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mars1129 [50]
3 years ago
5

You are comparing distillation column designs at 1 atm and 3 atm total pressure for a particular separation. You have the same f

eed in both cases, and the feed is a saturated liquid in both cases. The product specifications do not change, and the same reflux ratio is used in both cases.
Compared to the design at 3 atm, the design at 1 atm will have: (Select one)a) more stages, fewer stages, the same number of stagesb) a larger diameter, a smaller diameter, the same diameterc) a higher reboiler temperature, a lower reboiler temp, the same reboiler temp
Engineering
1 answer:
Anestetic [448]3 years ago
8 0

Answer:

more number of stages will be required , a smaller diameter , a higher reboiler temperature

Explanation:

It is stated in the question that the product specifications and the reflux ratio do not change.

The pressure in a distillation column varies:

  • directly as the diameter of the distillation column
  • inversely as the number of stages required for the design
  • inversely as the temperature of the reboiler

Based on the theory stated above, the distillation column design at a pressure of 1 atm requires a more number of stages than the design with 3 atm, it has a smaller diameter than the design at 3 atm. It also has a higher reboiler temperature than the design with 3 atm.

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Engineers are designing a cylindrical air tank and are trying to determine the dimensions of the tank. The proposed material for
lana66690 [7]

Answer:

The length of tank is found to be 0.6 m or 600 mm

Explanation:

In order to determine the length, we need to find a volume for the tank.

For this purpose, we use ideal gas equation:

PV  = nRT

n = no. of moles = m/M

Therefore,

PV = (m/M)(RT)

V = (mRT)/(MP)

where,

V = volume of air = volume of container

m = mass of air = 4.64 kg

R = General Gas Constant = 8.314 J/mol.k

T = temperature of air = 10°C + 273 = 283 K

M = molecular mass of air = 0.02897 kg/mol

P = Pressure of Air = 20 MPa = 20 x 10^6 N/m²

V = (4.64 kg)(8.314 J/mol.k)(283 k)/(0.02897 kg/mol)(20 x 10^6 N/m²)

V = 0.01884 m³

Now, the volume of cylindrical tank is given as:

V = 0.01884 m³ = π(Diameter/2)²(Length)

Length = (0.01884 m³)(4)/π(0.2 m)²

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4 0
3 years ago
Calculate the viscosity(dynamic) and kinematic viscosity of airwhen
nikitadnepr [17]

Answer:

(a) dynamic viscosity = 1.812\times 10^{-5}Pa-sec

(b) kinematic viscosity = 1.4732\times 10^{-5}m^2/sec

Explanation:

We have given temperature T = 288.15 K

Density d=1.23kg/m^3

According to Sutherland's Formula  dynamic viscosity is given by

{\mu} = {\mu}_0 \frac {T_0+C} {T + C} \left (\frac {T} {T_0} \right )^{3/2}, here

μ = dynamic viscosity in (Pa·s) at input temperature T,

\mu _0= reference viscosity in(Pa·s) at reference temperature T0,

T = input temperature in kelvin,

T_0 = reference temperature in kelvin,

C = Sutherland's constant for the gaseous material in question here C =120

\mu _0=4\pi \times 10^{-7}

T_0 = 291.15

\mu =4\pi \times 10^{-7}\times \frac{291.15+120}{285.15+120}\times \left ( \frac{288.15}{291.15} \right )^{\frac{3}{2}}=1.812\times 10^{-5}Pa-swhen T = 288.15 K

For kinematic viscosity :

\nu = \frac {\mu} {\rho}

kinemic\ viscosity=\frac{1.812\times 10^{-5}}{1.23}=1.4732\times 10^{-5}m^2/sec

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Answer:

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Explanation:

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The correct answer is B
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