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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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Multiple Choice
hichkok12 [17]

Answer:

Geological engineers can help determine if the soil and structural stability in the building's location are satisfactory.

Explanation:

Geologic Engineering refers to a job which applies<em> geology to engineering</em>. This type of job focuses on the investigation of sites in response to<em> soil, rock </em>and <em>groundwater.</em> They help design the major structures for engineering works, thus, it is also their role <u><em>to know whether the building's soil and structural stability are satisfactory.</em></u> They determine <em>how much of the structure is a safe load for the soil it is standing upon.</em> They also test the strength of both rock and soil at different depths. This will help them know whether the location will be suitable for the skyscraper.

4 0
3 years ago
. A Carnot heat pump is to be used to heat a house and maintain it at 22 °C in winter. When the outdoor temperature remains at 3
max2010maxim [7]

Answer:

The Carnot heat pump must work 3.624 hours per day to keep the temperature constant inside the house.

Explanation:

The net heat daily loss of the house is:

Q_{losses} = \left(76000\,\frac{kJ}{h}\right)\cdot (24\,h)

Q_{losses} = 1.824\times 10^{6}\,kJ

In order to keep the house warm, given heat must be equal to heat losses:

Q_{H} = Q_{losses}

Besides, the Coefficient of Performance for a Carnot heat pump is:

COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}

Where,

T_{L} - Temperature of the cold reservoir (Outdoors), measured in Kelvin.

T_{H} - Temperature of the hot reservoir (House), measured in Kelvin.

Given that T_{L} = 276.15\,K and T_{H} = 295.15\,K, the Coefficient of Performance is:

COP_{HP} = \frac{295.15\,K}{295.15\,K-276.15\,K}

COP_{HP} = 15.534

For a real heat machine, the Coefficient of Performance is determined by the following expression:

COP_{HP} = \frac{Q_{H}}{W}

Where:

Q_{H} - Heat received by the house, measured in kilojoules.

W - Work consumed by the Carnot heat pump, measured in kilojoules.

The daily work consumed is now cleared in the previous expression:

W = \frac{Q_{H}}{COP_{HP}}

W = \frac{1.824\times 10^{6}\,kJ}{15.534}

W = 117419.853\,kJ

The working time is calculated by dividing this result by input power. That is:

\Delta t = \frac{W}{\dot W}

\Delta t = \left(\frac{117419.853\,kJ}{9\,kW} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}\right)

\Delta t = 3.624\,h

The Carnot heat pump must work 3.624 hours per day to keep the temperature constant inside the house.

8 0
3 years ago
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Answer:

thats a good question that i have been trying to figure out. idk but people seem to have their own gods

Explanation:

6 0
3 years ago
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