Time in Hours with Base 10 given Fine Sand Solution

STEP 0: Pre-Calculation Summary
Formula Used
Time = (2.303/0.5)*log((Depression Head 1/Depression Head 2),10)
t = (2.303/0.5)*log((h1/h2),10)
This formula uses 1 Functions, 3 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.
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
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 = (2.303/0.5)*log((h1/h2),10) --> (2.303/0.5)*log((27/10),10)
Evaluating ... ...
t = 10.6777629061637
STEP 3: Convert Result to Output's Unit
38439.9464621895 Second -->10.6777629061637 Hour (Check conversion here)
FINAL ANSWER
10.6777629061637 10.67776 Hour <-- Time
(Calculation completed in 00.019 seconds)

Credits

Created by Suraj Kumar
Birsa Institute of Technology (BIT), Sindri
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Meerut Institute of Engineering and Technology (MIET), Meerut
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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 with Base 10 given Fine Sand Formula

Time = (2.303/0.5)*log((Depression Head 1/Depression Head 2),10)
t = (2.303/0.5)*log((h1/h2),10)

What is fine soil ?

Fine-grained soil is described depend on its dry strength, dilatancy, dispersion and plasticity. It has good load-bearing qualities. It has good-load bearing qualities when dry; however, it possesses little or no load-bearing strength if it is wet.

How to Calculate Time in Hours with Base 10 given Fine Sand?

Time in Hours with Base 10 given Fine Sand calculator uses Time = (2.303/0.5)*log((Depression Head 1/Depression Head 2),10) to calculate the Time, Time in Hours with Base 10 given Fine Sand 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 with Base 10 given Fine Sand using this online calculator? To use this online calculator for Time in Hours with Base 10 given Fine Sand, enter Depression Head 1 (h1) & Depression Head 2 (h2) and hit the calculate button. Here is how the Time in Hours with Base 10 given Fine Sand calculation can be explained with given input values -> 0.002966 = (2.303/0.5)*log((27/10),10).

FAQ

What is Time in Hours with Base 10 given Fine Sand?
Time in Hours with Base 10 given Fine Sand calculates value of time in hours when we have prior information of other parameters used and is represented as t = (2.303/0.5)*log((h1/h2),10) or Time = (2.303/0.5)*log((Depression Head 1/Depression Head 2),10). 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 with Base 10 given Fine Sand?
Time in Hours with Base 10 given Fine Sand calculates value of time in hours when we have prior information of other parameters used is calculated using Time = (2.303/0.5)*log((Depression Head 1/Depression Head 2),10). To calculate Time in Hours with Base 10 given Fine Sand, you need Depression Head 1 (h1) & Depression Head 2 (h2). With our tool, you need to enter the respective value for 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 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 = 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 = ((Cross-Sectional Area*2.303)/Constant)*log((Depression Head 1/Depression Head 2),10)
  • Time = (2.303*Cross-Sectional Area*Constant Depression Head*log((Depression Head 1/Depression Head 2),10))/Discharge
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