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Time in Hours given Time at 1st and 2nd Instance since Pumping Started Formula

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Time in Hours can be defined as the ongoing and continuous sequence of events that occur in succession, from the past through the present to the future, measured in hours. Check FAQs

t hour = \frac{2.303 \cdot Q \cdot \log ((\frac{t 2sec}{t1}), 10)}{4 \cdot π \cdot Δs}

t_hour - Time in Hours?Q - Discharge?t_2sec - Time of Drawdown (t2) in Wells?t1 - Time of Drawdown (t1) in Wells?Δs - Difference in Drawdowns?π - Archimedes' constant?

Time in Hours given Time at 1st and 2nd Instance since Pumping Started Example

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Here is how the Time in Hours given Time at 1st and 2nd Instance since Pumping Started equation looks like with Values.

Here is how the Time in Hours given Time at 1st and 2nd Instance since Pumping Started equation looks like with Units.

Here is how the Time in Hours given Time at 1st and 2nd Instance since Pumping Started equation looks like.

0.1546 = \frac{2.303 \cdot 1.01 \cdot \log ((\frac{62}{58.7}), 10)}{4 \cdot 3.1416 \cdot 0.014}

0.1546h = \frac{2.303 \cdot 1.01m³/s \cdot \log ((\frac{62s}{58.7s}), 10)}{4 \cdot 3.1416 \cdot 0.014m}

0.1546 = \frac{2.303 \cdot 1.01 \cdot \log ((\frac{62}{58.7}), 10)}{4 \cdot 3.1416 \cdot 0.014}

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Time in Hours given Time at 1st and 2nd Instance since Pumping Started Solution

Follow our step by step solution on how to calculate Time in Hours given Time at 1st and 2nd Instance since Pumping Started?

FIRST Step Consider the formula

t hour = \frac{2.303 \cdot Q \cdot \log ((\frac{t 2sec}{t1}), 10)}{4 \cdot π \cdot Δs}

Next Step Substitute values of Variables

∴

t hour = \frac{2.303 \cdot 1.01m³/s \cdot \log ((\frac{62s}{58.7s}), 10)}{4 \cdot π \cdot 0.014m}

Next Step Substitute values of Constants

∴

t hour = \frac{2.303 \cdot 1.01m³/s \cdot \log ((\frac{62s}{58.7s}), 10)}{4 \cdot 3.1416 \cdot 0.014m}

Next Step Prepare to Evaluate

∴

t hour = \frac{2.303 \cdot 1.01 \cdot \log ((\frac{62}{58.7}), 10)}{4 \cdot 3.1416 \cdot 0.014}

Next Step Evaluate

∴

t hour = 556.606390179623s

Next Step Convert to Output's Unit

∴

t hour = 0.154612886161006h

LAST Step Rounding Answer

∴

t hour = 0.1546h

Time in Hours given Time at 1st and 2nd Instance since Pumping Started Formula Elements

Variables

Constants

Functions

Time in Hours

Time in Hours can be defined as the ongoing and continuous sequence of events that occur in succession, from the past through the present to the future, measured in hours.

Symbol: t_hour

Measurement: TimeUnit: h

Note: Value should be greater than 0.

Formulas to find Time in Hours

Discharge

Discharge is the rate of flow of a fluid in unsteady flow.

Symbol: Q

Measurement: Volumetric Flow RateUnit: m³/s

Note: Value can be positive or negative.

Formulas to find Discharge

Time of Drawdown (t2) in Wells

Time of Drawdown (t2) in Wells is the total period for the difference between the pumping water level and the static (non-pumping) water level for wells.

Symbol: t_2sec

Measurement: TimeUnit: s

Note: Value should be greater than 0.

Formulas to find Time of Drawdown (t2) in Wells

Time of Drawdown (t1) in Wells

Time of Drawdown (t1) in Wells is the total period for the difference between the pumping water level and the static (non-pumping) water level for wells.

Symbol: t1

Measurement: TimeUnit: s

Note: Value should be greater than 0.

Formulas to find Time of Drawdown (t1) in Wells

Difference in Drawdowns

Difference in Drawdowns between the two wells.

Symbol: Δs

Measurement: LengthUnit: m

Note: Value should be greater than 0.

Formulas to find Difference in Drawdowns

Archimedes' constant

Archimedes' constant is a mathematical constant that represents the ratio of the circumference of a circle to its diameter.

Symbol: π

Value: 3.14159265358979323846264338327950288

log

Logarithmic function is an inverse function to exponentiation.

Syntax: log(Base, Number)

Other formulas in Time of Flow category

Go Time given Formation Constant S

t days = \frac{S c}{\frac{4 \cdot u \cdot T}{{(d radial)}^{2}}}

Go Time in Days given Radial Distance

t days = \frac{S c}{\frac{2.25 \cdot T}{{(d radial)}^{2}}}

Go Time at 1st Instance since Pumping Started given Discharge

t1 = \frac{t2}{10^{\frac{Δs}{\frac{2.303 \cdot Q}{4 \cdot π \cdot t seconds}}}}

Go Time at 2nd Instance since Pumping Started given Discharge

t2 = t1 \cdot 10^{\frac{Δs}{\frac{2.303 \cdot Q}{4 \cdot π \cdot t seconds}}}

How to Evaluate Time in Hours given Time at 1st and 2nd Instance since Pumping Started?

Time in Hours given Time at 1st and 2nd Instance since Pumping Started evaluator uses Time in Hours = (2.303*Discharge*log((Time of Drawdown (t2) in Wells/Time of Drawdown (t1) in Wells),10))/(4*pi*Difference in Drawdowns) to evaluate the Time in Hours, Time in Hours given Time at 1st and 2nd Instance since Pumping Started formula is defined as a measure of the time elapsed in hours since pumping started, considering the unsteady flow context, where the time at the first and second instances are known, enabling the calculation of the time in hours. Time in Hours is denoted by t_hour symbol.

How to evaluate Time in Hours given Time at 1st and 2nd Instance since Pumping Started using this online evaluator? To use this online evaluator for Time in Hours given Time at 1st and 2nd Instance since Pumping Started, enter Discharge (Q), Time of Drawdown (t2) in Wells (t_2sec), Time of Drawdown (t1) in Wells (t1) & Difference in Drawdowns (Δs) and hit the calculate button.

FAQs on Time in Hours given Time at 1st and 2nd Instance since Pumping Started

What is the formula to find Time in Hours given Time at 1st and 2nd Instance since Pumping Started?

The formula of Time in Hours given Time at 1st and 2nd Instance since Pumping Started is expressed as Time in Hours = (2.303*Discharge*log((Time of Drawdown (t2) in Wells/Time of Drawdown (t1) in Wells),10))/(4*pi*Difference in Drawdowns). Here is an example- 4.3E-5 = (2.303*1.01*log((62/58.7),10))/(4*pi*0.014).

How to calculate Time in Hours given Time at 1st and 2nd Instance since Pumping Started?

With Discharge (Q), Time of Drawdown (t2) in Wells (t_2sec), Time of Drawdown (t1) in Wells (t1) & Difference in Drawdowns (Δs) we can find Time in Hours given Time at 1st and 2nd Instance since Pumping Started using the formula - Time in Hours = (2.303*Discharge*log((Time of Drawdown (t2) in Wells/Time of Drawdown (t1) in Wells),10))/(4*pi*Difference in Drawdowns). This formula also uses Archimedes' constant and Logarithmic Inverse (log) function(s).

Can the Time in Hours given Time at 1st and 2nd Instance since Pumping Started be negative?

No, the Time in Hours given Time at 1st and 2nd Instance since Pumping Started, measured in Time cannot be negative.

Which unit is used to measure Time in Hours given Time at 1st and 2nd Instance since Pumping Started?

Time in Hours given Time at 1st and 2nd Instance since Pumping Started is usually measured using the Hour[h] for Time. Second[h], Millisecond[h], Microsecond[h] are the few other units in which Time in Hours given Time at 1st and 2nd Instance since Pumping Started can be measured.

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Time in Hours given Time at 1st and 2nd Instance since Pumping Started Formula

Time in Hours given Time at 1st and 2nd Instance since Pumping Started Example

Time in Hours given Time at 1st and 2nd Instance since Pumping Started Solution

Time in Hours given Time at 1st and 2nd Instance since Pumping Started Formula Elements

Other formulas in Time of Flow category

How to Evaluate Time in Hours given Time at 1st and 2nd Instance since Pumping Started?

FAQs on Time in Hours given Time at 1st and 2nd Instance since Pumping Started

Variable Name