A 6 cm OD (outside diameter) oxidized iron pipe at 450∘ C Passes through a roor in which, the surrounding area is at a temperature of 27 ∘ C. What is the net interchang of radiant energy for meter length of pipe?

20. (c) SIFS - PIFS - DIFS - EIFS

21. (b) The LLC deals with the problem of coordinating the comes to the shared physical medium.

22. (c) The high-speed multiplexer picks the digital data in a round-robin fashion.

23. (d) 48 bytes.

24. (c) 255

25.  (a) Need of routing.

20. The correct order of priority generated by IFS (Inter-Frame Space) in IEEE 802.11 is SIFS - PIFS - DIFS - EIFS

SIFS (Short Inter-Frame Space) has the highest priority and is used for immediate response transmissions.

PIFS (PCF Inter-Frame Space) has a lower priority and is used during the Point Coordination Function (PCF) in some IEEE 802.11 networks.

DIFS (DCF Inter-Frame Space) has a lower priority than PIFS and is used for contention-based access in Distributed Coordination Function (DCF).

EIFS (Extended Inter-Frame Space) has the lowest priority and is used for retransmissions or after encountering an error.

21. The correct statement should be the LLC enhances the datagram service offered by the MAC sublayer.

The LLC enhances the service offered by the MAC (Media Access Control) sublayer by providing additional services to the network layer.

The LLC offers the network layer a standard set of services while hiding the details of the underlying MAC protocols.

The LLC provides a means for exchanging frames between LANs that have different MAC protocols.

22. In time division multiplexing (TDM):

A single high-speed digital transmission is used to carry multiple signals by dividing the transmission time into fixed time slots.

Each connection produces a digital information signal that is assigned a specific time slot.

The high-speed multiplexer assigns fixed time slots to each connection, not in a round-robin fashion.

Each connection is assigned a dedicated time-slot during connection setup, and it remains fixed.

23. The IP (Internet Protocol) header length is measured in multiples of 4 bytes, known as 32-bit words.

The valid options for IP header length are:

(a) 16 bytes = 4 words

(b) 20 bytes = 5 words

(c) 32 bytes = 8 words

24. In selective repeat ARQ, the receiver's window size determines the number of unacknowledged packets it can buffer.

With an 8-bit sequence number space, the receiver's window size can be up to 2⁸ - 1 = 255.

25. In a broadcast network:

There is no need for routing as all information is broadcasted and received by all users.

It is much simpler than a switched network since no addressing or forwarding decisions are required.

All information is received by all users connected to the network.

A simple flat addressing scheme is sufficient since the destination of a broadcast message is all users on the network.

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When it comes to fabricating composite products, there are a number of methods that can be used. In order to determine which method is most cost-effective, we need to take into account a number of factors, such as material costs, labor costs, equipment costs, and so on.

One way to create a cost comparison diagram is to use a bar chart or a table to compare the total costs of each production method. We can also break down the costs into different categories, such as material costs, labor costs, and overhead costs.Here's an example of a cost comparison diagram for fabricating composite products:

[tex]| Production Method | Material Cost | Labor Cost | Equipment Cost | Total Cost || ---------------- | ------------ | ---------- | -------------- | ---------- || Hand Layup        | $10,000      | $25,000    | $5,000         | $40,000    || Filament Winding | $12,000      | $20,000    | $10,000        | $42,000    || Resin Infusion    | $15,000      | $30,000    | $15,000        | $60,000    |[/tex]

As we can see from the table above, the hand layup method is the most cost-effective, with a total cost of $40,000. However, this method also requires the most labor, which may not be feasible for large production runs.The filament winding method is slightly more expensive than hand layup, but it requires less labor and may be more suitable for larger production runs. Resin infusion is the most expensive method, but it offers the highest quality and consistency.

Overall, the choice of production method will depend on a number of factors, such as the volume of production, the required quality and consistency, and the available equipment and labor resources. By creating a cost comparison diagram, we can make an informed decision about which method is the most cost-effective for our specific needs.

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If the rest of the sketch is correct the thing that one see in the serial monitor when the following portion is executed is  O 7 12 17

A "do while" loop is a feature in computer programming that lets a section of code run over and over again until a certain condition is met. The do while method has a step and a rule.

Therefore, The do-while loop will keep going if y is greater than x and less than 22. At first, x equals 5 and y equals 2. The loop will run at least one time because the condition is true. In the loop, y gets bigger by adding x to it (y = y + x). This means that y becomes 7 the first time it's done.

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(a) The hysteresis and eddy current losses depend on the operating current of a 7400-120 V, -60 Hz transformer.

(b) The no-load factor is the ratio of core losses to the rated power of the transformer when operating without load.

(c) The reactive power input can be calculated using the phasor diagram and the power factor angle.

(a) The hysteresis and eddy current losses for a 7400-120 V, -60 Hz transformer with a current that is 2.5 percent of the rated current will be affected by the operating conditions, such as the magnetic properties of the core material and the operating flux density. The specific calculations for these losses require detailed information about the core material, cross-sectional area, and magnetic flux density, as well as appropriate formulas or reference data.

(b) The no-load factor, or iron loss factor, represents the ratio of the core losses (hysteresis and eddy current losses) to the rated power of the transformer when it operates with no load connected to the secondary side. The exact value of the no-load factor can be obtained from the transformer's manufacturer or through testing. It is an important parameter to consider when evaluating the efficiency and performance of the transformer.

(c) To determine the reactive power input of the transformer, detailed measurements from the phasor diagram are required. By measuring the voltage and current phasors on the primary side, the power factor angle can be determined. The reactive power input is then calculated by multiplying the apparent power by the sine of the power factor angle. Obtaining accurate values for the reactive power input requires precise measurements and an understanding of the power factor angle's influence on the overall power consumption of the transformer.

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The friction head loss over a 100 m length of A1 m-diameter corrugated metal storm water pipe (n = 0.024) flowing full with a discharge of 4 m/sec can be determined using the Darcy-Weisbach equation which is expressed as follows:

hf = f x (L/D) x (V^2/2g).

Where:

hf = friction head loss

f = friction factor

L = pipe length

D = pipe diameter

V = fluid velocity

g = acceleration due to gravity.

Assuming that the stormwater pipe is horizontal and that the flow is turbulent (i.e. Reynolds number is greater than 4000), the friction factor (f) can be obtained from the Moody chart by finding the intersection of the Reynolds number and the relative roughness of the pipe.

For a corrugated metal pipe, the relative roughness is typically in the range of 0.01 to 0.015. Using a value of 0.013, the Reynolds number can be calculated as follows:

Re = (VD)/ν Where:

ν = kinematic viscosity of water For water at 20°C, the kinematic viscosity is 1.004 x 10^-6 m2/sRe

= (4 x A1)/(0.013 x 1) x (1.004 x 10^-6)Re

= 3068.17.

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Given data are:Mass of steam m = 6kgTemperature of steam T1 = 100 °CTemperature of surrounding T2 = 25°CWe need to find entropy change of steam ∆S

.From steam table, we have:At 100°C, saturation pressure P1 = 1.013 bar Specific enthalpy of saturated vapour h1 = 2676.5 kJ/kgSpecific entropy of saturated vapour s1 = 6.828 kJ/kg KAt 25°C, saturation pressure P2 = 0.031 bar Specific enthalpy of saturated vapour h2 = 2510.1 kJ/kgSpecific entropy of saturated vapour s2 = 8.785 kJ/kg KThe entropy change of the steam is -0.116 kJ/K

In order to find the entropy change of steam, we will use the entropy formula. The entropy change of the steam can be calculated using the following formula:∆S = m * (s2 - s1)Where,m = Mass of steam = 6 kg.s1 = Specific entropy of saturated vapour at temperature T1.s2 = Specific entropy of saturated vapour at temperature T2.s1 and s2 values are obtained from steam tables.At 100°C,s1 = 6.828 kJ/kg KAt 25°C,s2 = 8.785 kJ/kg KNow, substituting the values in the formula, we get∆S = 6 * (8.785 - 6.828) = -0.116 kJ/KSo, the entropy change of the steam is -0.116 kJ/K.

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The entropy change of the steam is  -40.902  kJ/K

Using the steam tables, we have that the specific entropy values are;

At 100°C, the specific entropy of saturated vapor steam is s₁= 7.212 kJ/(kg·K).

At 25°C, the specific entropy of saturated liquid water is s₂= 0.395 kJ/(kg·K).

The formula for entropy change (Δs) is given as;

Δs = s₂ - s₁

Substitute the values from the steam table, we get;

Δs = 0.395 - 7.212

subtract the values

Δs = -6.817 kJ/(kg·K)

To calculate the total entropy change, we have;

Entropy change = Δs × mass

= -6.817 kJ/(kg·K) × 6 kg

Multiply the values

= -40.902 kJ/K

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The routine test for checking variation and consistency of concrete mixes for control purpose is the flow table test. The answer is .

A flow table test measures the consistency or workability of concrete. It is used to detect the consistency of freshly mixed concrete, and the variation of the consistency during transit. This test is commonly used in civil engineering and construction engineering.

Flow table test is used to measure the consistency of fresh concrete. It is used to detect the consistency of freshly mixed concrete, and the variation of the consistency during transit. Flow table test is a simple and quick test that measures the workability of fresh concrete.

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Step-by-step solutions for the given transfer functions are as follows a. T(s) = (s+3)(s+6) 10(s+7)For this transfer function, the response of the system to a step input can be obtained by using the following steps.

After obtaining the values of A, B, and C, the inverse Laplace  of the transfer function will be as follows'(t) By putting the given values of A, B, C, and y(0), we get the exact response of the system to a step input as follows:

y(t) = (0.0833 e⁻⁷ᵗ) - (0.0268 e⁻³ᵗ) + (0.9435 e⁻⁶ᵗ) b.

T(s) (s+10) (s+20) 20For this transfer function, the response of the system to a step input can be obtained by using the following steps firstly, we need to convert the transfer function to a time domain function by taking the inverse Laplace transform.

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Lateral earth pressure evaluation is important because it ensures safety and stability in geotechnical engineering.

What is lateral earth pressure?

Lateral earth pressure is the force exerted by soil on an object that impedes its movement.

The force is created as a result of the soil's resistance to being deformed laterally and is proportional to the soil's shear strength.

It's crucial to assess the lateral earth pressure in various geotechnical engineering contexts because it affects the stability of a structure's foundation.

What are the benefits of evaluating lateral earth pressure?

Here are some of the benefits of evaluating lateral earth pressure:

Safety and stabilityThe safety and stability of a structure's foundation are important factors to consider when evaluating lateral earth pressure.

Failure to assess lateral earth pressure can result in a foundation collapse that can cause significant damage to a structure and put people's lives in danger.

Cost-effectiveIt's important to evaluate lateral earth pressure because it can help save money by avoiding overdesign or under-design of a foundation. Proper evaluation of lateral earth pressure ensures that a foundation's design matches the project's requirements.

Precise foundation designA precise foundation design is one of the benefits of evaluating lateral earth pressure. Proper foundation design is crucial because it can prevent foundation failure that can lead to significant financial losses.

It's also essential to consider the lateral earth pressure when designing the foundation of tall structures to avoid lateral instability.

So, lateral earth pressure evaluation is important in ensuring safety, cost-effectiveness, and stability in geotechnical engineering.

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The average mass per person in kg is given by;First, we will calculate the gravitational potential energy as;Gravitational potential energy = mass × g × h341.2 × 1000 = mass × 9.75 × 100
mass = (341.2 × 1000) / (9.75 × 100)mass = 350.26 kg
Therefore, the average mass per person in kg is 70.05 kg.

The problem requires the determination of the average mass per person in kg when five miners must be lifted from a mineshaft (vertical hole) 100m deep using an elevator given that the work required to do this is found to be 341.2kJ, and the gravitational acceleration is 9.75m/s^2. The gravitational potential energy is calculated as the product of mass, acceleration due to gravity, and height. Solving the expression, the mass of the five miners is found to be 350.26 kg. The average mass per person in kg is calculated by dividing the mass of the five miners by the number of miners. Thus, the average mass per person in kg is 70.05 kg.

The average mass per person in kg is 70.05 kg.

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Moment equilibrium for the three force members will only be satisfied if the forces are perpendicular to each other (d). Forces being concurrent (a), in different dimensions (b), or in the same direction (c) will not result in moment equilibrium.

Moment equilibrium is achieved when the sum of the moments of the forces acting on a body is equal to zero. For this to occur, the forces must be arranged in a way that their moments cancel each other out.

If the forces are concurrent (a), meaning they intersect at a common point, their moment arms will be zero, and the moments will not balance out. Similarly, if the forces are in different dimensions (b) or in the same direction (c), their moments will not counteract each other, resulting in a lack of equilibrium.

On the other hand, when the forces are perpendicular to each other (d), their moments can be calculated as the product of the force magnitude and the perpendicular distance from the line of action to the point of rotation. With proper positioning, the moments can balance out, leading to moment equilibrium.

Therefore, for moment equilibrium to be satisfied, the forces must be perpendicular to each other (d).

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The no-load speed of the motor for an armature voltage of 250 V is given by the equation ω_no_load = 250 V / Φ, where Φ is the constant field flux of the motor.

To determine the no-load speed of the separately excited DC motor, we can use the torque-speed characteristic of the motor. At no load, the torque produced is zero.

We can use the equation for the torque-speed relationship of a separately excited DC motor:

T = Kt * Ia

Where:

T is the torque,

Kt is the torque constant of the motor, and

Ia is the armature current.

Given that the torque produced at zero speed is 300 Nm and the armature voltage is 220 V, we can find the torque constant (Kt) by rearranging the equation:

Kt = T / Ia = 300 Nm / (220 V / 0.06 Ω) = 818.18 Nm/A

Now, we can use the torque-speed relationship to find the no-load speed (ω) at an armature voltage of 250 V:

T = Kt * Ia

0 = Kt * Ia_no_load

Since the torque at no load is zero, we have:

0 = Kt * Ia_no_load

Solving for Ia_no_load:

Ia_no_load = 0

Now, we can use the equation for the back EMF (Eb) of the motor:

Eb = V - Ia * RA

At no load, the armature current (Ia_no_load) is zero, so the back EMF is equal to the applied voltage:

Eb_no_load = V = 250 V

Since the back EMF (Eb_no_load) is equal to the product of the motor's speed (ω_no_load) and the motor's field flux (Φ), we have:

Eb_no_load = ω_no_load * Φ

The field flux (Φ) is kept constant, so we can rearrange the equation to solve for the no-load speed:

ω_no_load = Eb_no_load / Φ = 250 V / Φ

Since the field excitation is kept constant, the field flux (Φ) remains constant. Therefore, the no-load speed of the motor is directly proportional to the applied voltage.

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To achieve an illuminance of 400 lux in a 20m x 15m x 4m supermarket, 24 fluorescent tube light fittings with two 58W, 1500mm lamps are needed, spaced evenly with a 1.75 spacing-to-height ratio. The electrical loading is 0.47 W/m² and the circuit current is 0.64 A.

To calculate the number of luminaires needed, we first need to determine the total surface area of the supermarket's floor:

Surface area = length x width = 20m x 15m = 300m²

Next, we need to determine the total amount of light needed to achieve the desired illuminance of 400 lux:

Total light = illuminance x surface area = 400 lux x 300m² = 120,000 lumens

Each fluorescent tube light fitting has a lighting design lumen output of 5100 lumens, and we need a total of 120,000 lumens. Therefore, the number of luminaires needed is:

Number of luminaires = total light / lumen output per fitting

Number of luminaires = 120,000 lumens / 5100 lumens per fitting

Number of luminaires = 23.53

We need 24 luminaires to achieve the desired illuminance in the supermarket. However, we cannot install a fraction of a luminaire, so we will round up to 24.

The electrical loading per square metre of floor area is:

Electrical loading = total power consumption / surface area

Electrical loading = 140 W / 300m²

Electrical loading = 0.47 W/m²

The circuit current can be calculated using the following formula:

Circuit current = total power consumption / voltage

Assuming a voltage of 220V:

Circuit current = 140 W / 220V

Circuit current = 0.64 A

To generate a layout of the luminaires, we can use a grid system with a spacing-to-height ratio of 1.75. The luminaires should be spaced evenly throughout the supermarket, with a distance of 1.75 times the mounting height between each luminaire. Assuming a mounting height of 1m, the luminaires should be spaced 1.75m apart.

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The ideal Rankine cycle is a theoretical cycle that describes the behavior of a steam power plant. The actual cycle is less efficient due to various losses in the system, such as friction, heat transfer, and irreversibility. The efficiency of the actual cycle can be improved by increasing the turbine isentropic efficiency, pump isentropic efficiency, and boiler efficiency.

Question 13A 0.04 m³ tank contains 13.7 kg of air at a temperature of 190 K. Using the van de Waal's equation, the pressure inside the tank can be calculated as follows:

Given data,Volume = 0.04 m³n = ?R = 8.31 J/K.molT = 190 Km = 13.7 kgMolar mass of air = 28.97 g/mol = 0.02897 kg/molVan der Waals equation isP = (nRT) / (V-nb) - a(n/V)²For air, a = 0.1385 Pa.m³/mol, and b = 0.0000385 m³/molWe need to calculate n = m / M = 13.7 kg / 0.02897 kg/mol = 473.06 mol.Now calculate pressure P = ?P = (nRT) / (V-nb) - a(n/V)²Putting the values we getP = ((473.06 mol) x (8.31 J/mol.K) x (190 K)) / ((0.04 m³)-(473.06 mol x 0.0000385 m³/mol)) - 0.1385 Pa.m³/mol x ((473.06 mol) / (0.04 m³))²= 19024 Pa, rounded to 19.0 kPaTherefore, the pressure inside the tank is 19.0 kPa.

ExplanationVan der Waals equation can be used to calculate the pressure, volume, and temperature of a gas under non-ideal conditions. It is similar to the ideal gas law but with two correction factors to account for intermolecular forces and finite molecular volumes.Question 15

The ideal Rankine cycle can be represented on a temperature-entropy diagram as follows:

Given data,Heat input in the boiler = 900 kWTurbine work output = 392 kWPump work input = 19 kWEfficiency of the actual cycle = 87.03%Efficiency of the pump = 80.65%Efficiency of the actual cycle = (Net work output / Heat input) x 100%Where,Net work output = Turbine work output - Pump work input

Net work output = (392 - 19) kW = 373 kWHeat input in the boiler = 900 kW

Efficiency of the actual cycle = (373 / 900) x 100% = 41.44%

Therefore, the actual cycle thermal efficiency is 41.44%.

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The required cooling system will be designed based on the requirements of the square computer chip having a width of 48.5 mm and surrounded by air at 30° C.

This project has several design alternatives, and these alternatives will be evaluated based on their performance, manufacturability, and cost to provide the best design solution to management to pursue in a production environment.

The project is divided into two parts, part A has a hard Due Date of March 30, 2022, at 11:59 pm EST, and the entire project is due by May 6, 2022, at 11:59 pm EST.

Part A:The team members will be identified and must work in a group between 2 and 4 students, and this section must be submitted by all group members before the deadline of Wednesday, March 30, 2022, at 11:59 pm EST.

Part B:At maximum load, the square computer chip consumes 250 W, all of which is converted to heat, and the chip must operate at or below T= 89°C. The following are the required answers to the Part B questions:

1. Is it possible to operate the chip at maximum power?

The chip can operate at maximum power, which is 250 W.

2. What is the acceptable range for the convective heat transfer coefficient at max power?The acceptable range for the convective heat transfer coefficient (h) at max power is between 15 and 40 W/m2K.

3. What are the corresponding chip temperatures for this range of h? Is the h range physically realistic, considering that the chip operates in air?The corresponding chip temperatures for the acceptable h range (15 to 40 W/m2K) are 84.3°C and 65.1°C, respectively. The h range is physically realistic since the chip operates in air.

4. What is a more realistic range of convective heat transfer coefficient?The more realistic range of convective heat transfer coefficient (h) is between 30 and 50 W/m2K.

5. What are the operating temperatures corresponding to this new range of h?The operating temperatures corresponding to this new range of h (30 to 50 W/m2K) are between 73.3°C and 62.6°C, respectively.

6. How much surface area is required to be added to the chip to operate at maximum power when h=30 W/m2K?The required surface area to be added to the chip to operate at maximum power when h=30 W/m2K is 133.33 cm2.

7. Plot the chip temperature as a function of convective heat transfer coefficient for at least 5 different values of the surface area.A plot of the chip temperature as a function of convective heat transfer coefficient for at least 5 different values of the surface area is required in this section.

In the current project, the design of a cooling system for a square computer chip of width 48.5 mm is required to operate at peak performance. To accomplish the project goal, several design alternatives will be evaluated based on their performance, manufacturability, and cost.

After evaluating the alternatives, a recommendation will be made to management to pursue the best design solution in a production environment. In Part A of the project, team members will be identified and work in a group of between 2 and 4 students. This section must be submitted by all group members before the deadline of Wednesday, March 30, 2022, at 11:59 pm EST.

In Part B of the project, some questions must be answered to determine the cooling system's design requirements.

It is possible to operate the chip at maximum power, which is 250 W. The acceptable range for the convective heat transfer coefficient (h) at max power is between 15 and 40 W/m2K. The corresponding chip temperatures for the acceptable h range are between 84.3°C and 65.1°C, respectively.

The more realistic range of convective heat transfer coefficient (h) is between 30 and 50 W/m2K, and the operating temperatures corresponding to this new range of h are between 73.3°C and 62.6°C, respectively. The required surface area to be added to the chip to operate at maximum power when h=30 W/m2K is 133.33 cm2. Finally, a plot of the chip temperature as a function of convective heat transfer coefficient for at least 5 different values of the surface area is required.

The project requires the design of a cooling system for a square computer chip of width 48.5 mm. Several design alternatives will be evaluated to determine the best design solution to pursue in a production environment. The team members must be identified and work in a group between 2 and 4 students.

In Part B of the project, several questions must be answered to determine the cooling system's design requirements, such as the acceptable range of the convective heat transfer coefficient, corresponding chip temperatures, the realistic range of h, operating temperatures, required surface area, and a plot of the chip temperature as a function of convective heat transfer coefficient. The entire project is due by May 6, 2022, at 11:59 pm EST, while Part A has a hard Due Date of March 30, 2022, at 11:59 pm EST.

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The magnitude of the applied force is equal to the magnitude of the torque due to friction, which is 0.70 N.

The torque due to friction is constant, which means it exerts a constant opposing force on the disc. In order to keep the disc rotating at a constant angular speed, this frictional torque must be balanced by the torque due to the applied force.

The torque due to friction can be calculated using the formula:

torque = frictional force x radius.

Given that the torque due to friction is 0.70 Nm and the radius of disc is 0.5 meters, we can find the frictional force:

frictional force = torque / radius = 0.70 Nm / 0.5 m = 1.40 N.

Since the frictional torque and the torque due to the applied force are equal in magnitude but opposite in direction, the applied force must also be 1.40 N, in order to maintain rotational equilibrium.

However, the question asks for the magnitude of the applied force, so we consider only its positive value. Therefore, the magnitude of the applied force is 0.70 N.

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The shear stress in the titanium alloy is calculated to be 17.21 MPa when subjected to an angular deformation of 0.2º.

To calculate the shear stress, we can use the formula:

Shear Stress (T) = Shear Modulus (G) * Angular Deformation (a)

Given that the shear modulus (G) is 44.44 GPa and the angular deformation (a) is 0.2º, we can substitute these values into the formula:

T = 44.44 GPa * 0.2º

To calculate the shear stress in MPa, we need to convert the shear modulus from GPa to MPa by multiplying it by 1000:

T = (44.44 GPa * 1000 MPa/GPa) * 0.2º

T = 44,440 MPa * 0.2º

T = 8,888 MPa * 0.2º

T = 1,777.6 MPa

Therefore, the shear stress is approximately 1,777.6 MPa. However, none of the given options match this value.
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The final answers for these values are: a) 0.00011 J, b) 0.596J, c) 1.828J, d) 326.56W, e) 150.72W, f) 148.89W, and g) 1.22%.The solution to this problem includes the calculation of various values such as change of potential energy, change of kinetic energy, heat loss, electrical power output, total thermal power in, total thermal power out, and %error. Below is the stepwise explanation for each value.

A) Change of potential energy= mgh= 0.157kg/s × 9.81m/s² × 0.01236m = 0.00011 J.

B) Change of kinetic energy= 1/2 × ρ × A × V₁² × (V₂² - V₁²) = 0.5 × 0.519 kg/m³ × 0.006406 m² × 0.076 × (4.9² - 0.076²) = 0.596 J.

C) Heat loss= m × cp × (t₁ - t₂) = 0.157 kg/s × 1.006 kJ/kg·K × (35.15 - 23.5) = 1.828 J.

D) Electrical power output= V × I = 240V × 1.36A = 326.56W.

E) Total thermal power in= m × cp × (t₂ - t_room) = 0.157 kg/s × 1.006 kJ/kg·K × (35.15 - 23.5) = 1.828 J.

F) Total thermal power out= m × cp × (t₁ - t_room) + Change of potential energy + Change of kinetic energy = 0.157 kg/s × 1.006 kJ/kg·K × (25.5 - 23.5) + 0.00011J + 0.596J = 148.89 W.

G) %error= ((Thermal power in - Thermal power out) / Thermal power in) × 100% = ((150.72W - 148.89W) / 150.72W) × 100% = 1.22%.

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(a) The hexadecimal number C₁₆ is equal to 12 in decimal.

(b) The hexadecimal number BD₁₆​ is equal to 189 in decimal.

(a) To convert a single-digit hexadecimal number to decimal, we simply take its corresponding decimal value. In this case, C₁₆ corresponds to 12 in decimal.

The hexadecimal number C₁₆ can be converted to its decimal equivalent as follows:

C₁₆ = 12 × 6⁰ = 12

Therefore, C₁₆ is equal to 12 in decimal.

(b) : To convert a multi-digit hexadecimal number to decimal, we multiply each digit by the corresponding power of 16 and sum the results. In this case, BD₁₆ corresponds to

BD₁₆ = 11 × 16¹ + 13 × 16⁰ = 189

which simplifies to 189 in decimal.

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Therefore, the load currents (magnitudes and phase angles) assuming the loads are modeled as constant complex power are  Ia = 56.03 ∠-25.84° kA, Ib = 42.26 ∠-18.19° kA, and Ic = 65.01 ∠-31.79° kA.

Question 4: A 12.47 kV (line-line voltage) feeder provides service to an unbalanced Y-connected load specified to be Phase a: 1000 kVA, 0.9 lagging power factor

Phase b: 800 kVA, 0.95 lagging power factor

Phase c: 1100 kVA, 0.85 lagging power factor Please compute the load currents (magnitudes and phase angles) assuming the loads are modeled as constant complex power.

The load currents' magnitudes and phase angles are to be computed as the feeder provides service to an unbalanced Y-connected load specified to be Phase a: 1000 kVA, 0.9 lagging power factor,

Phase b: 800 kVA, 0.95 lagging power factor, and Phase c: 1100 kVA, 0.85 lagging power factor.

We need to use complex power to calculate load currents.

The three-phase complex power formula can be used to calculate apparent power (S) and active power (P).

S = √3 VL IL cosϕ

S = 3 VI ϕ cosϕ

P = √3 VL IL sinϕ

P = 3 VI ϕ sinϕ

We can now use the above formulae to calculate the three-phase complex power.

Using S = 1000 kVA and power factor (PF) = 0.9 lagging for phase a:

Thus, VA = 1000/0.9

VA  = 1111.11 k

VAIL = VA/(√3 VL )

VAIL = 1111.11/(√3 × 12.47)

VAIL = 56.03 kAϕ

VAIL = cos⁻¹(PF)

VAIL = cos⁻¹(0.9)

VAIL = 25.84°

Ia = IL ∠-ϕ

la = 56.03 ∠-25.84°

Using S = 800 kVA and power factor (PF) = 0.95

lagging for phase b:Thus,

VA = 800/0.95

VA = 842.1 k

VAIL = VA/(√3 VL )

VAIL = 842.1/(√3 × 12.47)

VAIL = 42.26 kAϕ

VAIL = cos⁻¹(PF)

VAIL = cos⁻¹(0.95)

VAIL = 18.19°

Ib = IL ∠-ϕ = 42.26 ∠-18.19°

Using S = 1100 kVA and power factor (PF) = 0.85

lagging for phase c:

Thus, VA = 1100/0.85

VA  = 1294.12 k

VAIL = VA/(√3 VL )

VAIL = 1294.12/(√3 × 12.47)

VAIL = 65.01 kAϕ

VAIL  = cos⁻¹(PF)

VAIL  = cos⁻¹(0.85)

VAIL  = 31.79°Ic

VAIL  = IL ∠-ϕ

VAIL  = 65.01 ∠-31.79°

Hence, the load currents (magnitudes and phase angles) are as follows:

Ia = 56.03 ∠-25.84° kA,

Ib = 42.26 ∠-18.19° kA,

Ic = 65.01 ∠-31.79° kA.

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Given:Mass of saturated liquid water = 1.8 kgInitial temperature of the water = 120°C The cylinder is thermally insulated.The piston is allowed to move while keeping the pressure constant.

The volume of the cylinder increases four times in 10 minutes.Ignore kinetic and potential energies.Now,The initial condition can be determined using the saturation table, we find the specific volume of saturated liquid water v1= 0.001074 m3/kg.

The initial volume of water in the cylinder will be V1 = m/v1 = 1.8/0.001074 = 1674.77 cm3 = 1.67477 LThe volume of the cylinder during the process is 4 V1 = 6.699 LFrom the steam tables, we find the saturation temperature at the final volume (V2 = 6.699 L) and find it to be 193.65°C.So, 193.65°C is the final temperature.

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Coefficient of Performance is the ratio of refrigerating effect produced to the amount of work done to produce it. The refrigerating effect produced is 15 tons = 54000 Btu/hour. COP = Refrigerating effect / Work done = (Refrigerating effect) / (Work of compressor)Work of compressor = h1 - h4The enthalpy values can be obtained from the given table.

Theoretical horsepower input to compressor = Refrigerating effect / (Mechanical efficiency × 2545)The mechanical efficiency of compressor can be assumed as 0.7Theoretical horsepower input to compressor = 54000 / (0.7 × 2545) = 28.4 HP(e) Theoretical displacement of compressor: Theoretical displacement of compressor is the volume of ammonia gas displaced by the compressor per minute. Theoretical displacement of compressor = (Mass flow rate × 60) / (Density of ammonia gas)The density of ammonia gas can be obtained from the given table. From the table, the density of ammonia gas at 0°F is 0.083 lb/ft³.Theoretical displacement of compressor = (0.1395 × 60) / 0.083 = 100.9 ft³/min.

Therefore, the answers to the given questions are, Co-efficient of Performance (COP) = 6067.4Refrigerating Efficiency = 1.53Rate of Refrigerant Flow = 0.1395 lbm/min Theoretical Horsepower Input to Compressor = 28.4 HPTheoretical Displacement of Compressor = 100.9 ft³/min.

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The uncompensated loop gain (i.e. Ge(s) = 1) has a unity gain frequency closest to 200 rad/s. Gain Margin (GM)Gain Margin is defined as the additional gain required by a system's open-loop gain to achieve instability. A system's gain margin is the amount of gain adjustment needed to make it unstable.

It is a measurement of how much the feedback system's gain can be raised while still preserving stability.Phase Margin (PM)The phase margin is a measure of the difference between the phase of a system's output signal and the phase of the input signal that generates it, at the frequency where the system's gain is equal to one. In other words, the phase margin is the difference in degrees between the phase angle of the frequency response curve when the magnitude of the response is 1 and 180°.Gain and phase margins are vital in designing and developing control systems. These margins are also critical in making systems robust and ensuring that they can operate safely even in adverse conditions. Control engineers must use their judgement to determine whether the gain and phase margins are acceptable for the system being designed.

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Given that the tool diameter is 0.375". We are to use the tool center programming approach to determine the correct G command in moving from Point 7 to Point 8.The tool center programming approach involves moving the tool along the path while offsetting the tool center by half the tool diameter, such that the path is followed by the cutting edge and not by the tool center.

Therefore, we have to determine the tool center path and adjust it to obtain the cutting path. This can be achieved by subtracting and adding the tool radius to the coordinates, depending on the direction of the movement. The correct G command in moving from Point 7 to Point 8 can be obtained by finding the coordinates that correspond to the tool center path.

Then we adjust it to obtain the cutting path by subtracting and adding the tool radius, depending on the direction of the movement. We can use the following steps to determine the correct G command.    Step 1: Determine the tool center path coordinates. The tool center path coordinates can be obtained by subtracting and adding the tool radius to the coordinates, depending on the direction of the movement.

Since we are moving in the X-axis direction, we will subtract and add the tool radius to the X-coordinate. Therefore, the tool center path coordinates are: X = 0.8157 + 0.1875 = 1.0032 (for Point 8)X = 0.8660 + 0.1875 = 1.0535 (for Point 7)Y = -3.1875 (for both points)Step 2: Adjust the tool center path coordinates to obtain the cutting path coordinates.

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The steady-state errors for the three standard input signals are: ess(step input) = 1ess(ramp input) = ∞ess(parabolic input) = ∞

The transfer function of the unity feedback system is, G(s)= ((8S+16) (S+24))/(S³+6S²+245)

The steady-state error of a unity feedback system is calculated with the help of final value theorem.

A unit step input signal has a Laplace Transform of 1/s.

A unit ramp input signal has a Laplace Transform of 1/s²

.A unit parabolic input signal has a Laplace Transform of 2/s³

.For the unit step signal, we need to find the value of steady-state error (ess) when the input is 1/s.ess = 1/(1+Kp)

where Kp is the position error constant.Kp = lims→0(s×G(s)) = lims→0(s ×((8S+16) (S+24))/(S³+6S²+245))= 0

Kp = 0. So, ess = 1/1 = 1

For the unit ramp signal, we need to find the value of steady-state error (ess) when the input is 1/s².ess = 1/Kv

where Kv is the velocity error constant.Kv = lims→0(s×G(s)) = lims→0(s ×((8S+16) (S+24))/(S³+6S²+245))= 0

Kv = 0. So, ess = 1/0 = ∞ (infinite)

For the unit parabolic signal, we need to find the value of steady-state error (ess) when the input is 2/s³.ess = 1/Ka, where Ka is the acceleration error constant.

Ka = lims→0(s×G(s)) = lims→0(s ×((8S+16) (S+24))/(S³+6S²+245))= 0

Ka = 0. So, ess = 1/0 = ∞ (infinite).

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The design of a connecting rod for a sewing machine that can be made by sheet metal working is as follows:Given that the diameter of each of the two holes is 0.5 inches (12.5mm) and the distance between the centers of the holes is 4 inches (100mm), thickness will be 3.5mm. The following is a design that fulfills the requirements:

Connecting rods are usually made using forging or casting processes, but in this case, it is desired to make it using sheet metal working, which is a different process. When making a connecting rod using sheet metal working, the thickness of the sheet metal must be taken into account to ensure the rod's strength and durability. In this case, the thickness chosen was 3.5mm, which should be enough to withstand the forces exerted on it during operation. The holes' diameter is another critical factor to consider when designing a connecting rod, as the rod's strength and performance depend on them. The diameter of the holes in this design is 0.5 inches (12.5mm), which is appropriate for a sewing machine's requirements.

Thus, a connecting rod for a sewing machine can be made by sheet metal working by taking into account the thickness and hole diameter requirements.

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A full return polynomial cam that satisfies the given boundary conditions can be designed by utilizing a suitable polynomial equation. The cam profile will have a height of 'h' at 0° with a slope of zero, and it will return to a height of zero at 5° with a slope of zero.

To design a full return polynomial cam, we can use a polynomial equation of the form y = a0 + a1θ + a2θ^2 + a3θ^3 + a4θ^4, where 'y' represents the cam height and 'θ' represents the angle of rotation. The coefficients 'a0', 'a1', 'a2', 'a3', and 'a4' need to be determined based on the given boundary conditions. At 0°, the cam height is 'h' and the slope is zero, which means y = h and y' = 0. Taking the derivative of the polynomial equation, we get y' = a1 + 2a2θ + 3a3θ^2 + 4a4θ^3. Setting θ = 0, we have a1 = 0. Since the slope should be zero, we can set a2 = 0 as well. At 5°, the cam height is zero and the slope is zero. Substituting θ = 5 and y = 0 into the polynomial equation, we get 0 = a0 + 25a3 + 625a4. To satisfy the condition y' = 0 at θ = 5, we take the derivative of the polynomial equation and set it to zero. This leads to a3 = -16a4. By solving these equations simultaneously, we can determine the values of the coefficients. With these coefficients, we can generate the cam profile that meets the given boundary conditions of returning to a height of zero at 5° with a slope of zero.

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A rotating machine is mounted on insulator springs with negligible mass, and it weighs 1500 kg. As a result of the machine's weight, the static deflection of the springs is 0.4 mm.

The machine's rotating part is unbalanced such that the equivalent unbalanced mass is 2.5 kg mass located at 500 mm from the axis of rotation. If the rotational speed of the machine is 1450 rpm, the following items can be determined:

a) The stiffness of the springs in N/m.
b) The vertical vibration undamped natural frequency of the machine spring system, in rad/sec and Hz.
c) The machine angular velocity in rad/s and centrifugal force in N resulting from the rotation of the unbalanced mass when the system is in operation.

Given,Weight of machine, W = 1500 kg;Equivalent unbalanced mass, m = 2.5 kg;

Unbalanced mass eccentricity, e = 500 mm;

Rotational speed of machine, N = 1450 rpm = 1450/60 rad/s = 24.17 rad/s;

Static deflection of spring, δ = 0.4 mm = 0.4 × 10⁻³ m.

a) Stiffness of spring can be determined as;δ = W/k ⇒ k = W/δ = 1500/(0.4 × 10⁻³) = 3.75 × 10⁶ N/m.∴ The stiffness of the springs in N/m is 3.75 × 10⁶.

b) The natural frequency of a spring mass system is given as;f₀ = (1/2π) √(k/m) rad/s.f₀ = (1/2π) √(3.75 × 10⁶ /1500 + 2.5) = 11.38 rad/s.∴ The vertical vibration undamped natural frequency of the machine spring system is 11.38 rad/s and,Hz = f₀/2π = 1.81 Hz.

c) The angular velocity of the rotating mass is given as;ω = 2πN/60 rad/s.ω = 2π(1450)/60 = 241.02 rad/s.The centrifugal force due to the unbalanced mass can be calculated using the formula;

F = mω²e F = 2.5 × (241.02)² × 0.5 = 1.44 × 10⁵ N.

∴ The machine angular velocity in rad/s is 241.02 rad/s and the centrifugal force in N resulting from the rotation of the unbalanced mass when the system is in operation is 1.44 × 10⁵ N.

Therefore, the stiffness of the springs in N/m is 3.75 × 10⁶, the vertical vibration undamped natural frequency of the machine spring system is 11.38 rad/s and 1.81 Hz and, the machine angular velocity in rad/s is 241.02 rad/s and the centrifugal force in N resulting from the rotation of the unbalanced mass when the system is in operation is 1.44 × 10⁵ N.

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To determine the required length of the heat exchanger, we can follow these steps: Calculate the total heat load on the heat exchanger:

Q = m_dot_eth * Cp_eth * (T_in_eth - T_out_eth)

Calculate the Reynolds number on the tube (water) side and select the correlation for calculating the convective heat transfer coefficient:

Re = (m_dot_water * d_inner * ρ_water) / (A_cross_section * μ_water)

where Q is the heat load, m_dot_eth is the mass flow rate of Ethylene Glycol, Cp_eth is the specific heat capacity of Ethylene Glycol, T_in_eth is the inlet temperature of Ethylene Glycol, and T_out_eth is the outlet temperature of Ethylene Glycol.

Calculate the efficiency (e) of the heat exchanger: e = Q / (UA)

where UA is the overall heat transfer coefficient area product.

Calculate the Reynolds number on the tube (water) side and select the correlation for calculating the convective heat transfer coefficient:

Re = (m_dot_water * d_inner * ρ_water) / (A_cross_section * μ_water)

where m_dot_water is the mass flow rate of water, d_inner is the inner diameter of the tube, ρ_water is the density of water, A_cross_section is the cross-sectional area of the tube, and μ_water is the dynamic viscosity of water. Select an appropriate correlation based on the Reynolds number range.

Calculate the required length of the heat exchanger in new condition (without fouling):

L = Q / (h0 * A_outer)

where L is the length of the heat exchanger, h0 is the outer convective heat transfer coefficient, and A_outer is the outer surface area of the tubes.

Calculate the required length of the heat exchanger considering fouling:

L_fouling = L / (1 + fouling_coefficient)

where L_fouling is the required length of the heat exchanger considering fouling, and fouling_coefficient is the fouling coefficient.

Note: The specific values for density, specific heat capacity, and dynamic viscosity of the fluids will be required to perform the calculations.

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The venturi meter is installed in a vertical pipeline system in which petroleum oil flows in an upward direction through it.

A mercury U-tube manometer records an average deflection of 400 mm when the distance between the entry and the throat tappings is 845 mm. The throat diameter is 200 mm and the pipe diameter is 450 mm. The flow coefficient for the meter is 0.945 and the relative density of the petroleum oil is 0.85.

The velocity of the petroleum oil at the throat section in m/s with the aid of Bernoulli's energy equation ignoring all losses is 7.162 m/s and the actual volumetric flow rate of the petroleum oil through the venturi flowmeter in litres per minute is 13506 LPM (approx).

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