^-2 to 10^-5 meters per second. Since the specific value for K is not given, we will assume a value of 10^-4 meters per second.
T = K * H = (10^-4 m/s) * (14.05 m) = 1.405 x 10^-3 cubic meters per second
Next, we need to calculate the value of u:
u = (r^2 * S) / (4Tt)
Since the distance between the pumping well and the observation well is given as 10.00 meters and the drawdown in the observation well is given as 9.52 meters, we can use these values to calculate u.
u = (r^2 * S) / (4Tt) = (10.00^2 * 9.52) / (4 * 1.405 x 10^-3 * 1066) = 0.168
Now we can solve for S, the required drawdown at a certain distance from the pumping well:
S = (Q / (4πT)) * W(u)
The well function W(u) can be looked up in tables or calculated using numerical methods. For simplicity, we will use an approximation for W(u) in this example.
W(u) is approximately equal to ln(u) + 0.5772 + C, where C is a constant. For the given value of u (0.168), we can calculate W(u):
W(u) = ln(0.168) + 0.5772 + C
To solve for C, we can use the fact that W(u) should be equal to zero when u is equal to one:
0 = ln(1) + 0.5772 + C
0 = 0 + 0.5772 + C
C = -0.5772
Now we can calculate W(u):
W(u) = ln(0.168) + 0.5772 - 0.5772 = ln(0.168)
Now we have all the values required to calculate the drawdown at a certain distance from the pumping well:
S = (Q / (4πT)) * W(u)
S = (0.0280 / (4π * 1.405 x 10^-3)) * ln(0.168)
Calculating this expression, we can find the value of S, which represents the drawdown at the distance of the observation well.