Question

In a single effect evaporator, how much steam (in kg) is required for evaporation of 1 kg of of water from a solution?

• A. 2 to 2.5
• B. 1 to 1.3
• C. 2 to 3
• D. 3 to 4

Hint:

To increase steam economy and reduce operating costs, multiple-effect evaporators are used.
In a triple-effect evaporator, \(1\text{ kg}\) of steam can evaporate approximately \(2.2\text{ to }2.5\text{ kg}\) of water, significantly increasing efficiency.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
An evaporator is utilized to concentrate liquid foods by boiling off water.
The performance of an evaporator is measured by its steam economy, defined as the mass of water evaporated per unit mass of steam supplied:
\[ \text{Economy} = \frac{\text{Mass of water evaporated}}{\text{Mass of steam consumed}} \]

Step 2: Key Formula or Approach:
For a single-effect evaporator, some heat is lost to the surroundings and utilized in preheating the feed.
Consequently, the steam economy is always slightly less than \(1.0\) (typically in the range of \(0.8\text{ to }0.9\)).
Therefore:
\[ \text{Steam required} = \frac{\text{Mass of water evaporated}}{\text{Economy}} \]

Step 3: Detailed Explanation:
To evaporate \(1\text{ kg}\) of water from a solution:
\[ \text{Steam required} = \frac{1\text{ kg}}{0.8\text{ to }0.9} \approx 1.1\text{ to }1.3\text{ kg} \]
Thus, we require approximately \(1.0\text{ to }1.3\text{ kg}\) of live steam.

Step 4: Final Answer:
The steam required is in the range of 1 to 1.3 kg, which corresponds to option (B).