Rough rice were dried at 50 °C and 60% relative humidity. The moisture content at time 3 and 5 hrs are 12.0 and 10.0%, respectively. If the drying rate constant be 0.49 per hour, c... Rough rice were dried at 50 °C and 60% relative humidity. The moisture content at time 3 and 5 hrs are 12.0 and 10.0%, respectively. If the drying rate constant be 0.49 per hour, calculate the equilibrium moisture content of the sample. Read more

Steps to Solve

1. Identify given values

The problem provides the following values:

• Initial moisture content ( M_1 = 12.0% ) at ( t_1 = 3 ) hours
• Final moisture content ( M_2 = 10.0% ) at ( t_2 = 5 ) hours
• Drying rate constant ( k = 0.49 ) per hour

2. Calculate the moisture content difference

Determine the change in moisture content between the two time points:

$$ \Delta M = M_1 - M_2 $$ $$ \Delta M = 12.0% - 10.0% = 2.0% $$

3. Determine the time interval

Calculate the time interval ( \Delta t ) between the points:

$$ \Delta t = t_2 - t_1 $$ $$ \Delta t = 5 \text{ hours} - 3 \text{ hours} = 2 \text{ hours} $$

4. Calculate the average drying rate

The average drying rate ( R ) can be calculated as follows:

$$ R = \frac{\Delta M}{\Delta t} $$ $$ R = \frac{2.0%}{2 \text{ hours}} = 1.0% \text{ per hour} $$

5. Apply the drying rate expression

In drying processes, we use the equation:

$$ R = k(M - M_e) $$

Where ( M_e ) is the equilibrium moisture content that we want to find. Rearranging gives:

$$ M - M_e = \frac{R}{k} $$

6. Calculate equilibrium moisture content

Substituting the known values from the first time point ( M_1 ):

$$ M_1 - M_e = \frac{1.0%}{0.49} $$ $$ M_e = M_1 - \frac{1.0%}{0.49} $$

Calculating the right side:

$$ M_e = 12.0% - \frac{1.0}{0.49} \approx 12.0% - 2.04% \approx 9.96% $$

Thus, we conclude that:

$$ M_e \approx 10.0% $$