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jok3333 [9.3K]
3 years ago
14

For some transformation having kinetics that obey the Avrami equation, the parameter n is known to have a value of 2. If, after

100 s, the reaction is 40% complete, how long (total time in seconds) will it take the transformation to go to 95% completion
Engineering
1 answer:
kotegsom [21]3 years ago
8 0

This question is incomplete, the complete question is;

For some transformation having kinetics that obey the Avrami equation, the parameter n is known to have a value of 2. If, after 100 s, the reaction is 40% complete, how long (total time in seconds) will it take the transformation to go to 95% completion

y = 1 - exp( -ktⁿ )

Answer: the time required for 95% transformation is 242.17 s

Explanation:

First, we calculate the value of k which is the dependent variable in Avrami equation

y = 1 - exp( -ktⁿ )

exp( -ktⁿ ) = 1 - y

-ktⁿ = In( 1 - y )

k = - In( 1 - y ) / tⁿ

now given that; n = 2, y = 40% = 0.40, and t = 100 s

we substitute

k = - In( 1 - 0.40 ) / 100²

k = - In(0.60) / 10000

k = 0.5108 / 10000

k = 0.00005108 ≈ 5.108 × 10⁻⁵

Now calculate the time required for 95% transformation

tⁿ = - In( 1 - y ) / k

t = [- In( 1 - y ) / k ]^1/n

n = 2, y = 95% = 0.95 and k = 5.108 × 10⁻⁵

we substitute our values

t = [- In( 1 - 0.95 ) / 5.108 × 10⁻⁵ ]^1/2

t = [2.9957 / 5.108 × 10⁻⁵]^1/2

t = [ 58647.22 ]^1/2

t = 242.17 s

Therefore the time required for 95% transformation is 242.17 s

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A large well-mixed tank of unknown volume, open to the atmosphere initially, contains pure water. The initial height of the solu
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Answer:

The exact time when the sample was taken is = 0.4167337 hr

Explanation:

The diagram of a sketch of the tank is shown on the first uploaded image

Let A denote the  first inlet

Let B denote the second inlet

Let C denote the single outflow from the tank

From the question we are given that the diameter of A is = 1 cm = 0.01 m

                              Area of  A is  = \frac{\pi}{4}(0.01)^{2} m^{2}

                                                    = 7.85 *10^{-5}m^{2}

Velocity of liquid through A = 0.2 m/s

  The rate at which the liquid would flow through the first inlet in terms of volume  = \frac{Volume of Inlet }{time} = Velocity * Area i.e is m^{2} * \frac{m}{s}   = \frac{m^{3}}{s}

             = 0.2 *7.85*10^{-5} \frac{m^{3}}{s}

  The rate at which the liquid would flow through the first inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

                              =  1039.8 * 0.2 * 7.85 *10^{-5} Kg/s

                              = 0.016324 \frac{Kg}{s}

From the question the diameter of B = 2 cm = 0.02 m

                                           Area of B = \frac{\pi}{4} * (0.02)^{2} m^{2} = 3.14 * 10^{-4}m^{2}

                                     Velocity of liquid through B = 0.01 m/s

The rate at which the liquid would flow through the first inlet in terms of volume  = \frac{Volume of Inlet }{time} = Velocity * Area i.e is m^{2} * \frac{m}{s}   = \frac{m^{3}}{s}

             = 3.14*10^{-4} *0.01 \frac{m^{3}}{s}

The rate at which the liquid would flow through the second inlet in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

                              = 1053 * 3.14*10^{-6} \frac{Kg}{s}

                              = 0.00330642 \frac{Kg}{s}

From the question The flow rate in term of volume of the outflow at the time of measurement is given as  = 0.5 L/s

And also from the question the mass of  potassium chloride  at the time of measurement is given as 13 g/L

So The rate at which the liquid would flow through the outflow in terms of mass of the liquid = mass of liquid × the rate of flow in terms of volume

                              = 13\frac{g}{L} * 0.5 \frac{L}{s}

                              =  \frac{6.5}{1000}\frac{Kg}{s}       Note (1 Kg = 1000 g)

                              = 0.0065 kg/s

Considering potassium chloride

         Let denote the  rate at which liquid flows in terms of mass as   as \frac{dm}{dt} i.e change in mass with respect to time hence

           Input(in terms of mass flow ) - output(in terms of mass flow ) = Accumulation in the Tank(in terms of mass flow )

         

      (0.016324 + 0.00330642) - 0.0065 = \frac{dm}{dt}

          \int\limits {\frac{dm}{dt} } \, dx  =\int\limits {0.01313122} \, dx

      => 0.01313122 t = (m - m_{o})

  From the question  (m - m_{o})  is given as = 19.7 Kg

Hence the time when the sample was taken is given as

               0.01313122 t = 19.7 Kg

      =>  t = 1500.2414 sec

            t = .4167337 hours (1 hour = 3600 seconds)

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