Question

An evaporator concentrating liquid food is producing 100 g of steam every minute. If the steam consumption of evaporator is 10 kg/h, what will be the steam economy of the evaporator?

• A. 0.1
• B. 1.0
• C. 1.6
• D. 0.6

Hint:

Always ensure both quantities are in the exact same unit (kg/h) before calculating the ratio.
An economy of less than 1.0 is typical for single-effect evaporators due to heat losses.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Steam economy is a performance metric of an evaporator.
It represents the ratio of the mass of water evaporated from the feed to the mass of external steam supplied to the system.

Step 2: Key Formula or Approach:
The steam economy is calculated as: \[ \text{Steam Economy} = \frac{\text{Rate of water evaporated (kg/h)}}{\text{Rate of steam consumed (kg/h)}} \]

Step 3: Detailed Explanation:
First, convert the rates to consistent units (kg/h):
The rate of water evaporated (steam produced): \[ m_v = 100 \text{ g/min} \] \[ m_v = 100 \text{ g/min} \times \left(\frac{1 \text{ kg}}{1000 \text{ g}}\right) \times \left(\frac{60 \text{ min}}{1 \text{ h}}\right) = 6 \text{ kg/h} \] The rate of external steam consumed: \[ m_s = 10 \text{ kg/h} \] Now, calculate the steam economy: \[ \text{Steam Economy} = \frac{m_v}{m_s} = \frac{6 \text{ kg/h}}{10 \text{ kg/h}} = 0.6 \]

Step 4: Final Answer:
The steam economy of the evaporator is 0.6.