1. Dalam alat ukur tegangan berubah terdapat contoh tanah dengan tuhkan 18 cm serta luas penampang 22cm^2. Hitunglah waktu yang dibutuhkan dari kondisi head 25 cm menjadi 10 cm , bila luas penampang pipa 2cm^2 Contoh tanah adalah heterogen yaitu: lapisan 1, koefesien permeabilitasnya 3times 10^-4cm/detik, tebal 6 cm; lapisan 2. koefesien permeabilitasnya 4times 10^-4 cm/detik, tebal 6 cm; lapisan 3. koefesien permeabilitasnya 6times 10^-4 cm/detik, tebal X cm . Dianggap aliran yang terjadi tegak lurus bidang gambar.

This problem involves calculating the time required for the water level in a permeameter to drop from 25 cm to 10 cm, given the soil properties and dimensions. We'll need to use Darcy's Law to solve this. However, the problem statement is incomplete; the thickness (X) of the third layer is missing. We'll solve it assuming a value for X, and then show how to adapt the solution if X is provided.

Assumptions:

* Steady-state flow: We assume the flow rate is constant during the time interval. This is a simplification, as the flow rate will decrease as the head decreases.
* Homogenous layers: While the problem states the soil is heterogeneous, we'll treat each layer as homogenous with the given permeability.
* Value of X: We'll assume a thickness (X) of 6 cm for the third layer for the sake of calculation. If a different value is provided, simply substitute that value into the final equation.


1. Calculate the equivalent permeability (k_eq) for the three layers:

For layers in series, the equivalent permeability is calculated as:

1/k_eq = (L1/k1) + (L2/k2) + (L3/k3)

Where:

* L1, L2, L3 are the thicknesses of layers 1, 2, and 3 respectively.
* k1, k2, k3 are the permeabilities of layers 1, 2, and 3 respectively.

Substituting the given values (assuming X = 6 cm):

1/k_eq = (6 cm / (3 x 10⁻⁴ cm/s)) + (6 cm / (4 x 10⁻⁴ cm/s)) + (6 cm / (6 x 10⁻⁴ cm/s))
1/k_eq = 20000 s/cm + 15000 s/cm + 10000 s/cm = 45000 s/cm
k_eq = 2.22 x 10⁻⁵ cm/s


2. Apply Darcy's Law:

Darcy's Law states: Q = k_eq * A * (h1 - h2) / L

Where:

* Q is the flow rate (volume/time)
* A is the cross-sectional area of the pipe (2 cm²)
* h1 is the initial head (25 cm)
* h2 is the final head (10 cm)
* L is the total thickness of the soil layers (6 cm + 6 cm + 6 cm = 18 cm)

Substituting the values:

Q = (2.22 x 10⁻⁵ cm/s) * (2 cm²) * (25 cm - 10 cm) / (18 cm)
Q ≈ 3.7 x 10⁻⁵ cm³/s


3. Calculate the volume of water drained:

The volume of water drained from the standpipe is given by:

V = A_standpipe * (h1 - h2)

Where A_standpipe is the cross-sectional area of the standpipe (22 cm²)

V = 22 cm² * (25 cm - 10 cm) = 330 cm³


4. Calculate the time required:

Time (t) = Volume (V) / Flow rate (Q)

t = 330 cm³ / (3.7 x 10⁻⁵ cm³/s)
t ≈ 8.92 x 10⁶ s

Converting seconds to hours:

t ≈ 8.92 x 10⁶ s * (1 min / 60 s) * (1 hr / 60 min) ≈ 2478 hours


Therefore, assuming a thickness of 6 cm for the third layer, the time required is approximately 2478 hours. If a different thickness (X) is given for the third layer, recalculate k_eq using the formula in step 1 and repeat steps 2-4 with the new value of k_eq. Remember that this calculation assumes steady-state flow, which is an approximation.