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2 years ago

Explain the process of fractional distillation (include cracking)

Engineering

1 answer:

Grace [21]2 years ago

5 0

Answer:

Explanation:

Fractional distillation is the separation of a mixture into its component parts, or fractions. Chemical compounds are separated by heating them to a temperature at which one or more fractions of the mixture will vaporize. It uses distillation to fractionate

it is the separation of a liquid mixture into fractions differing in boiling point (and hence chemical composition) by means of distillation, typically using a fractionating column.

cracking allows large hydrocarbon molecules to be broken down into smaller, more useful hydrocarbon molecules. Fractions containing large hydrocarbon molecules are heated to vaporise them. They are then either: heated to 600-700°C.

I hope this was helpful

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1 year ago

A cylindrical tank is required to contain a gage pressure 520 kPa . The tank is to be made of A516 grade 60 steel with a maximum

enot [183]

Answer:

t= 4.5 mm

Explanation:

Given that

P = 520 KPa ( gauge)

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d= 2.6 m

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3 years ago

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2 years ago

The structure of a house is such that it loses heat at a rate of 4500kJ/h per °C difference between the indoors and outdoors. A

adelina 88 [10]

Answer:

15.24°C

Explanation:

The quality of any heat pump pumping heat from cold to hot place is determined by its coefficient of performance (COP) defined as

$COP=\frac{Q_{in}}{W}$

Where Q_{in} is heat delivered into the hot place, in this case, the house, and W is the work used to pump heat

You can think of this quantity as similar to heat engine's efficiency

In our case, the COP of our heater is

$COP_{heater} = \frac{\frac{4500\ kJ}{3600\ s} *(T_{house}-T_{out})}{4\ kW}$

Where T_{house} = 24°C and T_{out} is temperature outside

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Which has COP of:

$COP_{carnot}=\frac{T_{house}}{T_{house}-T_{out}}$

So we equate the COP of our heater with COP of Carnot heater

$\frac{1.25 *(T_{house}-T_{out})}{4}=\frac{T_{house}}{T_{house}-T_{out}}$

Rearrange the equation

$\frac{1.25}{4}(24-T_{out})^2-24=0$

Solve this simple quadratic equation, and you should get that the lowest outdoor temperature that could still allow heat to be pumped into your house would be

15.24°C

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