Specific Capacity under Unsteady Drawdown Conditions Formula

GoSpecific Capacity and Discharge into Well Relationship Formula

GoSpecific capacity per unit well area of aquifer Formula

Fx Copy

LaTeX Copy

The Specific Capacity is referred as amount of water furnished under a standard unit head. Check FAQs

K s = \frac{1}{(\frac{1}{4} \cdot π \cdot T) \cdot (\ln (2.25 \cdot T \cdot \frac{t}{{R w}^{2}} \cdot S)) + C 2 \cdot Q f}

K_s - Specific Capacity?T - Transmissivity?t - Time from start of pump?R_w - Radius of the Pumping Well?S - Storage Coefficient?C₂ - Well Constant C2?Q_f - Flow Discharge?π - Archimedes' constant?

Specific Capacity under Unsteady Drawdown Conditions Example

With values

With units

Only example

Here is how the Specific Capacity under Unsteady Drawdown Conditions equation looks like with Values.

Here is how the Specific Capacity under Unsteady Drawdown Conditions equation looks like with Units.

Here is how the Specific Capacity under Unsteady Drawdown Conditions equation looks like.

0.0145 = \frac{1}{(\frac{1}{4} \cdot 3.1416 \cdot 11) \cdot (\ln (2.25 \cdot 11 \cdot \frac{50}{6^{2}} \cdot 1.2)) + 0.05 \cdot 30}

0.0145 = \frac{1}{(\frac{1}{4} \cdot 3.1416 \cdot 11m²/s) \cdot (\ln (2.25 \cdot 11m²/s \cdot \frac{50min}{{6m}^{2}} \cdot 1.2)) + 0.05 \cdot 30m³/s}

0.0145 = \frac{1}{(\frac{1}{4} \cdot 3.1416 \cdot 11) \cdot (\ln (2.25 \cdot 11 \cdot \frac{50}{6^{2}} \cdot 1.2)) + 0.05 \cdot 30}

Tip: Use pencil () icon above to edit Input Values and Units.

Calculate

Recalculate

Solution

Copy

Reset

You are here -

Home » Engineering » Civil » Engineering Hydrology » Specific Capacity under Unsteady Drawdown Conditions

Specific Capacity under Unsteady Drawdown Conditions Solution

Follow our step by step solution on how to calculate Specific Capacity under Unsteady Drawdown Conditions?

FIRST Step Consider the formula

K s = \frac{1}{(\frac{1}{4} \cdot π \cdot T) \cdot (\ln (2.25 \cdot T \cdot \frac{t}{{R w}^{2}} \cdot S)) + C 2 \cdot Q f}

Next Step Substitute values of Variables

∴

K s = \frac{1}{(\frac{1}{4} \cdot π \cdot 11m²/s) \cdot (\ln (2.25 \cdot 11m²/s \cdot \frac{50min}{{6m}^{2}} \cdot 1.2)) + 0.05 \cdot 30m³/s}

Next Step Substitute values of Constants

∴

K s = \frac{1}{(\frac{1}{4} \cdot 3.1416 \cdot 11m²/s) \cdot (\ln (2.25 \cdot 11m²/s \cdot \frac{50min}{{6m}^{2}} \cdot 1.2)) + 0.05 \cdot 30m³/s}

Next Step Convert Units

∴

K s = \frac{1}{(\frac{1}{4} \cdot 3.1416 \cdot 11m²/s) \cdot (\ln (2.25 \cdot 11m²/s \cdot \frac{3000s}{{6m}^{2}} \cdot 1.2)) + 0.05 \cdot 30m³/s}

Next Step Prepare to Evaluate

∴

K s = \frac{1}{(\frac{1}{4} \cdot 3.1416 \cdot 11) \cdot (\ln (2.25 \cdot 11 \cdot \frac{3000}{6^{2}} \cdot 1.2)) + 0.05 \cdot 30}

Next Step Evaluate

∴

K s = 0.0144910574614259

LAST Step Rounding Answer

∴

K s = 0.0145

Specific Capacity under Unsteady Drawdown Conditions Formula Elements

Variables

Constants

Functions

Specific Capacity

The Specific Capacity is referred as amount of water furnished under a standard unit head.

Symbol: K_s

Measurement: NAUnit: Unitless