True or false sugar melting an example of oxidation and water evaporating
1 answer:
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Answer:
the amounts and compositions of the overflow V1 and underflow L1 leaving the stage are 75kg and 125kg respectively.
Explanation:
Let state the given parameters;
Let A= solvent (hexane)
B= solid(inert soiid)
C= solvent(oil)
= mass of solvent + mass of oil (i.e A+C)
<u>Feed Phase:</u>
Total feed (i.e slurry of flakes soybeans)= 100kg
B= mass of solid =75 kg
F= mass of solvent + mass of oil (i.e A+C)
= 25kg
Mass ratio of oil to solution =
mass of oil (C) =25 × 0.1 wt = 2.5kg
mass of hexane in feed = 25 × 0.9 =22.5kg + 2.5 =25kg
therefore =
= 0.1
mass ratio of solid to solution =
=3
<u>Solvent Phase:</u>
C= Mass of oil= 0(kg)
A= Mass of hexane = 100kg
mass of solutions = A+C = 0+100kg
solvent= 100kg
<u>Underflow:</u>
underflow = L₁ = (unknown) ???
L₁ = E₁ + B
the value of N for the outlet and underflow is 1.5 kg
i.e N₁ =
solution in underflow E₁ = Mass (A+C)
<u>Overflow:</u>
Overflow = V₁ = (unknown) ???
solution in overflow V₁ = Mass (A+C)
This is because, B = 0 in overflow
Solid Balance: (since the solid is inert, then is said to be same in feed & underflow).
solid in feed = solid in underflow = 75
75= E₁ × N₁
75 = E₁ × 1.5
E₁ = 50kg
Underflow L₁ = E₁ × B
= 50 + 75
=125kg
The Overall Balance: Feed + Solvent = underflow + overflow
100 + 100 = 125 + V₁
V₁ = 75kg
Answer:
a. 123.9°C
b.
c.
Explanation:
Hello, I'm attaching a picture with the numerical development of this exercise.
a. Since the steam is overheated vapour, the specific volume is gotten from the corresponding table. Then, as it became a saturated vapour, we look for the interval in which the same volume of state 1 is, then we interpolate and get the temperature.
b. Now, at 80°C, since it is about a rigid tank (constant volume for every thermodynamic process), the specific volume of the mixture is 0.79645 m^3/kg as well, so the specific volume for the liquid and the vapour are taken into account to get the quality of 0.234.
c. Now,since this is an isocoric process, the heat transfer per kg of steam is computed as the difference in the internal energy, considering the initial condition (showed in a. part) and the final one computed here.
** The thermodynamic data were obtained from Cengel's thermodynamics book 7th edition.
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