Answer:
genorator- electrical to mechanical
motor- mechanical to electrical
what is the Economic
Discuss the ethics of the circumstances that resulted in the Columbia shuttle disaster. Considering the predictions that were made years before the disaster, as well as the reliability of the Binomial distribution and its implications, what could or should the engineers associated with the program have done differently
Explanation:
This is not so much a mathematical issue as a case study, because the response will inevitably require us to test the special Columbic shuttle disaster scenario. I would suggest that you read this in detail and present the points accordingly. Here I give as many points as I think are relevant.
The failure of a space program is definitely a complex situation, more than a simple binomial distribution. It's definitely not as simple as repeating the flip of a coin. There are several coherent factors and situations that govern the overall coordination and execution of such an event. The problem is, those who are running a project like this are still making a trade off,It is never the case that they sealed the lid on any chance of failure between multiple parameters. You try to do something, but often, as is the case above, the potentially dangerous situation is impossible or uncontrollable. Since the root cause of failure, which is dried out tiles that can not withstand heat and water, it appears that owing to the constant use of the shuttle the head architects have not foreseen this.
For an Ac signal with peak voltage Vp equal to 60V, the same power would be delivered to a load with a dc voltage of
Answer:
30√2 ≈ 42.426 volts
Explanation:
The RMS value of the signal is the DC equivalent. For a sine wave, the mean of the square is half the square of the peak value. Then the Root-Mean-Square is ...
RMS = √((Vp)^2/2) = Vp/√2
For Vp = 60 volts, the equivalent DC voltage is ...
V = (60 volts)/√2 = 30√2 volts ≈ 42.426 volts
6. For the following waste treatment facility, chemical concentration (mg/gal) within the tank can be considered uniform. The initial chemical concentration inside the tank was 5 mg/gal, the concentration of effluent coming in is 20 mg/gal. The volume of the tank is 600 gallons. The fluid coming in rate is equal to fluid going out is equal to 60 gal/min. (a) Establish a dynamic model of how the concentration of chemical inside the tank increases over time. Specify the units for your variables. (Hint: establish a mass balance of the waste chemical. The concentration inside the tank = concentration going out) (b) Find the Laplace transformation of the concentration (transfer function) for this step input of 20 mg/gal.
Answer:
mass of chemical coming in per minute = 60 × 20 = 1200 mg/min
at a time t(min) , M = mass of chemical = 1200 × t mg
concentration of chemical = 1200t / 600 = 2t mg / gallon
Explanation:
Since it is only fluid that is leaving and not chemical.
A sheet of steel 4.4 mm thick has nitrogen atmospheres on both sides at 1200°C and is permitted to achieve a steady-state diffusion condition. The diffusion coefficient for nitrogen in steel at this temperature is 5.9 × 10^(-11) m^2/s, and the diffusion flux is found to be 4.7 × 10^(-7) kg/m^2.s. Also, it is known that the concentration of nitrogen in the steel at the high-pressure surface is 4.5 kg/m^3.
How far into the sheet from this high-pressure side will the concentration be 2.7 kg/m^3? Assume a linear concentration profile.
Answer:
0.544×10–³
Explanation:
Please see the attached file for the solution
Determine the drag on a small circular disk of 0.02-ft diameter moving 0.01 ft/s through oil with a specific gravity of 0.89 and a viscosity 10000 times that of water. The disk is oriented normal to the upstream velocity. By what percent is the drag reduced if the disk is oriented parallel to the flow?
Answer:
33.3%
Explanation:
Given that:
specific gravity (SG) = 0.89
Diameter (D) = 0.01 ft/s
Density of oil [tex]\rho= SG\rho _{h20} = 0.89 * 1.94=1.7266\frac{sl}{ft^3}[/tex]
Since the viscosity 10000 times that of water, The reynold number [tex]R_E=\frac{\rho VD}{\mu} =\frac{1.7266*0.01*0.01}{0.234}=7.38*10^{-4}[/tex]
Since RE < 1, the drag coefficient for normal flow is given as: [tex]C_{D1}=\frac{24.4}{R_E}= \frac{20.4}{7.38*10^{-4}}=2.76*10^4[/tex]
the drag coefficient for parallel flow is given as: [tex]C_{D2}=\frac{13.6}{R_E}= \frac{13.6}{7.38*10^{-4}}=1.84*10^4[/tex]
Percent reduced = [tex]\frac{D_1-D_2}{D_2} *100=\frac{2.76-1.84}{3.3}=33.3[/tex] = 33.3%
Consider a Carnot heat pump cycle executed in a steady-flow system in the saturated mixture region using R-134a flowing at a rate of 0.264 kg/s. The maximum absolute temperature in the cycle is 1.15 times the minimum absolute temperature, and the net power input to the cycle is 5 kW. If the refrigerant changes from saturated vapor to saturated liquid during the heat rejection process, determine the ratio of the maximum to minimum pressures in the cycle.
Answer:
7.15
Explanation:
Firstly, the COP of such heat pump must be measured that is,
[tex]COP_{HP}=\frac{T_H}{T_H-T_L}[/tex]
Therefore, the temperature relationship, [tex]T_H=1.15\;T_L[/tex]
Then, we should apply the values in the COP.
[tex]=\frac{1.15\;T_L}{1.15-1}[/tex]
[tex]=7.67[/tex]
The number of heat rejected by the heat pump must then be calculated.
[tex]Q_H=COP_{HP}\times W_{nst}[/tex]
[tex]=7.67\times5=38.35[/tex]
We must then calculate the refrigerant mass flow rate.
[tex]m=0.264\;kg/s[/tex]
[tex]q_H=\frac{Q_H}{m}[/tex]
[tex]=\frac{38.35}{0.264}=145.27[/tex]
The [tex]h_g[/tex] value is 145.27 and therefore the hot reservoir temperature is 64° C.
The pressure at 64 ° C is thus 1849.36 kPa by interpolation.
And, the lowest reservoir temperature must be calculated.
[tex]T_L=\frac{T_H}{1.15}[/tex]
[tex]=\frac{64+273}{1.15}=293.04[/tex]
[tex]=19.89\°C[/tex]
the lowest reservoir temperature = 258.703 kpa
So, the pressure ratio should be = 7.15
Have you ever had an ice cream headache that’s when a painful sensation resonates in your head after eating something cold usually ice cream on a hot day this pain is produced by the dilation of a nerve center in the roof of your mouth the nerve center is overreacting to the cold by trying to hit your brain ice cream headaches have turned many smiles to frowns identify the structure
Answer:
Cause and effect
Explanation:
pls mark brainliest
The structure that makes or turned many smiles to frowns can be regarded as compare/contrast.
What is compare contrast?The term compare/contrast is a common terms. The act of comparing is known to be depicting the similarities, and contrasting is said to be showing differences that exist between two things.
Conclusively, from the above, we can see that it is a compare/contrast scenario as it talks about the effects of taking ice cream. It went from smiles to frowns.
See option below
cause/effect
descriptive
compare/contrast
sequence/process
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Consider this example of a recurrence relation. A police officer needs to patrol a gated community. He would like to enter the gate, cruise all the streets exactly once, and then leave by the same gate. What information would you need to determine a Euler circuit and a Hamilton circuit
Answer:
the police officer cruise each streets precisely once and he enters and exit with the same gate.
Explanation:
NB: kindly check below for the attached picture.
The term ''Euler circuit'' can simply be defined as the graph that shows the edge of K once in a finite way by starting and putting a stop to it at the same vertex.
The term "Hamiltonian Circuit" is also known as the Hamiltonian cycle which is all about a one time visit to the vertex.
Here in this question, the door is the vertex and the road is the edge.
The information needed to detemine a Euler circuit and a Hamilton circuit is;
"the police officer cruise each streets precisely once and he enters and exit with the same gate."
Check attachment for each type of circuit and the differences.
the police officer cruise each streets precisely once and he enters and exit with the same gate.
A steady green traffic light means
Answer:
Its C. you may proceed, but only if the path is clear
Explanation:
I just gave Quiz and its correct
A flat site is being considered for a new school that will have a steel frame and brick façade. The steel columns will have a maximum load of 250 kips, and the planned column support will consist of a 6 foot by 6 foot square footing placed 2 feet below the ground surface (to the bottom of the footing). Subsurface conditions consist of a 15-foot-thick layer of uniform silty sand (unit weight = 122 pcf, soil modulus = 160,000 psf) over stiff clay (unit weight = 118 pcf, soil modulus = 230,000 psf). Groundwater is deep.
a) Sketch the problem (freehand, not to scale) and state any necessary assumptions.
b) Calculate the immediate (elastic) settlement.
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem
A refrigerator uses refrigerant-134a as the working fluid and operates on the ideal vapor-compression refrigeration cycle except for the compression process. The refrigerant enters the evaporator at 120 kPa with a quality of 34 percent and leaves the compressor at 70°C. If the compressor consumes 450 W of power, determine (a) the mass flow rate of the refrigerant, (b) the condenser pressure, and (c) the COP of the refrigerator
Answer:
(a) 0.0064 kg/s
(b) 800 KPa
(c) 2.03
Explanation:
The ideal vapor compression cycle consists of following processes:
Process 1-2 Isentropic compression in a compressor
Process 2-3 Constant-pressure heat rejection in a condenser
Process 3-4 Throttling in an expansion device
Process 4-1 Constant-pressure heat absorption in an evaporator
For state 4 (while entering compressor):
x₄ = 34% = 0.34
P₄ = 120 KPa
from saturated table:
h₄ = hf + x hfg = 22.4 KJ/kg + (0.34)(214.52 KJ/kg)
h₄ = 95.34 KJ/kg
For State 1 (Entering Compressor):
h₁ = hg at 120 KPa
h₁ = 236.99 KJ/kg
s₁ = sg at 120 KPa = 0.94789 KJ/kg.k
For State 3 (Entering Expansion Valve)
Since 3 - 4 is an isenthalpic process.
Therefore,
h₃ = h₄ = 95.34 KJ/kg
Since this state lies at liquid side of saturation line, therefore, h₃ must be hf. Hence from saturation table we find the pressure by interpolation.
P₃ = 800 KPa
For State 2 (Leaving Compressor)
Since, process 2-3 is at constant pressure. Therefore,
P₂ = P₃ = 800 KPa
T₂ = 70°C (given)
Saturation temperature at 800 KPa is 31.31°C, which is less than T₂. Thus, this is super heated state. From super heated property table:
h₂ = 306.9 KJ/kg
(a)
Compressor Power = m(h₂ - h₁)
where,
m = mass flow rate of refrigerant.
m = Compressor Power/(h₂ - h₁)
m = (0.450 KJ/s)/(306.9 KJ/kg - 236.99 KJ/kg)
m = 0.0064 kg/s
(b)
Condenser Pressure = P₂ = P₃ = 800 KPa
(c)
The COP of ideal vapor compression cycle is given as:
COP = (h₁ - h₄)/(h₂ - h₁)
COP = (236.99 - 95.34)/(306.9 - 236.99)
COP = 2.03
The Ph diagram is attached
g In the above water treatment facility, chemical concentration (mg/gal) within the tank can be considered uniform. The initial chemical concentration inside the tank was 0 mg/gal, the concentration of effluent coming in is 10 mg/gal. The volume of the tank is 10,000 gallons. The fluid coming in rate is equal to fluid going out is equal to 50 gal/min. Establish a dynamic model of how the concentration of the chemical inside the tank increases over time.
Answer:
0.05 mg / gallon
Explanation:
mass of chemecila coming in per minute = 50*10 = 500 mg/min
at a time t min , M = mass of chemical = 500*t mg
conecntartion of chemecal = 500t/10000 = 0.05 mg / gallon
Two Electric field vectors E1 and E2 are perpendicular to each other; obtain its base
vectors.
Answer:
<E1, E2>.
Explanation:
So, in the question above we are given that the Two Electric field vectors E1 and E2 are perpendicular to each other. Thus, we are going to have the i and the j components for the two Electric Field that is E1 and E2 respectively. That is to say the addition we give us a resultant E which is an arbitrary vector;
E = |E| cos θi + |E| sin θj. -------------------(1).
Therefore, if we make use of the components division rule we will have something like what we have below;
x = |E2|/ |E| cos θ and y = |E1|/|E| sin θ
Therefore, we will now have;
E = x |E2| i + y |E1| j.
The base vectors is then Given as <E1, E2>.
The supplement file* that enclosed to this homework consists Time Versus Force data. The first column in the file stands for time (second) and the 2nd stands for force (Volt), respectively. This data were retrieved during an impact event. In this test, an impactor strikes to a sample. A force-ring sensor that attached to the impactor generates voltage during collision. A data acquisition card gathers the generated signals.
• Take the mass of the impactor as 30 kg and strike velocity as 2.0 m/s.
• Pick the best numerical technique, justify your choice.
• All results must be in SI units.
a) Determine a factor that will be used in the conversion from V to N. The calibration data that supplied by manufacturer as below:
• Take the mass of the impactor as 30 kg and strike velocity as 2.0 m/s.
• Pick the best numerical technique, justify your choice.
• All results must be in SI units.
a) Determine a factor that will be used in the conversion from V to N. The calibration data that supplied by manufacturer as below:• Take the mass of the impactor as 30 kg and strike velocity as 2.0 m/s.
• Pick the best numerical technique, justify your choice.
• All results must be in SI units.
a) Determine a factor that will be used in the conversion from V to N. The calibration data that supplied by manufacturer as below:
b) Compute the impulse of this event.
c) Obtain acceleration, velocity and displacement histories by using Newton’s second law of motion.
d) Compute the absorbed energy during collision.
Answer:
A.) 1mv = 2000N
B.) Impulse = 60Ns
C.) Acceleration = 66.67 m/s^2
Velocity = 4 m/s
Displacement = 0.075 metre
Absorbed energy = 60 J
Explanation:
A.) Using a mathematical linear equation,
Y = MX + C
Where M = (2000 - 0)/( 898 - 0 )
M = 2000/898
M = 2.23
Let Y = 2000 and X = 898
2000 = 2.23(898) + C
2000 = 2000 + C
C = 0
We can therefore conclude that
1 mV = 2000N
B.) Impulse is the product of force and time.
Also, impulse = momentum
Given that
Mass M = 30kg
Velocity V = 2 m/s
Impulse = M × V = momentum
Impulse = 30 × 2 = 60 Ns
C.) Force = mass × acceleration
F = ma
Substitute force and mass into the formula
2000 = 30a
Make a the subject of formula
a = 2000/30
acceleration a = 66.67 m/s^2
Since impulse = 60 Ns
From Newton 2nd law,
Force = rate of change in momentum
Where
change in momentum = -MV - (- MU)
Impulse = -MV + MU
Where U = initial velocity
60 = -60 + MU
30U = 120
U = 120/30
U = 4 m/s
Force = 2000N
Impulse = Ft
Substitute force and impulse to get time
60 = 2000t
t = 60/2000
t = 0.03 second
Using third equation of motion
V^2 = U^2 + 2as
Where S = displacement
4^2 = 2^2 + 2 × 66.67S
16 = 4 + 133.4S
133.4S = 10
S = 10/133.4
S = 0.075 metre
D.) Energy = 1/2 mV^2
Energy = 0.5 × 30 × 2^2
Energy = 15 × 4 = 60J
An insulated, vertical piston–cylinder device initially contains 10 kg of water, 6 kg of which is in the vapor phase. The mass of the piston is such that it maintains a constant pressure of 200 kPa inside the cylinder. Now steam at 0.5 MPa and 350°C is allowed to enter the cylinder from a supply line until all the liquid in the cylinder has vaporized. Determine (a) the final temperature in the cylinder and (b) the mass of the steam that has entered.
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
A heat recovery device involves transferring energy from the hot flue gases passing through an annular region to pressurized water flowing through the inner tube of the annulus. The inner tube has inner and outer diameters of 24 and 30 mm and is connected by eight struts to an insulated outer tube of 60-mm diameter. Each strut is 3 mm thick and is integrally fabricated with the inner tube from carbon steel (k 50 W/m K). Consider conditions for which water at 300 K flows through the inner tube at 0.161 kg/s while flue gases at 800 K flow through the annulus, maintaining a convection coefficient of 100 W/m2 K on both the struts and the outer surface of the inner tube. What is the rate of heat transfer per unit length of tube from gas to the water?
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
is sampled at a rate of to produce the sampled vector and then quantized. Assume, as usual, the minimum voltage of the dynamic range is represented by all zeros and the maximum value with all ones. The numbers should increase in binary order from bottom to top. Find the bit combination used to store each sample when rounded to the nearest integer between and (clipping may occur). Note: A partially-correct answer will not be recognized. You must answer all three correctly on the same
Answer:
d[0] = 11111111
d[1] = 11011101
d[2] = 1111011
Explanation:
Assume that the number of bits is 8. The voltage range input is -8 to 7 volts. The range is thus 15V, and the resolution is 15/2^8 = 0.0586 volts. We will first add +8 to the input to convert it to a 0-15v signal. Then find the equivalent bit representation. For 7.8 volts, the binary signal will be all 1's, since the max input voltage for the ADC is 7 volts. For 4.95, we have 4.95+8 = 12.95 volts. Thus, N = 12.95/0.0586 = 221. The binary representation is 11011101. For -0.8, we have -0.8 + 8 = 7.2. Thus, N = 7.2/0.0586 = 123. The binary representation is 1111011.
Thus,
d[0] = 11111111
d[1] = 11011101
d[2] = 1111011
Design a circuit that will tell whether a given month has 31 days in it. The month is specified by a 4-bit input, A3:0. For example, if the inputs are 0001, the month is January, and if the inputs are 1100, the month is December. The circuit output, Y, should be HIGH only when the month specified by the inputs has 31 days in it. However, if the input is not a valid month (such as 0 or 13) then the output is don’t care (can be either 0 or 1).
Answer:
see attachments
Explanation:
A Karnaugh map for the output is shown in the first attachment. The labeled and shaded squares represent the cases where Y = HIGH. The associated logic can be simplified to
Y = A3 xor A0
when the don't care at 1110 gives an output of HIGH.
__
The second attachment shows a logic diagram using a 4:1 multiplexer to do the decoding. A simple XOR gate would serve as well. If AND-OR-INV logic is required, that would be ...
Y = Or(And(A3, Inv(A0)), And(A0, Inv(A3)))
The spherical pressure vessel has an inner diameter of 2 m and a thickness of 10 mm. A strain gauge having a length of 20 mm is attached to it, and it is observed to increase in length by 0.012 mm when the vessel is pressurized. Determine the pressure causing this deformation, and find the maximum in-plane shear stress, and the absolute maximum shear stress at a point on the outer surface of the vessel. The material is steel, for which Est
Question:
The spherical pressure vessel has an inner diameter of 2 m and a thickness of 10 mm. A strain gage having a length of 20 mm is attached to it, and it is observed to increase in length by 0.012 mm when the vessel is pressurized. Determine the pressure causing this deformation, and find the maximum in-plane shear stress, and the absolute maximum shear stress at a point on the outer surface of the vessel. The material is steel, for which Eₛₜ = 200 GPa and vₛₜ = 0.3.
Answer:
See explanation below
Explanation:
Given:
d = 2m = 2*10³ = 2000
thickness, t = 10 mm
Length of strain guage = 20 mm
i) Let's calculate d/t
[tex] \frac{d}{t} = \frac{2000}{10} = 200 [/tex]
Since [tex] \frac{d}{t}[/tex] is greater than length of strain guage, the pressure vessel is thin.
For the minimum normal stress, we have:
[tex] \sigma max= \frac{pd}{4t} [/tex]
[tex] \sigma max= \frac{2000p}{4 * 20} [/tex]
= 50p
For the minimum normal strain due to pressure, we have:
[tex] E_max= \frac{change in L}{L_g} [/tex]
[tex] = \frac{0.012}{20} = 0.60*10^-^3[/tex]
The minimum normal stress for a thin pressure vessel is 0.
[tex] \sigma _min = 0 [/tex]
i) Let's use Hookes law to calculate the pressure causing this deformation.
[tex] E_max = \frac{1}{E} [\sigma _max - V(\sigma _initial + \sigma _min)] [/tex]
Substituting figures, we have:
[tex] 0.60*10^-^3 = \frac{1}{200*10^9} [50p - 0.3 (50p + 0)] [/tex]
[tex] 120 * 10^6 = 35p [/tex]
[tex] p = \frac{120*10^6}{35}[/tex]
[tex] p = 3.429 * 10^6 [/tex]
p = 3.4 MPa
ii) Calculating the maximum in-plane shear stress, we have:
[tex] \frac{\sigma _max - \sigma _int}{2}[/tex]
[tex] = \frac{50p - 50p}{2} = 0 [/tex]
Max in plane shear stress = 0
iii) To find the absolute maximum shear stress at a point on the outer surface of the vessel, we have:
[tex] \frac{\sigma _max - \sigma _min}{2}[/tex]
[tex] = \frac{50p - 0}{2} = 25p [/tex]
since p = 3.429 MPa
25p = 25 * 3.4 MPa
= 85.71 ≈ 85.7 MPa
The absolute maximum shear stress at a point on the outer surface of the vessel is 85.7 MPa
(a) Show how two 2-to-1 multiplexers (with no added gates) could be connected to form a 3-to-1 MUX. Input selection should be as follows: If AB = 00, select I0 If AB = 01, select I1 If AB = 1− (B is a don’t-care), select I2 (b) Show how two 4-to-1 and one 2-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs. (c) Show how four 2-to-1 and one 4-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs
Answer:
Explanation:
a) Show how two 2-to-1 multiplexers (with no added gates) could be connected to form a 3-to-1 MUX. Input selection should be as follows: If AB = 00, select I0 If AB = 01, select I1 If AB = 1− (B is a don’t-care), select I2
We are to show how Two-2-to -1 multiplexers could be connected to form 3-to-1 MUX
If AB = 00 select [tex]I_o[/tex]
If AB = 01 select [tex]I_1[/tex]
If AB = 1_(B is don't care), select [tex]I_2[/tex]
However, the truth table is attached and shown in the first file below.
Also, the free- body diagram for 2- to - 1 MUX is shown in the second diagram attached below.
b) We are show how two 4-to-1 and one 2-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs.
The perfect illustration showing how they are connected in displayed in the third free-body diagram attached below.
Where ; [tex]I_o , I_1, I_2, I_3, I_4, I_5, I_6, I_7[/tex] are the inputs of the multiplexer and Z is the output.
c) Show how four 2-to-1 and one 4-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs.
For four 2-to-1 and one 4-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs, we have a perfect illustration of the diagram in the last( which is the fourth) diagram attached below.
Where ; [tex]I_o , I_1, I_2, I_3, I_4, I_5, I_6, I_7[/tex] are the inputs of the multiplexer and Z is the output
This question is a multiplexer which is a topic in digital circuit.
Multiplexer is a type of combination circuit that consist of a maximum of [tex]2^n[/tex] data inputs 'n' selection lines and single output line. One of these data inputs will be connected to the output based on the values of selection lines. Another name for multiplexers is MUX.
If we have 'n' selection lines, we will get [tex]2^n[/tex] possible combinations zero and ones. Each combination will select a maximum of only one data input.
a)
Two 2-to-1 multiplexers to form a 3-to-1 MUX.
If AB = 00, select [tex]I_o[/tex]
If AB = 01, select [tex]I_1[/tex]
If AB = 1- (B is don't care) select I
The truth table for the above scenario is in the attached document below.
Figure 1 and 2 represents the solution to this question.
b).
Two 4-to-1 multiplexers and one 2-to-1 multiplexers and one 2-to-1 multiplexers are used to form an 8-to-1 MUX.
In the attached diagram, figure 3 shows a comprehensive detail of how it is structured.
Where [tex]I_o[/tex] to [tex]I_7[/tex] are the inputs of the multiplexer and Z is the output.
c) Four 2-to-1 multiplexers and one 4-to -1 multiplexer are used to form 8-to-1 MUX.
In the attached diagram, figure 4 shows how it is structured.
We would see that [tex]I_o[/tex] to [tex]I_7[/tex] are the inputs of the multiplexer and Z is the output in the system.
Learn more about multiplexers here;
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An air conditioning unit is used to provide cooling during summer for a house. If the air conditioner provides 450 kW cooling by using 150 kW electrical power, determine the coefficient of performance (COP) of the air conditioner. The outside temperature and inside temperature are 40 and 20°C, respectively. Using the inequality of Clausius determine if the cycle is possible. Determine the COP of an air conditioner working based on the Carnot cycle between the same temperature difference. Compare the COPs of the Carnot and actual air conditioners and comment on them based on your answer for the previous part (the inequality of
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
Create an abstract class DiscountPolicy. It should have a single abstract method computeDiscount that will return the discount for the purchase of a given number of a single item. The method has two parameters, count and itemCost. 2. Derive a class BulkDiscount from DiscountPolicy, as described in the previous exercise. It should have a constructor that has two parameters, minimum and percent. It should define the method computeDiscount so that if the quantity purchased of an item is more than minimum, the discount is percent percent. 3. Derive a class BuyNItemsGetOneFree from DiscountPolicy, as described in Exercise 1. The class should have a constructor that has a single parameter n. In addition, the class should define the method computeDiscount so that every nth item is free. For example, the following table gives the discount for the purchase of various counts of an item that costs $10, when n is 3: count 1 2 3 4 5 6 7 Discount 0 0 10 10 10 20 20
4. Derive a class CombinedDiscount from DiscountPolicy, as described in Exercise 1. It should have a constructor that has two parameters of type DiscountPolicy. It should define the method computeDiscount to return the maximum value returned by computeDiscount for each of its two private discount policies. The two discount policies are described in Exercises 2 and 3. 5. Define DiscountPolicy as an interface instead of the abstract class described in Exercise 1.
Answer:
Java Code was used to define classes in the abstract discount policy,The bulk discount, The buy items get one free and the combined discount
Explanation:
Solution
Code:
Main.java
public class Main {
public static void main(String[] args) {
BulkDiscount bd=new BulkDiscount(10,5);
BuyNItemsGetOneFree bnd=new BuyNItemsGetOneFree(5);
CombinedDiscount cd=new CombinedDiscount(bd,bnd);
System.out.println("Bulk Discount :"+bd.computeDiscount(20, 20));
System.out.println("Nth item discount :"+bnd.computeDiscount(20, 20));
System.out.println("Combined discount :"+cd.computeDiscount(20, 20));
}
}
discountPolicy.java
public abstract class DiscountPolicy
{
public abstract double computeDiscount(int count, double itemCost);
}
BulkDiscount.java
public class BulkDiscount extends DiscountPolicy
{
private double percent;
private double minimum;
public BulkDiscount(int minimum, double percent)
{
this.minimum = minimum;
this.percent = percent;
}
at Override
public double computeDiscount(int count, double itemCost)
{
if (count >= minimum)
{
return (percent/100)*(count*itemCost); //discount is total price * percentage discount
}
return 0;
}
}
BuyNItemsGetOneFree.java
public class BuyNItemsGetOneFree extends DiscountPolicy
{
private int itemNumberForFree;
public BuyNItemsGetOneFree(int n)
{
itemNumberForFree = n;
}
at Override
public double computeDiscount(int count, double itemCost)
{
if(count > itemNumberForFree)
return (count/itemNumberForFree)*itemCost;
else
return 0;
}
}
CombinedDiscount.java
public class CombinedDiscount extends DiscountPolicy
{
private DiscountPolicy first, second;
public CombinedDiscount(DiscountPolicy firstDiscount, DiscountPolicy secondDiscount)
{
first = firstDiscount;
second = secondDiscount;
}
at Override
public double computeDiscount(int count, double itemCost)
{
double firstDiscount=first.computeDiscount(count, itemCost);
double secondDiscount=second.computeDiscount(count, itemCost);
if(firstDiscount>secondDiscount){
return firstDiscount;
}else{
return secondDiscount;
}
}
}
Sludge wasting rate (Qw) from the solids residence time (Thetac = mcrt) calculation. Given the following information from the previous problem. The total design flow is 15,000 m3/day. Theoretical hydraulic detention time (Theta) = 8 hours. The NPDES limit is 25 mg/L BOD/30 mg/L TSS.
Assume that the waste strength is 170 mg/L BOD after primary clarification.
XA=MLSS = 2200 mg/L,
Xw = Xu = XR = 6,600 mg/L,
qc = 8 days.
Make sure you account for the solids in the discharge.
What volume of sludge (Qw=m3/day) is wasted each day from the secondary clarifiers?
Answer:
The volume of sludge wasted each day from the secondary classifiers is Qw = 208.33 m^3 / day
Explanation:
Check the file attached for a complete solution.
The volume of the aeration tank was first calculated, V = 5000 m^3 / day.
The value of V was consequently substituted into the formula for the wasted sludge flow. The value of the wasted sludge flow was calculated to be Qw = 208.33 m^3 / day.
Problem Statement: Air flows at a rate of 0.1 kg/s through a device as shown below. The pressure and temperature of the air at location 1 are 0.2 MPa and 800 K and at location 2 the pressure and temperature are 0.75 MPa and 700 K. The surroundings are at 300 K and the surface temperature of the device is 1000 K. Determine the rate that the device performs work on its surroundings if the rate of heat transfer from the surface of the device to the environment is 1 kW. Justify your answer. Note that the flow direction for the air is not specified so you need to consider all possibilities for the direction of the airflow. Assume that the air is an ideal gas, that R
Answer:
The answer is "+9.05 kw"
Explanation:
In the given question some information is missing which can be given in the following attachment.
The solution to this question can be defined as follows:
let assume that flow is from 1 to 2 then
Q= 1kw
m=0.1 kg/s
From the steady flow energy equation is:
[tex]m\{n_1+ \frac{v^2_1}{z}+ gz_1 \}+Q= m \{h_2+ \frac{v^2_2}{2}+ gz_2\}+w\\\\\ change \ energy\\\\0.1[1.005 \times 800]-1= 0.01[1.005\times 700]+w\\\\w= +9.05 \ kw\\\\[/tex]
If the sign of the work performed is positive, it means the work is done on the surrounding so, that the expected direction of the flow is right.
How does a car batteray NOT die?
Answer:
bye hooking plugs up to it to amp it up
What is Postflow used to protect?
Answer:
The idea is to protect the puddle while it cools
Explanation:
Find a negative feedback controller with at least two tunable gains that (1) results in zero steady state error when the input is a unit step (1/s). (and show why it works); (2) Gives a settling time of 4 seconds; (3) has 10% overshoot. Use the standard 2nd order approximation. Plot the step response of the system and compare the standard approximation with the plot.
Answer:
Gc(s) = [tex]\frac{0.1s + 0.28727}{s}[/tex]
Explanation:
comparing the standard approximation with the plot attached we can tune the PI gains so that the desired response is obtained. this is because the time requirement of the setting is met while the %OS requirement is not achieved instead a 12% OS is seen from the plot.
attached is the detailed solution and the plot in Matlab
Air at 100°F, 1 atm, and 10% relative humidity enters an evaporative cooler operating at steady state. The volumetric flow rate of the incoming air is 1765 ft3/min. Liquid water at 68°F enters the cooler and fully evaporates. Moist air exits the cooler at 70°F, 1 atm. There is no significant heat transfer between the device and its surroundings and kinetic and potential energy effects can be neglected. Determine the mass flow rate at which liquid enters, in lb(water)/min.
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
A thin‐walled tube with a diameter of 12 mm and length of 25 m is used to carry exhaust gas from a smoke stack to the laboratory in a nearby building for analysis. The gas enters the tube at 200°C and with a mass flow rate of 0.006 kg/s. Autumn winds at a temperature of 15°C blow directly across the tube at a velocity of 2.5 m/s. Assume the thermophysical properties of the exhaust gas are those of air
Estimate the average heat transfer coefficient for the exhaust gas flowing inside the tube.
Answer:
The average heat transfer coefficient for the exhaust gas flowing inside the tube, h = 204.41 W/m^2 - K
Explanation:
The detailed solution is attached as files below.
However, the steps followed are highlighted:
1) The average temperature was calculated as 380.5 K
2) The properties of air at 380.5 K was highlighted
3) The Prandti number was calculated. Pr = 0.693
4) The Reynold number was calculated, Re = 28716.77
5) The Nusselt umber was calculated, Nu = 75.94
6) From Nu = (hD)/k , the average heat transfer coefficient, h, was calculated and a value of 204.41 W/m^2 - K was gotten.
