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Lena [83]
3 years ago
14

Conditions of special concern: i. Suggest two reasons each why distillation columns are run a.) above or b.) below ambient press

ure. Be sure to state clearly which explanation is for above and which is for below ambient pressure. ii. Suggest two reasons each why reactors are run at a.) elevated pressures and/or b.) elevated temperatures. Be sure to state clearly which explanation is for elevated pressure and which is for elevated temperature
Engineering
1 answer:
lutik1710 [3]3 years ago
7 0

Solution :

Methods for selling pressure of a distillation column :

a). Set, \text{based on the pressure required to condensed} the overhead stream using cooling water.

  (minimum of approximate 45°C condenser temperature)

b). Set, \text{based on highest temperature} of bottom product that avoids decomposition or reaction.

c). Set, \text{based on available highest } not utility for reboiler.

Running the distillation column above the ambient pressure because :

The components to be distilled have very high vapor pressures and the temperature at which they can be condensed at or below the ambient pressure.

Run the reactor at an evaluated temperature because :

a). The rate of reaction is taster. This results in a small reactor or high phase conversion.

b). The reaction is endothermic and equilibrium limited increasing the temperature shifts the equilibrium to the right.

Run the reaction at an evaluated pressure because :

The reaction is gas phase and the concentration and hence the rate is increased as the pressure is increased. This results in a smaller reactor and /or higher reactor conversion.

The reaction is equilibrium limited and there are few products moles than react moles. As increase in pressure shifts the equilibrium to the right.

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A light aircraft with a wing area of 200 ft^2 and a weight of 2000 lb has a lift coefficient of 0.39 and a drag coefficient of 0
Gnoma [55]

Answer: power required to maintain level flight=82.20hp

Explanation:

Given

Area = 200 ft^2

Weight = 2000 lb

Cl( Lift coefficient)= 0.39

Cd( Drag coefficient) = 0.06  

The density ρ of air at standard atmospheric  pressure = 2.38 X 10^-3 slugs/ft^3

For Equilibrium to be maintained during flight conditions, the lift force must be balanced by the weight of the aircraft such that

Lift force  = Weight of aircraft

(1/2)ρAU²Cl= W

1/2X 2.38 X 10^-3 X 200 X U² X 0.39 = 2000

U²= 2000 X 2 / 2.38 X 10^-3 X 200 X 0.39

U=\sqrt{21,547.08}

Velocity, U= 146.7892ft/s

Drag force of the velocity can be deduced from the formulae

Cd= Drag force(D) /1/2 ρU²A

Drag force=1/2 ρU²ACd

D=1/2 x (2.38 X 10^-3 slugs/ft^3) x (146.7892ft/s)² x 200 ft^2 x 0.06

D=307.69

Drag force= 308lb

power required to maintain level flight is given as

P = Drag force x Velocity = D x U

=308lb X  146.7892ft/s

=45,211.0736lb.ft/s

Changing to hp we have that

1 Horsepower, hp = 550 ft lbf/s

??=45,211.0736lb.ft/s

45,211.0736lb.ft/s/ 550 ft lbf/s= 82.20hp

6 0
3 years ago
An air compressor of mass 120 kg is mounted on an elastic foundation. It has been observed that, when a harmonic force of amplit
kupik [55]

Answer:

equivalent stiffness is 136906.78 N/m

damping constant is 718.96 N.s/m

Explanation:

given data

mass = 120 kg

amplitude = 120 N

frequency = 320 r/min

displacement = 5 mm

to find out

equivalent stiffness and damping

solution

we will apply here frequency formula that is

frequency ω = ω(n) √(1-∈ ²)      ......................1

here  ω(n) is natural frequency i.e = √(k/m)

so from equation 1

320×2π/60 = √(k/120) × √(1-2∈²)

k × ( 1 - 2∈²) = 33.51² ×120

k × ( 1 - 2∈²) = 134752.99    .....................2

and here amplitude ( max ) of displacement is express as

displacement = force / k  ×  (  \frac{1}{2\varepsilon \sqrt{1-\varepsilon ^2}})

put here value

0.005 = 120/k   ×  (  \frac{1}{2\varepsilon \sqrt{1-\varepsilon ^2}})  

k ×∈ × √(1-2∈²) = 1200       ......................3

so by equation 3 and 2

\frac{k\varepsilon \sqrt{1-\varepsilon^2})}{k(1-2\varepsilon^2)} = \frac{12000}{134752.99}

\varepsilon^{2} - \varepsilon^{4}  = 7.929 * 10^{-3} - 0.01585 * \varepsilon^{2}

solve it and we get

∈ = 1.00396

and

∈ = 0.08869

here small value we will consider so

by equation 2 we get

k × ( 1 - 2(0.08869)²) = 134752.99

k  = 136906.78 N/m

so equivalent stiffness is 136906.78 N/m

and

damping is express as

damping = 2∈ √(mk)

put here all value

damping = 2(0.08869) √(120×136906.78)

so damping constant is 718.96 N.s/m

7 0
3 years ago
Does anyone know what this is​
sammy [17]

Answer:

Looks like mold that got frosted over

Explanation:

4 0
2 years ago
Read 2 more answers
A rectangular open box, 25 ft by 10 ft in plan and 12 ft deep weighs 40 tons. Sufficient amount of stones is placed in the box a
dimulka [17.4K]

Answer:

44.95 tonnes

Explanation:

According to principle of buoyancy the object will just sink when it's weight is more than the weight of the liquid it displaces

It is given that empty weight of box = 40 tons

Let the mass of the stones to be placed be = M tonnes

Thus the combined mass of box and stones = (40+M) tonnes..........(i)

Since the box will displace water equal to it's volume V we have volume of box = 25ft*10ft*12ft= 3000ft^{3}

Volume= 84.95m^{3}

Since 1ft^{3} =0.028m^{3}

Now the weight of water displaced = Weight =\rho \times Volumewhererho is density of water = 1000kg/m^{3}

Thus weight of liquid displaced = \frac{84.95X1000}{1000}tonnes=84.95 tonnes..................(ii)

Equating i and ii we get

40 + M = 84.95

thus Mass of stones = 44.95 tonnes

3 0
3 years ago
Products that are in the process of being manufactured but are not yet complete are called:
vazorg [7]

Answer:

Those products are generally called Work in Process WIP

Explanation:

Work in process (WIP), or work in progress (WIP), goods in process, or in-process inventory in a manufacturing industry/company refer to the company's partially finished goods waiting for completion and eventual sale or the value of these items.

These items are either just being produced or require further processing (like purification, separation, packaging or handling) in a queue or a buffer storage.

6 0
3 years ago
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