Answer:
the diameter of the stand pipe is 24.456mm
Step-by-step explanation:
Given that:
Height of the soil sample = 90 mm
Cross section area of the soil sample = 6000 m²
The head fell from 500 mm to 300 mm
i,e
![H_1 = 500 \ mm](https://img.qammunity.org/2021/formulas/engineering/college/aolhnbznv59qmf3td1qjad1gn60q597rko.png)
![H_2 = 300 \ mm](https://img.qammunity.org/2021/formulas/engineering/college/vjb60gfjbchel7opp1bk2d5osw2p1mwxf5.png)
Time = 1500 s
The permeability of the soil (K) was 2.4x10⁻³ mm/s.
The objective is to determine the diameter of the stand pipe. (in mm)
Using the formula :
![K = ((2.303 )*a*L)/(A*t) log((H_1)/(H_2))](https://img.qammunity.org/2021/formulas/engineering/college/qnxmmjdncftuvrezfghjxqs55aqg18ybgx.png)
![2.4*10^(-3)= ((2.303 )*a*(90))/((6000 )*(1500 )) log((500)/(300 ))](https://img.qammunity.org/2021/formulas/engineering/college/11felfphi7ltynkv8qu7oy1hnc1jlx7zrt.png)
![2.4*10^(-3)= ((207.27 *a))/(9000000) *0.22185](https://img.qammunity.org/2021/formulas/engineering/college/r65it43dyswvrm75iqq31aq8i9peuwdokm.png)
![2.4*10^(-3)*9000000= (207.27 *a)}*0.22185](https://img.qammunity.org/2021/formulas/engineering/college/uxm0fmhv24cif8kt1c69evju2pmr8imelu.png)
21600 = 45.983 a
a = 21600/45.983
a = 469.74 mm
Recall thar:
a = π/4 (d²)
So;
469.74 = 0.7854 ((d²)
d² = 469.74/0.7854
d² = 598.09
d = √598.09
d = 24.456 mm
Hence ; the diameter of the stand pipe is 24.456mm