let us consider the filter shown below: a. determine the type of the filter shown below using qualitative analysis. b. determine the filter transfer function. c. determine the filter cut-off frequency

The most common times to perform a seasonal check on heating equipment are in the **fall** before the heating season begins and in the **spring** after the heating season ends.

Performing a seasonal check on heating equipment in the fall allows for early detection of any issues or malfunctions before the colder months when the heating system will be heavily relied upon. It involves inspecting and servicing components such as filters, burners, ignitors, thermostats, and ductwork to ensure they are clean, functioning properly, and ready for the heating season ahead.

Similarly, conducting a seasonal check in the spring helps assess the condition of the heating equipment after a prolonged period of use. This check ensures that the system is properly shut down for the warmer months and identifies any necessary maintenance or repairs that may be needed before the next heating season.

By performing seasonal checks in both the fall and spring, homeowners can maintain the efficiency, reliability, and safety of their heating equipment, extending its lifespan and avoiding potential heating emergencies during the colder months.

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A modern jet airplane equipped with inboard and outboard ailerons plus roll control spoilers is cruising at its normal cruise Mach number. In this scenario, option A is correct.

Both the inboard and outboard ailerons are active, and the spoilers may also be active. The inboard and outboard ailerons work together to control the roll of the aircraft, while the spoilers help to increase drag and reduce lift, allowing for more precise control of the aircraft's roll. So, both of these control surfaces would be active during normal cruise.

Roll control spoilers are also typically installed on modern jet airplanes to assist with roll control. These spoilers can be deployed on one wing to create more drag and lift, which causes the aircraft to roll in the opposite direction. In cruise flight, the spoilers may be used to assist the ailerons in controlling the roll of the aircraft, but they are not usually the primary means of roll control.

The correct option is A.

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Therefore, if the input piston moves 10 inches, the output piston will move 1.11 inches. This shows that the hydraulic press can magnify force and generate high-pressure output with a relatively small input force.

A hydraulic press is a device that utilizes the principle of Pascal's Law to multiply force. According to this law, pressure exerted at one point in a confined fluid is transmitted equally to all other points in the container. In this case, the input cylinder has a diameter of 3 inches and the output cylinder has a diameter of 9 inches.
The formula to calculate the movement of the output piston is based on the ratio of the areas of the input and output cylinders. This means that the output piston will move a distance that is directly proportional to the ratio of the area of the output cylinder to the area of the input cylinder.
Using the formula: Output force = Input force × (Area of output piston/Area of input piston)
We can rearrange the formula to find the distance that the output piston moves, which is:
Distance of output piston = Input distance × (Area of input piston/Area of output piston)
Substituting the values, we get:
Distance of output piston = 10 inches × (π × (3 in)^2)/(π × (9 in)^2)
Distance of output piston = 10 inches × (9/81)
Distance of output piston = 1.11 inches
Therefore, if the input piston moves 10 inches, the output piston will move 1.11 inches. This shows that the hydraulic press can magnify force and generate high-pressure output with a relatively small input force.

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How to Clean Hand Tools:

Cleaning Power Tools:

Answer:

True

Explanation:

Given data:The nitrogen compressor adiabatically compresses 600 ft3/s of N2 from 15 psia and 77°F to 300 psia and 900°F. The nitrogen then passes through a heat exchanger, where it is cooled at constant pressure to an outlet temperature of 150°F.

Assume a calorically perfect gas with specific heats evaluated at the mean temperatures for each process.Power Input (HP) to the compressor We know that, Hence,The power input to the compressor is 4760 HP.Rate of heat transfer (BTU/hr) removed from the nitrogen in the heat exchanger From the ideal gas law,

PV = nRTTherefore, the rate of heat transfer (Q) is:

Q = nCp(T2 – T1) … … (ii)Using the relationship

Q = pV/RT Cp (T2 – T1) … … (iii)For the first process, using

Cp = (7/2) RThus,  Using

Cp = (5/2) RThus, Substituting the given values in equation (iii), we getThe rate of heat transfer (Q) removed from the nitrogen in the heat exchanger is -377.24 × 106 BTU/hr.

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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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When troubleshooting an engine for too rich mixture to allow the engine to idle, the possible cause could be that there is a misadjusted carburetor.

This means that the air/fuel mixture is not adjusted properly and is allowing too much fuel into the engine, causing it to run rich. Another possible cause is that the fuel system may be clogged or dirty, causing the fuel to not be delivered properly to the engine. This can cause the engine to idle poorly and run rich.There are a few things that can be done to troubleshoot this issue.

The first step is to check the carburetor and make sure that it is adjusted properly. The air/fuel mixture should be adjusted so that the engine is running at the correct RPM and there is no hesitation or misfire. If the carburetor is misadjusted, it will need to be corrected.

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Answer:Falling or flying objects on a worksite can expose workers to relatively minor injuries, such as cuts and abrasions, as well as more serious injuries, such as concussions or blindness. Working beneath scaffolds or other areas where overhead work is being performed puts workers at risk from falling objects. Flying objects become a concern when workers are using power tools or performing tasks that involve pushing, pulling or prying.

Explanation:General

Always wear hard hats when work is being performed overhead or when other work conditions call for it.

Stack materials securely to prevent them from sliding, falling or collapsing.

Overhead work

Secure all tools and materials to prevent them from falling on people below.

Use toe boards or guardrails on scaffolds to prevent objects from falling. Alternately, use debris nets or catch platforms to grab falling objects.

Machine use

When working with machines or power tools that can produce flying particles, wear safety glasses, goggles or face shields.

Inspect tools prior to use, and be sure all guards are in place and in good working condition.

Allow only properly trained workers to use power-actuated tools.

Cranes/hoists

Whenever possible, avoid working under moving loads.

Erect barricades and post warning signs at hazardous work zones.

Inspect cranes and hoists prior to use to ensure all components are in good working order, including wire rope, lifting hooks and chains.

Never exceed the lifting capacity of cranes and hoists.

Compressed air

Reduce compressed air for cleaning to 30 psi, and always use proper personal protective equipment and guarding.

Never clean clothing with compressed air.

Answer:

  forever (no solution)

Explanation:

If the figure you're working with is the one shown below, no amount of mixing will reduce the solid content to 35%.

The vessel contains 40% solids.

The incoming feed is ...

  (100 kg/h)×(60% solids) = 60 kg solids/h

out of a total influx of material of ...

  (100 kg/h +70 kg/h) = 170 kg/h

That means the solids content of the inflow is ...

  (60 kg)/(170 kg) = 0.352941 ≈ 35.3%

The solids content cannot ever be less than 35.3%. The problem has no solution.

_____

We suggest you discuss this question with your teacher to see how they would solve it.

It is False to state that in the MVC Architecture the functionality of the system is organized into services, with each service delivered from a separate server. The above scenario occurs within Client-Server Architecture.

Model-view-controller is a software architecture style for building user interfaces that divides the underlying program logic into three interrelated pieces. This is done to isolate internal information representations from how data is communicated to and accepted by the user.

The Model-View-Controller (MVC) architectural pattern, on the other hand, divides an application into three key logical components: the model, the view, and the controller. Every one of these components is designed to handle particular parts of an application's development.

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Heat energy is the known to be a product of the movement of tiny particles called atoms, molecules or ions. The true statement is by Technician A.

Internal combustion engine is dependent on the heat of combustion so as to make torque to move the vehicle and power the system.

A lot of heat made during combustion is not often used productively and therefore need to be removed to avoid overheating of the engine.

The heat energy that is not used for is wasted in three ways: They are:

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195.96 degrees C and  -59.35 kW is the temperature of the steam after the flashing process and the power output from the turbine if the pressure of the steam at the exit of the flash chamber is 1 MPa.

To solve this problem, we need to apply the energy balance and the steam table.

First, we need to determine the state of the geothermal water before the flashing process. Since it enters the flash chamber as a saturated liquid, we can use the steam table to find its properties at the given temperature of 230 degrees C:

h1 = hf + x * hfg = 834.46 kJ/kg (from the steam table)

where h1 is the enthalpy of the geothermal water, hf is the enthalpy of the saturated liquid at 230 degrees C, hfg is the enthalpy of vaporization at 230 degrees C, and x is the quality of the water (which is 0 since it is a saturated liquid).

Next, we need to find the state of the steam after the flashing process. We know that the pressure at the exit of the flash chamber is 1 MPa, and we can assume that the process is adiabatic (no heat transfer). Using the steam table, we can find the enthalpy and quality of the steam at this pressure:

hf = 191.81 kJ/kg (from the steam table)

hfg = 1984.4 kJ/kg (from the steam table)

hg = hf + hfg = 2176.21 kJ/kg

x = (h1 - hf) / hfg = 0.314

where hg is the enthalpy of the saturated vapor at 1 MPa.

Therefore, the temperature of the steam after the flashing process can be found by interpolation:

Tg = 230 + x * (Tsat(1 MPa) - 230) = 230 + 0.314 * (184.97 - 230) = 195.96 degrees C

where Tsat(1 MPa) is the saturation temperature at 1 MPa (from the steam table).

Finally, we can use the steam table again to find the enthalpy of the steam at the exit of the turbine:

hf = 96.83 kJ/kg (from the steam table)

hfg = 2434.4 kJ/kg (from the steam table)

hg = hf + x * hfg = 835.63 kJ/kg

where x is the quality of the steam, which is given as 5%.

Therefore, the power output from the turbine can be calculated as:

P = m * (h1 - hg) = 50 * (834.46 - 835.63) = -59.35 kW

The negative sign indicates that the turbine is consuming power instead of generating power. This is because the quality of the steam at the exit of the turbine is only 95%, which means that there is some moisture content that needs to be removed. To improve the power output, we can use a moisture separator or a reheater to increase the quality of the steam.

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Correct question:

In a single-flash geothermal power plant, geothermal water enters the flash chamber (a throttling valve) at 230 dgrees C as a saturated liquid at a rate of 50 kg/s. The steam resulting from the flashing process enters a turbine and leaves at 20 kPa with a moisture content of 5%. Determine the temperature of the steam after the flashing process and the power output from the turbine if the pressure of the steam at the exit of the flash chamber is 1 MPa.

Answer:

Explanation:

It really all depends, it varies from 375 to 830, you can't mark one as " Lamborghinis have this much hp always " seeing it fluctuates so much car to car

Answer:

a)  Cr = Co - Fx / D

b)   ΔC / Δx = ( CR - Cr )  / ( xR - xRo )

Explanation:

A) Derive an expression for the profile c(r) inside the tissue

F = DΔC / X  = D ( Co - Cr ) / X   ------ 1

where : F = flux , D = drug diffusion coefficient

            X = radial distance between Ro and R

Hence : Cr = Co - Fx / D

B) Express the diffusive flux at outer surface of the balloon

Diffusive flux at outer surface =  ΔC / Δx = CR - Cr / xR - xRo

Answer: true

Explanation:

Answer:

Projectile motion is a form of motion experienced by a launched object.

Answer:

Projectile motion is the motion of a body which experiences both vertical and horizontal motions ( trajection ) from point of flight up to the point of landing.

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.

Let's take this step by step.

a. The process capability ratio is given as Cp = 392. However, I believe there may be a misunderstanding or typo in the original question. Process capability ratio typically falls within the range 0-3. The formula for the process capability ratio (Cp) is:

Cp = (USL - LSL) / (6σ)

Where:

- USL is the upper specification limit (in this case, 2.9 ounces)

- LSL is the lower specification limit (in this case, 2.1 ounces)

- σ is the standard deviation (in this case, 0.34 ounces)

Let's recalculate the Cp given these inputs:

Cp = (2.9 - 2.1) / (6 * 0.34) = 0.800

This indicates that Leah's Toys' process is currently capable of producing balls within the tolerance limits about 80% of the time, assuming a normal distribution of weights. There may be a misunderstanding with the provided Cp of 392.

b. If Leah's Toys wants to meet the tolerance limits 99.7% of the time, then they would need to reduce the standard deviation such that the output falls within +/- 3σ (3 standard deviations from the mean). This is also known as achieving a "Six Sigma" level of quality.

We can rearrange the Cp equation to solve for σ:

σ = (USL - LSL) / (6 * Cp)

Assuming a Cp of 1.0 (which represents a process that meets tolerance limits 99.73% of the time under a normal distribution), we find:

σ = (2.9 - 2.1) / (6 * 1.0) = 0.13 ounces

This is the standard deviation Leah's Toys would need to achieve to meet the tolerance limits 99.7% of the time.

c. If Leah's Toys invests in process improvements and lowers the standard deviation to 0.13 ounces, then the new process capability ratio (Cp) would be:

Cp = (2.9 - 2.1) / (6 * 0.13) = 1.026

This means Leah's Toys could achieve Six Sigma quality levels (99.7% of products within specification limits) with this new standard deviation. Six Sigma is often represented by a Cp or Cpk (which takes into account mean shift) of 1.5 or more, but in a perfect process centered between the limits, a Cp of 1.0 represents 99.73% within limits, which aligns with your 99.7% target.

Answer:

I can help you with your calculation.

To add and subtract angles in degrees, minutes, seconds (DMS) form, you need to follow these steps12:

Align the angles so that the degrees, minutes, and seconds are in the same column.

Add or subtract the seconds first. If the result is more than 60 or less than 0, adjust the minutes accordingly.

Add or subtract the minutes next. If the result is more than 60 or less than 0, adjust the degrees accordingly.

Add or subtract the degrees last.

Using this method, your calculation can be done as follows:

64º26’18’’ + 195º57’12,75’’ – 100º55’35’’ = 64º26’18’’ + 195º57’12.75’’ – 100º55’35’’ = 159º83’30.75’’ – 100º55’35’’ = 159º83’(30.75 - 35)’’ – 100º55’0’’ = 159º82’55.75’’ – 100º55’0’’ = 159º(82 - 55)’55.75’’ – 100º0’0’’ = 159º27’55.75’’ – 100º0’0’’ = (159 - 100)º27’55.75’’ = 59º27’55.75’’

Therefore, the answer is 59º27’55.75’’. I hope this helps!

Explanation:

Answer:

yes answer d is correct

Communication Process are actions that create physical solutions to problems. The correct option is d.

Human existence and organisational survival both depend on effective communication. It is a process of generating and disseminating thoughts, facts, opinions, and sentiments from one place, individual, or group to another. The Management function of Directing depends on effective communication.

The sending party, message encoding, channel selection, message receipt by the recipient, and message decoding are all aspects of the communication process.

Feedback is when the recipient communicates something back to the original sender. These procedures are actions that result in tangible fixes for issues.

Thus, the correct option is d.

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Answer:

The Dependent Variable is the Effect, Its value depends on changes from the Independent Variable

Hope this Helps

The Kano model is a framework used to classify customer requirements into different categories based on their impact on customer satisfaction.

It helps businesses understand which features or attributes of a product or service are essential for customer satisfaction and which ones are more likely to delight or dissatisfy customers.

Must-be attributes are basic requirements that customers expect to be fulfilled. In the context of an EV charging station, these would include fundamental features such as reliable and safe charging, compatibility with different EV models, and ease of use. If a charging station lacks these must-be attributes, it would significantly diminish customer satisfaction and render the offering non-competitive.

One-dimensional attributes are features that directly impact customer satisfaction in a linear manner. They are typically stated explicitly by customers and their presence or absence can be easily measured. In the case of an EV charging station, one-dimensional attributes may include factors like charging speed. etc.

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Methane, represented by CH4, is a compound with one carbon and four hydrogen atoms, and it is a non-polar molecule. It means that the CH4 molecule is symmetric in shape, and the charge distribution on the atoms is even and symmetrical as well.

The CH4 molecule's boiling point is -161.5°C, which means that at a temperature of 195 K, methane is in its liquid state. Methane's critical temperature and critical pressure are 190.6 K and 46 atm, respectively. It means that the methane molecule remains in its gaseous state at pressures below 46 atm and temperatures above 190.6 K.Repulsive forces are one of the intermolecular forces that exist between the molecules of a substance. They arise due to the overlapping of electron clouds of two or more molecules when they come close to each other.

The repulsive force becomes more significant when the molecules' distance is too close, and the electron clouds start overlapping with each other. The repulsive force increases with an increase in pressure, which causes the molecules to move closer to each other and, as a result, causes the electron clouds to overlap more significantly.When the pressure on the CH4 molecule increases beyond 300 atm, the repulsive forces between the CH4 molecules increase drastically, making it difficult for the molecules to remain in close proximity. Therefore, it is true that repulsive interactions become important above 300 atm for methane at 195 k.

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The tatal resistance of the series-parallel circuit with two resistor connected in parallel which combination is connected in series to a single resistor is 9 ohms.

This can be defined as the opposition to current flow in a circuit.

To calculate the total resistance, first we need to find the total resistance of the parallel resistor.

For parallel,

Where:

R' = Total resistance of the parallel resistor.

From the question,

Given:

Substitute these values into equation 1

Finally, we combine the effective parallel resistance in series to the single resistance to the the total resistance of the circuit.

Where:

From the question,

Substitute these values into equation 2

Hence, the total resistance of the circuit is 9 ohms.

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Answer:

35.92 kpsi

Explanation:

Given data:

diameter of the steel bar d = 0.875 in

Area A = πd^2/4 = π(0.875)^2/4

length L = 15.0 ft

Load P = 21.6 kip

Modulus of elesticity E = 29×10^6 Psi

Assume we are asked to determine axial stress in the bar which is given as

\(\sigma = Load, P/ Area, A\)

\(\sigma = 4P/\pi d^2\)

substitute the value

\(\sigma = \frac{4\times 21.6}{\pi \times (0.875)^2} \\=35.92\ kpsi\)

Answer:

The dimension of the  square cross section is = 30mm * 30mm

Explanation:

Before proceeding with the calculations convert MPa to Kpsi

Sut ( ultimate strength ) = 770 MPa * 0.145 Kpsi/MPa = 111.65 Kpsi

the fatigue strength factor of Sut at 10^3 cycles for Se = Se' = 0.5 Sut

at 10^6 cycles" for 111.65 Kpsi  = f ( fatigue strength factor) = 0.83

To calculate the endurance limit  use  Se' = 0.5 Sut      since Sut < 1400 MPa

Se'( endurance limit ) = 0.5 * 770 = 385 Mpa

The surface condition modification factor

Ka = 57.7 ( Sut )^-0.718

Ka = 57.7 ( 770 ) ^-0.718

Ka = 0.488

Assuming the size modification factor (Kb) = 0.85 and also assuming all modifiers are equal to one

The endurance limit at the critical location of a machine part can be expressed as :

Se = Ka*Kb*Se'

Se = 0.488 * 0.85 * 385 = 160 MPa

Next:

Calculating the constants to find the number of cycles

α = \(\frac{(fSut)^2}{Se}\)

α =\(\frac{(0.83*770)^2}{160}\)  =  2553 MPa

b = \(-\frac{1}{3} log(\frac{fSut}{Se} )\)

b = \(-\frac{1}{3} log (\frac{0.83*770}{160} )\)  = -0.2005

Next :

calculating the fatigue strength using the relation

Sf = αN^b

N = number of cycles

Sf = 2553 ( 10^4) ^ -0.2005

Sf = 403 MPa

Calculate the maximum moment of the beam

M = 2000 * 0.6 = 1200 N-m

calculating the maximum stress in the beam

∝ = ∝max = \(\frac{Mc}{I}\)

Where c = b/2 and   I = b(b^3) / 12

hence ∝max = \(\frac{6M}{b^3}\)  =  6(1200) / b^3   =  7200/ b^3   Pa

note: b is in (meters)

The expression for the factor of safety is written as

n = \(\frac{Sf}{\alpha max }\)

Sf = 403, n = 1.5 and ∝max = 7200 / b^3

= 1.5 = \(\frac{403(10^6 Pa/Mpa)}{7200 / B^3}\)   making b subject of the formula in other to get the value of b

b = 0.0299 m * 10^3 mm/m

b = 29.9 mm therefore b ≈ 30 mm

since  the size factor  assumed is near the calculated size factor using this relation : de = 0.808 ( hb)^1/2

the dimension = 30 mm by 30 mm

The company should go with the out-house treatment option, as it has a lower annual worth value and will result in lower costs over the 11-year period.

GivenDataAn oil refinery finds that it is necessary to treat the waste liquids from a new process before discharging them into a stream. The treatment will cost $40,000 the first year, but process improvements will allow the costs to decline by $4,000 each year.An outside company will process the wastes for the fixed price of $20,000/year throughout the 11 year period, payable at the beginning of each year.MARR = 7%FormulaAnnual Worth (AW) = (P/A, i%, n)Annual Worth (AW) = Present Worth (PW) + Future Worth (FW)Where,P = Initial Cost (Present Worth)A = Capital Recovery Factori = InterestRaten = Life of the ProjectCalculationFirst of all, we calculate the AW of in-house treatment. The cash outflow would be $40,000 in year 0, then $36,000 ($40,000 – $4,000) in year 1, then $32,000 ($36,000 – $4,000) in year 2, and so on until year 10, and the cash inflow would be $0 as there is no salvage value.Annual Worth (AW) = (P/A, i%, n)Present Worth (PW) = $40,000Future Worth (FW) = $0Capital Recovery Factor (CRF) = (i(1 + i)n)/((1 + i)n – 1) = (0.07(1 + 0.07)11)/((1 + 0.07)11 – 1) = 0.122053Annual Worth (AW) = (P/A, i%, n)= ($40,000/0.122053)= $327,814.53Therefore, the AW of in-house treatment is $327,814.53.Now, we calculate the AW of out-house treatment. The cash outflow would be $20,000 in each year from year 0 to year 10, and the cash inflow would be $0 as there is no salvage value.Annual Worth (AW) = (P/A, i%, n)Present Worth (PW) = $20,000Capital Recovery Factor (CRF) = (i(1 + i)n)/((1 + i)n – 1) = (0.07(1 + 0.07)11)/((1 + 0.07)11 – 1) = 0.122053Annual Worth (AW) = (P/A, i%, n)= ($20,000/0.122053)= $163,907.27Therefore, the AW of out-house treatment is $163,907.27. AW in-house treatment = $327,814.53 AW out-house treatment = $163,907.27.

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