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Flura [38]
3 years ago
14

An evaporation–crystallization process of the type described in Example 4.5-2 is used to obtain solid potassium sulfate from an

aqueous solution of this salt. The fresh feed to the process contains 19.6 wt% K2SO4. The wet filter cake consists of solid K2SO4 crystals and a 40.0 wt% K2SO4 solution, in a ratio 10 kg crystals/kg solution. The filtrate, also a 40.0% solution, is recycled to join the fresh feed. Of the water fed to the evaporator, 45.0% is evaporated. The evaporator has a maximum capacity of 175 kg water evaporated/s.Calculate the maximum production rate of solid K2SO4, the rate at which fresh feed must be supplied to achieve this production rate.

Chemistry
1 answer:
timofeeve [1]3 years ago
7 0

Answer:

feed = 220.77 kg/s; maximum production rate of solid crystal = 416 kg/s; the rate of supplying fresh feed to obtain the production rate = 1.6

Explanation:

Material or mass balance can be used to estimate the mass flow rates of all the streams in the diagram shown in the attached file.

Overall balance: M_{1} = 175 + 11M_{2}

Water: 0.804M_{1} = 175 + 0.6M_{2}

Using substitution method, we have:

M_{1} = 220.77 kg/s

M_{2}  = 4.16 kg/s

The maximum production rate of solid crystal is 10M_{2} = 10*4.16 = 416 kg/s

Around evaporator:

0.45M_{5} = 175

M_{5} = 175/0.45 = 389 kg/s

Around the mixing point:

M_{1} + M_{3} = M_{4} + M_{5}

Solid crystal: 0.196M_{1} + 0.4M_{3} = M_{4}

Using the last two equations, we can obtain:

0.804M_{1} + 0.6M_{3} = M_{5}

0.6M_{3} = 389 - 0.804*220.77 = 389 - 177.5 = 211.5

M_{3} = 211.5/0.6 = 352.5 kg/s

The rate of supplying fresh feed to obtain the production rate is:

\frac{M_{3}}{M_{1}} = 352.5/220.77 = 1.6

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Explanation:

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P1T2 = P2T1

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<em />

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P1 = 1000torr

T2 = ? -Incógnita-

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T1 = 273K -Temperatura del hielo fundido = 0°C = 273K

1000torrT2 = 100torr273

T2 = 27.3K

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<h3>-245.7°C es la temperatura del gas bajo 100 torr</h3>

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3 0
3 years ago
Determine how many carbon dioxide (CO2) molecules are produced if 8.45 x 1023 molecules of water (H2O) are produced 2 C2H6 (g) +
Andreas93 [3]
Ill get back to you on this
5 0
2 years ago
0.415 g of an unknown triprotic acid are used to make a 100.00 mL solution. Then 25.00 mL of this solution is transferred to an
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Answer:

Explanation:

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final burette reading = 39.7 mL

volume of NaOH used = 39.7 - 1.81 = 37.89 mL .  

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No of moles of triprotic acid = 3.89 x 10⁻³ / 3

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8 0
3 years ago
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steel, plastics

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3 0
3 years ago
Determine the ph of a 0.18 m h2co3 solution. carbonic acid is a diprotic acid whose ka1 = 4.3 × 10-7 and ka2 = 5.6 × 10-11. dete
oee [108]
Answer is: ph value is 3.56.
Chemical reaction 1: H₂CO₃(aq) ⇄ HCO₃⁻(aq) + H⁺(aq); Ka₁ = 4,3·10⁻⁷.
Chemical reaction 2: HCO₃⁻(aq) ⇄ CO₃²⁻(aq) + H⁺(aq); Ka₂ = 5,6·10⁻¹¹.
c(H₂CO₃) = 0,18 M.
[HCO₃⁻] = [H⁺<span>] = x.
</span>[H₂CO₃] = 0,18 M - x.
Ka₁ = [HCO₃⁻] · [H⁺] / [H₂CO₃].
4,3·10⁻⁷ = x² / (0,18 M -x).
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pH = -log[H⁺] = -log(0,000293 M).
pH = 3,56; second Ka do not contributes pH value a lot.


5 0
3 years ago
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