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butalik [34]
3 years ago
6

For some transformation having kinetics that obey the Avrami equation , the parameter n is known to have a value of 1.1. If, aft

er 114 s, the reaction is 50% complete, how long (total time) will it take the transformation to go to 87% completion
y = 1 - exp(-kt^n)
Engineering
1 answer:
ivanzaharov [21]3 years ago
7 0

Answer:

total time  = 304.21 s

Explanation:

given data

y = 50% = 0.5

n = 1.1

t = 114 s

y = 1 - exp(-kt^n)

solution

first we get here k value by given equation

y = 1 - e^{(-kt^n)}   ...........1    

put here value and we get

0.5 = 1 - e^{(-k(114)^{1.1})}    

solve it we get

k = 0.003786  = 37.86 × 10^{4}

so here

y = 1 - e^{(-kt^n)}

1 - y  =  e^{(-kt^n)}

take ln both side

ln(1-y) = -k × t^n  

so

t = \sqrt[n]{-\frac{ln(1-y)}{k}}    .............2

now we will put the value of y = 87% in equation  with k and find out t

t = \sqrt[1.1]{-\frac{ln(1-0.87)}{37.86*10^{-4}}}

total time  = 304.21 s

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Answer:

Power output, P_{out} = 178.56 kW

Given:

Pressure of steam, P = 1400 kPa

Temperature of steam, T = 350^{\circ}C

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Mass flow rate, \dot{m} = 0.1 kg.s^{- 1}

Diameter of exhaust pipe, d_{h} = 15 cm = 0.15 m

Pressure at exhaust, P' = 50 kPa

temperature, T' =  100^{\circ}C

Solution:

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\dot{m} = \frac{Av_{i}}{V_{1}}

0.1 = \frac{\frac{\pi d^{2}}{4}v_{i}}{0.2004}

0.1 = \frac{\frac{\pi 0.08^{2}}{4}v_{i}}{0.2004}

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Now, eqn for compressible fluid:

\rho_{1}v_{i}A_{1} = \rho_{2}v_{e}A_{2}

Now,

\frac{A_{1}v_{i}}{V_{1}} = \frac{A_{2}v_{e}}{V_{2}}

\frac{\frac{\pi d_{i}^{2}}{4}v_{i}}{V_{1}} = \frac{\frac{\pi d_{e}^{2}}{4}v_{e}}{V_{2}}

\frac{\frac{\pi \times 0.08^{2}}{4}\times 3.986}{0.2004} = \frac{\frac{\pi 0.15^{2}}{4}v_{e}}{3.418}

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P_{out} = -\dot{m}W_{s}

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Answer:

Python code is explained below

Explanation:

CODE(TEXT): -

9A.py:-

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9C.py: -

#inputs are strings by default applying a split() function to a string splits it into a list of strings

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How many meters per second is 100 meters and 10 seconds
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Answer:

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Read 2 more answers
One kilogram of water fills a 150-L rigid container at an initial pressure of 2 MPa. The container is then cooled to 40∘C. Deter
lukranit [14]

The pressure of water is 7.3851 kPa

<u>Explanation:</u>

Given data,

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m = 1 Kg

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P_{2} = P_{Sat} = 7.3851 kPa

P_{2} = 7.3851 kPa

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