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Question

The kinematic viscosity of glycerin and kerosene are 1.2 times and 0.95 times of that of water, respectively. Glycerin and kerosene flow through two identical porous media having same hydraulic gradient. Assuming Darcy's law is valid for the porous media, the ratio of flow rate of kerosene to that of glycerin is

The correct answer is
1.263

Understanding the Problem

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.

Applying 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:

  • \( Q \) is the volumetric flow rate
  • \( k \) is the permeability of the porous medium
  • \( A \) is the cross-sectional area
  • \( \mu \) is the dynamic viscosity of the fluid
  • \( \frac{\Delta h}{L} \) is the hydraulic gradient (\( i \))

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.

Calculating the Flow Rate Ratio

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:

  • \( Q_G = (kA i) \frac{1}{\nu_G} \)
  • \( Q_K = (kA i) \frac{1}{\nu_K} \)

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} $

Substituting Given Values

We are given:

  • \( \nu_G = 1.2 \times \nu_{water} \)
  • \( \nu_K = 0.95 \times \nu_{water} \)

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.

Final Answer Determination

The calculated ratio \( \frac{Q_K}{Q_G} \) is approximately 1.263, which matches Option D.

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Important Questions from Darcy Law

  1. A homogeneous isotropic confined aquifer of uniform thickness $30 \text{ m}$ has hydraulic conductivity of $5 \text{ m/day}$ and porosity of $0.3$. There are two observation wells X and Y along a radial line from a fully penetrating pumping well at $100 \text{ m}$ and $200 \text{ m}$ distance, respectively. The well is pumped at a uniform rate to produce steady drawdown of $5 \text{ m}$ at X and $3 \text{ m}$ at Y. If a non-reactive pollutant enters at the observation well Y, then the time taken by the pollutant (under advection) to reach the observation well X is ________ days. (rounded off to two decimal places)
  2. 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 ________

  3. 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]

  4. In a field test of a geological formation of permeable soil (porosity = 20%), the hydraulic gradient was found to be 2%. The actual seepage velocity of the flow was found to be 0.0025 m/s. Assume that the flow is in the laminar regime. The permeability (K) of the aquifer is ________ m/s (rounded off to three decimal places).
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