Answer:
2k20
Explanation:
4k ✈
The displacement volume of an internal combustion engine is 3 liters. The processes within each cylinder of the engine are modeled as an air-standard Diesel cycle with a cutoff ratio of 2.5. The state of the air at the beginning of compression is fixed by P1=95kPa, T1=22oC, and V1 = 3.17 liters.
Determine the net work per cycle, in kJ, the power developed by the engine, in kW, and the thermal efficiency, if the cycle is executed 1000 times per min.
Answer:
1) The power developed by the engine is 14705.7739 kW
2) The thermal efficiency is approximately 61.5%
Explanation:
The given parameters are;
P₁ = 95 kPa
T₁ = 22°C
V₁ = 3.17 liters
The cutoff ratio = 2.5
Displacement volume = 3 liters
The number of times the cycle is executed per minute = 1000 times per minute
We have;
The displacement volume = V₁ - V₂ = 3 l
V₁ = 3.17 l
V₂ = 3 - 3.17 = 0.17 l
Compression ratio = V₁/V₂ = 3.17/0.17 ≈ 18.65
P₂/P₁ = P₂/(95 kPa) = (V₁/V₂)^(k) = 18.65^1.4
P₂ = (95×18.65^(1.4)) ≈ 5710.5 kPa
T₂/T₁ = (V₁/V₂)^(k - 1)
T₂/(295 K)= (18.65)^(1.4 - 1)
T₂ = 295 * (18.65)^(1.4 - 1) = 950.81 K
The cutoff ratio = V₃/V₂ = 2.5
T₃ = T₂ × V₃/V₂ = 2.5 * 950.81 K = 2377.025 K
[tex]Q_{in}[/tex] = [tex]C_p[/tex]×(T₃ - T₂) = 1.006 × (2377.025 - 950.81) = 1,434.77 kJ/kg
T₄ = T₃ × (V₃/V₄)^(k-1) =
Therefore,
[tex]T_4 = T_3 \times \left (\dfrac{r_c}{r} \right )^{k - 1} = 2377.025 \times \left( \dfrac{2.5}{18.65} \right )^{1.4 - 1} \approx 1064 \ K[/tex]
T₄ ≈ 1064 K
[tex]Q_{out}[/tex] = [tex]-C_v \times (T_4 - T_1)[/tex]
[tex]C_v = C_p/k = 1.006/1.4 \approx 0.7186 \ kJ/kg[/tex]
∴ [tex]Q_{out}[/tex] = 0.7186×(1064 - 295) = 552.6034 kJ/kg
1) The net work = [tex]Q_{in}[/tex] - [tex]Q_{out}[/tex] = 1,434.77 kJ/kg - 552.6034 kJ/kg ≈ 882.17 kJ/kg
The number of cycle per minute = 1000 rpm
The number of cycle per minute = 1000 rpm/60 = 16.67 cycles per second
The power developed by the engine = The number of cycles per second × The net work of the engine
Therefore;
The power developed by the engine = 16.67 cycles/second × 882.17 kJ/kg
The power developed by the engine = 14705.7739 kW
2) Efficiency, [tex]\eta _{th}[/tex], is given as follows;
[tex]\eta _{th} = \dfrac{Q_{in}-Q_{out}}{Q_{in}} \times 100 = 1 - \dfrac{Q_{out}}{Q_{in}} \times 100= 1 - \dfrac{552.6034}{1434.77}\times 100 \approx 61.5\%[/tex]
Therefore, the thermal efficiency ≈ 61.5%.
Solve the compound inequality. 3x − 4 > 5 or 1 − 2x ≥ 7
Which of the following was an effect of world war 2 on agricultural industry
Answer:
Option C..Farmers saught new technology to help with the workload
hope this helped you
please mark as the brainliest (ㆁωㆁ)
- The four leading causes of death in the
construction industry include electrical
incidents, struck-by incidents, caught-in or
caught-between incidents, and
a. vehicular incidents
b. falls
C. radiation exposure
d. chemical burns
Help me for this question
A standard carbon resistor has a gold band to indicate + 5% tolerance. If its resistance is 3,500 , what are the upper and lower limits for its resistance? OA . 3495 - 3505 2 OB. 3300 Q - 3600 0 OC. 3325 N - 3675 OD 3450 - 35500
Answer:
C. 3325 Ω - 3675 Ω
Explanation:
5% of 3500 Ω is ...
0.05 × 3500 = 175
The lower limit is this amount less than the nominal value:
3500 -175 = 3325
The upper limit is the nominal value plus the tolerance:
3500 +175 = 3675
The lower and upper limits are 3325 Ω and 3675 Ω, respectively.
12. You need to be at the lift controls whenever the lift is in motion A) True B)False
the answer to that would be a) true
for high-volume production runs, machining parts from solid material might not be the best choice of manufacturing operations because
Answer:
There are actually multiple types of processes a manufacturer uses, and those can be grouped into four main categories: casting and molding, machining, joining, and shearing and forming.
Explanation:
Explain how engineering and science-related?
Answer:
While scientists study how nature works, engineers create new things, such as products, websites, environments, and experiences. Because engineers and scientists have different objectives, they follow different processes in their work.
Explanation:
What allows negative feedback to control a system
Answer:
Negative feedback control of the amplifier is achieved by applying a small part of the output voltage signal at Vout back to the inverting ( – ) input terminal via the feedback resistor
Answer:
The system has parts that sense the amount of output
Explanation:
Apex cirtified
How can you safely lift and support a vehicle
Answer:
Place the jack under the part of the vehicle that it should contact when raised. If you're using jack stands, place them near the jack. If you place your jack incorrectly, you can injure your car. To find the proper place to position the jack for your particular vehicle, check your owner's manual.
An eyewash station needs to be located no more than 10 seconds or less than 55 feet away from a work area.
Answer: True is correct
A third-degree burn may look charred and black or dry and white.
Explanation:
This ball is technology too! It can be
rolled, kicked, or thrown. Is that a need
or a want?
That is an example of a want because you don't need the ball.
Steam enters a two-stage adiabatic turbine at 8 MPa and 5008C. It expands in the first stage to a state of 2 MPa and 3508C. Steam is then reheated at constant pressure to a temperature of 5008C before it is routed to the second stage, where it exits at 30 kPa and a quality of 97 percent. The work output of the turbine is 5 MW. Assuming the surroundings to be at 258C, determine the reversible power output and the rate of exergy destruction within this turbine.
Answer:
1) The exergy of destruction is approximately 456.93 kW
2) The reversible power output is approximately 5456.93 kW
Explanation:
1) The given parameters are;
P₁ = 8 MPa
T₁ = 500°C
From which we have;
s₁ = 6.727 kJ/(kg·K)
h₁ = 3399 kJ/kg
P₂ = 2 MPa
T₂ = 350°C
From which we have;
s₂ = 6.958 kJ/(kg·K)
h₂ = 3138 kJ/kg
P₃ = 2 MPa
T₃ = 500°C
From which we have;
s₃ = 7.434 kJ/(kg·K)
h₃ = 3468 kJ/kg
P₄ = 30 KPa
T₄ = 69.09 C (saturation temperature)
From which we have;
h₄ = [tex]h_{f4}[/tex] + x₄×[tex]h_{fg}[/tex] = 289.229 + 0.97*2335.32 = 2554.49 kJ/kg
s₄ = [tex]s_{f4}[/tex] + x₄×[tex]s_{fg}[/tex] = 0.94394 + 0.97*6.8235 ≈ 7.563 kJ/(kg·K)
The exergy of destruction, [tex]\dot X_{dest}[/tex], is given as follows;
[tex]\dot X_{dest}[/tex] = T₀ × [tex]\dot S_{gen}[/tex] = T₀ × [tex]\dot m[/tex] × (s₄ + s₂ - s₁ - s₃)
[tex]\dot X_{dest}[/tex] = T₀ × [tex]\dot W[/tex]×(s₄ + s₂ - s₁ - s₃)/(h₁ + h₃ - h₂ - h₄)
∴ [tex]\dot X_{dest}[/tex] = 298.15 × 5000 × (7.563 + 6.958 - 6.727 - 7.434)/(3399 + 3468 - 3138 - 2554.49) ≈ 456.93 kW
The exergy of destruction ≈ 456.93 kW
2) The reversible power output, [tex]\dot W_{rev}[/tex] = [tex]\dot W_{}[/tex] + [tex]\dot X_{dest}[/tex] ≈ 5000 + 456.93 kW = 5456.93 kW
The reversible power output ≈ 5456.93 kW.
You have a piece of film paper that is 3 in x 5 in. You fix it inside the back of a pinhole camera with a focal length of 5.5 in. You want to use it to take a picture of your team’s mascot – a giant guinea pig that just barely fits in a 4 ft. tall cube. The picture will be taken directly in front, from a stool that places the aperture 2 ft. above the ground. You have to determine how far away the camera must be from your mascot to get a good portrait that fills up the whole film paper, without cutting any part of him off. How far apart should the camera and the mascot be to take the portrait? Show your work.
Answer:
In order to take a portrait, the distance of the mascot from the camera should be approximately 7.33 feet
Explanation:
The size of the film paper = 3 in. × 5 in.
The focal length of the camera = 5.5 in.
The height and width of the guinea pig = 4 ft.
The height of the aperture above the ground = 2 ft.
Therefore, we have;
Magnification = Height of image/(Height of object)
Withe the 3 in. wide film, we have;
Magnification = 3 in./(4 ft.) = 3 in./(48 in.) = 0.0625
Magnification = Length of camera/(Distance of object from pin hole)
∴ Length of camera/(Distance of object from pin hole) = 0.0625
Length of camera = Focal length of the camera = 5.5 in.
Therefore;
5.5 in./(Distance of object from pin hole) = 0.0625
Distance of object from pin hole = 5.5/0.0625 = 88 inches = 7.33 ft
Therefore, the camera should be approximately 7.33 ft. from the mascot to take a portrait.
In this exercise we have to use the magnification knowledge to calculate the distance that the photograph should be taken, thus we have to:
Distance of the mascot from the camera should be approximately 7.33 feet
To calculate the best distance to take the photo, we have that some information must be taken into account such as:
Size of the film paper: [tex](3)*(5) in[/tex] Focal length of the camera: [tex]5.5 in[/tex] Height and width of the guinea pig: [tex]4 ft[/tex] Height of the aperture above the ground: [tex]2 ft[/tex]
Therefore, we have that the formula of magnification is:
[tex]Magnification = Height \ of \ image/(Height \ of \ object)[/tex]
With the 3 in wide film, we have;
[tex]Magnification = 3 in/(4 ft) \\= 3 in/(48 in) = 0.0625 in[/tex]
Rewriting the magnification formula as:
[tex]Magnification = Length \ of \ camera/(Distance \ of \ object \ from \ pin \ hole)[/tex]
Substituting the values already known we have the equation will be matched as:
[tex]Length\ of \ camera/(Distance\ of\ object \ from \ pin \ hole) = 0.0625\\Length \ of \ camera = Focal \ length \ of \ the \ camera = 5.5 in.[/tex]
Therefore;
[tex]5.5 /(Distance \ of \ object\ from \ pin\ hole) = 0.0625 in\\Distance \ of \ object\ from \ pin \ hole = 5.5/0.0625\\ = 88 inches = 7.33 ft[/tex]
See more about distance at brainly.com/question/989117
What is the period if the clock frequency is 3.5 GHz?
Answer:
Period = 0.2857 nanoseconds
Explanation:
We are told that frequency = 3.5 GHz
This is simply 3.5 × 10^(9) Hz
Now, from wave equations, Period is given by the formula;
Period = 1/frequency
Thus;
Period = 1/(3.5 × 10^(9))
Period = 0.2857 × 10^(-9) seconds
From conversions, we can simplify the answer.
1 second = 10^(-9) nanoseconds
Thus, 0.2857 × 10^(-9) seconds = 0.2857 × 10^(-9) × 10^(-9) nanoseconds = 0.2857 nanoseconds
Two wooden members of uniform rectangular cross section are joined using a simple glued scarf splice. The maximum allowable shearing stress and maximum allowable normal stress in the glued splice is 50 MPa and 100 MPa, respectively. The cross-section area of the glued member is 400 mm2. (a) What should the value of the angle be to achieve maximum load Fmax? (b) What is the magnitude of the maximum load Fmax?
Answer:
a). α = 26.57
b). Maximum load is 50 .kN
Explanation:
a).
The normal force is given by
N = σ A cosec β
where, σ is the normal stress
A is the cross sectional area
Similarly, shear force is given by
S= τ A cosec β
where, τ is the shearing stress
Now from the figure,
tan β = S/N
= τ/σ
Therefore, [tex]$\beta = \tan^{-1}(2)$[/tex] = 63.43
α = 90 - β = 26.57
b).
The normal force is given by
[tex]$N=(100\times 10^6)(400\times 10^{-6}) \text{ cosec}\ 63.43$[/tex]
[tex]$N=44.78\times 10^3$[/tex] N
We have
[tex]$\Sigma F_y=0$[/tex]
∴ N - F sin β = 0
⇒ F = N / sin β
= [tex]$\frac{44.72\times 10^3}{\sin(63.43)} = 50\times 10^3 N$[/tex]
Similarly,
The shear force is given by
S = τ A cosec β
= [tex]$(50\times 10^6)(400\times 10^{-6}) \text{ cosec}\ 63.43 = 22.36\times 10^3 N$[/tex]
[tex]$\Sigma F_x=0$[/tex]
∴ S - F cos β = 0
⇒ F = S / cos β
[tex]$\frac{22.36\times 10^3}{\cos(63.43)} = 49.99\times 10^3 N$[/tex]
Therefore, force is 50 kN.
Which items are NOT found on a
door?*
5 points
Cladding
Moulding
Weatherstrip
Check Strap
Striker
All of the above
None of the above
Answer:
None of the above cause thats what i put
parallel circuits???
PLEASE GIVE BRAINLIST
A parallel circuit has two or more paths for current to flow through. Voltage is the same across each component of the parallel circuit. The sum of the currents through each path is equal to the total current that flows from the source.
HOPE THIS HELPED
To create an effective study schedule, a student must
guess the amount of time needed for studying.
O write down all activities and commitments.
O allow the same amount of time each day for studying.
O budget periods of time to study daily, even if it is interrupted.
Answer:
B) write down all activities and commitments.
Explanation:
also i found the answer on quizlet hope this works:))
Answer:
write down all activities and commitments.
Explanation:
i had got the same question on my unit test and it was the answer it was right
A sign that has a white background with a
green panel with white lettering is a
a. general information sign
b. safety instruction sign
c. caution sign
d. danger sign
Answer:
A
Explanation:
general information sign
safety is orange
caution is yellow
danger is red
A picture of a ____ is the label pictogram on a chemical that warms of a product's acute toxicity
Answer:skull and crossbones
Explanation:
Which one is NOT a benefit of Shine? 1. Less production downtime 2.Happier employees 3. Improved quality 4. Inventory reduction 5. Customer satisfaction
Answer:
Inventory reduction
Explanation:
Shine is a term that deals with the general cleaning and overall maintenance of the workplace. It is used as a part of an organizational theme involving five steps which are often referred to as 5S and it is basically defined as the mainstay of the visual workplace. They are Sort, Set in order, Shine, Standardize, and Sustain.
Hence, in this situation, considering the option that is not a benefit of Shine the correct answer is "Inventory reduction." This is because Shine is not about reducing the inventory levels. It is Sort and Set-in-Order will help reduce inventory.
The answer choice which is NOT a benefit of Shine is:
D. Inventory reduction
According to the given question, we are asked to show the answer choice which is not a direct benefit of Shine and how Shine is used to increase productivity in an office space.
Shine is a term which is used to refer to the general cleanliness of an office space and is a part of organization which is used to improve quality, make the employees happier, and have a less production downtime, etc
As a result of this, we can see that inventory reduction is not a benefit of Shine.
Therefore, the correct answer is option D
Read more here:
https://brainly.com/question/22061432
While out on the International Space Station, an engineer was able to gather a sample of a new type of unidentified rock. What knowledge will the engineer use to predict the potential of this new material?
Answer:
The engineer will conduct a variety of tests, including chemical, mechanical, electrical, and physical examinations, to determine the potential of the new material.
Explanation:
They will need to test the material, this will also help to determine its malleability.
Hope this helps!
You are designing an airplane to carry liquid cargo that will slosh and move side to side in its container. This could make the plane unstable. What type of airfoil would you use for the wing? Why. Answer in a full sentence or more
1. Looking at the case study provided under the Companion Material section, what is the main problem that is addressed in this case study? Maintenance workers painted over the pipes. Engineers lost all of their work because of the flood. The water pipes broke and flooded the space center. There was no paperwork for the maintenance work done on the pipes.
Incomplete question. However, I provided information that could assist you in identifying the main problem or issue addressed in any case study.
Explanation:
First, note that a case study is simply a learning aid that allows one to learn from a real-life scenario.
To determine the main problems of a case study one needs to:
Read the case as many times as possible to become familiar with the message been expressed. For example, by highlighting or underlining the most important facts it can help you to discover the main problem or issue. Check for any facts provided in the case study, by so doing you can identify the most important problems.Thus, by taking these few steps you may be able to determine the main problem in that case study.
Key length is designed to provide desired factor of safety
a. True
b. False
Answer: true
Explanation:
A key is a machine element that us used to connect the element of a rotating machine to a shaft. It should be noted that the key hinders the relative rotation that may take place between the two parts.
Key length is designed to provide desired factor of safety. It should also be noted that the factor of safety shouldn't be much and the key length is typically limited to the hub length.
Wheel grinders need be equipped with an
Answer:
wheel guard
Explanation:
to protect our hands and reduce spark
Which statement demonstrates the most scientific observation?
A company that produces footballs uses a proprietary mixture of ideal gases to inflate their footballs. If the temperature of 230 grams [g] of gas mixture in a 15-liter [L] tank is maintained at 465 degrees Rankine [°R] and the tank is pressurized to 135 pound-force per square inch [psi], what is the molecular weight of the gas mixture in units of grams per mole
Answer:
The molecular weight of the gas mixture is 35.38 g/mol.
Explanation:
The molecular weight of the gas can be found using the following equation:
[tex] M = \frac{m}{n} [/tex]
Where:
m: is the mass = 230 g
n: is the number of moles
First, we need to find the number of moles using Ideal Gas Law:
[tex] PV = nRT [/tex]
Where:
P: is the pressure = 135 psi
V: is the volume = 15 L
R: is the gas constant = 0.082 L*atm/(K*mol)
T: is the temperature = 465 °R (K = R*5/9)
[tex]n = \frac{PV}{RT} = \frac{135 psi*\frac{1 atm}{14.6959 psi}*15 L}{0.082 L*atm/(K*mol)*465*(5/9) K} = 6.50 moles[/tex]
Finally, the molecular weight of the gas is:
[tex] M = \frac{m}{n} = \frac{230 g}{6.50 moles} = 35.38 g/mol [/tex]
Therefore, the molecular weight of the gas mixture is 35.38 g/mol.
I hope it helps you!