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Explanation:

Give data:

(S_c) Ultimate primary consolidation settlement = 50mm

(S_i) Consolidation settlement after 300 day = 10mm

(U) degree of consolidation at 10mm settlement

\(\rm U_{1}=\frac{S_{\mathrm{i}} \times 100}{S_{c}}=\frac{10}{50} \times 100=20 %\)

(t₁) time to achieve 10mm settlement = 300 days

Given, (U₂) degree of consolidation = 50%

The time factor is given by

\(\rm \therefore \quad T_{v}=\frac{C_{v} t}{d^{2}}\)

T_v ∝ t

where, (T_v) time factor and (t) time needed to achieve a particular degree of consolidation

\(\rm \frac{T_{v 1}}{T_{v 2}}=\frac{t_{1}}{t_{2}}\)

\(\rm \therefore \quad T_{v}=\frac{\pi}{4}\left(U^{2}\right)\) for U ≤ 60% 

So, 

\(\rm \frac{\left(U_{1}\right)^{2}}{\left(U_{2}\right)^{2}}=\frac{t_{1}}{t_{2}}\)

\(\rm \frac{(20)^{2}}{(50)^{2}}=\frac{300}{t_{2}}\)

(t₂) time needed to achieve 50% degree of consolidation = 1875 days

Hence, additional time needed = 1875 – 300 = 1575 days

Solution:

To calculate the eigenvalues of the given matrix:

Given Data:

\( A = \begin{bmatrix} 2 & 1 & 1 \ 1 & 4 & 1 \ 1 & 1 & 2 \end{bmatrix} \)

Step 1: Write the characteristic equation

\( \text{det}(A - \lambda I) = 0 \)

Subtract \( \lambda \) along the diagonal elements:

\( \text{det} \begin{bmatrix} 2-\lambda & 1 & 1 \ 1 & 4-\lambda & 1 \ 1 & 1 & 2-\lambda \end{bmatrix} = 0 \)

Step 2: Expand the determinant

Expand using the first row:

\( (2-\lambda) \cdot \text{det} \begin{bmatrix} 4-\lambda & 1 \ 1 & 2-\lambda \end{bmatrix} - 1 \cdot \text{det} \begin{bmatrix} 1 & 1 \ 1 & 2-\lambda \end{bmatrix} + 1 \cdot \text{det} \begin{bmatrix} 1 & 4-\lambda \ 1 & 1 \end{bmatrix} = 0 \)

Simplify the determinant step-by-step to get the cubic equation:

\( -\lambda^3 + 9\lambda^2 - 26\lambda + 30 = 0 \)

Step 3: Solve the cubic equation

Using numerical methods or a solver, the roots of the equation are:

\( \lambda_1 = 5, , \lambda_2 = 1, , \lambda_3 = 2 \)

The eigenvalues of the matrix are 5, 1, and 2.

Explanation:

Now,

N_o = Number of pollutant present in waste water intially i.e. at t = 0

N_t = Number of pollutant present in waste water at time, ‘t’

Treatment of activated carbon to remove pollutant in mixed batch reactor follows 1^st order kinetics.

i.e.  N_t = N_oe^-kt …(i)

As, removal efficiency at any time ‘t’ = 95%

i.e. \(\rm \frac{N_{o}-N_{t}}{N_{o}} \times 100=95 \)

So, \(\rm \frac{N_{t}}{N_{o}}=0.05\)

On putting in eq.

(i) So, 0.05 = e^–0.38t

On solving,

t = 7.88 days ≈ 7.9 days

Explanation:

Natural alkalinity = 4 mg/l as CaCO₃

Alkalinity req = \(\rm \left(\frac{450 \times 10^{3} \mathrm{~g} / \text { day }}{25 \times 10^{6} l / \mathrm{d} \times \text { eq wt of } \mathrm{CO}\left(\mathrm{HCO}_{3}\right)_{2}} \times \text { eq wt of } \mathrm{CaCO}_{3}\right)\) \(\times 10^{3} \frac{\mathrm{mg}}{\mathrm{l}} \mathrm{as\ CaCO}_{3}\)

\(=\frac{450 \times 10^{3} \times 10^{3}}{25 \times 10^{6} \times 81} \times 50\)

\(=\frac{450 \times 50}{25 \times 81}=11.11 \mathrm{mg} / \mathrm{l\ as\ CaCO}_{3}\)

Alkalinity needed to be provided through outside source is

= (11.11 – 4) = 7.11 mg/l as CaCO₃ 

= \(\frac{7.11}{50}\) milieq/l

CaO required is

= \(\frac{7.11}{50}\) milieq/l

\(=\left[\frac{7.11}{50} \times \text { eq. wt\ of } \mathrm{CaO}\right] \frac{\mathrm{mg}}{\mathrm{l}} \)

\(\rm =\left[\frac{7.11}{50} \times 28\right] \frac{\mathrm{mg}}{\mathrm{l}} \times 25 \times 10^{6} l / \mathrm{d}\)

= 99.54 kg/d

Explanation:

Vehicle Damage Factor:

\(\text { VDF }=\mathrm{EWLF} =\left(\frac{\text { Axle load }}{\text { Standard axle load }}\right)^{4} \)

\(=\left(\frac{12000}{8160}\right)^{4}=4.677 \)

A = P(1 + r)^x

= 800(1.04)⁵ = 973.322 

The cumulative standard axle (in msa) for the pavement design is

\(\rm C S A =\frac{365 A\left((1+r)^{n}-1\right)}{r} L D F \times V D F \) 

where assuming LDF to be 1

CSA \(=\frac{365 \times 973.322\left[1.04^{15}-1\right]}{0.04} \times 1 \times 4.677\)

CSA = 33.27 msa

Explanation:

When the cell pressure was increased to 30 kPa

Δσ₃ = (30 – 20) = 10 kPa

Δu₁ = 1 kPa

The value of Skempton’s pore pressure parameter B for the soil sample is

\(\rm B=\frac{\Delta \mathrm{u}_{1}}{\Delta \sigma_{3}}=\frac{1}{10}=0.1\)

B = 0.1

Explanation:

Considering PQR

P_R + 4 × (0.8γw) = 10γw

⇒ PR = (10 3.2)γw = 6.8γw

Now, 

Considering force prism QRST 

\(\rm F_{1}=\frac{(6.8+10)}{2} \gamma_{w} \times 4 \times 5=168 \gamma_{w}=1680 \mathrm{kN}\)

Considering force prism PQTU 

\(\rm F_{2}=\frac{(10+16)}{2} \gamma_{w} \times 6 \times 5=3900 \mathrm{kN}\)

So, F = F₁ + F₂ = 1680 + 3900 = 5580 kN

Explanation:

Seepage is occuring upto the ground surface 

Given, (γ_sat) saturated unit weight = 20 kN/m³

(γW) unit weight of water = 10 kN/m³

(β) slope angle = 30° 

(ϕ) angle of shearing resistance of soil = ?

The water table is at the ground surface & seepage is occurring so, 

(FOS) factor of safety = \(\rm \frac{\gamma_{\text {sub }}}{\gamma_{\text {sat }}} \frac{\tan \phi}{\tan \beta}\)

For critical conditions, 

\(\rm \text { FOS }=1=\left(\frac{\gamma_{\mathrm{sat}}-\gamma_{\mathrm{w}}}{\gamma_{\mathrm{sat}}}\right) \times \frac{\tan \phi}{\tan \beta}\)

\(1=\left(\frac{20-10}{20}\right) \times \frac{\tan \phi}{\tan 30^{\circ}}\)

The angle of shearing resistance (ϕ) = 49.106° = 49°

Explanation:

We know that 

\(\rm \frac{d x}{d t}+P x=Q\)

\(\rm \Rightarrow \quad \frac{\mathrm{dm}}{\mathrm{dt}}+\mathrm{Pm} =Q \)

\(\rm P =\frac{1}{6000+t}\)

Q = 0.01 

Integration factor

(I.F.) = \(\rm e^{\int P\ d t}=e^{\int \frac{1}{6000+t} d t}\)

= e^ln(6000+t) = 6000 + t

⇒ m.(I.F.) = ∫Q.(I.F.)dt + C

m(6000 + t) = ∫0.01(6000 + t)dt + C

\(\rm m(6000+t)=0.01\left[6000 t+\frac{t^{2}}{2}\right]+C\)

at t = 0, m = 0

[Since there is Fresh water in tank initially]

⇒ C = 0

⇒ \(\mathrm{m}=\frac{0.01\left(6000 \mathrm{t}+\frac{\mathrm{t}^{2}}{2}\right)}{(6000+\mathrm{t})}\)

From volume balance

Accumulation = Inflow – Ourtflow

2 × 2 ×(3 × 0.25)m3 = \(\rm (2-1) \frac{l}{5} \times \mathrm{t}_{\mathrm{sec}} \times 10^{-3} \frac{\mathrm{~m}^{3}}{l}\)

⇒ t = 3000 sec. 

Hence, at 3000 sec. i.e. when tank is 75% full.

\(\mathrm{m}=\frac{0.01\left(6000 \times 3000+\frac{(3000)^{2}}{2}\right)}{6000+3000}\)

m = 0.01[2000 + 500]

m = 25 gram

Explanation:

As per given data:

T = 200 yrs

\(\lambda=\frac{1}{200}\)

Probability of non occurrence in 50 years

= e^-λt  = 0.7788

Probability of occurrence atleast once

= 1 – 0.7788 = 0.2212 ≃ 22.12%