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Triss [41]
2 years ago
10

For binary flash distillation, we discussed in class that there are 8 variables (F, ZA, V, ya, L, XA, P and T) and 4 equations d

erived from VLE and mass balances. Thus, we typically require 4 of these variables to be given so that we can obtain a unique solution to the problem. Let's say, your manager tells you that he has a feed mixture with 2 components (given F, za) and he requires you to come up with a flash column that can produce a certain desired amount of Vapor product (thus V, ya are specified). Identity of both components is known and all VLE data has been provided to you. Has the manager given you enough data? If yes, give a step-by-step description of how would you go about designing the flash column (basically find P and T)? If no, why?
Engineering
1 answer:
Keith_Richards [23]2 years ago
8 0

Answer:

yes

Explanation:

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irinina [24]

Answer:

False I'm pretty sure sorry If its wrong

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Water flows steadily through the pipe as shown below, such that the pressure at section (1) and at section (2) are 300 kPa and 1
steposvetlana [31]

Answer:

The velocity at section is approximately 42.2 m/s

Explanation:

For the water flowing through the pipe, we have;

The pressure at section (1), P₁ = 300 kPa

The pressure at section (2), P₂ = 100 kPa

The diameter at section (1), D₁ = 0.1 m

The height of section (1) above section (2), D₂ = 50 m

The velocity at section (1), v₁ = 20 m/s

Let 'v₂' represent the velocity at section (2)

According to Bernoulli's equation, we have;

z_1 + \dfrac{P_1}{\rho \cdot g} + \dfrac{v^2_1}{2 \cdot g} = z_2 + \dfrac{P_2}{\rho \cdot g} + \dfrac{v^2_2}{2 \cdot g}

Where;

ρ = The density of water = 997 kg/m³

g = The acceleration due to gravity = 9.8 m/s²

z₁ = 50 m

z₂ = The reference = 0 m

By plugging in the values, we have;

50 \, m + \dfrac{300 \ kPa}{997 \, kg/m^3 \times 9.8 \, m/s^2} + \dfrac{(20 \, m/s)^2}{2 \times 9.8 \, m/s^2} = \dfrac{100 \ kPa}{997 \, kg/m^3 \times 9.8 \, m/s^2} + \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}50 m + 30.704358 m + 20.4081633 m = 10.234786 m + \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}

50 m + 30.704358 m + 20.4081633 m - 10.234786 m = \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}

90.8777353 m = \dfrac{v_2^2}{2 \times 9.8 \, m/s^2}

v₂² = 2 × 9.8 m/s² × 90.8777353 m

v₂² = 1,781.20361 m²/s²

v₂ = √(1,781.20361 m²/s²) ≈ 42.204308 m/s

The velocity at section (2), v₂ ≈ 42.2 m/s

3 0
2 years ago
A metal bar has a 0.6 in. x 0.6 in. cross section and a gauge length of 2 in. The bar is loaded with a tensile force of 50,000 l
Aleks [24]

Answer:

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The two boxcars A and B have a weight of 20 000 Ib and 30 000 Ib, respectively. If they coast freely down the incline when the b
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Answer:

Answer for the question :

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is explained in the attachment.

Explanation:

Download pdf
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Galina-37 [17]

Answer:

Explanation:

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