Question

Effluent from drainage system with 8 dSm\(^{-1}\) EC is discharged in a canal water having 0.3 dS m\(^{-1}\) EC. If discharge of drainage effluent and canal water is 30 and 900 m\(^3\) h\(^{-1}\), respectively then what will be EC of downstream canal water?

• A. 4.15 dS m\(^{-1}\)
• B. 0.55 dS m\(^{-1}\)
• C. 0.3 dS m\(^{-1}\)
• D. 1.55 dS m\(^{-1}\)

Hint:

To quickly verify your answer:
Because the clean canal water flow (\(900 \text{ m}^3\text{h}^{-1}\)) is much larger than the salty drainage flow (\(30 \text{ m}^3\text{h}^{-1}\)), the final salinity must be much closer to the clean water level (\(0.3 \text{ dS/m}\)) than the salty effluent level (\(8 \text{ dS/m}\)).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
When two streams with different salt concentrations merge, we can calculate the downstream salinity using a simple conservation of mass (mass balance) equation.
This assumes the salt dissolves and mixes completely without undergoing chemical precipitation.
Key Formula or Approach:
The mass balance equation for the mixture is: \[ Q_{\text{down}} \times \text{EC}_{\text{down}} = (Q_e \times \text{EC}_e) + (Q_c \times \text{EC}_c) \] where:
- \(Q_e\) and \(\text{EC}_e\) are the discharge and electrical conductivity of the drainage effluent.
- \(Q_c\) and \(\text{EC}_c\) are the discharge and electrical conductivity of the canal water.
- \(Q_{\text{down}}\) and \(\text{EC}_{\text{down}}\) are the combined downstream values, where \(Q_{\text{down}} = Q_e + Q_c\).

Step 2: Detailed Explanation:
Let's list the given values:
- Drainage Effluent: \(Q_e = 30 \text{ m}^3\text{h}^{-1}\), \(\text{EC}_e = 8 \text{ dS/m}\)
- Canal Water: \(Q_c = 900 \text{ m}^3\text{h}^{-1}\), \(\text{EC}_c = 0.3 \text{ dS/m}\)
First, calculate the total downstream discharge: \[ Q_{\text{down}} = 30 + 900 = 930 \text{ m}^3\text{h}^{-1} \] Next, set up the mass balance equation to solve for \(\text{EC}_{\text{down}}\): \[ 930 \times \text{EC}_{\text{down}} = (30 \times 8) + (900 \times 0.3) \] \[ 930 \times \text{EC}_{\text{down}} = 240 + 270 \] \[ 930 \times \text{EC}_{\text{down}} = 510 \] \[ \text{EC}_{\text{down}} = \frac{510}{930} \approx 0.5484 \text{ dS/m} \] Rounding to two decimal places gives \(0.55 \text{ dS m}^{-1}\).

Step 3: Final Answer:
The electrical conductivity of the downstream canal water is 0.55 dS m\(^{-1}\).