Chemical Engineering - Heat Transfer - Discussion

Discussion Forum : Heat Transfer - Section 7 (Q.No. 29)
29.
1000 kg of wet solids are to be dried from 60% to 20% moisture (by weight). The mass of moisture removed in kg is
520
200
400
500
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
12 comments Page 1 of 2.

Vishnu Kushwaha said:   6 years ago
The correct Answer is D.

Initially water is 60% means 600kg.&400 kg is solid.
&finally 20%.

Therefore;
X/(x+400)=0.2 solve it you will get;
X=100, Now water remaining is 100 and evaporated water is =600-100=500kg.
(2)

Shailesh kumar said:   8 years ago
Outlet condition 1000 * .2=200,
Inlet condition 1000 * .6=600.
So moisture removed =Outlet-inlet.=600-200=400kg moisture removed.
(1)

Dboipai said:   9 years ago
Answer is D: after removing 500 kgs is water from 600 kgs we are left with 100 kgs of water and 400 kgs of solid. Then the % of water in the system will be 100/500 = 20%.

Philip said:   8 years ago
It should be D. 1000kg has a mass of water included in it and it is stated that initially, it has 60% water content thus 600kg of water.
(1)

Rana said:   5 years ago
@All.

D is the right option.

It is a kind of dryer. Apply overall material balance and then component balance (solid).
(1)

Ramsajan said:   2 years ago
(D) is right because M.B of solid 1000*40/100 = x * 80/100,
x = 500.
Then moisture in kg 1000 - 500 = 500kg.
(3)

Shahid said:   1 decade ago
m(w-we) = mf.
1000(0.6-0.2) = mf.
400 = mf.
(2)

S.roy said:   6 years ago
Yes, Option D is correct. I too agree.
(1)

Asmah sitotaw said:   5 years ago
Anyone, Explain the right answer here.

TANMAY SHAH said:   2 years ago
The Right Answer should be option D.
(2)


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