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iVinArrow [24]
3 years ago
5

The following liquids are stored in a storage vessel at 1 atm and 25°C. The vessels are vented with air. Determine whether the e

quilibrium vapor above the liquid will be flammable. The liquids are:________.
a. Acetone
b. Benzene
c. Cyclohexane
d. Toluene Problem
Engineering
1 answer:
defon3 years ago
4 0

Answer:

The liquids are TOLUENE because the equilibrum vapor above it will be flammable ( D )

Explanation:

Liquids stored at : 1 atm , 25⁰c  and they are vented with air

Determining whether the equilibrum vapor above the liquid will be flammable

We can determine this by using Antoine equation to calculate saturation vapor pressure also apply Dalton's law to determine the volume % concentration of air and finally we compare answer to flammable limits to determine which liquid will be flammable

A) For acetone

using the Antoine equation to calculate saturation vapor pressure

In(P^{out} ) = A - \frac{B}{C + T}

values gotten appendix E ( chemical process safety (3rd edition) )

A = 16.6513

B = 2940.46

C = -35.93

T = 298 k      input values into Antoine equation

therefore ; p^{out} = 228.4 mg

calculate volume percentage using Dalton's law

= V% = (saturation vapor pressure / pressure ) *100

         = (228.4 mmHg / 760 mmHg) * 100 = 30.1%

The liquid is not flammable because its UFL = 12.8%

B) For Benzene

using the Antoine equation to calculate saturation vapor pressure

In(P^{out} ) = A - \frac{B}{C + T}

values gotten appendix E ( chemical process safety (3rd edition) )

A= 15.9008

B = 2788.52

C = -52.36

T = 298 k   input values into the above equation

p^{out} = 94.5 mmHg

calculate volume percentage using Dalton's law

V% = (saturation vapor pressure / pressure ) *100

      = (94.5 / 760 ) * 100 = 12.4%

Benzene is not flammable under the given conditions because its UFL =7.1%

C) For cyclohexane

using the Antoine equation to calculate saturation vapor pressure

In(P^{out} ) = A - \frac{B}{C + T}

values gotten appendix E ( chemical process safety (3rd edition) )

A = 15.7527

B = 2766.63

c = -50.50

T = 298 k

solving the above equation using the given values

p^{out} = 96.9 mmHg

calculate volume percentage using Dalton's law

V% = (saturation vapor pressure / pressure ) *100

      = ( 96.9 mmHg /760 mmHg) * 100 = 12.7%

cyclohexane not flammable under the given conditions because its UFL= 8%

D) For Toluene

using the Antoine equation to calculate saturation vapor pressure

In(P^{out} ) = A - \frac{B}{C + T}

values gotten from appendix E ( chemical process safety (3rd edition) )

A = 16.0137

B = 3096.52

C = -53.67

T = 298 k

solving the above equation using the given values

p^{out} = 28.2 mmHg

calculate volume percentage using Dalton's law

V% = (saturation vapor pressure / pressure ) *100

     = (28.2 mmHg / 760 mmHg) * 100 = 3.7%

Toluene is flammable under the given conditions because its UFL= 7.1%

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Answer:

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Explanation:

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We know that

1 inch = 0.0254 m

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We know that density of Hg =13.6\times 10^3\ kg/m^3

P = ρ g h

P(atm) = 13.6 x 1000 x 9.81 x 0.739 Pa

P(atm) = 13.6  x 9.81 x 0.739 KPa

P(atm) =98.54 KPa

Now

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solmaris [256]

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For a given set of input values, a NAND gate produces the opposite output as an OR gate with inverted inputs.A. True
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Answer:

B.

Explanation:

For a given set of input values, A NAND gate produces exactly the same values as an OR gate with inverted inputs.

The truth table  for a NAND gate with 2 inputs is as follows:

0 0    1

0 1     1

1  0    1

1  1    0

The  truth table for an OR gate, is as follows:

0 0    0

0 1     1

1  0    1

1  1     1

If we add two extra columns for inverted inputs, the truth table will be this one:

0 0    1  1       1

0 1     1  0      1

1  0    0  1      1

1  1     0  0     0

which is the same as for the NAND gate, not the opposite, so the statement is false.

This means that the right choice is B.

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Answer:

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