ENGINEERING COLLEGE

For some transformation having kinetics that obey the Avrami equation (Equation 10.17), the parameter n is known to have a value of 2.1. If, after 146 s, the reaction is 50% complete, how long (total time) will it take the transformation to go to 86% completion?

Answers

Answer 1

Answer:

Answer:

t = 25.10 sec

Explanation:

we know that Avrami equation

$Y = 1 - e^(-kt^n)$

here Y is percentage of completion of reaction = 50%

t is duration of reaction = 146 sec

so,

$0.50 = 1 - e^(-k^146^2.1)$

$0.50 = e^(-k306.6)$

taking natural log on both side

ln(0.5) = -k(306.6)

$k = 2.26* 10^(-3)$

for 86 % completion

$0.86 = 1 - e^{-2.26* 10^(-3) * t^(2.1)}$

$e^{-2.26* 10^(-3) * t^(2.1)} = 0.14$

$-2.26* 10^(-3) * t^(2.1) = ln(0.14)$

$t^(2.1) = 869.96$

t = 25.10 sec

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Air expands adiabatically through a nozzle from a negligible initial velocity to a final velocity of 300 m/s, what is the temperature drop of the air, if air is assumed to be an ideal gas for which CP = (7/2)R?

Answers

The temperature drop of air if air is assumed to be an ideal gas for which C_p = ⁷/₂R is; Δt = 1546 K

We are given;

Final velocity; v₂ = 300 m/s

C_p = ⁷/₂R

At constant pressure, the change in enthalpy is;

Δh = C_p × Δt

Now, from first law of thermodynamics;

h₂ + (v₂²/2) = h₁ + (v₁²/2)

We are told initial velocity is negligible and as such v₁ = 0 m/s

Thus;

h₂ + (v₂²/2) = h₁ + 0

(h₁ - h₂) = (v₂²/2)

Thus; Δh = v₂²/2

Finally;

C_p × Δt = v₂²/2

Δt = v₂²/2/(C_p)

Δt = (300²/2)/(⁷/₂R)

where R is ideal gas constant = 8.314 Kj/kg.mol

Thus;

Δt = (300²/2)/(⁷/₂ × 8.314)

Δt = 1546 K

Read more at; brainly.com/question/24188841

Answer:

ΔH+U²/2=0

and

ΔH= $C{p$ ×ΔT

∴to get the temperature drop of air, you make ΔT subject of the formula

ΔT=-U²/2Cp

=-300²/2× $(7)/(2)$ ×8.314

∴ΔT=-1546K

Explanation:

Which property of real numbers is shown below?3 + ((-5) + 6) = (3 + (-5)) + 6
commutative property of addition
identity property of multiplication
associative property of addition
commutative property of multiplication

Answers

The property of realnumbers is shown below is associative property of addition. The correct option is C.

What is associative property of addition?

According to the associativeproperty of addition, you can arrange the addends in several ways without changing the result.

According to the commutative property of addition, you can rearrange the addends without altering the result.

When more than two numbers are added together or multiplied, the outcome is always the same, regardless of how the numbers are arranged.

This is known as the associativeproperty. As an illustration, 2 (7 6) = (2 7) 6. 2 + (7 + 6) = (2 + 7) + 6.

Thus, the correct option is C.

For more details regarding associative property, visit:

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Answer:

Explanation:

The pressure in an automobile tire depends on thetemperature of the air in the tire. When the air temperature is25°C, the pressure gage reads 210 kPa. If the volume of the tire is 0.025 m3, determine the pressure rise in the tire whenthe air temperature in the tire rises to 50°C. Also, determinethe amount of air that must be bled off to restore pressure toits original value at this temperature. Assume the atmosphericpressure to be 100 kPa.

Answers

Answer:

The pressure rise in the tire when the air temperature in the tire rises to 50°C is 337.43 KPa.

The amount of air that must be bled off to restore pressure 0.007 Kg

Explanation:

Knowing

T1 = 25°C = 298 K

T2 = 50°C = 323 K

volume of the tire = V = 0.025 $m^(3)$

P = 210 kPa (gage)

Pabs = 210 + 101 = 311 KPa

Before the temperature rise

P1 V1 = m1 R1 T1

m1 = $(P1 V1)/(R1 T1) = (310 * 10^(3) * 0.25 )/(287 - 298) = 0.091 Kg\n \n$

After the temperature rise

P2 = $(m2 * R * T2)/(V2) = (0.091 *287*323 )/(0.025) = 337.43 KPa$

after bleeding the pressure and the volume returns to its first value

P1 = P2 and V1 = V2

then

$(m2 * R * T2)/(V) = (m1 * R * T1)/(V)$

m2 = $(m1*T1)/(T2)$

m2 = $(0.091*298)/(332) = 0.084 Kg\n\n$

mbleed = m1 - m2 --> mbleed = 0.91 - 0.84 = 0.007 Kg

P2 = 337.43 KPa

mbleed = 0.007 Kg

The process in which the system pressure remain constant is called a)-isobaric b)-isochoric c)-isolated d)-isothermal

Answers

Answer:

Isobaric process

Explanation:

The process in which the system pressure remain constant is called is called isobaric process. The word "iso"means same and baric means pressure.

At constant pressure, the work done is given by :

$W=p* \Delta V$

Where

W is the work done by the system

p is the constant pressure

$\Delta V$ is the change in volume

So, the correct option is (c) " isobaric process ".

Which of the following expressions causes an implicit conversion between types? Assume variable x is an integer, t is a float, and name is a string.Group of answer choices7.5 + (x / 2)x + 2 * x"Hello, " + str(name)print(str(t))

Answers

x + 2 * x is the correct option. The above-selected option demonstrates implicit conversion, which is an automated type of conversion. Thus, option B is correct.

The series of conversions are necessary to change the type of a function call's argument to that of the parameter with the same name in the function declaration is known as an implicit conversion sequence. For each parameter, the compiler tries to identify an implicit conversion sequence.

If both user-defined conversion sequences A and B contain the same user-defined conversion function or constructor, and if the second standard conversion sequence of A is superior to the second standard conversion sequence of B, then user-defined conversion sequence A is preferable to user-defined conversion sequence B.

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Required information NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. The gas-turbine cycle of a combined gas–steam power plant has a pressure ratio of 8. Air enters the compressor at 290 K and the turbine at 1400 K. The combustion gases leaving the gas turbine are used to heat the steam at 15 MPa to 450°C in a heat exchanger. The combustion gases leave the heat exchanger at 247°C. Steam expands in a high-pressure turbine to a pressure of 3 MPa and is reheated in the combustion chamber to 500°C before it expands in a low-pressure turbine to 10 kPa. The mass flow rate of steam is 17 kg/s. Assume isentropic efficiencies of 100 percent for the pump, 85 percent for the compressor, and 90 percent for the gas and steam turbines.Determine the rate of total heat input.

Answers

Answer:

h1 = 290.16kj/kg

P = 1.2311

Prandil expression at 8

P=p1/p7×pr

=8(1.2311)

=9.85

Enthalpy state at 8 corresponding to 9.85

h1 = 526.13kj/kg

Now prandtl state at 9 that correspond to 1400k.

h9 = 1515.42kj/kg

Pr = 450.5

Prandtl expression at state 10

P= p10/p9×pr

=1/8(450.5)

=56.31

Enthalpy at state 10 corresponding to prandtl 56.31

h10 = 860.39kj/kg

At 520k

h11 = 523.63kj/kg

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