Question

A batch of 5 kg of food product has a moisture content of 150% dry basis. Calculate how much water must be removed from this product to reduce its moisture content to 20% wet basis.

Answer 1

Explanation:

The 20 kg of food at a moisture content of 80 % wet basis is dried to 50 % wet basis. Calculate the amount of water removedF = 20 kg of raw product

Answer 2

Given:

mass of dry food product=5kg

moisture content(dry basis)=150%

moisture content(wet basis)=20%

To find:

amount of water that must be removed

Solution:

We know that,

$Moisture content(dry basis)=\frac{mass of water}{mass of dry solid}$

Therefore we can write,

$150=\frac{mass of water}{5}$×$100$

⇒$mass of water=\frac{750}{100}$

⇒mass of water=7.5

So, when we calculate moisture on dry basis the mass of water is 7.5kg.

Now we know,

$Moisture content(wet basis)=\frac{mass of water}{mass of moist solid}$

⇒$Moisture content(wet basis)=\frac{mass of water}{mass of water+mass of dry solid}$

Therefore, we can write,

$20=\frac{mass of water}{mass of water+5}$×$100$

Let the mass of water be 'x'.

⇒$(1-0.2)x=1$

⇒$0.8x=1$

⇒$x=\frac{1}{0.8}$

⇒$x=1.25$

The percent decrease$=\frac{7.5-1.25}{7.5}$×$100$

=$0.8333$×$100$

=$83.33$%

Hence, 6.25kg of water must be reduced or 83.33% of water must be reduced.