SSC JE Civil Engineering 30-Oct-2020 Shift 2 Paper

Concept:
Depth of water Held in Root zone: For ease in calculation water present in voids of soil needs to be expressed as the depth of water.
Let root zone depth = ‘D’ in m
Specific wt. of soil (dry) = ${\mathrm{\gamma }}_{\mathrm{d}}$
The cross-sectional area of soil considered = A
Equivalent depth of water stored in voids of soil = ‘d’.
We know, ∴ ${\mathrm{F}}_{\mathrm{c}}=\frac{\mathrm{Ad}{\mathrm{\gamma }}_{\mathrm{w}}}{\mathrm{A}\mathrm{D}{\mathrm{\gamma }}_{\mathrm{d}}}$
∴ ${\mathrm{F}}_{\mathrm{c}}=\frac{\mathrm{d}}{\mathrm{D}}\frac{{\mathrm{\gamma }}_{\mathrm{w}}}{{\mathrm{\gamma }}_{\mathrm{d}}}$
⇒ d = $\frac{{\mathrm{\gamma }}_{\mathrm{d}}}{{\mathrm{\gamma }}_{\mathrm{w}}}\times \mathrm{D}\times {\mathrm{F}}_{\mathrm{c}}$
It is the depth of water stored in the root zone for full field capacity.
But this entire depth of water cannot be extracted by plants. Hence available moisture will be
${\mathrm{d}}_{\mathrm{available}}$ = $\frac{{\mathrm{\gamma }}_{\mathrm{d}}}{{\mathrm{\gamma }}_{\mathrm{w}}}\times \mathrm{D}\times \left[{\mathrm{F}}_{\mathrm{c}}-\mathrm{\varphi }\right]$
Where ϕ = permanent wilting point.
Depth of Readily available moisture (d’):
$\mathrm{d}=\frac{{\mathrm{\gamma }}_{\mathrm{d}}}{{\mathrm{\gamma }}_{\mathrm{w}}}\times \mathrm{D}\times \left({\mathrm{F}}_{\mathrm{c}}-{\mathrm{M}}_{\mathrm{0}}\right)$
Frequency of Irrigation:
The frequency of irrigation is decided based on the maximum allowable deficit and rate of consumptive use.
${\mathrm{f}}_{\mathrm{w}}=\frac{\mathrm{d}}{{\mathrm{C}}_{\mathrm{u}}∕\mathrm{day}}$
Where, ${\mathrm{C}}_{\mathrm{u}}$ = Consumptive use/day
Calculation:
${\mathrm{F}}_{\mathrm{c}}$ = 29 % = 0..29, ϕ = 10% = 0.10, D = 0.8 m,
Relative density of soil $\left(\frac{{\mathrm{\rho }}_{\mathrm{d}}}{{\mathrm{\rho }}_{\mathrm{w}}}\right)=1.302$
Frequency of irrigation = 10 days, Optimum moisture $\left({\mathrm{M}}_0\right)=17\%$
Depth of Readily available moisture (d′) = $\frac{{\mathrm{\rho }}_{\mathrm{d}}}{{\mathrm{\rho }}_{\mathrm{w}}}\times \mathrm{D}\times \left[{\mathrm{F}}_{\mathrm{c}}-{\mathrm{M}}_0\right]$
$\therefore \mathit{d}\prime =1.302\times 0.8\times \left[0.29-0.17\right]$
∴ d' = 0.12499 m = 124.99 mm
∴ Frequency of Irrigation = $\frac{\mathrm{d}\prime }{{\mathrm{C}}_{\mathrm{u}}}$ = $\frac{124.99}{{\mathrm{C}}_{\mathrm{u}}}$ = 10 days
∴ Cu = 12.49 mm = 1.25 cm.