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Chapter 6: Problem 2

In the USA the unit gpd/sqft (gallon per day per square foot) is sometimes used to measure the hydraulic conductivity \(k\), and the specific discharge q. What is the relation with the SI-unit \(\mathrm{m} / \mathrm{s}\) ?

Short Answer

Expert verified
Answer: 1 gpd/sqft ≈ 4.719 x 10^-8 m/s

Step by step solution

01

Identify unit conversions

To convert from gpd/sqft to m/s, we need to convert gallons to cubic meters and square feet to square meters. 1 gallon = \(0.00378541\) cubic meters 1 square foot = \(0.092903\) square meters
02

Convert gpd/sqft to cubic meters per day per square meter

Start by converting gallons per day to cubic meters per day (multiply by 0.00378541): 1 gpd = \(1 * 0.00378541 = 0.00378541\) cubic meters per day Next, convert square feet to square meters (multiply by 0.092903): 1 sqft = \(1 * 0.092903 = 0.092903\) square meters Now, divide cubic meters per day by square meters: \(\frac{0.00378541\, \mathrm{m}^3 /\mathrm{day}}{0.092903\, \mathrm{m}^2} = \frac{0.00378541}{0.092903}\, (\frac{\mathrm{m}^3}{\mathrm{m}^2 \cdot \mathrm{day}})\)
03

Convert cubic meters per day per square meter to m/s

Finally, convert from days to seconds: 1 day = 86400 seconds Divide the fraction obtained in step 2 by 86400 to get m/s: \(\frac{0.00378541}{0.092903}\, (\frac{\mathrm{m}^3}{\mathrm{m}^2 \cdot \mathrm{day}}) * \frac{1 \, \mathrm{day}}{86400 \, \mathrm{s}} = \frac{0.00378541}{0.092903 * 86400}\, (\frac{\mathrm{m}^3}{\mathrm{m}^2 \cdot \mathrm{s}})\) Simplify the fraction to obtain the conversion factor between gpd/sqft and m/s: \(\frac{0.00378541}{0.092903 * 86400} \approx 4.719 \times 10^{-8}\, (\frac{\mathrm{m}}{ \mathrm{s}})\)
04

Conclusion

The relation between gpd/sqft and m/s for both the hydraulic conductivity (k) and the specific discharge (q) is: 1 gpd/sqft \(\approx 4.719 \times 10^{-8}\, \frac{\mathrm{m}}{ \mathrm{s}}\)

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Most popular questions from this chapter

A certain soil has a hydraulic conductivity \(k=5 \mathrm{~m} / \mathrm{d}\). This value has been measured in summer. In winter the temperature is much lower, and if it supposed that the viscosity \(\mu\) then is a factor \(1.5\) as large as in summer, determine the value of the hydraulic conductivity in winter.

In geohydrology the unit \(\mathrm{m} / \mathrm{d}\) is often used to measure the hydraulic conductivity \(k\). What is the relation with the SI-unit \(\mathrm{m} / \mathrm{s}\) ?
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