Chemical Engineering - Heat Transfer - Discussion
Discussion Forum : Heat Transfer - Section 2 (Q.No. 24)
24.
It is desired to concentrate a 20% salt solution (20 kg of salt in 100 kg of solution) to a 30% salt solution in an evaporator. Consider a feed of 300 kg/min at 30°C. The boiling point of the solution is 110°C, the latent heat of vaporisation is 2100 kJ/kg and the specific heat of the solution is 4 kJ/kg.K. The rate at which the heat has to be supplied in (kJ/min) to the evaporator is
Discussion:
15 comments Page 1 of 2.
Nagesh said:
7 years ago
Q= (m Cp dt) + (m x latenet heat).
Latent heat * amount of water evaporates.
So according to that,
Latent heat is given 21000.
And by material balance, 100 kg of water evaporates.
So,
21000*100_______(1)
And. MCp"T is heat rate required for heating the solution of 300 kg. Cp is given. Latenet heat is (110-30)=80.
So
300*4*80=96000_____&(2).
By adding (1)And (2).
= 3.06 * 10^5.
Latent heat * amount of water evaporates.
So according to that,
Latent heat is given 21000.
And by material balance, 100 kg of water evaporates.
So,
21000*100_______(1)
And. MCp"T is heat rate required for heating the solution of 300 kg. Cp is given. Latenet heat is (110-30)=80.
So
300*4*80=96000_____&(2).
By adding (1)And (2).
= 3.06 * 10^5.
(7)
Honey said:
1 decade ago
First sensible heat is provided to rise temp up to 110 deg celsius.
(i.e. 300*4*(110-30) = 96000).
Amount of water evaporate is obtained by material balance of salt
(i.e. 100 kg water evaporate so heat require 100*2100 = 210000).
Total heat required = 210000+96000 = 306000. Option A.
(i.e. 300*4*(110-30) = 96000).
Amount of water evaporate is obtained by material balance of salt
(i.e. 100 kg water evaporate so heat require 100*2100 = 210000).
Total heat required = 210000+96000 = 306000. Option A.
Alpesh Solanki said:
9 years ago
Dear all,
First find the liquor flow rate.
Wt fra in feed * feed rate = wt fra in exit * liquor flow rate.
.2 * 300 = .3 * m',
m' = 200kg/min.
Now, find out heat requirement.
By using the formula,
=> (300 * 4 * 80) + (300 - 200) * 2100,
= 306000.
First find the liquor flow rate.
Wt fra in feed * feed rate = wt fra in exit * liquor flow rate.
.2 * 300 = .3 * m',
m' = 200kg/min.
Now, find out heat requirement.
By using the formula,
=> (300 * 4 * 80) + (300 - 200) * 2100,
= 306000.
(2)
Naveen solanki said:
6 years ago
100kg/min water is evaporating as per material balance we find out that 200 Kg/min output remains but we get 300Kg/min.
So, remaining was evaporated which is 300-200= 100 kg/min.
So, remaining was evaporated which is 300-200= 100 kg/min.
(2)
Shan Rana said:
1 decade ago
Q= (m Cp dt) + (m x latenet heat).
dt = 80.
Q = (300x4x80) + (300 x 2100).
= 96000 + 630000.
= 726000.
= 7.26 x 10^5.
Correct Answer is C.
dt = 80.
Q = (300x4x80) + (300 x 2100).
= 96000 + 630000.
= 726000.
= 7.26 x 10^5.
Correct Answer is C.
Emma said:
10 years ago
@Honey, you added to term of different unit. Both masses ought to be in terms of rate.
Vinit pandya said:
1 decade ago
I think in formula m*Cp*dt+ m*latent heat.
Both the m are mass flow rates.
Both the m are mass flow rates.
Akshat said:
6 years ago
@Naveen Solanki.
Please, can you explain material balance?
Please, can you explain material balance?
(1)
Richard said:
7 years ago
@Honey.
Your units aren't consistent. Please explain.
Your units aren't consistent. Please explain.
Akash said:
7 years ago
@Alpesh Solanki.
Can you explain (300 - 200) * 2100?
Can you explain (300 - 200) * 2100?
(1)
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