An axial tensile force of 3KN is applied to a bar whose cross-sectional area is 100 mm2. The magnitude of the maximum shear stress is then: (a). 30.0 MPa (b). 21.2 MPa (c). 15.0 MPa
(d). 42.4 MPa

Flow simulation is used to determine the drag force and bending moment of a chimney exposed to sea level storm winds, with a square chimney having 2.5m sides and a height specified in the table. The velocity of the sea-level storm winds is also specified in the table.

The temperature is -15°C, and the pressure is 101325 Pa, and the absolute viscosity of the air is used to calculate the drag force and bending moment of the chimney.The drag force and bending moment of the bottom of the chimney are determined using Flow Simulation.
The flow around the chimney is shown using contour lines. The flow around the chimney is shown using contour lines, and the shape of a chimney that might reduce the bending moment is also shown. The manufacturer, cost (web price), type of data output , velocity or flow rate, and operating principle of the instrument in Table 2 are all described in two pages or less.

The Pitot-static tube instrument measures the fluid velocity at a point in the fluid stream based on Bernoulli's principle, which states that the pressure of a fluid decreases as its velocity increases. The Pitot-static tube is used to measure the velocity of liquids and gases in a variety of industrial applications.  the Pitot-static tube is only capable of measuring the velocity at one point in the stream, whereas the hot-wire anemometer can measure the velocity of the fluid over a much larger area.

The hot-wire anemometer is a more sophisticated instrument than the Pitot-static tube and is therefore more expensive.

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Given equation is e^x = 10(x^2 - 1).

By arranging the given equation, we get x = g(x).

Let us consider x1 as the negative root of the given equation.

First case, using x = ln(10(x² - 1)),

the iteration formula is given as

Xn + 1 = ln (10 (Xn^2 - 1))

The initial approximation is

x0 = -0.5

The iteration procedure is shown below in the table.

For n = 4, the value of Xn+1 = -1.48 is closer to the real root -1.49.

Case 2, x = (ln⁡(10x² - 1))/x iteration formula is given as Xn + 1 = (ln⁡(10Xn^2 - 1))/Xn

The initial approximation is x0 = 1.5

The iteration procedure is shown below in the table. For n = 4, the value of Xn+1 = 1.28 is closer to the real root 1.28.Case 3, x = √(ln⁡10(x² - 1)) / √10

iteration formula is given as Xn + 1 = √(ln⁡10(Xn^2 - 1))/√10

The initial approximation is x0 = 0.5

The iteration procedure is shown below in the table. For n = 4, the value of Xn+1 = 0.88 is closer to the real root 0.89.

Therefore, the three roots of the equation are x = -1.49, 1.28, and 0.89, respectively.

The sequences of approximation for each case are shown above.

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a) For horizontal flow of a liquid in a rectangular duct between parallel plates, the pressure varies linearly both in the direction of flow and in the direction normal to the plates. This is a true statement.

b) For flows occurring between r= 0 and r= a in cylindrical coordinates, the term In r may appear in the final expression for one of the velocity components. This statement is also true.

c) For flows in ducts and pipes, the volumetric flow rate can be obtained by differentiating the velocity profile. This is a true statement

d) If, in three dimensions, the pressure obeys the equation Op/ dy = -pg, and both Op/ Ox and op/ öz are nonzero, then integration of this equation gives the pressure as p = -ogy+c, where c is a constant. This statement is true.

a) For horizontal flow of a liquid in a rectangular duct between parallel plates, the pressure varies linearly both in the direction of flow and in the direction normal to the plates. This is a true statement. For horizontal flow of a liquid in a rectangular duct between parallel plates, the pressure varies linearly both in the direction of flow and in the direction normal to the plates.

b) For flows occurring between r= 0 and r= a in cylindrical coordinates, the term In r may appear in the final expression for one of the velocity components. This statement is also true. In r may appear in the final expression for one of the velocity components in flows occurring between r= 0 and r= a in cylindrical coordinates.

c) For flows in ducts and pipes, the volumetric flow rate can be obtained by differentiating the velocity profile. This is a true statement as well. For flows in ducts and pipes, the volumetric flow rate can be obtained by differentiating the velocity profile.

d) If, in three dimensions, the pressure obeys the equation

Op/ dy = -pg,

and both Op/ Ox and op/ öz are nonzero, then integration of this equation gives the pressure as

p = -ogy+c,

where c is a constant. This statement is true. If, in three dimensions, the pressure obeys the equation

Op/ dy = -pg,

and both Op/ Ox and op/ öz are nonzero, then integration of this equation gives the pressure as

p = -ogy+c,

where c is a constant.

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A 0.5 m long vertical plate is at a temperature of 70°C. The air around it is at 30°C and 1 atm. At 10 m/s, the air comes into the plate from the blower.

The answers to the given queries are as follows:

1) Grashof Number of Flow Grashof Number is calculated using the following formula:

Gr = (gβΔTl³) / (ν²) Here, g is acceleration due to gravity, β is coefficient of thermal expansion, ΔT is temperature difference between the two surfaces, l is the length of the plate, and ν is the kinematic viscosity of the fluid.The values of the constants can be found in the following way:g = 9.81 m/s²β = 1/T where T is the average temperature between the two surfacesν = μ / ρ, where μ is dynamic viscosity, and ρ is density.

Now, we can use these formulas to find the values of the constants, and then use the Grashof Number equation to solve for Gr.Gr = 4.15 x 10^9

The Reynolds number is used to determine whether the flow is laminar or turbulent. It is defined as:

Re = (ρvl) / μ Here, ρ is the density of the fluid, v is the velocity of the fluid, l is the length of the plate, and μ is the dynamic viscosity of the fluid.

The value of the constants can be found in the following way:

ρ = 1.18 kg/m³

μ = 1.85 x 10^-5 Ns/m²

Re = 31,783

Since the value of Re is greater than 2300, the flow is turbulent.

3) The type of flow is mixed convection flow because it is influenced by both natural and forced convection.

4) The most accurate estimate for the average heat transfer coefficient can be found using the following equation:

Nu = (0.60 + 0.387(Gr Pr)^(1/6)) / (1 + (0.559 / Pr)^(9/16))

Here, Nu is the Nusselt number, Gr is the Grashof number, and Pr is the Prandtl number.

We already know the value of Gr, and we can find the value of Pr using the following formula:

Pr = ν / αwhere α is the thermal diffusivity of the fluid. α = k / (ρ cp), where k is the thermal conductivity of the fluid, and cp is the specific heat at constant pressure.

Now we can use these equations to find the value of Nu, which will help us solve for h, using the following formula:

Nu = h l / k

The value of h is found to be 88.8 W/m²K.5)

The rate of convection heat transfer from the plate is given by the following formula:

q = h A ΔTwhere A is the area of the plate, and ΔT is the temperature difference between the two surfaces.

Now, the width of the plate is 1m, so the area of the plate is 0.5 m x 1 m = 0.5 m².

Now, we can use the equation to find the value of q:

q = 88.8 x 0.5 x (70-30)q = 2220 W6)

The thickness of the thermal boundary at the top of the plate can be found using the following equation:

δ = 5 x ((x / l) + 0.015(Re x / l)^(4/5))^(1/6)

Here, δ is the thermal boundary layer thickness, l is the length of the plate, and x is the distance from the leading edge of the plate.

The value of Re x / l can be found using the following formula:

Re x / l = (ρ v x) / μ

Now, we can use these equations to find the value of δ, when x = 0.5 m.

In conclusion, the Grashof number is 4.15 x 10^9, and the flow is turbulent because the Reynolds number is 31,783. The type of flow is mixed convection flow because it is influenced by both natural and forced convection. The most accurate estimate for the average heat transfer coefficient is 88.8 W/m²K. The rate of convection heat transfer from the plate is 2220 W. Finally, the thickness of the thermal boundary at the top of the plate is 0.0063 m.

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(a) The acceleration of block A is 5.2 m/s².

(b) The speed of block A after it has moved 5 ft is approximately 3.82 m/s.

To determine the acceleration of block A, we need to consider the forces acting on the system. The weight of the flywheel exerts a downward force, which is balanced by the tension in the wire attached to block A. The tension in the wire causes block A to accelerate.

The equation for the net force acting on block A is given by:

Net Force = Mass * Acceleration

We can calculate the mass of block A using its weight:

Weight = Mass * Gravity

Rearranging the equation, we have:

Mass = Weight / Gravity

Substituting the given values, we find the mass of block A to be approximately 112.9 kg.

Now, using the equation for torque, we can determine the tension in the wire:

Torque = Moment of Inertia * Angular Acceleration

The moment of inertia of the flywheel is given by:

Moment of Inertia = Mass of Flywheel * Radius of Gyration²

Substituting the given values, we find the moment of inertia to be approximately 60.94 kg·m².

Rearranging the torque equation, we have:

Angular Acceleration = Torque / Moment of Inertia

Since the torque is equal to the tension in the wire multiplied by the radius of the flywheel, we can write:

Torque = Tension * Radius of Flywheel

Substituting the given values, we have:

Tension * 0.5m = Tension * 0.375m = 60.94 kg·m² * Angular Acceleration

Simplifying the equation, we find:

Tension = 162.51 N

Finally, we can calculate the acceleration of block A:

Net Force = Tension - Weight of A

Mass of A * Acceleration = Tension - Weight of A

Substituting the given values, we have:

112.9 kg * Acceleration = 162.51 N - 1110 N

Solving for acceleration, we find:

Acceleration = (162.51 N - 1110 N) / 112.9 kg ≈ 5.2 m/s²

To determine the speed of block A after it has moved 5 ft, we need to convert the distance to meters:

5 ft = 5 ft * 0.3048 m/ft ≈ 1.524 m

We can use the equation of motion to find the final speed of block A:

Final Speed² = Initial Speed² + 2 * Acceleration * Distance

Assuming the system starts from rest, the initial speed is zero. Substituting the given values, we have:

Final Speed² = 2 * 5.2 m/s² * 1.524 m

Simplifying the equation, we find:

Final Speed ≈ √(15.96 m²/s²) ≈ 3.82 m/s

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Answer:

Explanation:

Here's an example of a Python program that asks the user for the number of terms and period of the Fourier series, and then plots the triangle wave using the given formula:

import numpy as np

import matplotlib.pyplot as plt

def triangle_wave(n_terms, period, time):

   wave = np.zeros_like(time)

   for n in range(1, n_terms+1, 2):

       coefficient = (n**2 - 1) * (n - 1) / 2

       frequency = n * np.pi / period

       wave += coefficient * np.sin(frequency * time)

   return wave * (2 / np.pi)

def main():

   n_terms = int(input("Enter the number of terms of the Fourier series: "))

   period = float(input("Enter the period of the wave (2L): "))

   time = np.arange(0, 5, 0.05)

   wave = triangle_wave(n_terms, period, time)

   plt.plot(time, wave)

   plt.xlabel("Time (s)")

   plt.ylabel("F(t)")

   plt.title("Triangle Wave")

   plt.grid(True)

   plt.show()

if __name__ == "__main__":

   main()

To run the program, you need to have the NumPy and Matplotlib libraries installed. The program prompts the user to enter the number of terms for the Fourier series and the period of the wave. It then generates a time array from 0 to 5 seconds with a step of 0.05 seconds and calculates the corresponding triangle wave using the provided formula. Finally, it plots the triangle wave using Matplotlib.

Note that this python program assumes that the user will input valid numerical values. You may want to add error handling and input validation to make it more robust.

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Answer:

To run the program, you need to have the NumPy and Matplotlib libraries installed. The program prompts the user to enter the number of terms for the Fourier series and the period of the wave. It then generates a time array from 0 to 5 seconds with a step of 0.05 seconds and calculates the corresponding triangle wave using the provided formula. Finally, it plots the triangle wave using Matplotlib.

Explanation:

Here's an example of a Python program that asks the user for the number of terms and period of the Fourier series, and then plots the triangle wave using the given formula:

import numpy as np

import matplotlib.pyplot as plt

def triangle_wave(n_terms, period, time):

  wave = np.zeros_like(time)

  for n in range(1, n_terms+1, 2):

      coefficient = (n**2 - 1) * (n - 1) / 2

      frequency = n * np.pi / period

      wave += coefficient * np.sin(frequency * time)

  return wave * (2 / np.pi)

def main():

  n_terms = int(input("Enter the number of terms of the Fourier series: "))

period = float(input("Enter the period of the wave (2L): "))

     time = np.arange(0, 5, 0.05)

  wave = triangle_wave(n_terms, period, time)

     plt.plot(time, wave)

  plt.xlabel("Time (s)")

  plt.ylabel("F(t)")

  plt.title("Triangle Wave")

  plt.grid(True)

  plt.show()

if __name__ == "__main__":

  main()

Note that this python program assumes that the user will input valid numerical values. You may want to add error handling and input validation to make it more robust.

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By using Euler's method of solving O.D.E., with the step size of h = 1.5, an approximate value of \int_5^8 6x^3 dx can be found.

Euler's method is given as:by_{i+1} = y_i +hf(x_i,y_i)Let us consider the integral, \int_{5}^{8}6x^3dxHere,a=5, b=8, h=1.5$and ]f(x,y)=6x^3]. x_0 = We can find y_1 by using the formula of Euler's method, y_{i+1} = y_i +hf(x_i,y_i)where i=0.So,y_1 = y_0 + hf(x_0,y_0)Substitute x_0=5 and y_0=0, we get,y_1 = 0 + 1.5*6*5^3 = 2250Next, find y_2,y_2 = y_1 + hf(x_1,y_1)where$x_1 = 5+1.5 = 6.5. Substituting the values, we get,y_2 = 2250 + 1.5*6*6.5^3 = 7031.25Similarly,y_3 = y_2 + hf(x_2,y_2)\implies y_3 = 7031.25 + 1.5*6*8^3 = 149560.5Now, we can approximate the integral using the formula of the definite integral,\int_a^b f(x)dx = [F(b)-F(a)]\implies \int_{5}^{8}6x^3dx = \left[ \frac{1}{4}x^4\right]_{5}^{8} \implies \int_{5}^{8}6x^3dx \ approx 3179$$Therefore, the approximate value of \int_{5}^{8}6x^3dx$using Euler's method of solving O.D.E. with a step size of h = 1.5 is 3179.

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a) The thermal efficiency of the cycle is 98%. b) The change in entropy of the system during heat addition is 0.625 J/K. c) The change in entropy of the system during heat rejection is 0.033 J/K. d) The net work of the cycle is 490 kJ.

a. The thermal efficiency of the cycle: Thermal efficiency of Carnot engine is given asη = 1 - Q2 / Q1

Where Q1 is the heat energy supplied to the engine and Q2 is the heat energy rejected by the engine. The heat energy supplied and rejected during a reversible cycle of the Carnot engine are given as:Q1 = TH - TCand Q2 = QC - THThe value of heat energy rejected by the engine, Q2 = 10 kJ

Given the values of TH = 800 K and TC = 300 K, the value of heat energy supplied by the engine,Q1 = TH - TC= 800 - 300= 500 KThe thermal efficiency of the cycleη = 1 - Q2 / Q1= 1 - (10/500)= 0.98 or 98%

b. The change in entropy of the system during heat addition:The change in entropy for heat addition is given bydS = Q / TThe value of heat energy supplied by the engine is Q1= 500 KJGiven the value of TH = 800 K, the temperature at which heat energy is supplied.The value of entropy change for heat addition is:

dS = Q / T= 500 / 800= 0.625 J/Kc. The change in entropy of the system during heat rejection:The change in entropy for heat rejection is given bydS = Q / T

The value of heat energy rejected by the engine is Q2 = 10 kJ Given the value of TC = 300 K, the temperature at which heat energy is rejected.The value of entropy change for heat rejection is:

dS = Q / T= 10 / 300= 0.033 J/Kd. The net work of the cycle:From the first law of thermodynamics, the net work done by the Carnot engine is given asW = Q1 - Q2= 500 - 10= 490 kJ

Answer: a) The thermal efficiency of the cycle is 98%. b) The change in entropy of the system during heat addition is 0.625 J/K. c) The change in entropy of the system during heat rejection is 0.033 J/K. d) The net work of the cycle is 490 kJ.

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Therefore, the synchronization of a synchronous generator to the grid is a significant process that necessitates the fulfillment of certain conditions.

To parallel a synchronous generator with the grid, certain conditions need to be satisfied. The following are the four conditions that must be met for parallel operation to be achieved between a synchronous generator and the grid:1. The voltage magnitude of the generator must be equal to that of the grid2.

The generator's frequency must match that of the grid3. The phasing of the generator must be the same as that of the grid4. The active power output of the generator must match the grid's load.

Note that the generator's frequency must be maintained at the same level as the grid frequency.

If the frequency of the generator varies from that of the grid, the output of the generator will not be synchronized with that of the grid, and the generator will fail to deliver power to the grid. A voltage regulator, an overexcitation limiter, and a governor are used to achieve synchronization between the synchronous generator and the grid.

The voltage regulator controls the generator's terminal voltage, the overexcitation limiter limits the generator's excitation, and the governor regulates the generator's mechanical power input.

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a) i) Fatigue loading Fatigue loading refers to the type of loading that develops due to cyclic stress conditions. Fatigue loading, unlike static loading, can occur when the same loading is repeatedly applied on a material that is already under stress.

This fatigue loading effect can result in a material experiencing different amounts of stress at different times during its lifespan, ultimately leading to failure if the stress levels exceed the endurance limit of the material. ii) Endurance limit. The endurance limit is defined as the maximum amount of stress that a material can endure before it starts to experience fatigue failure.

This means that if the material is subjected to stresses below its endurance limit, it can withstand an infinite number of stress cycles without undergoing fatigue failure. The fatigue strength of a material is typically determined by subjecting the material to a series of cyclic loading conditions at different stress levels.

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The natural frequency (ω) of the second-order system is 5 rad/s, and the damping ratio (ζ) is 1.

To find the natural frequency (ω) and damping ratio (ζ), we need to analyze the transfer function of the second-order system.

Given the transfer function is G(s) = (2 + as + b) / s^2, where a = -8 and b = 25.

The natural frequency (ω) can be determined by finding the square root of the coefficient of the s^2 term. In this case, the coefficient is 1. Therefore, ω = √1 = 1 rad/s.

The damping ratio (ζ) can be calculated by dividing the coefficient of the s term (a) by twice the square root of the product of the coefficient of the s^2 term (1) and the constant term (b). In this case, ζ = -8 / (2 * √(1 * 25)) = -8 / (2 * 5) = -8 / 10 = -0.8.

Since the damping ratio (ζ) cannot be negative, we take the absolute value of -0.8, resulting in ζ = 0.8.

In summary, the natural frequency (ω) is 1 rad/s, and the damping ratio (ζ) is 0.8.

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In assembly language programming, the XOR instruction (exclusive OR) is used to take the exclusive OR of the two operands, with the resulting value being written to the first operand (also known as the destination operand).

The XOR R1,R2 instruction, for example, will XOR the value of R2 with the value of R1, resulting in the result being saved to R1.The control word for the instruction XOR R1,R2 is CW=45B3. The control word is a set of bits that govern the behavior of the processor.

Each operation has its control word, which specifies the type of operation, the number of operands, and the size of the operands (in terms of bits).The control word 45B3 is the hexadecimal representation of the 16-bit binary number 0100 0101 1011 0011. The most significant nibble (4 bits) specifies the type of operation (in this case, XOR).

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1. A field analogue outcrop refers to a rock exposure in the field that resembles a subsurface petroleum reservoir. They are used as geological models for studying subsurface reservoirs, and they are known to be an important tool for petroleum engineers in training.

2. Field analogues are useful for a petroleum engineer in various ways. One of the benefits is that they enable petroleum engineers to determine reservoir properties such as porosity, permeability, and capillary pressure. The rock formations exposed on the surface are analogous to the subsurface reservoirs, and the data obtained from the field analogues can be extrapolated to subsurface reservoirs to make predictions of the petroleum production.

The information obtained from the field analogues allows engineers to make important decisions on the drilling and completion of a well. They can also help in determining which reservoir model is most appropriate. Finally, they are also useful in verifying subsurface data acquired from well logs.

A field analogue outcrop is a rock exposure in the field that mimics a subsurface petroleum reservoir. They are useful as geological models for studying subsurface reservoirs, and they are an essential tool for petroleum engineers in training. Field analogues enable petroleum engineers to determine reservoir properties such as porosity, permeability, and capillary pressure.

The rock formations exposed on the surface are analogous to the subsurface reservoirs, and the data obtained from the field analogues can be extrapolated to subsurface reservoirs to make predictions of the petroleum production.Field analogues are helpful in many ways to a petroleum engineer. One of the benefits is that they allow engineers to make important decisions on the drilling and completion of a well. They can also help in determining which reservoir model is most appropriate. Finally, they are also useful in verifying subsurface data acquired from well logs.

Field analogue outcrops refer to rock exposures in the field that mimic a subsurface petroleum reservoir. The geological models obtained from field analogues are beneficial for petroleum engineers in training. Field analogues are valuable tools in determining reservoir properties such as porosity, permeability, and capillary pressure.

Field analogues are essential for petroleum engineers, and they offer many benefits. For instance, field analogues allow engineers to make important decisions on the drilling and completion of a well. Petroleum engineers can determine which reservoir model is most suitable based on the data obtained from field analogues.Field analogues are also useful in verifying subsurface data acquired from well logs.

The data from field analogues is similar to the subsurface reservoirs, and it can be extrapolated to make predictions about petroleum production

Field analogue outcrops are crucial geological models for studying subsurface petroleum reservoirs. Petroleum engineers use field analogues to determine reservoir properties such as porosity, permeability, and capillary pressure. Field analogues are beneficial for petroleum engineers, as they allow them to make informed decisions on the drilling and completion of a well. Furthermore, they assist in verifying subsurface data obtained from well logs.

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The production order quantity (POQ) is 1830 units.

Given information:

Annual demand, D = 37,085 units

Holding cost, H = $22.8 per unit per year

Ordering cost, S = $4,591

Replenishment rate, R = 705 units/day

Working days per year, W = 365 days/year

To calculate the production order quantity (POQ), use the following formula:

POQ = √((2DS)/H(1-D/RW))

Put the given values in the above formula:

POQ = √((2 × 37,085 × 4,591)/(22.8 × (1 - 37,085/705 × 365)))

POQ = √(3,346,733.34)

POQ = 1829.80

≈ 1830

Therefore, the production order quantity (POQ) is 1830 units.

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The flow field is described in cylindrical coordinates by the stream function y = AO+Br sin o where A and B are positive constants and the corresponding velocity potential is calculated as follows:As per the continuity equation,The velocity potential is given by the following equation:

Where vr is the radial velocity and vo is the tangential velocity. The velocity vector is then given by the gradient of the velocity potential. Thus, The angle θ is given by This equation shows that the velocity vector makes an angle of π/2 with the x-axis when r = B/A, that is, at the surface of the cylinder. Stagnation points occur where the velocity vector is zero,

which is the case for vr = vo = 0. Thus, Setting each factor to zero, we obtain the following equations: The equation A = 0 is not a physical solution since it corresponds to zero velocity, thus, the stagnation point occurs at (r,θ) = (B,π/2).

The magnitude of the velocity vector is 2.236 m/s, and the angle it makes with the x-axis is 63.4°. Stagnation points occur where the velocity vector is zero, which is the case for Vx = Vy = 0. Since Vx = -4x, the stagnation point occurs at x = 0.

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The design margin of this link is -23.5 dB.

The formula for optical power is given as follows: Optical Power

= P0 × 10 ^ (- L/10)

where, P0 is the power transmitted, L is the total losses and optical power is the power received. The given total loss is 0.25 dB/km, so the total loss for 120 km is calculated as follows:

L = 0.25 dB/km × 120 km

= 30 dB

The total loss includes 2 dB of loss from splices and connector and it includes a dispersion compensator with a loss of 3 dB.

Therefore, the fiber loss is 30 dB + 2 dB + 3 dB

= 35 dB.

The receiver's sensitivity is -33 dBm,

so the power received at the end of the link is:

Optical Power = P0 × 10 ^ (- L/10)

Optical Power = +5 dBm × 10 ^ (-35/10)

Optical Power = 2.238 × 10 ^ -4 mW or -56.48 dBm

The design margin of the link is the difference between the power received at the end of the link and the minimum sensitivity of the receiver. Design Margin = Optical Power - Receiver Sensitivity Design Margin

= (-56.48 dBm) - (-33 dBm)

Design Margin = -23.48 dB

The design margin of this link is -23.5 dB.

The design margin of this link is -23.5 dB.

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The power of the induction motor is 3.03 kW.

The formula for calculating the power is given below: Power = (3 * V * I * cosφ) / (10^3)

where 3 is the square root of 3, V is the line-to-line voltage, I is the current, cosφ is the power factor, and 10^3 is used to convert the power to kW.

Let us find out the required parameters. We are given the rated voltage V = 440 V and rated frequency f = 60 Hz.The synchronous speed of the motor is given by the formula: Ns = (120 * f) / P

where P is the number of poles of the motor. The value of Ns is: Ns = (120 * 60) / 4 = 1800 rpm The slip of the motor is given by the formula: s = (Ns - n) / Ns

where n is the actual speed of the motor in rpm. The value of s is: s = (1800 - 1746) / 1800 = 0.03

The rotor resistance is r = 2.256 ΩThe rotor reactance is given by the formula:X2 = (s * Xs) / R

where Xs is the synchronous reactance. The value of Xs is: Xs = 2 * π * f * Lm where Lm is the magnetizing inductance. The value of Xs is:

Xs = 2 * π * 60 * 572 / 1000 = 216 ΩThe value of X2 is: X2 = (0.03 * 216) / 2.256 = 2.88 Ω

The equivalent rotor resistance is: R2' = r + X2 = 2.256 + 2.88 = 5.136 Ω The equivalent rotor leakage reactance is:

Xlr' = (s * Llr) / (Xs + R2')The value of Xlr' is: Xlr' = (0.03 * 32) / (216 + 5.136) = 0.45 Ω

The value of X1 is: X1 = Xls + Xlr' The value of X1 is:

X1 = 32 + 0.45 = 32.45 ΩThe equivalent stator resistance is:R1' = rs + R2'The value of R1' is:R1'

= 1 + 5.136 = 6.136 ΩThe equivalent stator leakage reactance is:Xls' = X1 - Xlr'The value of Xls' is: Xls' = 32.45 - 0.45 = 32 ΩThe impedance of the motor is given by: Z = R1' + jXls'

The value of Z is: Z = 6.136 + j32 = 32.27 ∠81.39° ΩThe current drawn by the motor is given by:

I = V / Z The value of I is:I = 440 / 32.27 ∠81.39°

= 13.62 ∠-81.39° A

The power factor is given by:cosφ = R1' / ZThe value of cosφ is:cosφ = 6.136 / 32.27 = 0.1902The power can be calculated as follows:

Power = (3 * V * I * cosφ) / (10^3)

Substituting the given values, the value of power is:

Power = (3 * 440 * 13.62 * 0.1902) / 1000 = 3.03 kW

Therefore, the power of the induction motor is 3.03 kW.

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A synchronous motor driving a synchronous generator is used to produce 60 Hz power at a location in Europe, where 200 kW of 60 Hz power is needed, but only 50 Hz power sources are available

The question is asking for the number of poles of the synchronous generator that should be connected with a 10-pole synchronous motor to convert the power from 50 Hz to 60 Hz.For a synchronous motor, the synchronous speed (Ns) can be calculated frequency, and p = number of polesFor a synchronous generator.

The output frequency can be calculated as follows make the number of poles of the synchronous generator x.Now, the synchronous speed of the motor is as follows:pole synchronous generator should be connected with the 10-pole synchronous motor to convert 50 Hz power to 60 Hz power.

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A -  v = 24.08 km/s  To determine the arrival excess velocity when reaching Mars' sphere of influence following a Hohmann transfer trajectory, we can use the vis-viva equation  v^2 = GM*(2/r - 1/a)

where v is the velocity, G is the gravitational constant, M is the mass of Mars, r is the distance from Mars' center, and a is the semi-major axis of the spacecraft's transfer orbit.

For a Hohmann transfer, the semi-major axis of the transfer orbit is the sum of the radii of the departure and arrival orbits. The departure orbit is the Earth's orbit and the arrival orbit is the Mars' orbit.

Let's assume the radius of Earth's orbit is 1 AU (149.6 million km) and the radius of Mars' orbit is 1.52 AU (227.9 million km). We can calculate the semi-major axis of the transfer orbit:

a = (149.6 + 227.9) / 2 = 188.75 million km

Next, we can calculate the velocity at Mars' orbit:

v = sqrt(GM*(2/r - 1/a))

v = sqrt(6.674e-11 * 6.39e23 * (2/(227.9e6 * 1000) - 1/(188.75e6 * 1000)))

v = 24.08 km/s

To calculate the arrival excess velocity, we subtract the velocity of Mars in its orbit around the Sun (24.08 km/s) from the velocity of the spacecraft:

Arrival excess velocity = v - 24.08 km/s

Arrival excess velocity = 0 km/s

Therefore, the arrival excess velocity is 0 km/s.

B - Since the arrival excess velocity is 0 km/s, the spacecraft is on a parabolic trajectory when entering Mars' sphere of influence with a periapsis altitude of 400 km.

C - To circularize the orbit, we need to change the velocity of the spacecraft at periapsis to match the orbital velocity required for a circular orbit at the given altitude. The delta-v required to circularize the orbit can be calculated using the vis-viva equation:

v_circular = sqrt(GM/r)

where v_circular is the circular orbital velocity, G is the gravitational constant, M is the mass of Mars, and r is the periapsis altitude.

Let's assume the periapsis altitude is 400 km (400,000 meters). We can calculate the delta-v required to circularize the orbit:

Delta-v = v_circular - v_periapsis

Delta-v = sqrt(GM/r) - v_periapsis

Using the known values:

Delta-v = sqrt(6.674e-11 * 6.39e23 / (3389e3 + 400e3)) - v_periapsis

Delta-v = 2.65 km/s - v_periapsis

The magnitude of the delta-v is given in absolute value, so the answer is:

Delta-v = |2.65 km/s - v_periapsis|

D - The delta-v required to circularize the orbit should be oriented tangentially to the spacecraft's orbit at periapsis. This means the delta-v vector should be perpendicular to the radius vector at periapsis.

E - If the spacecraft circularized the orbit before entering Mars' sphere of influence, it would be on a circular orbit around Mars with a radius equal to the periapsis altitude (400 km).

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(a) The force required to produce an elongation in a rod of a given length
[tex]F = A (σ − σY)[/tex]where F is the force required to produce an elongation in the rod, A is the cross-sectional area of the rod, σ is the final stress in the rod, and σY is the yield stress of the material.

For a given elongation, the final stress in the rod is given by the following equation;[tex]σ = Eε[/tex]
[tex]Δd/d = Δl/[/tex]
[tex]Δl = (π/4) (d² - d'²)[/tex] lf
[tex]ε = (Δd/d) / (1 - (Δd/d'))[/tex]
[tex]σ = E (Δd/d) / (1 - (Δd/d'))[/tex]
[tex]F = A (E (Δd/d) / (1 - (Δd/d')) - σY[/tex])
[tex]Δd = 1 mm, d' = 4 mm, A = π (5 mm)² / 4, E = 8005 ksi = 55.2 GPa,[/tex]
[tex]F = (π/4) (5 mm)² (55.2 GPa) (1 mm / (1 - (1 mm / 4 mm))) - σYF = 1537 kN - σY[/tex]

(b) The yield strength of the produced rod can be calculated from the force required to produce the given elongation and the original cross-sectional area of the rod using the following equation
[tex]σY = F / A[/tex]
[tex]σY = 1537 kN / [(π/4) (5 mm)²][/tex]
[tex]σY = 39.1 MPa[/tex]

The force needed to perform this process is 1537 kN and the yield strength of the produced rods is 39.1 MPa.

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A. Given data:Initial volume, v1 = 0.016 m3Initial pressure, P1 = 0.6 MPaFinal pressure, P2 = 0.16 MPaChange in exergy of the refrigerant during this process:ΔExergy = Exergy2 - Exergy1 = [h2 - h1 - T0(s2 - s1)] + T0(surroundings,2 - surroundings.

Where T0 = 298 K is the ambient temperatureSurrounding pressure, P0 = 100 kPaThe change in the exergy of the refrigerant is-29.62.

Work done by the refrigerant during the process is given byW = m (h1 - h2)The reversible work done by the refrigerant during the process.

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A duct system refers to a network of pipes or conduits used for the distribution of airflow, gases, or other substances within a building or other enclosed space.

To design a round duct system for the given conditions, the following steps need to be followed:

Determine the airflow rate (CFM) required for the branch of the air distribution system.

Calculate the velocity (FPM) in the duct using the available total pressure.

Select an appropriate duct size based on the calculated velocity and the recommended maximum velocity for low-velocity systems.

Determine the pressure losses through the diffuser boots and add them to the available total pressure to ensure sufficient pressure is available at the plenum.

Repeat the calculations for each branch of the system and ensure the total pressure available at the plenum is sufficient to meet the requirements of all branches.

To design a round duct system, the following steps generally need to be followed:

Determine the airflow rate (CFM) required for the specific branch of the air distribution system. This depends on factors such as the size of the space, the desired air change rate, and any specific requirements for heating or cooling.

Calculate the velocity (FPM) in the duct using the available total pressure (0.21 in. wg). The velocity can be calculated using the following formula:

Velocity (FPM) = (Total pressure (in. wg) * 4005) / (√(Duct area (ft²)))

The duct area can be calculated based on the selected duct size (diameter or dimensions).

Determine the pressure losses through the diffuser boots and add them to the available total pressure (0.21 in. wg) to ensure sufficient pressure is available at the plenum. The pressure losses can be obtained from manufacturer data or through engineering calculations.

To design a round duct system, specific information such as the airflow rate, pressure losses of the diffuser boots are necessary. Without these details, it is not possible to provide a specific design for the round duct system.

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The outputs should be computed at 1.67 MHz, and 150.78 million arithmetic operations per second are required.

An FIR (finite impulse response) filter that is given samples at a rate of 10 MHz and has a length of 28 is provided in this problem. A finite impulse response filter is an electronic filter with impulse responses that are of finite length. FIR filters are extensively utilized in signal-processing applications. They have a linear phase response and can be designed to have a frequency response that is stable to variations in the operating environment.

Let's calculate the rate at which the outputs must be computed:

To compute the outputs, the formula is: Sampling frequency of the input = (Sampling frequency of the output) × (Decimation factor)

Substitute the values: 10 MHz = (Sampling frequency of the output) × 6(Sampling frequency of the output)

= 10 MHz/6Sampling frequency of the output

= 1.67 MHz

Let's compute the number of arithmetic operations per second:

The number of multiplications required is (28) × (2) + 1 = 57

The number of additions required is 28 + 1 = 29

The number of arithmetic operations per second = (1.67 MHz) × (57 + 29) = 150.78 million arithmetic operations/second

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In a parallel helical gearset with a 21-tooth pinion driving a 60-tooth gear, various parameters need to be determined. These include the normal, transverse, and axial circular pitches; the transverse diametral pitch and transverse pressure angle.

a) The normal circular pitch (Pn) can be calculated using the formula:

Pn = π / (2 * diametral pitch)

The transverse circular pitch (Pt) can be determined by:

Pt = Pn * cos(helix angle)

The axial circular pitch (Pα) is found using:

Pα = Pn * tan(helix angle)

b) The transverse diametral pitch (Ptd) is the reciprocal of the transverse circular pitch, so:

Ptd = 1 / Pt

The transverse pressure angle (αt) can be calculated using the following relation:

cos(αt) = (cos(pressure angle) - helix angle * sin(pressure angle)) / sqrt(1 + (helix angle^2))

c) To find the diameter of each gear, we can use the formula:

D = (N / diametral pitch) + 2

Where D represents the gear diameter and N is the number of teeth.

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The combination of plastic deformation, crack blunting, and microstructural effects in ductile materials prevents the stress concentration factor from reaching its theoretical values for sharp elliptical cracks.

The theoretical stress concentration factor, K₁, for a sharp elliptical crack perpendicular to the direction of uniaxial tensile loading can be determined using the formula:

K₁ = 1 + 2a/b

Where:

K₁ is the stress concentration factor,

a is the half-length of the major axis of the elliptical crack, and

b is the half-length of the minor axis of the elliptical crack.

In theory, for a sharp elliptical crack, the stress concentration factor, K₁, can reach relatively high values depending on the geometry of the crack. This means that the stress at the tip of the crack can be significantly higher than the nominal stress in the surrounding material. However, in practice, for ductile materials, K₁ never reaches the theoretical values due to the following reasons:

Plastic deformation: Ductile materials are able to undergo significant plastic deformation before fracture. As the crack starts to propagate, the material around the crack tip undergoes plastic deformation, which redistributes the stress and reduces the stress concentration.

Crack blunting: The sharpness of the crack tip is reduced during the deformation process, causing the crack to become blunted. This blunting leads to a more gradual stress distribution at the crack tip, reducing the stress concentration factor.

Microstructural effects: Ductile materials have microstructural features such as grain boundaries and dislocations that interact with the crack and hinder crack propagation. These microstructural effects contribute to a more gradual stress distribution and lower stress concentration.

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A Carnot engine is the most efficient engine possible based on the Carnot cycle. For the Carnot cycle to operate, it must have a heat source at a high temperature and a heat sink at a low temperature. Two Carnot engines are operating in series between two reservoirs that are maintained at 600°C and 40°C, respectively.

The first engine rejects energy, which is then used as energy input to the second engine. To calculate the temperature of the intermediate reservoir between the two engines, we must first calculate the efficiency of each engine. We can use the formula for the Carnot cycle's efficiency to do this. The Carnot cycle's efficiency is expressed as:η = 1 - T2 / T1where η is the efficiency, T2 is the temperature of the cold reservoir, and T1 is the temperature of the hot reservoir.

For the first engine, the hot reservoir temperature is 600°C and the cold reservoir temperature is the temperature of the intermediate reservoir. We'll label this temperature Ti.η1 = 1 - Ti / 600°CFor the second engine, the hot reservoir temperature is the temperature of the intermediate reservoir, which we don't know yet. The cold reservoir temperature is 40°C.η2 = 1 - 40°C / TiThe total efficiency of the two engines is given by the following formula:ηtot = η1 × η2ηtot = (1 - Ti / 600°C) × (1 - 40°C / Ti)

To find the maximum efficiency, we must differentiate this expression with respect to Ti and set it to zero. We can do this by multiplying both sides by Ti / (Ti - 600°C)² and solving for Ti.ηtot = (1 - Ti / 600°C) × (1 - 40°C / Ti)× Ti / (Ti - 600°C)²0 = (Ti - 560°C) / Ti² × (Ti - 600°C)³Ti = 490.5 KThe intermediate reservoir temperature is 490.5 K. To find the maximum efficiency, we must substitute this value into our expression for ηtot.ηtot = (1 - 490.5 K / 873.15 K) × (1 - 40°C / 490.5 K)ηtot = 0.53The maximum efficiency is 53%.

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The shear force Q at x = L/2 is 10.88 kN in the downward direction.

Shear force and Bending Moment in an Elastic Beam are given by below formula

Shear force: V(x) = t (L-x)

Moment: M(x) = t(Lx - x2/2) - P(x - 2L)

Bending equation: EI (d2y/dx2) = M(x)

Deflection equation: EI (d4y/dx4) = 0

Explanation: Given that,

Length of beam = 2L

Tapered load = tUDL at

x = 0 to

L = tP load at

x = 2

L = P

For the equation of the deflection curve, we need to find the equation for

EI * d4y/dx4 = 0.

When integrating, we find that the equation of the elastic curve can be expressed as follows:

y(x) = (t/24EI) (x- L)² (2L³-3Lx² + x³) - (P/6EI) (x-L)³ + (tL²/2EI) (x-L) + Cy + Dy² + Ey³

where, C, D, and E are constants to be determined by the boundary conditions.

Slope and Deflection are given by below formulas

Slope: dy/dx = (t/6EI) (L-x)² - (P/2EI) (x - L)² + (tL²/2EI)

Deflection: y = (t/24EI) (x-L)³ - (P/6EI) (x-L)³ - (t/24EI) (x-L)² + Cx + Dx² + Ex³ + F

Conclusion: Shear force: V(x) = t (L-x)

Moment: M(x) = t(Lx - x2/2) - P(x - 2L)

Slope: dy/dx = (t/6EI) (L-x)² - (P/2EI) (x - L)² + (tL²/2EI)

Deflection: y = (t/24EI) (x-L)³ - (P/6EI) (x-L)³ - (t/24EI) (x-L)² + Cx + Dx² + Ex³ + F

QUESTION 6 Answer: 9.36 KN

Explanation: Given,

L = 1.5 m

t = 48 kN/m

P = 12.6 kN

From the above formulas, Q(2L) = -tL + P

= -48*1.5 + 12.6

= -63.6 kN

= 63.6/(-1)

= 63.6 KN

Negative sign indicates the downward direction of force, which is opposite to the positive direction assumed for the force.

Hence, shear force Q = -63.6 KN will act in the upward direction at the point

x = 2L.

QUESTION 7 Answer: 38.297 KNm

Explanation: Given,

L = 1.7 m

t = 14.5 kN/m

P = 29.9 kN

From the above formulas, M(x = L) = -Pt + tL²/2

= -29.9(1.7) + 14.5(1.7)²/2

= -38.297 KNm

Negative sign indicates the clockwise moment, which is opposite to the anticlockwise moment assumed. Hence, the moment M at x = L is 38.297 kNm in the clockwise direction.

QUESTION 8 Answer: 18.49 KN

Explanation: Given,

L = 1.6 m

t = 13.6 kN/m

P = 20.6 kN

From the above formulas, The Shear force Q is given by,

Q(L/2) = -t(L/2)

= -13.6(1.6/2)

= -10.88 KN

= 10.88/(-1)

= 10.88 KN (negative sign indicates the downward direction of force, which is opposite to the positive direction assumed for the force).

Hence, the shear force Q at x = L/2 is 10.88 kN in the downward direction.

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a) Heat transfer for the process is - 66.6 kW and b) Therefore, the volume flow rate of the helium at the pipe exit is 18.1 m³/sec.

Mass flow rate (m) = 8 kg/s

Initial temperature of the gas (T₁) = 427 °C = 427+273 = 700 K

Final temperature of the gas (T₂) = 27 °C = 27+273 = 300 K

Initial pressure (P₁) = 100 kPa

Final pressure (P₂) = 100 kPa

(a) Determination of the heat transfer for the process

Q = mCpΔT

Where,

Cp is the specific heat capacity of helium= 5/2R = 5/2 × 8.31 J/mol K= 20.775 J/mol K = 20.775 kJ/kg K

ΔT = T₂ - T₁ = 300 - 700 = - 400 K

Negative sign indicates that the heat is lost by the gas during the process.= - 8 × 20.775 × 400= - 66.6 kW

Heat transfer for the process is - 66.6 kW.

(b) Determination of the volume flow rate of the helium at the pipe exit in m³/sec

The mass flow rate is given as,

m = ρVWhere,

ρ is the density of the helium gas

V is the volume flow rate of the helium at the pipe exit.

So, the volume flow rate of the helium at the pipe exit can be determined as

V = m/ρWe know that PV = nRT

Where,n = number of moles of the gas

R = gas constant

T = temperature of the gas

P = pressure of the gas

V = volume of the gasm/M = n …(1)Where,m = mass of the gas

M = molecular mass of the gas

We can write the density of the gas asρ = m/V = (m/M) (M/V) = (m/M) (P/RT) …(2)

On combining (1) and (2), we have

ρ = Pm/RTMM = molecular weight of helium gas = 4 g/mol = 0.004 kg/mol= P/m × RT= 100 × 10³/ (8 × 0.004) × (8.31) × (700)ρ = 0.442 kg/m³

Volume flow rate of the helium at the pipe exit, V = m/ρ= 8/0.442= 18.1 m³/sec

Therefore, the volume flow rate of the helium at the pipe exit is 18.1 m³/sec.
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This function has four essential prime implicants, which are m1, m3, m6, and m11.

F(w, x, y, z) = m1 + m3 + m6 + m11

(b) F(A, B, C, D) = Σ(0, 2, 3, 5, 7, 8, 10, 11, 14, 15)

The K-map of the function is shown below. This function has three essential prime implicants, which are m1, m3, and m5.

F(A, B, C, D) = m1 + m3 + m5

(c) F(A, B, C, D) = Σ(1, 3, 4, 5, 10, 11, 12, 13, 14, 15)

The K-map of the function is shown below. This function has four essential prime implicants, which are m0, m2, m6, and m7.

F(A, B, C, D) = m0 + m2 + m6 + m7(d) F(w, x, y, z) = Σ(0, 1, 4, 5, 6, 7, 9, 11, 14, 15)

This function has three essential prime implicants, which are m2, m4, and m6.

F(w, x, y, z) = m2 + m4 + m6(

e) F(A, B, C, D) = Σ(0, 1, 3, 7, 8, 9, 10, 13, 15)

This function has three essential prime implicants, which are m0, m3, and m5.

F(A, B, C, D) = m0 + m3 + m5

(f) F(w, x, y, z) = Σ(0, 1, 2, 4, 5, 6, 7, 10, 15)

This function has three essential prime implicants, which are m0, m2, and m7.F(w, x, y, z) = m0 + m2 + m7Thus, the given Boolean functions are simplified by first finding the essential prime implicants.

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The moment of inertia of the rotating parts of an electric motor is 12 kg m, and it is connected to another shaft that has a moment of inertia of 24 kg m by means of a disk clutch while running at 1200 rev/min.

We have to find the common speed of rotation immediately after slip has ceased and the time taken for the two shafts together to regain the initial speed of 1200 revs/min.
(a) To find the common speed of rotation immediately after slip has ceased, we can apply the law of conservation of angular momentum, which states that the initial angular momentum equals the final angular momentum.
The initial angular momentum of the electric motor is given by,
Initial angular momentum = Moment of inertia × Angular velocity
= 12 kgm × (1200 rev/min × 2π/60)
= 1507.96 kgm²/s

Let ω be the common angular velocity of the electric motor and the other shaft after slip has ceased.
The final angular momentum is given by,
Final angular momentum = (Moment of inertia of electric motor + Moment of inertia of other shaft) × Angular velocity
= (12 kgm + 24 kgm) × (ω)
= 36 ω kgm²/s

By the law of conservation of angular momentum, we have,
Initial angular momentum = Final angular momentum
Therefore, 1507.96 = 36 ω
Hence, the common speed of rotation immediately after slip has ceased is:
ω = 1507.96/36 = 41.89 rev/min.

(b) To find the time taken for the two shafts together to regain the initial speed of 1200 revs/min, we can use the equation:
T = (Final angular momentum - Initial angular momentum)/Torque
The final angular momentum is 1507.96 kgm²/s, which is the same as the initial angular momentum since the two shafts are rotating at 1200 rev/min.
The torque applied is 160 Nm.
Substituting the values into the above equation, we have:
T = (1507.96 - 1507.96)/160 = 0 s

Therefore, it will take no time for the two shafts together to regain the initial speed of 1200 revs/min.

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