Time in Hours given Specific Capacity of Open Well with Base 10 Calculator

✖Specific Capacity is capacity of an open well.ⓘ Specific Capacity [K_a]			+10% -10%
✖Depression Head is the difference of level of waters table and the water level in the well when pumping stopped.ⓘ Depression Head [h_d]			+10% -10%
✖Depression Head in Well 2 is depression head in the well at a time T after the pumping stopped.ⓘ Depression Head in Well 2 [h_w2]			+10% -10%

✖Time refers to the ongoing and continuous sequence of events that occur in succession, from the past through the present to the future.ⓘ Time in Hours given Specific Capacity of Open Well with Base 10 [t]

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Time in Hours given Specific Capacity of Open Well with Base 10 Solution

STEP 0: Pre-Calculation Summary

Formula Used

Time = (2.303/Specific Capacity)*log((Depression Head/Depression Head in Well 2),10)
t = (2.303/K_a)*log((h_d/h_w2),10)
This formula uses 1 Functions, 4 Variables

Functions Used

log - Logarithmic function is an inverse function to exponentiation., log(Base, Number)

Variables Used

Time - (Measured in Second) - Time refers to the ongoing and continuous sequence of events that occur in succession, from the past through the present to the future.
Specific Capacity - (Measured in Meter per Second) - Specific Capacity is capacity of an open well.
Depression Head - (Measured in Meter) - Depression Head is the difference of level of waters table and the water level in the well when pumping stopped.
Depression Head in Well 2 - (Measured in Meter) - Depression Head in Well 2 is depression head in the well at a time T after the pumping stopped.

STEP 1: Convert Input(s) to Base Unit

Specific Capacity: 2 Meter per Hour --> 0.000555555555555556 Meter per Second (Check conversion here)
Depression Head: 27 Meter --> 27 Meter No Conversion Required
Depression Head in Well 2: 10 Meter --> 10 Meter No Conversion Required

STEP 2: Evaluate Formula

Substituting Input Values in Formula

t = (2.303/K_a)*log((h_d/h_w2),10) --> (2.303/0.000555555555555556)*log((27/10),10)

Evaluating ... ...

t = 9609.98661554736

STEP 3: Convert Result to Output's Unit

9609.98661554736 Second -->2.66944072654093 Hour (Check conversion here)

FINAL ANSWER

2.66944072654093 ≈ 2.669441 Hour <-- Time

(Calculation completed in 00.004 seconds)

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< Discharge from Well Calculators

Discharge from Well given Specific Capacity

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Discharge from Well given Specific Capacity for Coarse Sand

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Discharge from Well given Specific Capacity for Fine Sand

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Time in Hours given Specific Capacity of Open Well with Base 10 Formula

LaTeX Go

Time = (2.303/Specific Capacity)*log((Depression Head/Depression Head in Well 2),10)
t = (2.303/K_a)*log((h_d/h_w2),10)

What is Specific Capacity?

Specific Capacity refers to the efficiency of a well in producing water. It is a measure of how much water (in volume) a well can yield per unit of drawdown (or drop in water level) over a specific period of time. The specific capacity is typically expressed in units such as gallons per minute per foot (gpm/ft) or liters per second per meter (L/s/m).

How to Calculate Time in Hours given Specific Capacity of Open Well with Base 10?

Time in Hours given Specific Capacity of Open Well with Base 10 calculator uses Time = (2.303/Specific Capacity)*log((Depression Head/Depression Head in Well 2),10) to calculate the Time, The Time in Hours given Specific Capacity of Open Well with Base 10 formula is defined as the calculation of 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 Specific Capacity of Open Well with Base 10 using this online calculator? To use this online calculator for Time in Hours given Specific Capacity of Open Well with Base 10, enter Specific Capacity (K_a), Depression Head (h_d) & Depression Head in Well 2 (h_w2) and hit the calculate button. Here is how the Time in Hours given Specific Capacity of Open Well with Base 10 calculation can be explained with given input values -> 0.000742 = (2.303/0.000555555555555556)*log((27/10),10).

FAQ

What is Time in Hours given Specific Capacity of Open Well with Base 10?

The Time in Hours given Specific Capacity of Open Well with Base 10 formula is defined as the calculation of value of time in hours when we have prior information of other parameters used and is represented as t = (2.303/K_a)*log((h_d/h_w2),10) or Time = (2.303/Specific Capacity)*log((Depression Head/Depression Head in Well 2),10). Specific Capacity is capacity of an open well, Depression Head is the difference of level of waters table and the water level in the well when pumping stopped & Depression Head in Well 2 is depression head in the well at a time T after the pumping stopped.

How to calculate Time in Hours given Specific Capacity of Open Well with Base 10?

The Time in Hours given Specific Capacity of Open Well with Base 10 formula is defined as the calculation of value of time in hours when we have prior information of other parameters used is calculated using Time = (2.303/Specific Capacity)*log((Depression Head/Depression Head in Well 2),10). To calculate Time in Hours given Specific Capacity of Open Well with Base 10, you need Specific Capacity (K_a), Depression Head (h_d) & Depression Head in Well 2 (h_w2). With our tool, you need to enter the respective value for Specific Capacity, Depression Head & Depression Head in Well 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 Specific Capacity, Depression Head & Depression Head in Well 2. We can use 1 other way(s) to calculate the same, which is/are as follows -

Time = (1/Specific Capacity)*log((Depression Head/Depression Head in Well 2),e)

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