You are an Associate Professional working in the Faculty of Engineering and a newly appointed technician in the Mechanical Workshop asks you to help him with a task he was given. The department recently purchased a new 3-phase lathe, and he is required to wire the power supply. The nameplate of the motor on the lathe indicated that it is delta connected with an equivalent impedance of (5+j15) £ per phase. The workshop has a balanced star connected supply and you measured the voltage in phase A to be 230 Đ0° V. (a) Discuss three (3) advantage of using a three phase supply as opposed to a single phase supply (6 marks) (b) Draw a diagram showing a star-connected source supplying a delta-connected load. Show clearly labelled phase voltages, line voltages, phase currents and line currents. (6 marks) (c) If this balanced, star-connected source is connected to the delta-connected load, calculate: i) The phase voltages of the load (4 marks) ii) The phase currents in the load (4 marks) iii) The line currents (3 marks) iv) The total apparent power supplied

Answer:

do the wam wam

Explanation:

Answer:

Most technicians install rebuilt or new power steering pumps rather than overhauling the defective unit in the shop.

True

Explanation:

It has been established that most technicians, instead of overhauling the defective power steering pumps, prefer to install rebuilt or revamped pumps or even new pumps when they could have overhauled the unit themselves.  Some are afraid that leakage could occur again.  They are not confident enough to do the overhauling in-house or some claim that they do not have the time and other tools to carry out the overhauling.  Acknowledgedly, overhauling a power steering pump is not a job for the novice.  They require experienced and skilled hands to do the hard work.

Option B is accurate since it indicates that after opening a two-pipe system, the system must be bled to remove any air in the fuel line. This is a false assertion.

There are distinct pipes in a two-pipe fuel delivery system for feeding fuel to the burner and returning surplus fuel to the tank. This method is frequently utilized in oil-fired heating systems.

When the system is first loaded or when it runs out of fuel, bleeding may be required to remove any air that has entered the fuel line. Bleeding entails removing all air from the system and ensuring that only fuel remains in the lines.

Therefore, option B is false because a properly functioning two-pipe system should not require bleeding every time it is opened.

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The complete question:

Which of the following statements regarding two-pipe fuel delivery systems is false?

A. A two-pipe system is self-priming.

B. If a two-pipe system is opened, the system must be bled to get rid of any air in the fuel line.

C. A two-pipe system’s fuel unit pumps more fuel oil in a given time than is burned in that time.

D. All of the above.

Before charging a system that has been opened for service, it's vital that one need to follow proper procedures to ensure safety and optimal performance of the system

Check for leaks before charging the system. Address leaks before vacuuming the system to remove air and moisture. Prevent contamination to avoid issues and component damage.

Charge system with correct refrigerant amount after evacuation. Use a charging scale for accuracy. Test system after charging. Check temperature, pressure, airflow, and component functionality. Follow safety guidelines and disposal procedures. Follow manufacturer's instructions for optimal performance and safety.

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The velocity vector of point C with respect to point B is (-1.5ωj^ + 2.598ωi^) ft/s.

Given:

Coordinate of C = (0, 0, 0) ft

Coordinate of B = (-3cos30°, 3sin30°) ft = (-2.598, 1.5) ft

r_BC = (3cos30° i^ - 3sin30° j^) ft = (2.598i^ - 1.5j^) ft (position vector for point C with respect to B)

ω_BC = ω_AB = ω (since link AB and link BC are connected and rotating together)

The velocity of point C with respect to point B is given by:

v_CB = ω_BC x r_BC

Since link BC rotates in the counter-clockwise direction, the direction of ω_BC is in the positive z direction, i.e., ω_BC = ωk^.

Substituting the values,

v_CB = ω_BC x r_BC

= ωk^ x (2.598i^ - 1.5j^) ft

= (-1.5ωj^ + 2.598ωi^) ft/s

Therefore, the velocity vector of point C with respect to point B is (-1.5ωj^ + 2.598ωi^) ft/s.

Given:

v_C = (6.5 ft/s) j^

v_CB = (2.6ω_BC j^ + 1.5ω_BC i^) ft/s

v_B = -ω_AB i^

The velocity of point C is given by:

v_C = v_CB + v_B

Substituting the given values,

(6.5 ft/s) j^ = (2.6ω_BC j^ + 1.5ω_BC i^) ft/s + (-ω_AB i^)

Equating the components of the vectors on both sides, we get:

2.6ω_BC = 0

1.5ω_BC - ω_AB = 0

Solving these equations, we get:

ω_BC = 2.5 ft/s

ω_AB = 1.5 ft/s

Substituting the value of ω_BC in v_CB, we get:

v_CB = (2.6 x 2.5 j^ + 1.5 x 2.5 i^) ft/s

= (3.25 i^ + 6.5 j^) ft/s

Substituting the values of v_CB and v_B in v_C, we get:

v_C = v_CB + v_B

= (3.25 i^ + 6.5 j^) ft/s + (-1.5 ω_AB i^) ft/s

= (3.25 - 1.5ω_AB) i^ + 6.5 j^ ft/s

= (3.25 - 1.5 x 1.5) i^ + 6.5 j^ ft/s

= 0.5 i^ + 6.5 j^ ft/s

Therefore, the velocity vector of point C is (0.5 i^ + 6.5 j^) ft/s.

Given:

ω_BC = 2.5 ft/s

ω_AB = 1.5 ft/s

From equation (2):

1.5ω_BC - ω_AB = 0

Rearranging the equation, we get:

ω_AB = 1.5ω_BC

Substituting the value of ω_BC, we get:

ω_AB = 1.5 x 2.5 rad/s

= 3.75 rad/s

Therefore, the angular velocity of link AB is 3.75 rad/s.

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Answer:

Borders, Dimensions Identified, Scale, Text Box with Date/Author/Title/Version Number

Explanation:

Technical drawings have many standards in order to maintain integrity and ensure that they can be used by engineers to produce what is mapped out in the drawing.

Every technical drawing needs to have borders and a text box with all information related to the drawing. This includes the date it was created, date it was updated, version number, title, and author.

Furthermore, each drawing should have necessary dimensions with a scale. At the same time, technical drawings should not be over-dimensioned or repetitive.

All of these things must be on a technical drawing to allow professional engineers to sign off on drawings.

there is no smoke its a electric train

To determine the convective and radiative heat fluxes to the wall at x = 0, we can use the following equations:

1. Convective heat flux:

q_conv = h * (T_s - T_∞)

2. Radiative heat flux:

q_rad = α * σ * (T_s^4 - T_∞^4)

where:

Given:

T_∞ = 20°C

T_sur = 40°C

h = 20 W/(m²·K)

α = 0.78

T_s = 24°C

Let's calculate the convective and radiative heat fluxes:

1. Convective heat flux:

q_conv = 20 * (24 - 20) = 80 W/m²

2. Radiative heat flux:

q_rad = 0.78 * 5.67 × 10^-8 * ((24 + 273.15)^4 - (20 + 273.15)^4) = 238.17 W/m²

Therefore, the convective heat flux to the wall at x = 0 is 80 W/m², and the radiative heat flux is 238.17 W/m².

In this scenario, the one-dimensional plane wall is exposed to both convective and radiative heat transfer. The convective heat transfer occurs due to the temperature difference between the surface of the wall (T_s) and the surrounding ambient temperature (T_∞), which is characterized by the convective heat transfer coefficient (h). The convective heat flux represents the amount of heat transferred per unit area through convection.

The radiative heat transfer, on the other hand, occurs due to the temperature difference and the emissivity/absorptivity of the wall's exposed surface (α). The radiative heat flux depends on the Stefan-Boltzmann constant (σ) and the fourth power of the temperatures involved (T_s and T_∞). The emissivity/absorptivity of the surface determines how efficiently it radiates and absorbs thermal radiation.

In this case, since the surroundings are considered "large," we assume that the temperature of the surroundings (T_sur) remains constant and unaffected by the wall's heat transfer. This assumption simplifies the analysis by focusing only on the heat exchange between the wall and the ambient environment.

By applying the equations for convective and radiative heat fluxes, we can calculate the respective contributions to the total heat transfer at the wall's exposed surface.

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1. Each Qwiklab takes one hour to complete. 2. The statement is true. 3. True, loading a functioning Linux or Windows OS instance can take some time. 4. The correct option is C i.e., Click the red "End Lab" button.

The operating system is the most important piece of software that operates on a computer. It controls the hardware, software, processes, and memory of the computer. Using this technique, you can communicate with the computer even if you don't comprehend its language.

The operating system is the most important element of computer software. It manages the memory and processes on the computer as well as all of the hardware and software. Additionally, it enables you to communicate with the computer without any prior computer language expertise.

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The pump horsepower required per mile of pipe is approximately 68.3 times the pump efficiency.

To estimate the pump horsepower required per mile of pipe, we can use the following formula:

P = Q * rho * H * L / (3960 * eff)

Where P is the pump horsepower, Q is the flow rate in barrels per day, rho is the density of the oil, H is the viscosity of the oil, L is the length of the pipe in feet, and eff is the pump efficiency.

Plugging in the given values, we get:

P = 590,000 * 1.67 * 1.11e-5 * 5280 / (3960 * eff)
P = 68.3 * eff

Therefore, the pump horsepower required per mile of pipe is approximately 68.3 times the pump efficiency.

As for the second part of the question, it is not unusual for crude oil to be transported at elevated temperatures to reduce its viscosity and make it easier to pump. However, given that the Keystone Pipeline is located in a cold environment, it is likely that the oil is cooled before it is delivered to its destination.

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Answer:

Some good ways and popular ways are:

1. Investing in data literacy.

2. Think about cybersecurity.

3. Choosing good and the right tools.

4. Establish metrics that matter.

Explanation:

These all are helpful.

(Hope this helped! Have a great day.)

Technician A is wrong.

Technician B is right.

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Answer:

0.00695 A

Explanation:

µ represents \(10^{-6}\). Multiply this by 6,950.

Small particles in your power steering system may Damage the hydraulic pump.

Check more about the reason for the above in the power steering system below:

If a person is said to have notice that there is  small black particles that can be seen inside of the power steering reservoir, there is therefore some chances that are these are some piece of the hose, instead of  the pump or rack.

Since High-temperature pulsations can be able to lead the power steering hoses to be destroyed  from the inside.

Therefore, based on the above, one can say that Small particles in your power steering system may Damage the hydraulic pump.

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Answer:

A

Explanation:

Answer:

A

Explanation:

The fuse breaks the circuit if a fault in an appliance causes too much current to flow. The fuse contains a piece of wire that melts easily. If the current going through the fuse is too great,the wire heats up until it melts and breaks the circuit

The quartiles divide a set of observations into four portions, each representing 25% of the observations, together with the minimum and maximum values of the data set. The interquartile range, a measurement of variation around the median, is calculated using quartiles.

Statistical file: {2, -6, 5}

Quartile Q1: -6

Quartile Q2: 2

Quartile Q3: 5.

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Answer:

See explanation below

Explanation:

Hypo-eutectoid steel has less than 0,8% of C in its composition.

It is composed by pearlite and α-ferrite, whereas Hyper-eutectoid steel has between 0.8% and 2% of C, composed by pearlite and cementite.

Ferrite has a higher tensile strength than cementite but cementite is harder.

Considering that hypoeutectoid steel contains ferrite at grain boundaries and pearlite inside grains whereas hypereutectoid steel contains a higher amount of cementite, the following properties are obtainable:

Hypo-eutectoid steel has higher yield strength than Hyper-eutectoid steel

Hypo-eutectoid steel is more ductile than Hyper-eutectoid steel

Hyper-eutectoid steel is harder than Hyper-eutectoid steel

Hypo-eutectoid steel has more tensile strength than Hyper-eutectoid steel.

When making a knife or axe blade, I would choose Hyper-eutectoid steel alloy because

1. It is harder

2. It has low cost

3. It is lighter

When making a die to press powders or stamp a softer metals, I will choose hypo-eutectoid steel alloy because

1. It is ductile

2. It has high tensile strength

3. It is durable

Answer:

(a)

Steels having carbon within 0.02% – 0.8% which consist of ferrite and pearlite are known as hypoeutectoid steel.

Steels having greater than 0.8% carbon but less than 2.0% are known as hypereutectoid steel.

(b)

The proeutectoid ferrite formed at a range of temperatures from austenite in the austenite+ferrite region above 726°C. The eutectoid ferrite formed during the eutectoid transformation as it cools below 726°C. It is a part of the pearlite microconstiutents . Note that both hypereutectoid and hypoeutectoid steels have proeutectoid phases, while in eutectoid steel, no proeutectoid phase is present.

Proeutectoid signifies is a phase that forms (on cooling) before the eutectoid austenite decomposes. It has a parallel with primary solids in that it is the first phase to crystallize out of the austenite phase. If the steel is hypoeutectoid it will produce proeutectoid ferrite and if it is hypereutectoid it will produce proeutectoid cementite.  The carbon concentration for both ferrites is 0.022 wt% C.

(c)

(i) Yield strength: The hypoeutectoid steel have good yield strength and hypereutectoid steels have little higher yield strengh.

(ii) Ductility: The hypoeutectoid steel is more ductile and the ductility has decreased by a factor of three for the eutectoid alloy. In hypereutectoid alloys the additional, brittle cementite on the pearlite grain boundaries further decreases the ductility of the alloy. The proeutectoid cementite restricts plastic deformation to the ferrite lamellae in the pearlite.

(iii) Hardness:  hypoeutectoid steels are softer and hypereutectoid steel contains low strength cementite at grain boundary region which makes it harder than hypoeutectoids.

(iv) Tensile strength: Grain boundary regions of hypereutectoid steel are high energy regions prone to cracking because of cementite in the grain boundaries, its tensile strength decreases drastically even though pearlite is present. Hypoeutectoid steel contains ferrite at grain boundaries and pearlite inside grains, so grain boundaries being the high energy state region, it has a higher tensile strength.

(d)

I would recommend hypereutectoid steel alloy to make a knife or ax blade

1- Hardness is required at the surface of the blades.

2- Ductility is not needed for such application.

3- Due to constant impact, the material will not easily yield to stress.

(e)

I would choose a hypoeutectoid steel alloy to make a steel that was easy to machine.

1- hypoeutectoid steel alloys have high machinability, hence better productivity

2- It will be used on softer metals, hence its fitness for the application

3- Certain amount of ductility is required which hypoeutectoid steel alloys possess.

Explanation:

See all together above

Answer:

Option B (Cold working) would be the correct alternative.

Explanation:

The other possibility isn't linked to the given scenario. Therefore the alternative above is the right one.

The first performance tool a technician would use to check if a computer has sufficient hardware resources for a specific application is a resource monitor.

A resource monitor is the ideal tool for a technician to assess hardware resource utilization on a computer. It provides real-time information about CPU usage, memory consumption, disk activity, and network traffic. By analyzing these metrics while the particular application is executing, the technician can determine if the computer has enough CPU power, memory, and disk space to handle the application's requirements.

If the resource monitor shows high CPU usage, excessive memory consumption, or heavy disk activity, it may indicate that the computer is struggling to meet the demands of the application. The technician can then consider upgrading the hardware or optimizing system settings to ensure smooth performance.

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cast.set("Mackenzie Foy", "Young Murph");

To add "Mackenzie Foy" to the cast map with the key "Young Murph", you can use the following code:

```javascript
let cast = new Map();
cast.set("Mackenzie Foy", "Young Murph");
```

Now, the cast map has the key-value pair ("Mackenzie Foy", "Young Murph").

JavaScript is a scripting language that enables you to create dynamically updating content, control multimedia, animate images, and pretty much everything else.Key differences between Java and JavaScript: Java is an OOP programming language while Java Script is an OOP scripting language. Java creates applications that run in a virtual machine or browser while JavaScript code is run on a browser only. Java code needs to be compiled while JavaScript code are all in text.

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The volumetric flow rate into the tank if the wine level remains constant at 2.5m is; 0.1232 m³/s

The formula for volumetric flow rate is;

V' = area * velocity

We are given;

Diameter; d = 15 cm = 0.15 m

Thus;

Area; A = (π/4)d²

A = (π/4)(0.15)²

A = 0.0176 m²

Velocity is; v = √(2gh)

v = √(2 * 9.8 * 2.5)

v = 7 m/s

Thus volumetric flow rate is;

V' = 0.0176 * 7

V' = 0.1232 m³/s

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Explanation:

commands to be and function arguments

The two (2) things that must be included in your function definition are: D. A function name and function variables.

A function can be defined as a set of statements that comprises executable codes and can be used in a software program to calculate a value or perform a specific task on a computer.

In Computer programming, there are two (2) things that must be included in a function definition and these include the following:

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The work done and the heat supplied by the process are equal to -2237.08 kJ and 2237.08 kJ, respectively.

The fluid of 0. 25kg at a pressure of 3 bar with specific internal energy and specific volume of 2544 kJ/kg and 0.6057 m^3/kg, respectively, is heated reversibly at constant pressure until the specific internal energy is 807 kJ/kg and the volume occupied is 0.2189 m^3.

Work done (W) = nRT (V1 - V2)

= 0.25 x 8.314 x 2544 (0.6057 - 0.2189)

= -2237.08 kJ

Heat supplied (Q) = U2 - U1

= 807 - 2544

= -2237.08 kJ

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The values of sum1 and sum2 if the operands in the expressions are evaluated left to right are; sum1 = 46 and sum2 = 48

In computer programming, the term operand is defined as the object of a mathematical operation or it is the object or quantity that is operated on.

A) If the operands in the expressions are evaluated left to right, then the values of sum1 and sum2 are;

sum1 = (10/2) + 41 = 46

sum2 = 41 + (14/2) = 48

B) If the operands in the expressions are evaluated right to left, then the values of sum1 and sum2 are;

sum1 = (14/2) + 41 = 48

sum2 = 41 + (10/2) = 46

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Answer:

Engineers work in a variety of fields to analyze, develop and evaluate large-scale, complex systems. This can mean and improve and maintaining current systems or creating brand new projects. Engineers will design and draft blueprints, visit systems in the field and manage projects.

Answer:

Engineers work in a variety of fields to analyze, develop and evaluate large-scale, complex systems.

Explanation:

It is not safe to operate the motor with a 10-hp shaft load.

Explanation :
Solution-
First, determine the rated speed and current of the motor:
Rated speed = 850 r/min
Rated current = 500 V / (0.1414 ohms + 0.0412 ohms + 0.0189 ohms) = 30.22 A
Next, determine the speed and current for a 10-hp shaft load:
Speed = 1700 r/min
Current = 10 hp / (75 hp * 0.902) = 0.1343 A
Finally, determine whether it is safe to operate the motor with the 10-hp shaft load using the magnetization curve. we can see that the current at 1700 r/min is 0.1264 A, which is less than the current for the 10-hp shaft load (0.1343 A).

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There may be a thrust bearing on either side of the engine.

In a crankshaft, rod bearings are the bearings that allow the connection rod to rotate on the crankshaft. Tech A and Tech B both provided information regarding the same topic. Tech A says that rod bearings are held in place by the torque of the rod bolts crushing the bearing into the big end of the rod bore. Tech B says that one of the connecting rod bearings acts as a thrust bearing for the engine. Let's discuss each one in brief:Tech A is correct: The correct answer is Tech A, as Rod bearings are held in place by the torque of the rod bolts crushing the bearing into the big end of the rod bore. The big end of the connecting rod has a machined half-moon-shaped hole in it, which is where the bearing shells are mounted. As torque is applied to the rod bolts, the conrod is pulled tight against the crankshaft, crushing the bearing shells between the rod and the crankshaft.Tech B is not completely accurate: Though it is true that some connecting rod bearings serve as thrust bearings, it is not entirely accurate. The thrust bearing is generally located between the front of the engine block and the rear of the crankshaft, and its purpose is to resist forward and backward crankshaft movement due to combustion pressures.

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Electrical symbols are the ones most frequently used in circuit diagrams. Amplifiers (shown by triangle shapes) in your circuit amplify the output signal. Your system's capacitors (parallel lines) store energy, but resistors (zigzag lines) lower current flow.

Electrical terminology uses the following five symbols: switch, wire, contactor, motor, and transformer. Any electrical sketch can use these icons.

A basic circuit has the three elements that are required to establish a working electric circuit, namely a source of voltage, a conductive route, and a resistor. The route an electric current takes is known as a circuit.

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Answer:

La diferencia media logarítimica de temperatura del intercambiador en paralelo es aproximadamente 75.466 ºC.

La diferencia media logarítmica de temperatura del intercambiador en contracorriente es aproximadamente 97.881 ºC.

Explanation:

De la teoría de Transferencia de Calor tenemos que un intercambiador de calor en paralelo presenta las siguientes dos características:

1) Tanto el fluido caliente como el fluido frío entran por el mismo lado.

2) Tanto el fluido caliente como el fluido frío salen por el mismo lado.

Mientras que el intercambiador de calor en contracorriente tiene que:

1) El fluido caliente y el fluido frío entran por lados opuestos.

2) El fluido caliente y el fluido frío salen por lados opuestos.

A continuación, anexamos los esquemas de perfil de cada intercambiador.

Ahora, la Diferencia Media Logarítimica de Temperatura (\(\Delta T_{lm}\)), medida en grados Celsius, queda definida como sigue:

\(\Delta T_{lm} = \frac{\Delta T_{1}-\Delta T_{2}}{\ln \frac{\Delta T_{1}}{\Delta T_{2}} }\) (Eq. 1)

Donde \(\Delta T_{1}\) y \(\Delta T_{2}\) son las diferencias de temperatura de los fluidos en cada extremo del intercambiador, medido en grados Celsius.

Procedemos a determinar esas diferencias y la Diferencia Media Logarítimica de Temperatura para cada configuración:

Intercambiador en paralelo

\(\Delta T_{1} = 180\,^{\circ}C-3\,^{\circ}C\)

\(\Delta T_{1} = 177\,^{\circ}C\)

\(\Delta T_{2} = 78\,^{\circ}C - 55\,^{\circ}C\)

\(\Delta T_{2} = 23\,^{\circ}C\)

\(\Delta T_{lm} = \frac{177\,^{\circ}C-23\,^{\circ}C}{\ln \frac{177\,^{\circ}C}{23\,^{\circ}C} }\)

\(\Delta T_{lm} \approx 75.466\,^{\circ}C\)

La diferencia media logarítimica de temperatura del intercambiador en paralelo es aproximadamente 75.466 ºC.

Intercambiador en contracorriente

\(\Delta T_{1} = 180\,^{\circ}C-55\,^{\circ}C\)

\(\Delta T_{1} = 125\,^{\circ}C\)

\(\Delta T_{2} = 78\,^{\circ}C-3\,^{\circ}C\)

\(\Delta T_{2} = 75\,^{\circ}C\)

\(\Delta T_{lm} = \frac{125\,^{\circ}C-75\,^{\circ}C}{\ln \frac{125\,^{\circ}C}{75\,^{\circ}C} }\)

\(\Delta T_{lm} \approx 97.881\,^{\circ}C\)

La diferencia media logarítmica de temperatura del intercambiador en contracorriente es aproximadamente 97.881 ºC.