Evaporation in Double-Effect Reverse-Feed Evaporators. A feed containing 2 wt % dissolved organic solids in water is fed to a do
uble-effect evaporator with reverse feed. The feed enters at 100°F and is concentrated to 25% solids. The boiling-point rise can be considered negligible as well as the heat of solution. Each evaporator has a 1000-ft2 surface area and the heat-transfer 700 btu/h -ft2-°F. The feed enters evaporator number 2 and steam at 100 psia is fed to number 1. The pressure iin the vapor space of evaporator number 2 is 0.98 psia. Assume that the heat capacity of all liquid solutions is that of liquid water. Calculate the feed rate F and the product rate Li of a solution containing 25% solids. (Hint: Assume a feed rate of, say, F 1000 lbm/h. Calculate the area. Then calculate the actual coefficients are U1 500 and U2- feed rate by multiplying 1000 by 1000/calculated area.) Ans. F- 133 800 lbm/h (60 691 kg/h), L,- 10700lb»/h (4853 kg/h)
1 answer:
Answer:
472,826 lb/hr
Explanation:
As per the given data:
Solids in feed= 2%
Solids in concentrate= 25%
HTA1 = HTA2 = 1000 ft^2
U1 = 500 Btu/ h ft^2 F & U2 = 700 Btu/ h ft^2 F
Overall material balance: Feed= Distillate + concentrate ----> Eq-1
Component balance: Feed * 0.02 = Distillate * 0 + concentrate * 0.25
Feed = 12.5 * concentrate ---> Eq-2
Boiling point rise = negligible, so solution & solvent vapor temperature will be same.
Assumed that the 1st effect is operating under atmospheric pressure (Boiling point - 212F).
As per the data:
Latent heat 212F = 300 Btu/lb
Latent heat 100F = 320 Btu/lb
As per material balance:
Vapor flowrate * latent heat = Overall HT coefficient * HTA * DT
1st effect: M-1= (500 * 1000 * (326-212)) / 300 = 190,000 lb/hr
2nd effect: M-2= (700*1000 * (212-100)) / 320 = 245,000 lb/hr
Distillate = M-1 + M-2 = 190,000+245,000 = 435,000 lb/hr
Substituting the above in Eq-1
Feed = 435,000 + concentrate
Substitute Eq-2 in the above
12.5 * concentrate = 435,000 + concentrate
Concentrate, L1 = 435,000/11.5 = 37826 lb /hr
Feed, F = 435,000 + 37826 = 472,826 lb/hr
1st effect operating pressure is not given, That may be the reason we are not getting the given answer. But procedure is right.
[ Find the figure in the attachment ]
You might be interested in
Answer:
For block A, a = 9.66 ft/s²
For block B, a = 15 ft/s²
Explanation:
A free body diagram for this force system is attached to this solution
Mass of block A = m₁ = 8 lb
Mass of block B = m₂ = 6 lb
Coefficient of kinetic friction = μ
Normal reaction on the blocks = N
Spring stiffness of the spring btw block A and B = k = 20 lb/ft
Compression of the spring = 0.2 ft
Analysing Block A first
The forces on block A include, the weight, normal reaction, frictional force and the elastic force due to the spring
Sum of forces in the y-direction = 0
So, the weight of the block = Normal reaction of the surface on the block
N = W = 8 lb
Sum of forces in the x-direction = maₓ
(k × x) - (μ × N) = maₓ
m = W/g = 8/32.2 = 0.248 lbm
(20×0.2) - (0.2 × 8) = (0.248) aₓ
aₓ = 9.66 ft/s²
The forces on block B include, the weight, normal reaction, frictional force and the elastic force due to the spring
Sum of forces in the y-direction = 0
So, the weight of the block = Normal reaction of the surface on the block
N = W = 6 lb
Sum of forces in the x-direction = maₓ
(k × x) - (μ × N) = maₓ
m = W/g = 6/32.2 = 0.186 lbm
(20×0.2) - (0.2 × 6) = (0.186) aₓ
aₓ = 15 ft/s²
