Steam enters an adiabatic turbine at 10 mpa and 500°c and leaves at 10 kpa with a quality of 90 percent. neglecting the changes in kinetic and potential energies, determine the mass flow rate required for a power output of 5 mw.

Answer: None of the above

Explanation:

Business process modeling refers to the graphical representation of the business processes of a company, which is vital in the identification of potential improvements.

Business pticess modelling can be done through graphing methods, like data-flow diagram, flowchart etc. It is vital as business managers can effectively and quickly communicate their ideas.

It also enhances the customization of business processes, enhances the competitive advantage and enhances the process communication as well.

Therefore, the answer to the question will be "None of the above".

It might be difficult to read damaged instruments, especially older analogue "round gauges" common in crash-landed aircraft.

These difficulties include:

Instruments and switches can offer useful information with regard to the gathering of compelling and single source evidence of crash dynamics.

Modern aircraft come with sophisticated cockpit voice recorders (CVR) and flight data recorders (FDR), which record audio and data, respectively, during flight.

Personal handheld digital devices discovered at the site can have information that is helpful to the accident investigation.

Thus, to ensure the authenticity and acceptability of the information gathered for accident investigation purposes, it is crucial to consult professionals and adhere to correct processes when recovering and analysing data from personal portable devices.

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Dual-element fuses' derating due to rising ambient temperatures The permitted ampacities for insulated copper are shown in the table below.

What is a dual element?

Ranges with a gas cooktop and an electric oven combine the best qualities of both. These are called dual fuel stoves. Quick heating and extensive temperature control are features of gas cooktops. Electric oven elements cycle in ways that enable the best temperature control, making them perfect for baking. Both a broiler and a baker are included in this range. But when cooking in the "Bake" mode, heat is largely generated by the lower element while also coming from the upper element. The Dual Zone burner contains two circular burners that produce both direct and indirect heat, and it is controlled by a single control knob. Food can be made to cook by positioning it inside the outer ring.

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For 5°C and For 35°C the backwash velocity can be between 1.5-2 times the minimum fluidization velocity.  backwash rate will be between 0.069 m/s and 0.092 m/s for For 5°C and 0.078 m/s and 0.104 m/s For 35°C.

The terminal velocity and minimum fluidization velocity of filter sand can be calculated using the following equations:

Terminal velocity (Vt):

Vt = [4 × (ρp - ρf) × g × dp²] / (3 × Cd × ρf)

Minimum fluidization velocity (Umf):

Umf = [((1 - ε) × g × dp³ × (ρp - ρf)) / (150 × μ × ε³)]¹/⁴

Where:

ρp is the density of the filter sand particles (assumed to be 2650 kg/m³)

ρf is the density of the fluid (water, assumed to be 1000 kg/m³)

g is the acceleration due to gravity (9.81 m/s²)

dp is the effective diameter of the filter sand (0.50 mm)

Cd is the drag coefficient (assumed to be 0.44 for a smooth sphere)

ε is the porosity of the filter bed (0.45)

μ is the dynamic viscosity of the fluid (1.787 x 10⁻³ Pa s at 5°C, and 1.138 x 10⁻³ Pa s at 35°C)

Substituting the given values, we get:

For 5°C:

Vt = [4 × (2650 - 1000) × 9.81 × (0.0005)²] / (3 × 0.44 × 1000)

= 0.037 m/s

Umf = [((1 - 0.45) × 9.81 × (0.0005)³ × (2650 - 1000)) / (150 × 1.787 × 10⁻³ × 0.45³)]¹/⁴

= 0.046 m/s

Backwash velocity can be between 1.5-2 times the minimum fluidization velocity, thus the appropriate backwash rate will be between 0.069 m/s and 0.092 m/s.

For 35°C:

Vt = [4 × (2650 - 1000) × 9.81 × (0.0005)²] / (3 × 0.44 × 1000)

= 0.042 m/s

Umf = [((1 - 0.45) × 9.81 × (0.0005)³ × (2650 - 1000)) / (150 × 1.138 x 10⁻³ × 0.45³)]¹/⁴

= 0.052 m/s

Backwash velocity can be between 1.5-2 times the minimum fluidization velocity, thus the appropriate backwash rate will be between 0.078 m/s and 0.104 m/s.

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(a) The percent thermal efficiency of the cycle is 33.3%.

(b) The back work ratio is 0.25.

(c) The net power developed is 4.5 kW.

Answer: y' = - x'

Explanation:

Let f(x) = 2x + y

then f'(x) = 2 + y'

Let f(y) = x - 2y

then f'(y) = x' - 2

Given:  f'(x) + f'(y) = 0

         2 + y' + x' - 2 = 0

                y' + x'= 0

                 y' = -x'

This can also be written as:       \(\dfrac{dy}{dx}=-\dfrac{d}{dx}\)

Answer:

D.  Resistance = 30 ohms

Explanation:

Using Ohm's law

V = I times R

Given:

V = 60 V

I = 2 A

Resistance = V / I = 60 V / 20 A

Resistance = 30 ohms

Answer:

shaft diameter = \(\sqrt[3]{0.3512}\) mm = 0.7055 mm

Explanation:

Ratio of inside diameter to outside diameter ( i.e. d/D )= 0.6

Shear stress of material ( Z ) ≤ 500 KPa

power transmitted by shaft ( P ) = 1.5MW of mechanical power

Revolution ( N ) = 1500 rev/min

Calculate shaft Diameter

Given that: P = \(\frac{2\pi NT}{60}\)  ---- 1

therefore; T = ( 1.5 *10^3 * 60 ) / ( 2\(\pi\) * 1500 )  = 9.554 KN-M

next

\(\frac{T}{I_{p} } = \frac{Z}{R}\)

hence ; T = Z\(_{p} *Z\)

attached below is the remaining part of the solution

Answer: Can't help

Explanation:

Answer:

What are you talking about? pls explain and then maybe I can help.

Answer:

The torque at the end of the beam is 2.5 Nm

Explanation:

Given;

length of beam, r = 0.5 m

applied force, F = 5 N

The torque at the end of the beam is given by;

τ = F x r

where;

τ  is the torque

F is applied force

r is length of the beam

τ = 5 x 0.5

τ = 2.5 Nm

Therefore, the torque at the end of the beam is 2.5 Nm

The new plant would require 4 forging cells.

The proportion is  10.07%.

Total available working time per year:

24 hours/day * 5 days/week * 50 weeks/year

= 6,000 hours/year

Total parts required including scrap:

800,000 forgings / (1 - 0.03)

= 800,000 / 0.97

≈ 824,742 forgings

Number of batches required:

824,742 forgings / 1,250 parts per batch

≈ 660 batches

Total forging time (excluding changeover time) for 824,742 parts:

1.5 minutes/part * 824,742 parts * (1 hour / 60 minutes)

≈ 20,618.55 hours

Total changeover time for 660 batches:

660 batches * 3.5 hours/changeover

≈ 2,310 hours

Total time required to produce 824,742 forgings, including changeovers:

20,618.55 hours + 2,310 hours

≈ 22,928.55 hours

Now, considering the availability of each cell during operation (96%):

Effective operation time required

= 22,928.55 hours / 0.96

≈ 23,883.49 hours

Now, we can determine the number of forging cells needed to meet production requirements:

Number of cells = Total time required / Total available working time

= 23,883.49 hours / 6,000 hours/year

≈ 3.98

Therefore, the new plant would require 4 forging cells.

B. Proportion of time spent in setup = 2,310 hours / 22,928.55 hours ≈ 0.1007

The proportion of time spent in setup for each batch is approximately 10.07%.

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The electronic components that could be used in making of energy saving switch are:​

Transistors as  well as other kinds of semiconductor devices are known to be tools or element that can be used in the making of switches. In its use or applications.

Note that the base or gate of a transistor, based  on the kind of transistor that is known to be in use, and it is one that is often employed as a form of control element to be able to switch on as well as switch off the current that often exist between the emitter as well as the collector or the source and that of the drain.

Note also that the electronic components that are able to  store energy in regards to electronic devices, are known to be capacitors and coils (inductors) and they often play a key function of temporarily saving  energy. One key role of a capacitor is to save an electric charge.

Therefore, The electronic components that could be used in making of energy saving switch are:​

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A monitor is implemented as a BCPL procedure, and any associated operations are declared as additional procedures inside the monitor. Prior to the procedure declarations, condition variables, initialized in the monitor body.

While smart TVs have special technology built in to improve picture quality, a smart monitor is smaller, so picture quality will still be sharp and clear. For games and watching movies, you also have a variety of viewing angles. A monitor is especially useful when there are several people involved in the adult's life or named in the Agreement. Meeting scheduling and information exchange can be facilitated by the monitor. If there is already tension or dispute between any of the persons involved, a monitor is also crucial. Monitors display data that has been processed by the computer's video card. Output is the term used to describe the processing outcome displayed on the monitor. The output gives you immediate feedback by displaying text and graphic pictures while you work or play.

A monitor to implement

monitor bounded_stack

buffer[max] // may contain at the most, max elements

full_slots = 0; // 0 < full_slots < max

condition notempty, notfull

deposit(x) {

if (full_slots == max) notfull.wait // or discard

stack.push(x)

full_slots = full_slots + 1

notempty.signal

}

remove(x) {

if (full_slots == 0) notempty.wait

stack.pop(x)

full_slots = full_slots - 1

notfull.signal

}

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Answer:

ABS: Antilock Braking System – Prevents the wheels from locking during emergency braking.

ACC: Adaptive Cruise Control – Monitors traffic ahead and reduces or increases the car's speed based on the flow of traffic.

AEB: ...

AWD: ...

BHP: ...

CVT: ...

DDI: ...

DFI:

Explanation:

hope it helpssss

Answer:

ABS: Antilock Braking System – Prevents the wheels from locking during emergency braking.

ACC: Adaptive Cruise Control – Monitors traffic ahead and reduces or increases the car's speed based on the flow of traffic.

AEB: ...

AWD: ...

BHP: ...

CVT: ...

DDI: ...

DFI:

Answer:

Under cold air standard conditions, the thermal efficiency of this cycle is 56.9 percent.

Explanation:

From Thermodynamics we remember that thermal efficiency of the ideal Otto cycle (\(\eta_{th}\)), dimensionless, is defined by the following formula:

\(\eta_{th} = 1-\frac{1}{r^{\gamma-1}}\) (Eq. 1)

Where:

\(r\) - Compression ratio, dimensionless.

\(\gamma\) - Specific heat ratio, dimensionless.

Please notice that specific heat ratio under cold air standard conditions is \(\gamma = 1.4\).

If we know that \(r = 8.2\) and \(\gamma = 1.4\), then thermal efficiency of the ideal Otto cycle is:

\(\eta_{th} = 1-\frac{1}{8.2^{1.4-1}}\)

\(\eta_{th} = 0.569\)

Under cold air standard conditions, the thermal efficiency of this cycle is 56.9 percent.

Saws, files, chisels, and planes belong to the category of hand tools, specifically woodworking hand tools. These tools are commonly used in woodworking and carpentry tasks to shape, cut, and refine wood.

Each tool serves a specific purpose:Saws: Saws are cutting tools with a toothed blade used to cut through various materials, including wood, metal, and plastic. In woodworking, hand saws such as crosscut saws, rip saws, and coping saws are commonly used.Files: Files are abrasive tools with a rough surface made of hardened steel. They are used to shape and smooth surfaces of various materials, including wood, metal, and plastic. Woodworking files typically have a specific shape or pattern, such as flat, half-round, or round.Chisels: Chisels are cutting tools with a sharp blade at the end of a handle. They are used to remove or carve out material from wood, creating precise cuts, mortises, or sculptural details. Chisels come in various shapes and sizes, including bench chisels, mortise chisels, and carving chisels.Planes: Planes are tools used to shape and smooth wooden surfaces. They consist of a cutting blade set at an angle within a flat base. By pushing or pulling the plane across the wood's surface, it shaves off thin layers to create a smooth and even finish. Different types of planes include bench planes, block planes, and shoulder planes.All these tools require manual operation and are essential for traditional woodworking techniques .

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Answer:

lol Jebus?

Explanation:

Answer:

addsa

sdadsa

Explanation:

dsads

Geo environmental engineering is a branch of engineering that deals with the interaction between the geologic and environmental aspects of a project. It involves an understanding of the geological environment and how it affects the project and the environment.

It encompasses soil mechanics, rock mechanics, hydrology, and environmental engineering. Geo environmental engineering is also concerned with understanding the behavior of soil under different conditions, including the multiphase behavior of soil. Soil behavior is of interest to geo environmental engineering because the presence of water or other liquids in the soil affects its strength, stiffness, and compressibility.

Multiphase behavior of soil involves the interaction between soil, water, and air in the soil. Soil in its natural state can contain different amounts of water and air, and the behavior of soil depends on the relative proportions of these different phases.

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The stress in the belt is calculated as follows: Stress in belt = 212.12 N / (14.14 mm x 10 mm) = 1.50 N/mm2.

Just "add 2" to your off-the-rack trouser size will yield the belt size. For instance, a 38-belt size will be a good bet if you wear 36" waist trousers. Most people will discover that this straightforward method works best when pants are worn at the customary height—near the natural waistline.

Power = (Tension in belt) x (Belt speed)

Tension in belt = (Power x 1000) / (Belt speed x Angle of lap x Coefficient of friction)

Substituting the given values, we get:

Tension in belt = (37,000 x 1000) / (1500 x 165 x 0.3) = 212.12 NEffective tension in belt = 15 N/mm x Belt width

Belt width = Effective tension in belt / 15 N/mmBelt width = 212.12 N / 15 N/mm = 14.14 mm

Stress in belt = Tension in belt / (Belt width x Belt thickness)

Substituting the given values, we get:

Stress in belt = \(212.12 N / (14.14 mm x 10 mm) = 1.50 N/mm^2\)

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Answer:

  any of AND, NOR, XOR

Explanation:

An AND gate will give a Low output for any input Low. The logic circuit could be an AND gate.

A NOR gate will give a Low output for any input High. The logic circuit could be a NOR gate.

An XOR gate will give a Low output for an even number of High inputs. The logic circuit could be an XOR gate.

The logic circuit could be any of ...

  AND or NOR or XOR

__

Additional comment

What you consider a "basic" gate is not defined here. All of these are catalog items. If you consider only AND, OR, and NOT to be the basic gates, then your answer is AND.

The graph of the function y = (sin x)/(x * 2-1) is given in the attached image.

Recall that a graph is a visual representation of data or depiction of the mathematical relationships between various factors in a mathematical function.

A function in math is a rule, or theorem defining a connection between one variable (the independent variable) and another (the dependent variable).

It is to be noted that functions are common in mathematics and are required for the formulation of physical connections in the sciences.

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Note that a milling machine can do all of the following tasks except: " ream a highly accurate hole with a fine finish" (Option D).

A milling machine is a tool that is used to machine flat and contoured surfaces, and it can also be used to rotate a multitoothed cutter into the workpiece to remove material. It is also capable of machining threads, gears, and spirals.

However, a milling machine is not typically used to ream holes, as reaming is a finishing operation that is usually done with a reamer, which is a specialized tool designed for this purpose. Reaming is used to create highly accurate holes with a fine finish, and it is typically done after the hole has been drilled or milled to its rough size.

There are several advantages to using a milling machine, including:

Versatility: Milling machines can be used to machine a wide range of parts and materials, including flat and contoured surfaces, threads, gears, and spirals.

Precision: Milling machines are capable of producing highly precise parts, as they can be accurately positioned and controlled using computer numerical control (CNC) technology, etc.

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Based on this information, it can be inferred that Plant X likely has better quality control for the asphaltic material compared to Plant Y.

Based on the results of the analysis, we can conclude that Type B tyres have a higher mean wear than Type A tyres.

A peaked bell-shaped distribution indicates that the data points are concentrated around the mean, with fewer extreme values. This suggests that Plant X has tighter quality control, resulting in more consistent asphaltic material. On the other hand, a flat distribution indicates a larger spread of values, potentially including more extreme values. This suggests that Plant Y has less consistent quality control.

The sample mean wear for Type A tyres is  9.6 and the sample mean wear for Type B tyres is 10.1. The sample standard deviation for Type A tyres is 0.7 and the sample standard deviation for Type B tyres is 0.7.

The 95% confidence interval for the difference between the means of the populations of the wear of the tyres is given by:

(9.6 - 10.1) ± 1.96 * √(0.7^2 / 5 + 0.7^2 / 5)

= -0.5 ± 0.84

= (-1.34, -0.66)

The p-value for the hypothesis test is given by:

p-value = 2 * (1 - NORMSDIST(0.5 / 0.84))

= 0.046

Since the p-value is less than 0.05, we can reject the null hypothesis and conclude that there is a significant difference between the means of the populations of the wear of the tyres.

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Technician a is correct. The car may veer off course due to loose ball joints. The propensity of a vehicle to veer from one side of the road to the other is known as wander.

The propensity of a vehicle to veer from one side of the road to the other is known as wander. Uneven tire pressure or mismatched tires are potential causes number one. Linkage binding or inadequate lubrication is a potential cause number two. The third potential factor is binding or inadequate lubrication of the steering gear.

The bottom line is that before you start to address poor steering performance, you must identify the underlying source of the issue. Oversteer and understeer have been discussed separately, but we've also included 10 other common steering issues with their likely causes below. You can stop cursing and start fixing your steering problems by locating potential problem areas.

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Answer:

Explanation:

Temperature of atmospheric air To = 25°C = 298 K

Free  stream velocity of air Vo = 25 m/s

Length and width of plate = 1m

Temperature of plate Tp = 125°C = 398 K

We know for air, Prandtl number Pr = 1

And for air, thermal conductivity K = 24.1×10?³ W/mK

Here, charectorestic dimension D = 1m

Given value of Reynolds number Re = 105

For laminar boundary layer flow over flat plate

= 3.402

Therefore, hx = 0.08199 W/m²K

So, heat transfer rate q = hx×A×(Tp – To)

                                          = 0.08199×1×(398 – 298)

In an ignition coil, the primary and secondary windings are responsible for generating and amplifying high voltage electrical energy that is required to create a spark in the engine's combustion chamber.

The primary winding is typically made up of a thicker gauge wire with fewer turns, whereas the secondary winding is made up of a thinner gauge wire with more turns.

In response to the given statement, Technician A says that the primary winding has more windings, whereas Technician B says that the secondary winding has more windings. However, both statements cannot be true at the same time. Therefore, one of the technicians is correct, and the other is wrong.

In reality, Technician B is correct. The secondary winding has more windings than the primary winding. The reason for this is that the ignition coil operates on the principle of electromagnetic induction, which states that a change in the magnetic field around a conductor induces an electrical current in that conductor. When a current flows through the primary winding, it generates a magnetic field around it. This magnetic field then induces a current in the secondary winding, which is connected to the spark plug.

The number of windings in the secondary winding is typically much greater than the number of windings in the primary winding. This is because the voltage needed to create a spark in the combustion chamber is much higher than the voltage generated by the battery. The secondary winding acts as a transformer, increasing the voltage generated by the primary winding to a level that is sufficient to create a spark in the engine.

In conclusion, Technician B is correct in stating that the secondary winding has more windings than the primary winding. The primary winding is typically made up of a thicker gauge wire with fewer turns, whereas the secondary winding is made up of a thinner gauge wire with more turns. This design allows the ignition coil to generate high voltage electrical energy needed to create a spark in the engine's combustion chamber.

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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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