Question

Length of soil sample for constant head permeability test is 30 cm, Area of specimen is 177cm2 constant head difference is 50 cm, total volume of water collected in 5 min is 350cm Calculate the hydraulic conductivity of the soil. ) 1.09x10 cm/sec )3.95x10cm/sec A) 2.85 x10'cm/sec )1.09x10 cm/sec

Please write correct option after completion of your answer

Answer 1

1)

Length of soil sample (L) = 30 cm

Cross sectional area of soil specimen (A) = 177 cm³

Constant head difference (h) = 50 cm

Volume of water collected (q) = 350 cm³

Duration of water collection (t) = 5 minutes.

Hydraulic conductivity (K) = ?

Flow rate of water (Q) = q / t = 350 / 5 = 70 cm³/min

We know that,

Q = K * * A for constant head test .................................................(1)

Where,

= Hydraulic gradient = h / L = 50 / 30 = 5 /3

Therefore from equation (1),

Q = K * (h/L) * A

$\Rightarrow$ 70 = K * (5/3) * 177

$\Rightarrow$ K = 0.237 cm/min

$\Rightarrow$ K = 0.237 / 60 cm/s [$\because$ 1 min = 60 sec]

$\therefore$K = 3.95 * 10^-3 cm/s

Hence option (C) is correct answer.

2)

Given,

Q_r = 8.7 m³/s

C_r = 6 mg/L

Q_e = 1.10 m³/s

C_e = 50 mg/L

K_d = 0.20 /day

Ultimate BOD (Y_u) = ?

Let the BOD concentration of river after effluent discharge = C_s and

New discharge of river after effluent mixture = Q_s = Q_r + Q_e = 8.7 + 1.1 = 9.8 m³/s

Therefore,

Q_s * C_s = Q_r * C_r + Q_e * C_e

$\Rightarrow$ 9.8 * C_s = 8.7 * 6 + 1.1 * 50

$\Rightarrow$ C_s = 10.939 mg/L

Therefore ultimate BOD of the river after effluent discharge is 10.939 mg/L

Hence option (A) is correct answer

3)

Given,

Error in measuring head = 1%

^i.e.,_{$\partial H/H = 0.01$}

Where, H = Head of water.

Let the coefficient of discharge of triangular weir (C_d) = 0.6

Discharge (Q) for triangular weir can be found using the following equation:

Q = (8/15) * C_d * H^(5/2) * (2 * g)^(1/2) * tan($\theta$/2)

Let the given weir is a 90^$\degree$ weir.

$\therefore$$\theta$/2 = 90 / 2 = 45^$\degree$

Therefore from the above equation,

Q = (8/15) * 0.60* H^(5/2) * (2 * 9.81)^(1/2) * tan(45)

$\Rightarrow$ Q = 1.4174 H^(5/2) .........................................................................................(1)

Differentiating the above equation with respect to head of water,

$\partial$Q / $\partial$H = 1.4174 * (5/2) * H^(3/2)

$\Rightarrow$$\partial$Q / $\partial$H = 3.5435 *  H^(3/2) .............................................................................(2)

Dividing equation (2) by equation (1),

{$\partial$Q / $\partial$H} / Q = {3.5435 *  H^(3/2)} / {1.4174 H^(5/2) }

$\Rightarrow$$\partial$Q / Q = (2.5 / H) * $\partial$H

$\Rightarrow$$\partial$Q / Q = 2.5 * ($\partial$H / H)

Given that, $\partial$H / H = 1%

Therefore,

$\partial$Q / Q = 2.5 * 1%

$\therefore$$\partial$Q / Q = 2.5 %

Therefore, error in measuring the discharge is 2.5 %

Hence option (D) is the correct answer.

4)

Given,

Rectangular channel

Bottom width (B) = 12 ft

Water depth (h) = 3 ft

Hydraulic depth (D) = ?

Hydraulic depth of a channel is the depth which is obtained by dividing the cross sectional area by the top water surface width.

It is nothing but the depth of an equivalent rectangular channel for the same cross sectional area and top surface width.

Mathematically,

D = A / T .........................................................................................(1)

Where,

A = Cross sectional area of the channel

T = Top surface width

As the given channel is a rectangular channel,

A = B * h = 12 * 3 = 36 ft²

For rectangular channel top and bottom width remains same.

$\therefore$ T = 12 ft

Therefore, from equation (1)

D = 36 / 12 = 3 ft

Therefore hydraulic depth for the given rectangular channel is 3 ft

Hence no options are matching with the answer.

[Note: For rectangular channel hydraulic depth is equal to the normal water depth. Hence no need to solve the problem in case of rectangular channel to find the hydraulic depth, directly answer can be put as the depth of water ]