Helium is used as the working fluid in a Brayton cycle with regeneration. The pressure ratio of the cycle is 8, the compressor inlet temperature is 300 K, and the turbine inlet temperature is 1800 K. The effectiveness of the regenerator is 75 percent. Determine the thermal efficiency and the required mass flow rate of helium for a net power output of 60 MW, assuming both the compressor and the turbine have an isentropic efficiency of (a) 100 percent and (b) 80 percent. The properties of Helium are cp = 5.1926 kJ/kg.K and k = 1.667.

Answers

Answer 1

Answer:

Explanation:

Find the temperature at exit of compressor

\(T_2=300 \times 8^{\frac{1.667-1}{1.667} }\\=689.3k\)

Find the work done by the compressor

\(\frac{W}{m} =c_p(T_2-T_1)\\\\=5.19(689.3-300)\\=2020.4kJ/kg\)

Find the actual workdone by the compressor

\(\frac{W}{m} =n_c(\frac{W}{m} )\\\\=1 \times 2020.4kJ/kg\)

Find the temperature at exit of the turbine

\(T_4=\frac{1800}{8^{\frac{1.667-1}{1.667} }} \\\\=787.3k\)

Find the actual workdone by the turbine

\(1 \times 5.19 (1800-783.3)\\=5276.6kJ/kg\)

Find the temperature of the regeneration

\(\epsilon = \frac{T_5-T_2}{T_4-T_2} \\\\0.75=\frac{T_5-689.3}{783.3-689.3} \\\\T_5=759.8k\)

Find the heat supplied

\(Q_i_n=c_p(T_3-T_5)\\\\=5.19(1800-759.8)\\\\=5388.2kJ/kg\)

Find the thermal efficiency

\(n_t_h=\frac{W_t-W_c}{Q_i_n} \\\\=\frac{5276.6-2020.4}{5388.2} \\\\n_t_h=60.4\)

60.4%

Find the mass flow rate

\(m=\frac{W_net}{P} \\\\\frac{60 \times 10^3}{5276.6-2020.4} \\\\=18.42\)

Find the actual workdone by the compressor

\(\frac{W_c}{m} =\frac{(\frac{W}{m} )}{n_c} \\\\=\frac{2020.4}{0.8} \\\\=2525.5kg\)

Find the actual workdone by the turbine

\(\frac{W_t}{m} =n_t(\frac{W}{m} )\\\\=0.8 \times5.19(1800-783.3)\\\\=4221.2kJ/kg\)

Find the temperature of the compressor exit

\(\frac{W_t}{m} =c_p(T_2_a-T_1)\\2525.5=5.18(T_2_a-300)\\T_2_a=787.5k\)

Find the temperature at the turbine exit

\(4221.2=5.18(1800-T_4_a)\\\\T_4_a=985k\)

Find the temperature of regeneration

\(\epsilon =\frac{T_5-T_2}{T_4-T_2}\\\\0.75=\frac{T_5-787.5}{985-787.5}\\\\T_5=935.5k\)

Answer 2

Answer:

a) 60.4%;  18.42 kg/s

b) 37.8% ;    35.4 kg/s

Explanation:

a) at an isentropic efficiency of 100%.

Let's first find the exit temperature of the compressor T2, using the formula:

\((r_p) ^k^-^1^/^k = \frac{T_2}{T_1}\)

Solving for T2, we have:

\( T_2 = 300 * (8)^1^.^6^6^7^-^1^/^1^.^6^6^7 = 689.3 K \)

Let's now find the work dine by the compressor.

\( \frac{W_c}{m} = c_p(T_2 - T_1) \)

\( \frac{W_c}{m} = 5.19(689.3 - 300) = 2020.4 KJ/kg\)

The actual work done by the compressor =

\( W_c = 1 * 2020.4 = 2020.4 KJ/kg \)

Let's find the temperature at the exit of the turbine, T4

\((r_p) ^k^-^1^/^k = \frac{T_3}{T_4}\)

Solving for T4, we have:

\(T_4 = \frac{1800}{(8)^1^.^6^6^7^-^1^/^1^.^6^6^7} = 783.3 K\)

Let's find the work done by the turbine.

\(\frac{W_t}{m} = c_p(T_3 - T_4)\)

\(\frac{W_t}{m} = 5.19(1800 - 783.3) = 5276.6 KJ/kg\)

The actual work done by the turbine:

= 1 * 5276.6 = 5276.6 KJ/kg

Let's find the regeneration temperature, using the formula:

\( e = \frac{T_r - T_2}{T_4 - T_2}\)

Substituting figures, we have:

\( 0.75 = \frac{T_r - 689.3}{783.3 - 689.3} \)

\( T_r = [0.75(783.3 - 689.3)] + 689.3 = 759.8 \)

Let's calculate the heat supplied.

\(Q = c_p(T_3 - T_r)\)

\( Q = 5.19(1800 - 759.8) \)

Q = 5388.2 kJ/kg

For thermal efficiency, we have:

\( n = \frac{W_t - W_c}{Q} \)

Substituting figures, we have:

\( n = \frac{5276.6 - 2020.4}{5388.2} = 0.604 \)

0.604 * 100 = 60.4%

For mass flow rate:

Let's use the formula:

\( m = \frac{W_n_e_t}{P} \)

Wnet = 60MW = 60*1000

\( m = \frac{60*10^3}{5276.6 - 2020.4} = 18.42 \)

b) at an isentropic efficiency of 80%.

Let's now find the work done by the compressor.

\( \frac{W_c}{m} = c_p(T_2 - T_1) \)

\( \frac{W_c}{m} = 5.19(689.3 - 300) = 2020.4 KJ/kg\)

The actual work done by the compressor =

\( W_c = \frac{2020.4}{0.8}= 2525.5 KJ/kg \)

Let's find the work done by the turbine.

\( \frac{W_t}{m} = c_p(T_3 - T_4) \)

\( \frac{W_t}{m} = 5.19(1800 - 787.5) = 5276.6 KJ/kg\)

The actual work done by the turbine:

= 0.8 * 5276.6 = 4221.2 KJ/kg

Let's find the exit temperature of the compressor T2, using the formula:

\(\frac{W_c}{m} = c_p(T_2 - T_1) \)

\( 2525.5 = 5.19(T_2 - 300) \)

Solving for T2, we have:

\( T_2 = \frac{2525.5 + 300}{5.19} = 787.5 \)

Let's find the temperature at the exit of the turbine, T4

\( \frac{W_t}{m} = c_p(T_3 - T_4) \)

\( 4221.2 = 5.19(1800 - T_4) \)

Solving for T4 we have:

\( T_4 = 958 K\)

Let's find the regeneration temperature, using the formula:

\( e = \frac{T_r - T_2}{T_4 - T_2}\)

Substituting figures, we have:

\( 0.75 = \frac{T_r - 787.5}{985 - 787.5} \)

\( T_r = [0.75(958 - 787.5)] + 787.5 = 935.5 K \)

Let's calculate the heat supplied.

\(Q = c_p(T_3 - T_r)\)

\( Q = 5.19(1800 - 935.5) \)

Q =  4486.2 kJ/kg

For thermal efficiency, we have:

\( n = \frac{W_t - W_c}{Q} \)

Substituting figures, we have:

\( n = \frac{4221.2 - 2525.2}{4486.2} = 0.378 \)

0.378 * 100 = 37.8%

For mass flow rate:

Let's use the formula:

\( m = \frac{W_n_e_t}{P} \)

Wnet = 60MW = 60*1000

\( m = \frac{60*10^3}{4221.2 - 2525.2} = 35.4 kg/s \)


Related Questions

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Answer:

The temperature gauge showing that the vehicle has been running warmer or has recently began to have issues from overheating is  an idication that your vehicle may be developing a cooling system problem.

Explanation:

A 20cm-long rod with a diameter of 0.250 cm is loaded with a 5000 N weight. If the diameter of the bar is 0.490 at this load, determine: I. the engineering stress and strain, and [2] II. the true stress and strain

Answers

If the diameter of the bar is 0.490 at this load, determine I. the engineering stress and strain, and [2] II. the true stress and strain is 1561. 84 MPa.

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Strain is a unitless degree of ways a great deal an item receives larger or smaller from an implemented load. Normal stress happens while the elongation of an item is in reaction to an everyday pressure (i.e. perpendicular to a surface), and is denoted via way of means of the Greek letter epsilon.

L = 20 cm d x 1 = 0.21 cmdx 2 = 0.25 cmF=5500 a) σ= F/A1= 5000/(π/4x(0.0025)^2)= 1018.5916 MPa lateral stress= Ad/d1= (0.0021-0.0025)/0.0025 = - 0.1 longitudinal stress (ɛ_l)= -lateral stress/v = -(-0.16)/0.3(assuming a poisson's ration of 0.3) ε_l=0.16/0.3 = 0.5333b) σ_true= σ(1+ ɛ_I)= 1018.5916(1+0.5333 = 1561.84 MPa.

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Grizzly bears in captivity frequently display abnormal behaviors, such as repetitive pacing, swaying from side-to-side, and biting the bars of their enclosure. In order to improve animal welfare, zoos and parks need to stimulate their captive bears to behave as closely as possible to their wild counterparts. Bear sanctuaries are an ideal place where grizzly bears can live out their lives in an enriched and safe environment that is as close their natural habitat as possible. Which of the following statements best expresses the main conclusion of the above argument?

Grizzly bears in captivity frequently display abnormal behaviors, such as repetitive pacing, swaying

Answers

Note that where Grizzly bears in captivity frequently display abnormal behaviors, such as repetitive pacing, swaying from side to side, and biting the bars of their enclosure, the statement that  best expresses the main conclusion of the above argument is: "Bears need to be in an environment as close as possible to their natural habitat." (Option 2)

What is the rationale for the above response?

Grizzly bears in captivity exhibit abnormal behaviors, which is a sign of poor animal welfare.

To improve their welfare, they need to be in an environment as close as possible to their natural habitat. Bear sanctuaries provide an ideal solution by providing an enriched and safe environment for the bears, mimicking their natural habitat. This helps to reduce abnormal behaviors and improve their overall quality of life.

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difference between velocity profile and velocity distribution​

Answers

Answer:

Profile is a graphical representation of velocity distribution

whether the exception is thrown or not, change the innerhtml of the wordcountbox object to the value of the charcount variable

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The following code fragment, whether or not an exception is thrown, should change the `innerHTML` of the `wordcountbox` object to the value of the `charcount` variable:

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The `innerHTML` property specifies the HTML content of an element. In the given question, the code is required to change the `innerHTML` of `wordcountbox` to the value of the `charcount` variable. The code to achieve this is very simple, as follows:

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The code should execute whether or not an exception is thrown.

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A specimen of steel 100mm diameter with a guage length of 100mm tested a distruction it has an extension of 0.50mm under a load of 70KN and the load at elastic limit is 10KN the maximum load is 140KN the total extension of fracture is 58mm and the diameter at the neck is 16mm . find the Stress of elastic limit, young modulus, percentage enlogation, percentage reduction in area , and ultimate tensile stress?​

Answers

The stress at elastic limit is 1.27 N/mm^2, the Young's modulus is 254 N/mm^2, the percentage elongation is 57.5%, the percentage reduction in area is 97.44%, and the ultimate tensile stress is 696.67 N/mm^2.

How to calculate the values

Stress = Force / Area

Young's modulus = Stress / Strain

Percentage elongation = (extension / gauge length) x 100%

Percentage reduction in area = [(original area - area at neck) / original area] x 100%

Ultimate tensile stress = Maximum load / Area

Diameter of specimen = 100 mm

Gauge length = 100 mm

Extension at 70 KN load = 0.50 mm

Load at elastic limit = 10 KN

Maximum load = 140 KN

Total extension at fracture = 58 mm

Diameter at neck = 16 mm

We can calculate the area of the specimen as follows:

Area = π/4 x d^2

Area = π/4 x (100 mm)^2

Area = 7853.98 mm^2

The stress at elastic limit can be calculated as:

Stress = Load / Area

Stress = 10 KN / 7853.98 mm^2

Stress = 1.27 N/mm^2

The Young's modulus can be calculated as:

Strain = Extension / Gauge length

Strain = 0.50 mm / 100 mm

Strain = 0.005

Stress = Load / Area

Load = Stress x Area

Load = 1.27 N/mm^2 x 7853.98 mm^2

Load = 9982.16 N

Young's modulus = Stress / Strain

Young's modulus = 1.27 N/mm^2 / 0.005

Young's modulus = 254 N/mm^2

The percentage elongation can be calculated as:

Percentage elongation = (extension / gauge length) x 100%

Percentage elongation = (58 mm - 0.50 mm) / 100 mm x 100%

Percentage elongation = 57.5%

The percentage reduction in area can be calculated as:

Original area = π/4 x (100 mm)^2 = 7853.98 mm^2

Area at neck = π/4 x (16 mm)^2 = 201.06 mm^2

Percentage reduction in area = [(original area - area at neck) / original area] x 100%

Percentage reduction in area = [(7853.98 mm^2 - 201.06 mm^2) / 7853.98 mm^2] x 100%

Percentage reduction in area = 97.44%

The ultimate tensile stress can be calculated as:

Area at neck = π/4 x (16 mm)^2 = 201.06 mm^2

Ultimate tensile stress = Maximum load / Area

Ultimate tensile stress = 140 KN / 201.06 mm^2

Ultimate tensile stress = 696.67 N/mm^2

Therefore, the stress at elastic limit is 1.27 N/mm^2, the Young's modulus is 254 N/mm^2, the percentage elongation is 57.5%, the percentage reduction in area is 97.44%, and the ultimate tensile stress is 696.67 N/mm^2.

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Another name for a load-center distribution system is a A. primary radial system. B. complex radial system. C. split-radial system. D. dual-radial system.

Answers

Answer:

A

Explanation:

Primary radial system

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Answers

Answer:

Explained below

Explanation:

Perfect Elastic Plastic in steel design is simply a method whereby the structural members are selected using the criteria of the overall ultimate capacity of the system. However, when safety is considered, the applied loads are usually increased by factors of safety as prescribed in the relevant steel design codes. Therefore, this model of design is just based on the yield capacity of the steel.

A 50-mm cube of the graphite fiber reinforced polymer matrix composite material is subjected to 125-kN uniformly distributed compressive force in the direction 2, which is perpendicular to the fiber direction (direction 1). The cube is constrained against expansion in direction 3. Determine:

a. changes in the 50-mm dimensions.
b. stresses required to provide constraints.

Answers

Answer:

hello some parts of your question is missing attached below is the missing part

answer :

A) Determine changes in the 50-mm dimensions

The changes are : 0.006mm compression  in y-direction

                               0.002 mm expansion in x and z directions

B) the stress required are evenly distributed

Explanation:

Given data :

50-mm cube of graphite fiber reinforced polymer matrix

subjected to 125-KN force in direction 2,

direction 2 is perpendicular to fiber direction ( direction 1 ) and cube is constrained against expansion in direction 3

A) Determine changes in the 50-mm dimensions

The changes are : 0.006mm compression  in y-direction

                               0.002 mm expansion in x and z directions

B) the stress required are evenly distributed

attached below is the detailed solution

A 50-mm cube of the graphite fiber reinforced polymer matrix composite material is subjected to 125-kN
A 50-mm cube of the graphite fiber reinforced polymer matrix composite material is subjected to 125-kN
A 50-mm cube of the graphite fiber reinforced polymer matrix composite material is subjected to 125-kN

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Answers

The rate of heat transfer to the water is 9.04 kW.

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The latent heat of vaporization of water is 2260 kJ/kg, which means that 2260 kJ of heat energy is required to convert 1 kg of water at 100°C to steam at the same temperature.

Therefore, the heat energy required to convert 2 kg of water to steam is 2 x 2260 kJ = 4520 kJ.

The time taken for this process is 30 minutes or 0.5 hours. Therefore, the rate of heat transfer to the water is 4520 kJ / 0.5 hours = 9040 kJ/hour or 9.04 kW.

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Answer:

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Answers

Answer:

Conic Sections

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A. Both Technicians A and B
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C. Technician A
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Answers

B.neither technicians A and B

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High CO2 levels can actually indicate inefficient engine operation, as it means that not all of the fuel in the engine is being burned and is being wasted as exhaust.

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Briefing:

1)here rotational velocity and acceleration of the rudder can be found by relative velocity and acceleration method with Coriolis component of acceleration

2)here let's draw a velocity diagram as follow with the scale of

2 cm=.1 m/s

velocity triangle is given by AB'B, where AB' shows slider velocity along horizontal, AB along the tangent to link 2 or guide, and velocity of pin  BB' inside the guide along the length of the guide, here we know the velocity of AB' but don't know velocity AB and BB' hence draw them along given direction point of intersection is B point, so from the diagram, I get that

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Vbb'=.9 cm=.045 m/s

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What quantity measures the effect of change?

control variable
independent variable
relative variable
dependent variable

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Answer:

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Answers

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A coral reef is in danger of being destroyed by a seaside construction project. Which of the following best relates to how an engineer would approach this problem?

Answers

Answer:Bio team

Explanation:

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Answers

The individual outputs of a typical AC output interface module usually have a maximum current rating of about:

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Exactly 5. 00 mL aliquots of a solution containing Phenobarbital were measured into 50. 00ml volumetric flasks and made basic with KOH.

The following volumes of a standard solution of Phenobarbital containing 2. 00 µg/ml of Phenobarbital were then introduced into each flask and the mixture was diluted to volume: 0. 00, 0. 500,1. 000, 1. 50, and 2. 00ml.

The fluorescence of each of these solutions was measured with a fluorometer, which gave volumes of 3. 26, 4. 80, 6. 41, 8. 02, and 9. 56 respectively.


(a) Pot the data.

(b) Using the plot from (a) calculate the concentration of Phenobarbital in the unknown.

(c) Derive a least-squares equation for the data.

(d) Find the concentration of Phenobarbital from the equation in (c)

(e) Calculate a standard deviation for the concentration obtained in (d)

Answers

The data can be plotted as a standard curve. The concentration of Phenobarbital in the unknown can be determined by finding the point on the curve where the fluorescence of the solution intersects the x-axis.

Pot the data

a) The data is plotted in form of a standard curve, which is a graph of the fluorescence of the solutions versus the volume of the standard solution added.

Using the plot from (a) calculate the concentration of Phenobarbital in the unknown.

b) By drawing a line of best fit through the data points, we can find the point where the line intersects the x-axis, the x-intercept, which will give us the concentration of phenobarbital in the unknown.

Derive a least-squares equation for the data.

c) The least-squares equation is a mathematical representation of the line of best fit. It is used to find the relationship between two variables by minimizing the sum of the squared residuals between the data points and the line.

Find the concentration of Phenobarbital from the equation in (c)

d) Using the equation obtained in c), we can calculate the concentration of phenobarbital in the unknown.

Calculate a standard deviation for the concentration obtained in (d)

e) By using the residuals of the line of best fit, we can calculate the standard deviation of the concentration obtained in d). This will give us an idea of how much the concentration of phenobarbital in the unknown deviates from the actual value.

The least-squares equation can be found by fitting a line to the data points. The concentration of Phenobarbital can then be determined using this equation. The standard deviation for the concentration can be calculated using the residuals from the line of best fit.

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The thickness of the oil reservoir is 10cm, the production rate of oil is 100m3, the well radius is 10cm, the permeability of the reservoir is 1um2, the porosity is 0.2, the viscosity of the oil is 4mPa.S, the density of oil is 850kg/m3. If the flow regime can be considered as linear flow, can this flow in the reservoir satisfy Darcy's law?

Answers

Answer:

The thickness of the oil reservoir is 10cm, the production rate of oil is 100m3, the well radius is 10cm, the permeability of the reservoir is 1um2, the porosity is 0.2, the viscosity of the oil is 4mPa.s, the density of oil is 850kg/m3. If the flow regime can be considered as linear flow, can this flow in the reservoir satisfy Darcy's

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Used building skyscrapers, Steel frame construction uses a metal cage to support a decorative exterior of another material such as glass, masonry, or stone.

A material? What do you mean by that?

The material(s) from which anything is created or composed: Stone is a hard substance. any part of the whole. fabric for an outfit made of textiles.

What materials are commonplace?

Any excavated material that is not "topsoil," "rock," or man-made hard material is referred to as "common material." Sample 1: "Common Material" refers to components utilized in the manufacture, testing, and packaging of devices for the buyer as well as for devices made by the supplier and its other clients.

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Draw the circuit diagram for 3 cells connected in series, two resistors in series and a switch.

If V = I x R, calculate l if one cell = 2V and the resistor = 4Ω.​

Answers

Answer:

See the image for solution

Draw the circuit diagram for 3 cells connected in series, two resistors in series and a switch. If V

Technician A says that shock absorbers should be replaced in pairs.
Technician B says that gas shocks should be bled before installation.
Who is right?
O A only
O Bonly
O Both A and B
O Neither Anor B
Previous
Exact answer

Answers

Answer:

The correct choice is C

Explanation A:

Shock Absorbers are best replaced in pairs if the car is relatively old. For new cars, however, one may elect to do the replacements as the need arises given that the other part is not worn out yet.

Explanation B:

It also recommended to bleed shock absorbers by moving the piston rod up and down several times. The reason is that sometimes, whilst in a horizontal position the hydraulic fluid and the gas get mixed up. This dampens the ability of the device to absorb shocks until the vehicle has been bounced up and down several times.

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the four mechanisms of adhesion, mechanical interlocking, diffusion, adsorption and electrostatic attraction respectively [

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The four mechanisms of adhesion are mechanical interlocking, diffusion, adsorption, and electrostatic attraction.


Mechanical interlocking: This mechanism involves physical interlocking of two surfaces, such as the interlocking of fibers in textiles or the interlocking of teeth in a zipper. The irregularities on the surface create a strong bond through intermolecular forces.

Diffusion: Diffusion occurs when molecules from one material penetrate the surface of another material. This mechanism is commonly observed in adhesives that consist of polymers. The adhesive molecules diffuse into the substrate, forming intermolecular bonds and creating adhesion.

Adsorption: Adsorption refers to the attraction of molecules from one material to the surface of another material. Van der Waals forces, dipole-dipole interactions, or chemical bonds can facilitate adsorption. Adhesives often rely on adsorption to adhere to substrates.

Electrostatic attraction: This mechanism involves the attraction between positively and negatively charged particles. It can occur between polar molecules or charged particles on the surfaces. Electrostatic attraction contributes to the adhesion of materials like plastics or rubber.

Adhesion can be achieved through various mechanisms. Mechanical interlocking relies on physical interlocking, diffusion involves the penetration of molecules into the substrate, adsorption occurs through the attraction of molecules to a surface, and electrostatic attraction relies on the attraction between charged particles. Understanding these mechanisms is crucial for developing effective adhesives and improving bonding techniques in various industries

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The four mechanisms of adhesion are mechanical interlocking, diffusion, adsorption, and electrostatic attraction. Each mechanism plays an important role in creating adhesion between two surfaces.

Mechanical interlocking is the process of creating a mechanical bond between two surfaces by physically interlocking them. This can be achieved by creating grooves or hooks in the surface of one of the materials and then pressing the two surfaces together. Diffusion is the process of atoms or molecules from one surface migrating into the other surface, creating a bond. This is typically seen in metal-to-metal bonding, where atoms from one surface diffuse into the other surface and create a solid-state weld. Adsorption is the process of molecules or atoms adhering to the surface of a material.

This can be achieved by creating a chemical bond between the surface of the material and the atoms or molecules. This type of bonding is often used in the creation of coatings and paints. Electrostatic attraction is the process of positively charged ions attracting negatively charged ions. This type of bonding is often seen in the bonding of two different materials, where one material has a positive charge and the other material has a negative charge.

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Determine the critical load if the bottom is fixed and the top is pinned. ewew = 1. 6 ×(10)3ksi×(10)3ksi ,σyσy = 5 ksiksi

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What is a Critical Load?

Critical load Fcr or buckling load is the value of load that causes the phenomenon of change from stable to unstable equilibrium state.

With that beign said, first it is neessary to calculate the moment of inercia about the x-axis:

\(Ix= \frac{db^3}{12}\\ Ix = \frac{2.(4)^3}{12} = 10.667in\)

Then it is necessary to calculate the moment of inercia about the y-axis:

\(Iy = \frac{db^3}{12}\\ Iy = \frac{4.(2)^3}{12} = 2.662in\)

Comparing both moments of inercia it is possible to assume that the minimun moment of inercia is the y-axis, so the minimun moment of inercia is 2662in.

And so, it is possible to calculate the critical load:

\(Pc\gamma = \frac{2046\pi ^2E.I}{L^2} \\Pc\gamma= \frac{2046.\pi ^2.(1,6.10^3.10^3).2662}{(10.12)^2} \\Pc\gamma= 5983,9db\)

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why is personal development necessary based activity success life and career​

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Personal development is a lifelong process that helps you assess your life goals and upskill to fulfill your potential. ... You may not always reach your objective but you lead a more fulfilling life because your purpose is defined. Personal development contributes to both satisfaction and success in life.

Engineered lumber should not be used for
A. finger-jointed studs.
B. plywood roof sheathing.
C. composite panel garage doors.
D. items that are in contact with concrete.

Answers

Answer:

Composite panel garage doors

Explanation:

Engineered lumber should not be used for composite panel garage doors. Hence, option C is correct.

What is a Pane lift garage door?

Horizontal panels are connected to form a single garage door in panel lift doors, sometimes referred to as sectional doors. These panels have wheels attached to them that are positioned in tracks on either side and guide the door through a reasonably sharp turn to reposition it horizontally above the garage door opening.

Yes, however it's crucial to have a pro take their place for the finest outcomes. It can be unsafe for anybody other than a professional to install garage doors due to their complexity and difficulty.

One of the most affordable solutions on the market is an aluminum garage door. They are available in a variety of design options that can complement the exterior style of your home. Low-cost aluminum garage doors.

Thus, option C is correct.

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