The question asks for the ratio of the flow rate of kerosene to glycerin flowing through identical porous media under the same hydraulic gradient. We are given the relationship between the kinematic viscosities of glycerin, kerosene, and water, and we need to apply Darcy's law.
Darcy's law relates the flow rate (Q) to the properties of the fluid and the porous medium. A common form of Darcy's law is:
$ Q = \frac{kA}{\mu} \frac{\Delta h}{L} $
Where:
We know that kinematic viscosity (\( \nu \)) is defined as dynamic viscosity (\( \mu \)) divided by density (\( \rho \)): \( \nu = \frac{\mu}{\rho} \). Therefore, \( \mu = \rho \nu \).
Substituting this into Darcy's law:
$ Q = \frac{kA}{\rho \nu} i $
Rearranging for the terms that are constant:
$ Q = (kA i) \frac{1}{\nu} $
Since the porous media are identical (\( k \) is the same) and the hydraulic gradient (\( i \)) is the same, the term \( (kA i) \) is constant for both fluids.
Let \( Q_K \) be the flow rate of kerosene and \( Q_G \) be the flow rate of glycerin.
Let \( \nu_K \) be the kinematic viscosity of kerosene and \( \nu_G \) be the kinematic viscosity of glycerin.
From Darcy's law, we have:
The ratio of the flow rate of kerosene to glycerin is:
$ \frac{Q_K}{Q_G} = \frac{(kA i) / \nu_K}{(kA i) / \nu_G} $
The constant term \( (kA i) \) cancels out:
$ \frac{Q_K}{Q_G} = \frac{\nu_G}{\nu_K} $
We are given:
Substitute these into the ratio:
$ \frac{Q_K}{Q_G} = \frac{1.2 \times \nu_{water}}{0.95 \times \nu_{water}} $
The \( \nu_{water} \) term cancels out:
$ \frac{Q_K}{Q_G} = \frac{1.2}{0.95} $
Calculating the value:
$ \frac{1.2}{0.95} \approx 1.263157... $
Rounding to three decimal places, the ratio is approximately 1.263.
The calculated ratio \( \frac{Q_K}{Q_G} \) is approximately 1.263, which matches Option D.
Two reservoirs having a difference of 20 m in their water surface elevations are connected through a confined aquifer (thickness = 5 m, length = 2 km, hydraulic conductivity $K = 3 \times 10^{-3}$ m/s, and porosity $\eta = 0.3$) as shown in the following figure (drawn not to scale). 
If the Reservoir 1 is contaminated by a pollutant then the time (in days, rounded off to one decimal place) taken by the pollutant to reach to the Reservoir 2 under advection will be ________
Two fully penetrating wells are dug 1.4 km apart in a homogenous confined aquifer. The difference in their piezometric levels is 4.0 m. The groundwater flow is steady and unidirectional. If the aquifer has a hydraulic conductivity of $3.5\ m\cdot day^{-1}$ and effective porosity of 40%, the time taken for water to move from one well to the other in days is _____________. [in integer]