Question

Orange juice is the feed introduced at a rate of 0.50 kg/s to a single effect evaporator. The concentration of the dilute juice is 10% total solids. The juice is concentrated to 40% total solids. Calculate the product mass flow rate.

• A. 0.125 kg/s
• B. 0.135 kg/s
• C. 0.145 kg/s
• D. 0.155 kg/s

Hint:

Always perform a "Solids Balance" for concentration problems. It is much simpler than a total mass balance because the solids don't evaporate!

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a standard mass balance problem in food engineering.
The principle involved is the Conservation of Mass, specifically the "Solids Balance."
In an evaporator, water is removed as vapor, but the total amount of solids entering the system in the feed must equal the total amount of solids leaving in the product.
Key Formula or Approach:
The formula for solids balance is:
\[ F \times X_f = P \times X_p \]
Where:
$F$ = Feed mass flow rate ($kg/s$)
$X_f$ = Mass fraction of solids in the feed
$P$ = Product mass flow rate ($kg/s$)
$X_p$ = Mass fraction of solids in the product

Step 2: Detailed Explanation:
Given data:
Feed rate ($F$) = 0.50 $kg/s$
Dilute concentration ($X_f$) = 10% = 0.10
Concentrated target ($X_p$) = 40% = 0.40
We need to find the product rate ($P$).
Using the solids balance equation:
\[ 0.50 \times 0.10 = P \times 0.40 \]
\[ 0.05 = P \times 0.40 \]
Now, solve for $P$:
\[ P = \frac{0.05}{0.40} \]
\[ P = \frac{5}{40} \]
\[ P = \frac{1}{8} \]
\[ P = 0.125 \text{ kg/s} \]
This means that for every 0.50 kg of juice entering, 0.125 kg comes out as concentrate and the remaining 0.375 kg ($0.50 - 0.125$) is evaporated as water vapor.

Step 3: Final Answer:
The product mass flow rate is 0.125 kg/s.