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 2 of 2.
Emma said:
10 years ago
@Honey, you added to term of different unit. Both masses ought to be in terms of rate.
Pavan said:
1 decade ago
From where you got m value?
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.
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.
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.
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