Time in Hours given Constant Depending upon Soil at Base with Base 10 Solution

STEP 0: Pre-Calculation Summary
Formula Used
Time = ((Cross-Sectional Area*2.303)/Constant)*log((Depression Head 1/Depression Head 2),10)
t = ((Acs*2.303)/K)*log((h1/h2),10)
This formula uses 1 Functions, 5 Variables
Functions Used
log - Logarithmic function is an inverse function to exponentiation., log(Base, Number)
Variables Used
Time - (Measured in Hour) - Time can be defined as the ongoing and continuous sequence of events that occur in succession, from the past through the present to the future.
Cross-Sectional Area - (Measured in Square Meter) - Cross-Sectional Area is the area of a two-dimensional shape that is obtained when a three-dimensional shape is sliced perpendicular to some specified axis at a point.
Constant - Constant depending on the soil at the base of the well.
Depression Head 1 - (Measured in Meter) - Depression Head 1 is the difference of level of waters table and the water level in the well when pumping stopped.
Depression Head 2 - (Measured in Meter) - Depression Head 2 is the difference of level of waters table and the water level in the well when pumping stopped.
STEP 1: Convert Input(s) to Base Unit
Cross-Sectional Area: 13 Square Meter --> 13 Square Meter No Conversion Required
Constant: 5 --> No Conversion Required
Depression Head 1: 27 Meter --> 27 Meter No Conversion Required
Depression Head 2: 10 Meter --> 10 Meter No Conversion Required
STEP 2: Evaluate Formula
Substituting Input Values in Formula
t = ((Acs*2.303)/K)*log((h1/h2),10) --> ((13*2.303)/5)*log((27/10),10)
Evaluating ... ...
t = 13.8810917780129
STEP 3: Convert Result to Output's Unit
49971.9304008463 Second -->13.8810917780129 Hour (Check conversion here)
FINAL ANSWER
13.8810917780129 13.88109 Hour <-- Time
(Calculation completed in 00.004 seconds)

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Birsa Institute of Technology (BIT), Sindri
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9 Recuperate Time Calculators

Time in Hours given Constant Depression Head and Area of Well
Go Time = (2.303*Cross-Sectional Area*Constant Depression Head*log((Depression Head 1/Depression Head 2),10))/Discharge
Time in Hours given Constant Depending upon Soil at Base
Go Time = (Cross-Sectional Area/Constant)*log((Depression Head 1/Depression Head 2),e)
Time in Hours given Constant Depending upon Soil at Base with Base 10
Go Time = ((Cross-Sectional Area*2.303)/Constant)*log((Depression Head 1/Depression Head 2),10)
Time in Hours given Clay Soil
Go Time = (1/0.25)*log((Depression Head 1/Depression Head 2),e)
Time in Hours given Fine Sand
Go Time = (1/0.5)*log((Depression Head 1/Depression Head 2),e)
Time in Hours given Coarse Sand
Go Time = log((Depression Head 1/Depression Head 2),e)
Time in Hours with Base 10 given Clay Soil
Go Time = (2.303/0.25)*log((Depression Head 1/Depression Head 2),10)
Time in Hours with Base 10 given Fine Sand
Go Time = (2.303/0.5)*log((Depression Head 1/Depression Head 2),10)
Time in Hours with Base 10 given Coarse Sand
Go Time = (2.303/1)*log((Depression Head 1/Depression Head 2),10)

Time in Hours given Constant Depending upon Soil at Base with Base 10 Formula

Time = ((Cross-Sectional Area*2.303)/Constant)*log((Depression Head 1/Depression Head 2),10)
t = ((Acs*2.303)/K)*log((h1/h2),10)

What is specific capacity ?

Specific capacity is a quantity that which a water well can produce per unit of drawdown. It is normally obtained from a step drawdown test.

How to Calculate Time in Hours given Constant Depending upon Soil at Base with Base 10?

Time in Hours given Constant Depending upon Soil at Base with Base 10 calculator uses Time = ((Cross-Sectional Area*2.303)/Constant)*log((Depression Head 1/Depression Head 2),10) to calculate the Time, Time in Hours given Constant Depending upon Soil at Base with Base 10 calculates value of time in hours when we have prior information of other parameters used. Time is denoted by t symbol.

How to calculate Time in Hours given Constant Depending upon Soil at Base with Base 10 using this online calculator? To use this online calculator for Time in Hours given Constant Depending upon Soil at Base with Base 10, enter Cross-Sectional Area (Acs), Constant (K), Depression Head 1 (h1) & Depression Head 2 (h2) and hit the calculate button. Here is how the Time in Hours given Constant Depending upon Soil at Base with Base 10 calculation can be explained with given input values -> 0.003856 = ((13*2.303)/5)*log((27/10),10).

FAQ

What is Time in Hours given Constant Depending upon Soil at Base with Base 10?
Time in Hours given Constant Depending upon Soil at Base with Base 10 calculates value of time in hours when we have prior information of other parameters used and is represented as t = ((Acs*2.303)/K)*log((h1/h2),10) or Time = ((Cross-Sectional Area*2.303)/Constant)*log((Depression Head 1/Depression Head 2),10). Cross-Sectional Area is the area of a two-dimensional shape that is obtained when a three-dimensional shape is sliced perpendicular to some specified axis at a point, Constant depending on the soil at the base of the well, Depression Head 1 is the difference of level of waters table and the water level in the well when pumping stopped & Depression Head 2 is the difference of level of waters table and the water level in the well when pumping stopped.
How to calculate Time in Hours given Constant Depending upon Soil at Base with Base 10?
Time in Hours given Constant Depending upon Soil at Base with Base 10 calculates value of time in hours when we have prior information of other parameters used is calculated using Time = ((Cross-Sectional Area*2.303)/Constant)*log((Depression Head 1/Depression Head 2),10). To calculate Time in Hours given Constant Depending upon Soil at Base with Base 10, you need Cross-Sectional Area (Acs), Constant (K), Depression Head 1 (h1) & Depression Head 2 (h2). With our tool, you need to enter the respective value for Cross-Sectional Area, Constant, Depression Head 1 & Depression Head 2 and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.
How many ways are there to calculate Time?
In this formula, Time uses Cross-Sectional Area, Constant, Depression Head 1 & Depression Head 2. We can use 8 other way(s) to calculate the same, which is/are as follows -
  • Time = (1/0.25)*log((Depression Head 1/Depression Head 2),e)
  • Time = (2.303/0.25)*log((Depression Head 1/Depression Head 2),10)
  • Time = (1/0.5)*log((Depression Head 1/Depression Head 2),e)
  • Time = (2.303/0.5)*log((Depression Head 1/Depression Head 2),10)
  • Time = log((Depression Head 1/Depression Head 2),e)
  • Time = (2.303/1)*log((Depression Head 1/Depression Head 2),10)
  • Time = (Cross-Sectional Area/Constant)*log((Depression Head 1/Depression Head 2),e)
  • Time = (2.303*Cross-Sectional Area*Constant Depression Head*log((Depression Head 1/Depression Head 2),10))/Discharge
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