5. Explain turn ratio' on control valve. What effect would this have on the control loop? 6. What effect would a valves installed gain have on the overall performance of a typical industrial closed loop control if it were to change? Use simple valve gain curves to elaborate.

b. Ductility property of steel is illustrated by its ability to be shaped into automobile fenders.

Metal will stretch when pulled as opposed to breaking because of its ductility. To put it another way, a material's ductile properties are its ability to undergo significant plastic deformation under tensile stress before rapture. Examples include steel, copper, nickel, and other ductile materials.

Metals' brittleness is what causes them to break under tensile tension without causing stretch. It also goes by the label of sudden failure. Examples of fragile materials include glass, iron, and others.

The ductility of a material, an important consideration in engineering and production, determines its capacity to withstand mechanical overload.

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When the source-to-film distance was increased from 60 in. to 120 in., the geometric unsharpness was improved. This means that the image on the radiograph will be sharper and clearer, making it easier to identify any defects or issues with the weld.

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Energy transfer is factor that affects unit operation this is because each processing of change requires energy transfer.

Unit operations involves reactions that lead to physical change or chemical transformation.

It include separating of mixtures which can be done by filtering, crystallization and polymerization.

Unit operation often results into changes that can be seen or visible changes.

Energy transfer is one the factors that affect unit operation.

Therefore, Energy transfer is factor that affects unit operation this is because each processing of change requires energy transfer.

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Two examples of judgment errors related to anchoring and adjustment are the anchoring effect in pricing and the anchoring effect in negotiations.

The anchoring effect in pricing occurs when consumers base their perception of a product's value on the initial price they see. For example, if a store initially sets a high price for a product and then offers a discount, consumers may perceive the discounted price as a great deal even if it is still higher than the product's actual value. The initial high price acts as an anchor, influencing the consumer's judgment.

The anchoring effect in negotiations occurs when the first offer made in a negotiation sets the reference point for further negotiations. For instance, if a seller sets a high initial asking price, it can anchor the buyer's perception of the item's value and influence subsequent offers. The buyer may adjust their offer based on the initial anchor, potentially leading to higher final prices.

In both examples, the initial anchor (the initial price or offer) influences subsequent judgments or decisions, demonstrating the anchoring and adjustment bias.

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The question is incomplete. The complete question is :

The hydrofoil boat has an A-36 steel propeller shaft that is 100 ft long. It is connected to an in-line diesel engine that delivers a maximum power of 2590 hp and causes the shaft to rotate at 1700 rpm . If the outer diameter of the shaft is 8 in. and the wall thickness is \($\frac{3}{8}$\)  in.

A) Determine the maximum shear stress developed in the shaft.

\($\tau_{max}$\) = ?

B) Also, what is the "wind up," or angle of twist in the shaft at full power?

\($ \phi $\) = ?

Solution :

Given :

Angular speed, ω = 1700 rpm

                              \($ = 1700 \frac{\text{rev}}{\text{min}}\left(\frac{2 \pi \text{ rad}}{\text{rev}}\right) \frac{1 \text{ min}}{60 \ \text{s}}$\)

                              \($= 56.67 \pi \text{ rad/s}$\)

Power \($= 2590 \text{ hp} \left( \frac{550 \text{ ft. lb/s}}{1 \text{ hp}}\right)$\)

          = 1424500 ft. lb/s

Torque, \($T = \frac{P}{\omega}$\)

                 \($=\frac{1424500}{56.67 \pi}$\)

                 = 8001.27 lb.ft

A). Therefore, maximum shear stress is given by :

Applying the torsion formula

\($\tau_{max} = \frac{T_c}{J}$\)

        \($=\frac{8001.27 \times 12 \times 4}{\frac{\pi}{2}\left(4^2 - 3.625^4 \right)}$\)

      = 2.93 ksi

B). Angle of twist :

     \($\phi = \frac{TL}{JG}$\)

         \($=\frac{8001.27 \times 12 \times 100 \times 12}{\frac{\pi}{2}\left(4^4 - 3.625^4\right) \times 11 \times 10^3}$\)

         = 0.08002 rad

         = 4.58°

The instruction that performs division of unsigned numbers is called DIV.

This instruction is used to perform division operations between two unsigned numbers in assembly language programming. The instruction only DIV is used to perform integer division in assembly language.

This instruction performs division of a 16-bit or 32-bit unsigned integer by another 8-bit, 16-bit, or 32-bit unsigned integer .The DIV instruction is used to perform the division operation on two unsigned numbers in assembly programming.

This instruction is a signed integer division instruction. It is used to divide a 16-bit or 32-bit unsigned integer by another 8-bit, 16-bit, or 32-bit unsigned integer and it generates the quotient in AX or DX and remainder in AH or DX.

If the dividend is 16-bit, then the divisor can be either 8-bit or 16-bit. If the dividend is 32-bit, then the divisor can be either 8-bit, 16-bit, or 32-bit.

The DIV instruction in assembly language is used in the following way:DIV destThe destination operand is divided by the AX register and the result is stored in the AX register. The DIV instruction performs unsigned division of two unsigned integers and it generates a quotient and remainder.

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Answer:

Concentric circles

Explanation:

Concentric circles are two or more circles which have the same center point. The region between two concentric circles is called an annulus.

The correct answer is a) Here is a diagram of the milling circuit with labels: The product from the SAG mill is fed to the two ball mills in parallel, with the product from the ball mills being combined and classified in cyclones. The underflow from the cyclones is recycled, and the overflow goes to the flotation circuit.

To calculate the mass rate of each size fraction in each stream, we can start with the total feed rate of 1,500 t/hr. Since 25% of the feed is below 75 microns, that means 75% of the feed is above 75 microns. Of that 75%, 50% will be broken down in the SAG mill to below 75 microns, while the remaining 50% will be broken down in the two ball mills in parallel. Therefore: Mass rate of particles above 75 microns in feed = 1,500 x 0.75 = 1,125 t/hr Mass rate of particles below 75 microns in feed = 1,500 x 0.25 = 375 t/hr Mass rate of particles above 75 microns in SAG mill product = 1,125 x 0.5 = 562.5 t/hr Mass rate of particles below 75 microns in SAG mill product = 1,125 x 0.5 + 375 = 937.5 t/hr Mass rate of particles above 75 microns in each ball mill product = 562.5 / 2 = 281.25 t/hr Mass rate of particles below 75 microns in each ball mill product = 375 / 2 = 187.5 t/hr Mass rate of particles above 75 microns in combined ball mill product = 562.5 t/hr Mass rate of particles below 75 microns in combined ball mill product = 375 t/hr + 187.5 t/hr = 562.5 t/hr Mass rate of particles above 75 microns in cyclone overflow = 562.5 t/hr x 0.95 = 534.38 t/hr (assuming a 5% bypass) Mass rate of particles below 75 microns in cyclone overflow = 562.5 t/hr x 0.05 + 375 t/hr x 0.95 = 393.75 t/hr (assuming a 5% bypass) Mass rate of particles above 75 microns in cyclone underflow/recycle = 562.5 t/hr x 0.05 = 28.13 t/hr Mass rate of particles below 75 microns in cyclone underflow/recycle = 375 t/hr x 0.05 + 187.5 t/hr = 205.63 t/hr To calculate the feed rate to each ball mill, we can use the mass rate of particles below 75 microns in the combined ball mill product, which is 562.5 t/hr. Since there are two ball mills in parallel, the feed rate to each ball mill will be: Feed rate to each ball mill = 562.5 t/hr / 2 = 281.25 t/hr Finally, to calculate the percentage passing 75 microns in the circuit product.

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When the mean velocity doubles, the head loss becomes 32 m per meter of pipe length. The head loss in a laminar flow of a liquid through a pipe can be calculated using the Darcy-Weisbach equation:

Δh = f * (L/D) * (V^2/2g)

Where:

Δh is the head loss

f is the friction factor

L is the pipe length

D is the pipe diameter

V is the velocity of the fluid

g is the acceleration due to gravity

Given:

Diameter of the pipe (D) = 2.5 cm = 0.025 m

Head loss (Δh) at velocity V = 0.5 m/s = 2 m/m

Velocity (V) = 2 * 0.5 m/s = 1 m/s

To find the head loss when the mean velocity doubles, we need to calculate the new head loss (Δh') using the same pipe diameter and new velocity.

Using the Darcy-Weisbach equation, we have:

Δh = f * (L/D) * (V^2/2g)

Since the pipe diameter and length remain the same, and the acceleration due to gravity (g) is constant, we can rewrite the equation as:

Δh = f * (V^2/2g)

Now, let's calculate the new head loss (Δh') at the doubled mean velocity:

Δh' = f * (V'^2/2g)

Where V' = 2 m/s (doubled mean velocity)

To find the relationship between Δh and Δh', we can use the following equation:

Δh / Δh' = (V^2 / V'^2)

Plugging in the given values:

2 m/m / Δh' = (0.5 m/s)^2 / (2 m/s)^2

2 m/m / Δh' = 0.25 / 4

2 m/m / Δh' = 0.0625

Now, rearranging the equation to solve for Δh':

Δh' = 2 m/m / 0.0625

Δh' = 32 m/m

Therefore, when the mean velocity doubles, the head loss becomes 32 m per meter of pipe length.

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The clock period of the pipeline is 2 units of time.

Given a 5-stage pipeline with stages taking 1, 2, 3, 1, and 1 units of time

The clock period of the pipeline is equal to 3 units of time.

For a pipeline with 'n' stages, the clock period is equal to the sum of the time taken by each stage divided by 'n'.

The time taken by each stage of the pipeline is given as:

Stage 1: 1 unit of time

Stage 2: 2 units of time

Stage 3: 3 units of time

Stage 4: 1 unit of time

Stage 5: 1 unit of time

Therefore, the total time taken by all the stages is 1 + 2 + 3 + 1 + 1 = 8 units of time.

The number of stages in the pipeline is 5. Hence, the clock period of the pipeline is:

Clock period = (1 + 2 + 3 + 1 + 1)/5= 8/5= 1.6 units of time.

However, the pipeline must have integer clock cycles. Therefore, the clock period is rounded up to the nearest integer.

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Answer:

A) return.

A return is a separate piece attached to the rear edge of a countertop that extends it vertically to meet the wall. It is used to create a finished look and to protect the wall from water and other spills that may occur on the countertop.

Answer: Apply battery-terminal grease to the terminals to help prevent corrosion. It's available at any auto parts store and usually comes in a little ketchup-like packet. Another great option is AMSOIL Heavy-Duty Metal Protector. It creates a protective coating on terminals that wards off corrosion.

Explanation:

In high-volume turboprop and turbojet engines with two-spool or split compressors, the low-pressure rotor will increase in speed as the compressor load decreases in the lower density air at high altitudes. This is due to the difference in air pressure and density at higher altitudes, which affects the engine's performance.

1)A two-spool or split compressor system features separate high-pressure and low-pressure rotors. These rotors work independently to compress air, which allows the engine to achieve higher efficiency and performance. This design also improves the engine's response to varying altitude and air density conditions.

2)At high altitudes, the air density decreases, leading to a reduction in the compressor load. As the load on the compressor decreases, the low-pressure rotor can increase in speed to maintain the desired compression ratio and engine performance. This increase in speed allows the engine to adapt to the changing conditions and continue operating efficiently.

3)In summary, high-volume turboprop and turbojet engines with two-spool or split compressors can adjust their performance at high altitudes due to the increased speed of the low-pressure rotor as the compressor load decreases in the lower density air. This design feature enhances the engine's adaptability and efficiency in various operating conditions.

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Answer:

True

Explanation:

All big companies are pretty much required in today's day and ages to complete these reports whether they truly believe it.

The running time of this algorithm is O(log n), where n is the number of elements in the array. This is because at each step, the search space is halved, leading to a logarithmic time complexity.

To determine if there exists an integer I in an array of increasing integers such that Ai = I, we can use a modified version of the binary search algorithm. Here's the algorithm:

Set the left pointer left to the first index of the array and the right pointer right to the last index of the array.

Repeat the following steps while left is less than or equal to right:

Calculate the middle index mid as (left + right) / 2.

If the value at index mid is equal to mid, return True as we have found an integer I such that Ai = I.

If the value at index mid is greater than mid, update right to mid - 1 as we only need to search in the left half of the array.

If the value at index mid is less than mid, update left to mid + 1 as we only need to search in the right half of the array.

If the above loop completes without finding a match, return False as there is no integer I such that Ai = I.

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Answer:

COP(heat pump) = 2.66

COP(Theoretical maximum) = 14.65

Explanation:

Given:

Q(h) = 200 KW

W = 75 KW

Temperature (T1) = 293 K

Temperature (T2) = 273 K

Find:

COP(heat pump)

COP(Theoretical maximum)

Computation:

COP(heat pump) = Q(h) / W

COP(heat pump) = 200 / 75

COP(heat pump) = 2.66

COP(Theoretical maximum) = T1 / (T1 - T2)

COP(Theoretical maximum) = 293 / (293 - 273)

COP(Theoretical maximum) = 293 / 20

COP(Theoretical maximum) = 14.65

Answer:

One major way I can think of is competition. Someone who is just starting in the career may have trouble competing with much larger companies or businesses. This may cause them to have to adapt to the tough environment, and if they do not do so, it may cause them to become unsuccessful in the pursuit of their career.

Explanation:

The state must put aside $66,415,794.83 to build the highway, assuming money is invested in an account that earns 3% per year.

Given that the project is a 3-year project, the cost of the project will be $21 million in year 1, and costs are expected to increase by 5% each year due to anticipated design modifications. The state regulations require that the money must be set aside at the beginning of the project to cover the entire project cost. We are to determine the amount of money that should be set aside to build the highway, assuming money is invested in an account that earns 3% per year.To determine the total cost of the project, we can use geometric progression. The first term is 21 and the common ratio is 1.05. Thus the total cost, T = 21 + 21 × 1.05 + 21 × 1.05²We know that the formula for the sum of a geometric progression is given as:S = (a₁(1 - rⁿ))/(1 - r)where;S = the sum of the terms,a₁ = the first term,r = the common ratio,n = the number of termsSubstituting the values of a₁, r and n in the equation above, we get;T = 21 + 21 × 1.05 + 21 × 1.05²= 21(1 + 1.05 + 1.05²)= 21 × (1 - 1.05³)/(1 - 1.05)= 21 × (1 - 1.15763)/(-0.05)= $66,415,794.83.

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The factor of safety for fatigue based on infinite life, using the Goodman criterion, is 256, and the number of cycles, using the Walker criterion to find the equivalent completely reversed stress, is 10⁶. The part will not yield.

FS = S160 / (σa + σm/σu)

Where

FS = S160 / (σa + σm/σu)

FS = 160 MPa / (60 MPa + 20 MPa/400 MPa)

FS = 160 MPa / (0.625)

FS = 256

N = (σa/S160)^(-1/b)

Where

N is the number of cycles

σa is the alternating stress

S160 is the completely adjusted endurance limit

b is the fatigue strength exponent

N = (σa/S160)^(-1/b)

N = (60 MPa/160 MPa)^(-1/0.1)

N = 10⁶

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What is Auto Cad?

Answer : AutoCAD is a commercial computer-aided design (CAD) and drafting software application. Developed and marketed by Autodesk, AutoCAD was first released in December 1982 as a desktop app running on microcomputers with internal graphics controllers.

This is the Answer for your question :3

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Answer:

Answer is "Increasing the number of base stations & reusing the channels"

Explanation:

Answer:

radio capacity may be increased in cellular concept by increasing the number of base stations and reusing the channels.

Answer:

look online

Explanation:

Answer:

6 Ω

Explanation:

given data :

Voltage ( v ) = 12cos ( 60t + 45° )

L = 0.1 H

calculate the inductor's impedance Z  

Z = jXL

  = jx = 60

  = L = 0.1

hence Z = 60 * 0.1 =  6 Ω

The thermal conductivities in the ξ and η directions, kξ and kη, are known for a flat plate of laminated material with laminations making an angle β with the surfaces. To determine the overall thermal conductivity of the plate, we need to consider the effective thermal conductivity in the direction of heat flow. This can be calculated using the rule of mixtures.

The effective thermal conductivity, keff, is given by the equation:

keff = (kξ cos²β + kη sin²β)

Here, cos²β represents the fraction of heat flow in the direction of kξ, and sin²β represents the fraction of heat flow in the direction of kη. By substituting the given values of kξ, kη, and β into the equation, you can calculate the effective thermal conductivity of the laminated plate.

Remember that this calculation assumes that heat flow is one-dimensional and perpendicular to the surfaces. Additionally, it is important to note that this equation applies only to flat plates of laminated material and may not be applicable to other geometries or materials.

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It is recommended to use an anti-seize compound on spark plugs installed in aluminum cylinder heads.

The anti-seize compound helps to prevent the spark plug from seizing in the cylinder head, which can make it difficult to remove the plug for maintenance or replacement.

This is particularly important for aluminum cylinder heads, as aluminum is a soft metal that is more prone to seizing than other materials. The anti-seize compound creates a barrier between the spark plug and the cylinder head, reducing the risk of corrosion and seizure.

However, it is important to use the compound sparingly and only on the threads of the spark plug, as excess compound can interfere with the electrical conductivity of the plug.

Overall, using an anti-seize compound can help to extend the life of spark plugs and make maintenance easier for those working on aluminum cylinder heads.

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