This question is incomplete, the complete question is;
An air-standard Carnot cycle is executed in a closed system between the temperature limits of 350 and 1200 K. The pressures before and after the isothermal compression are 150 and 300 kPa, respectively. If the net work output per cycle is 0.5 kJ, determine (a) the maximum pressure in the cycle, (b) the heat transfer to air, and (c) the mass of air. Assume variable specific heats for air.
Answer:
a) the maximum pressure in the cycle is 30.01 Mpa
b) the heat transfer to air is 0.7058 KJ
c) mass of Air is 0.002957 kg
Explanation:
Given the data in the question;
We find the relative pressure of air at 1200 K (T1) and 350 K ( T4)
so from the "ideal gas properties of air table"
Pr1 = 238
Pr4 = 2.379
we know that Pressure P1 is only maximum at the beginning of the expansion process,
so
now we express the relative pressure and pressure relation for the process 4-1
P1 = (Pr2/Pr4)P4
so we substitute
P1 = (238/2.379)300 kPa
P1 = 30012.6 kPa = 30.01 Mpa
Therefore the maximum pressure in the cycle is 30.01 Mpa
b)
the Thermal heat efficiency of the Carnot cycle is expressed as;
ηth = 1 - (TL/TH)
we substitute
ηth = 1 - (350K/1200K)
ηth = 1 - 0.2916
ηth = 0.7084
now we find the heat transferred
Qin = W_net.out / ηth
given that the net work output per cycle is 0.5 kJ
we substitute
Qin = 0.5 / 0.7084
Qin = 0.7058 KJ
Therefore, the heat transfer to air is 0.7058 KJ
c)
first lets express the change in entropy for process 3 - 4
S4 - S3 = (S°4 - S°3) - R.In(P4/P3)
S4 - S3 = - (0.287 kJ/Kg.K) In(300/150)kPa
= -0.1989 Kj/Kg.K = S1 - S2
so that; S2 - S1 = 0.1989 Kj/Kg.K
Next we find the net work output per unit mass for the Carnot cycle
W"_netout = (S2 - S1)(TH - TL)
we substitute
W"_netout = ( 0.1989 Kj/Kg.K )( 1200 - 350)K
= 169.065 kJ/kg
Finally we find the mass
mass m = W_ net.out / W"_netout
we substitute
m = 0.5 / 169.065
m = 0.002957 kg
Therefore, mass of Air is 0.002957 kg
which of the following activities can help expand engineers' creative thinking capabilities?
web design
team building exercises
journal keeping
trial and error
Answer:
team building exercises
Explanation:
Answer: its not team building excercize
Explanation:
Corrections for curvature and refraction (c r) are applied to: ____________.
a. benchmark elevations (BM)
b. both foresight rod readings (FS) and backsight rod readings (BS)
c. foresight rod readings (FS)
d. backsight rod readings (BS)
Answer:
b.
Explanation:
Foresight rod reading is any measurement taken at a given sight to determine a particular elevation. The backsight rod readings are usually used to a point of certain elevation. It is added to the elevation to determine the height of the instrument. These are phenomena are used in differential leveling and are applied to corrections for curvature and refraction.
What time ----–- the train arrve?
On a day when the barometer reads 755 mmhg, a tire pressure gage reads 204 kPa. The absolute pressure in the tire is:
Answer:
2.29mHg
Explanation:
We first do a conversion of gage pressure to mmHg
= 204(10³) / 13595(9.807)
= 204000/133326.165
= 1.53mHg
755 mmhg = 0.755
The pressure is therefore
P = p(atmosphere) + Of
= 1.53 + 0.755
= 2.29 mHg
This is the absolute pressure on this tire. So this answers this question.
The absolute pressure in the tire is equal to 2,285.13 mmHg.
Given the following data:
Atmospheric pressure of barometer = 755 mmHg.Gauge pressure of tire = 204 kPa.To calculate the absolute pressure in the tire:
First of all, we would convert the value of gauge pressure in kPa to mmHg.
Conversion:
1 kPa = 7.50062 mmHg
204 kPa = [tex]204 \times 7.50062 = 1530.13\;mmHg[/tex]
The formula for absolute pressure.Mathematically, absolute pressure is given by this formula:
[tex]P_{abs} = P_{atm} + P_{ga}[/tex]
Where:
[tex]P_{atm}[/tex] is the atmospheric pressure.[tex]P_{ga}[/tex] is the gauge pressure.Substituting the parameters into the formula, we have;
[tex]P_{abs} = 755 + 1530.13[/tex]
Absolute pressure = 2,285.13 mmHg.
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One - tenth kilogram of air as an ideal gas with k= 1.4 executes a carnot refrigeration cycle as shown i fig. 5,16, the isothermal expansion occurs at - 23C with a heat transfer to the air of 3.4 kj. The isothermal compression occurs at 27C to a final volume of 0.01m. Using the results of prob. 5.80 adapted to the case, Determine (a) the pressure, in Kpa, at each of the four principal states (b) the work, in KJ for each of the four processes (c) the coefficient of performance
Answer:
Hello your question is incomplete attached below is the missing part
a) p1 = 454.83 kPa, p2 = 283.359 Kpa , p3 = 536.423 kpa , p4 = 860.959kPa
b) W12 = 3.4 kJ, W23 = -3.5875 KJ, W34 = -4.0735 KJ, W41 = 3.5875 KJ
c) 5
Explanation:
Given data:
mass of air ( m ) = 1/10 kg
adiabatic index ( k ) = 1.4
temperature for isothermal expansion = 250K
rate of heat transfer ( Q12 ) = 3.4 KJ
temperature for Isothermal compression ( T4 ) = 300k
final volume ( V4 ) = 0.01m ^3
a) Calculate the pressure, in Kpa, at each of the four principal states
from an ideal gas equation
P4V4 = mRT4 ( input values above )
hence P4 = 860.959kPa
attached below is the detailed solution
b) Calculate work done for each processes
attached below is the detailed solution
C) Calculate the coefficient of performance
attached below is detailed solution
(TCO 1) The cutoff frequency of a low pass filter is usually identified by its _____. Group of answer choices
Answer:
-3 dB point
Explanation:
A filter is a circuit deigned to pass or amplify certain frequencies while also preventing other frequencies from passing. There are different types of filter depending on the frequency range such as the low pass filter, high pass filter, bandstop filter, bandpass filter.
A low pass filter is a filter that allows the passage of frequencies below the cut off frequency and blocks frequencies above the cut off frequency. The cutoff frequency of a low pass filter is the frequency at the -3dB point.
A new construction firm completed its first project of a residential building. But a few months later, the building developed cracks, and eventually the entire building crumbled to the ground. What caused the building to fall?
A. Poor air conditioning
B. Weak walls
C. External force
D. Furniture
Answer:
it is weak walls
Explanation:
The primary energy source for the controller in a typical control system is either brainlythe primary energy source for the controller in a typical control system is either
Answer:
a pneumatic or electric power
Explanation:
The primary energy source for the controller in a typical control system is either "a pneumatic or electric power."
This is because a typical control system has majorly four elements which include the following:
1. Sensor: this calculates the controlled variable
2. Controller: this receives and process inputs from the sensor to the controlled device as output
3. Controlled device: this tweak the controlled variable
4. Source of energy: this is the energy used to power the control system. It could be a pneumatic or electric power
At what angle should you place your elbows while at your workstation? Above 120 degrees At 60 degrees Below 90 degrees Between 90 and 110 degrees
Answer:
Between 90 and 110 degrees
Explanation:
Ideally, the angle should not be less than 90° so that your elbows could be parallel to the workstation. This will in turn mean that the shoulders will be relaxed and it will make it a lot less stressful to keep the wrists in a straight manner.
Looking at the options, there is no exact figure of 90°, however, it could be between 90° to 110° because the difference is not going to be much in terms of making the shoulders relaxed.
Answer:
90 degrees i did the test and i also looked it up to make sure i was right.
Explanation:
how much metal can be removed from a cracked drum to restore surface
No amount of metal removal will restore a cracked surface, rather, there is a need to completely change the metal drum.
What is a metal drum?This means a from of cylindal metal container that is used for shipping or storage of liquids.
When its surface is cracked, then, no amount of metal removal will restore a cracked surface, rather, there is a need to completely change the metal drum.
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Animation can occur before an action
Answer:Animation can occur before an action. Prepares the audience for the action. ... A pose or action should clearly communicate to the audience the attitude, mood, reaction or idea of the character. The effective use of long, medium or close up shots, as well as camera angles helps tell the story.
Explanation:May i have brainlist plz only if u wanna give me brainlist though have an nice day and stay safe.
DİGİTAL LOGİC DESİGN
Answer:
Uh- dude if you think I'm gonna download that, think twice...
Convert 5m/h to yds/week
918.63517 is the answer.
ASAE 1060 Steel wire (1 mm diameter) is coated with copper to form a composite with a diameter of 2mm. Use the following properties for parts a, b, and c of question 2: The elastic modulus of copper is 110 GPa The yield stress of the copper is 140 MPa The coefficient of thermal expansion of the copper is 17 times 10^-6/degree C. The elastic modulus of steel is 205 GPa The yield stress of the steel is 280 MPa The coefficient of thermal expansion of the copper is 10 times 10^-6/degree C Determine: a. The elastic modulus of the composite b. The maximum force that the composite will carry before either material yields c. The coefficient of thermal expansion of the composite material.
Answer:
a) [tex]E_{m}[/tex] = 133.75 Gpa
b) Fnet = 560 N
c) thermal expansion of the composite material = 14.31 [tex]10^{-6 }[/tex] / °C
Explanation:
Solution:
a) Elastic Modulus of the composite:
Area of steel wire = [tex]\frac{\pi }{4}[/tex] x ([tex]0.001^{2}[/tex]) = 0.8 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]
Area of Copper wire = [tex]\frac{\pi }{4}[/tex] x ([tex]0.002^{2}[/tex]) - 0.8 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]
Area of Copper wire = 2.4 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]
Young's Modulus of Composite mixture:
[tex]E_{m}[/tex] = [tex]F_{st}[/tex][tex]E_{st}[/tex] + [tex]F_{Cu}[/tex][tex]E_{Cu}[/tex] Equation 1
here,
[tex]F_{st}[/tex] = Stress in Steel
[tex]F_{Cu}[/tex] = Stress in Copper.
We know that,
F = P/A
F is inversely proportional to Area, so if area is large, stress will less and vice versa. So, Take
Ratio for area of steel = [tex]\frac{0.8. 10^{-6} }{(0.8 + 2.4) .10^{-6} }[/tex]
Ratio for area of steel = [tex]\frac{0.8}{3.2 }[/tex] = 0.25
Similarly, for Copper,
Ratio for area of copper = [tex]\frac{2.4. 10^{-6} }{(0.8 + 2.4) .10^{-6} }[/tex]
Ratio for area of copper = [tex]\frac{2.4 }{3.2}[/tex] = 0.75
Put these values in equation 1:
[tex]E_{m}[/tex] = [tex]F_{st}[/tex][tex]E_{st}[/tex] + [tex]F_{Cu}[/tex][tex]E_{Cu}[/tex]
[tex]E_{m}[/tex] = (0.25) [tex]E_{st}[/tex] + (0.75)[tex]E_{Cu}[/tex]
We are given that,
[tex]E_{st}[/tex] = 205 Gpa
[tex]E_{Cu}[/tex] = 110 Gpa
So,
[tex]E_{m}[/tex] = (0.25) (205 Gpa) + (0.75) (110 GPa)
[tex]E_{m}[/tex] = 51.25GPa + 82.5 Gpa
Hence, the Elastic Modulus of the composite will be:
[tex]E_{m}[/tex] = 133.75 Gpa
b) maximum force:
Fnet = Fst + Fcu
We know that F = (Yield Stress x Area)
F = fst x Ast + fcu x Acu
And we are given that,
Yield stress of Steel = 280 Mpa
Yield stress of Copper = 140 Mpa
And,
Ast = 0.8 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]
Acu = 2.4 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]
Just plugging in the values, we get:
F = (280 Mpa) (0.8 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]) + (140 Mpa) (2.4 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex])
F = 224 + 336
Fnet = 560 N ( because Mpa = [tex]10^{6}[/tex] N/[tex]m^{2}[/tex])
So, it means the composite will carry the maximum force of 560N
c) Coefficient of Thermal Expansion:
Strain on both material is same upon loading so,
(ΔL/L)st = (ΔL/L)cu
by thermal expansion equation:
([tex]\alpha .[/tex]ΔT + [tex]\frac{F}{A}[/tex][tex]. \frac{1}{Est}[/tex]) = [tex]\alpha .[/tex]ΔT + [tex]\frac{F}{A}[/tex][tex]. \frac{1}{Ecu}[/tex])
Where [tex]\alpha[/tex] = Coefficient of Thermal expansion
Here, fst = -fcu = F
and ΔT = 1°
So,
Plugging in the values, we get.
( 10 x [tex]10^{-6}[/tex] x (1) + [tex]\frac{F}{0.8.10^{-6} } . \frac{1}{205 . 10^{9} }[/tex] ) = ( 17 x [tex]10^{-6}[/tex] x (1) + [tex]\frac{-F}{2.4.10^{-6} } . \frac{1}{110 . 10^{9} }[/tex] )
Solving for F, we get:
F = 0.71 N
Here,
fst = F = 0.71 N (Tension on Heating)
fcu = -F = 0.71 N ( Compression on Heating )
So, the combined thermal expansion of the composite material will be:
(ΔL/L)cu = ( 17 x [tex]10^{-6}[/tex] x (1°) + [tex]\frac{-0.71}{2.4.10^{-6} } . \frac{1}{110 . 10^{9} }[/tex] )
(ΔL/L)cu = ( 17 x [tex]10^{-6}[/tex] x (1°) - 2.69 x [tex]10^{-6}[/tex]
combined thermal expansion of the composite material = 14.31 [tex]10^{-6 }[/tex] / °C
A certain power-supply filter produces an output with a ripple of 100 mV peak-to-peak and a dc value of 20 V. The ripple factor is
Answer: the ripple factor is 0.005
Explanation:
Given the data in the question;
we know that expression of ripple factor is;
r = Vr(pp) / Vdc
where Vr(pp) the peak to peak is ripple voltage ( 100mv = 0.1 V)
and Vdc is the dc value of the filter output voltage ( 20 V)
so we substitute our given values;
r = 0.1 / 20
r = 0.005
Therefore; the ripple factor is 0.005
The ripple factor of the given power supply filter is; γ = 0.005
We are given;
Ripple peak to peak voltage; V_r,pp = 100 mV = 0.1 V
DC value of the filter output voltage; V_dc = 20 V
Now, formula for the ripple factor is given as;
γ = V_rms/V_dc
Now, our ripple peak to peak voltage is also known as the rms value of ripple voltage at the output. Thus;
γ = 0.1/20
γ = 0.005
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Which of these construction materials does the government restrict because of toxicity?
A.
lead
B.
silica
C.
concrete
D.
cement
Answer:
A
Explanation:
It is also a toxic material
Answer:
The correct answer is A. Lead.
Explanation:
I got it right on the Plato test.
The fan pressure differential gage on an air handler reads 12 cm H2O. What is this pressure differential in kiloPascals
Answer:
[tex]1.18\ \text{kPa}[/tex]
Explanation:
g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
h = Height of reading = 12 cm
[tex]\rho[/tex] = Density of water = [tex]1000\ \text{kg/m}^3[/tex]
Pressure due to height difference is given by
[tex]P=\rho gh\\\Rightarrow P=1000\times 9.81\times 12\times 10^{-2}\\\Rightarrow P=1177.2\ \text{Pa}=1.1772\ \text{kPa}\approx 1.18\ \text{kPa}[/tex]
The pressure differential is [tex]1.18\ \text{kPa}[/tex].
According to the question,
Height of reading, [tex]h = 12 \ cm[/tex]Density of water, [tex]\rho = 1000 \ kg/m^3[/tex]Acceleration due to gravity, [tex]g = 9.8 \ m/s^2[/tex]The formula will be:
→ [tex]P = \rho g h[/tex]
By putting the values, we get
[tex]= 1000\times 9.8\times 12\times 10^{-2}[/tex]
[tex]= 1177.2 \ Pa[/tex]
[tex]= 1.18 \ kPa[/tex]
Thus the answer above is correct.
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Technician A says that tailor-rolled parts may be used for collision energy managements.
Technician B says that tailor-welded parts are aluminum and steel parts joined together. Who is right?
A Only
B only
Both A and B
Neither A nor B
At a high school science fair, Connor won first place for his replica of the Golden Gate Bridge. Connor liked the
project so much, he now wants to design bridges as a career. Which will best position Connor to do that?
earning a bachelor's degree in Civil Engineering from a four-year university, completing an internship, and
seeking a job at a private firm
earning a doctoral degree in Civil Engineering and seeking a job in the public works department of a state or
federal government
O completing an internship with a state or federal government before earning an associate degree in Civil
Engineering from a technical school
working at an entry-level job at a private engineering firm before earning a degree in Civil Engineering from a
four-year university
The correct answer is A. Earning a bachelor's degree in Civil Engineering from a four-year university, completing an internship, and seeking a job at a private firm.
Explanation:
In the U.S. and many countries, the best to start a career is to enroll in a formal educational program at a university or college. This helps students learn concepts, theories, methods, etc. they need for their profession. Moreover, a degree such as a bachelor's degree is required by employers. In this context, the first step for Connor is to earn a bachelor's degree in Civil Engineering.
Besides this, an internship is recommended after earning a degree because this is the way students can gain real-life work experience, which is considered positive by employers. This means the next step should be an internship.
Finally, Connor can seek a job to design bridges and other buildings because after the degree and internship he will have the experience and knowledge required by employers and by the job.
The bulk density of a compacted soil specimen (Gs = 2.70) and its water content are 2060 kg/m^3 and 15.3%, respectively. If the specimen is soaked in a bucket of water for sev-eral days until it is fully saturated, what should the saturated density be?
Answer:
the saturated density should be
Explanation:
A cylindrical metal specimen of initial diameter d0 =14 mm, initial length L0=53 mm, strain hardening exponent n=0.31, strength coefficient K=665 MPa and reduction of area at fracture Arf=0.53 - Determine the true strain at maximum load and true strain at fracture - Determine the length of specimen at maximum load and diameter at fracture - Determine the maximum load force - Determine the Load (F) on the specimen when a true strain Ɛt=0.25 is applied during the tension test
Answer:
a) Ef = 0.755
b) length of specimen( Lf )= 72.26mm
diameter at fracture = 9.598 mm
c) max load ( Fmax ) = 52223.24 N
d) Ft = 51874.67 N
Explanation:
a) Determine the true strain at maximum load and true strain at fracture
True strain at maximum load
Df = 9.598 mm
True strain at fracture
Ef = 0.755
b) determine the length of specimen at maximum load and diameter at fracture
Length of specimen at max load
Lf = 72.26 mm
Diameter at fracture
= 9.598 mm
c) Determine max load force
Fmax = 52223.24 N
d) Determine Load ( F ) on the specimen when a true strain et = 0.25 is applied during tension test
F = 51874.67 N
attached below is a detailed solution of the question above
Describe the differences between case hardening and through hardening, insofar as engineering applications of metals are concerned.
Answer:
The answer is below
Explanation:
Case hardening is a form of steel hardening that is applied on mild steel with a high temperature of heat.
It results in material forming a hard surface membrane, while the inner layer is soft.
It is mostly used for universal joints, construction cranes, machine tools, etc.
On the other hand, Through hardening is a form of steel hardening in engineering that involves heat treatment of carbon steel.
It increases the hardness and brittleness of the material.
It is often used for axles, blades, nuts and bolts, nails, etc.
A 30*30 cm cross section concrete pile with length of 18 m, the carrying capacity of the pile at
the pile tip is 9445 KN/m2 and the average unit skin resistance is 25 kN/m2. The ultimate load
carrying capacity of this pile is:
Select one:
0 2231 KN
2111 KN
1390 KN
1300 KN
Answer:
b
Explanation:
Trust me. I am not the smartest though...
The speed of sound is 1150 ft/s convert to mile/h
Answer:
784.090909mph
Explanation:
1ft/s=0.681818 mph
1150ft/s=0.681818 x 1150=784.090909
A separately excited DC motor turns at 1200 r/min when the armature is connected to 115V source. Calculate the armature voltage required so that the motor runs at 1500 r/min. At 100 r/min.
Answer:
The armature voltage required to run the motor at 1500 RPM is 143.75 V
The armature voltage required to run the motor at 100 RPM is 9.58 V
Explanation:
Given;
angular speed of the DC motor, ω = 1200 RPM
source voltage, V = 115 V
The armature voltage required to run the motor at 1500 RPM;
[tex]V_a = 115V \ \times \ \frac{1500 \ RPM}{1200 \ RPM} \\\\V_a = 143.75 \ V[/tex]
The armature voltage required to run the motor at 100 RPM;
[tex]V_a = 115V \ \times \ \frac{100 \ RPM}{1200 \ RPM} \\\\V_a = 9.58 \ V[/tex]
