Answer:
Explanation:
Given that:
f = 250 Hz
[tex]\delta[/tex]= 2%
[tex]f_n[/tex]= 600 Hz
[tex]\zeta[/tex] = 0.5 to 1.5 increment by 0.05
[tex]F = A sin (Xt)[/tex]
For 250 Hz = 250 cycle/sec
[tex]X = 2 \pi t 250[/tex]
[tex]X = 500 \pi t[/tex]
[tex]X = Asin (500 \pi t)[/tex]
[tex]\omega = 250 \\ \\ \omega_n = 600[/tex]
M = 0.98 , 1.02
[tex]M_{(w)}} = \sqrt{[1-(\frac{w}{w_n})^2 + ( 2\zeta \frac{w}{w_n})^2}[/tex]
[tex]\frac{1}{M} = [1-(\frac{w}{w_n})^2]+(2 \zeta \frac{w}{w_n})^2[/tex]
[tex]\zeta = \frac{w_n}{2w}\sqrt{\frac{1}{M^2}-(1-(\frac{w}{w_n})^2)^2}[/tex]
[tex]\zeta = \frac{600}{2(250)}\sqrt{\frac{1}{0.98^2}-(1-(\frac{250}{600})^2)^2}[/tex]
[tex]\zeta = 0.7183[/tex]
At 0.7183 value of damping ratio the error value was 2% at 0.98 value of M
[tex]\frac{1}{M} = [1-(\frac{w}{w_n})^2]+(2 \zeta \frac{w}{w_n})^2[/tex]
[tex]\zeta = \frac{w_n}{2w}\sqrt{\frac{1}{M^2}-(1-(\frac{w}{w_n})^2)^2}[/tex]
[tex]\zeta = \frac{600}{2(250)}\sqrt{\frac{1}{1.02^2}-(1-(\frac{250}{600})^2)^2}[/tex]
[tex]\zeta = 0.6330[/tex]
At 0.6330 value of damping ratio the error value was 2% at 1.02 value of M.
Hence, the damping ratio [tex]\zeta[/tex] of the transducer must be placed between 0.6330 to 0.7183
Oxygen combines with nitrogen in the air to form NOx at about
A. 1500
B. 2500
C.50
D.000
Oxygen combines with nitrogen in the air to form NO at about 2500 celsius.
What combines nitrogen and oxygen in the air?
The enormous energy of lightning breaks nitrogen molecules and enables their atoms to combine with oxygen in the air forming nitrogen oxides. These dissolve in rain, forming nitrates, that are carried to the earth.
At these high temperatures, nitrogen and oxygen from the air combine to produce nitrogen monoxide. One nitrogen molecule (N2) reacts with one oxygen molecule (O2) to make two nitrogen monoxide molecules (NO).
Learn more about NO here:
https://brainly.com/question/27548777
#SPJ2
Experiment: With the battery voltage set to 15 volts, measure the current in a parallel circuit with 1, 2, 3, and 4 light bulbs. (In each case, place the ammeter next to the battery.) Use Ohm’s law to calculate the total resistance of the circuit. Record results below. Is this right?
Answer:
No
Explanation:
We expect current to be proportional to the number of identical bulbs. The total resistance is the ratio of voltage to current, so will be inversely proportional to the number of bulbs.
The current readings look wrong in that the first bulb caused the current to be 1 A, but each additional bulb increased it by 2 A. If that is what happened, the bulbs were not identical. That may be OK, but we expect the point of the experiment is to let you see the result described above.
In any event, the total resistance is not calculated properly. It should be the result of dividing voltage (15 V) by current.
Answer:
No, it is not right.
Explanation:
Your table is not consistent with bulbs of the same resistance.
Current comes from a measurement, but resistance comes from a calculation.
I presume that the measured currents are correct.
Ohm's Law states that the current flowing in a circuit is directly proportional to the voltage.
We usually write it as
V/I = R
1. One bulb in circuit
[tex]R = \dfrac{V}{I} = \dfrac{\text{15 V}}{\text{1 A}}= \mathbf{15 \, \Omega}[/tex]
2. Two bulbs
[tex]R = \dfrac{V}{I} = \dfrac{\text{15 V}}{\text{3 A}} = \mathbf{5 \, \Omega}[/tex]
3. Three bulbs
[tex]R = \dfrac{V}{I} = \dfrac{\text{15 V}}{\text{5 A}} = \mathbf{3 \, \Omega}[/tex]
4. Four bulbs
[tex]R = \dfrac{V}{I} = \dfrac{\text{15 V}}{\text{7 A}} = \mathbf{2.1 \, \Omega}[/tex]
In a steam power plant, the temperature of the burning fuel is 1100 °C, and cooling water is available at 15 °C. Steam leaving the boiler is at 2 MPa and 700 °C, and the condenser produces a saturated liquid at 50 kPa. The steam lines are well insulated. The turbine and pump operate reversibly and adiabatically. Some of the mechanical work generated by the turbine is used to drive the pump.
a. Draw a T-s diagram of this cycle.
b. What is the net work obtained in the cycle per kg steam generated in the boiler?
c. How much heat is discarded in the condenser per kg steam generated in the boiler?
d. What fraction of the work generated by the turbine is used to operate the pump?
e. How much heat is absorbed in the boiler per kg steam generated?
Answer:
b. 1655.7 KJ/kg ( net work produced )
c. 2324.86 KJ/kg
d. 0.25539 --- 25.5%
e. 3980.63 KJ/kg
Explanation:
Given:-
Condenser exit parameters:
P1 = 50 KPa , saturated liquid
Boiler exit / Turbine exit parameters:
P3 = 2 MPa
T3 = 1100°C
Solution:-
- Adiabatic and reversible processes for pump and turbine are to be applied
- Assume changes in elevation heads within the turbomachinery to be negligible.
- Assume steady state conditions for fluid flow and the use of property tables will be employed.
Isentropic compression of water in pump:
Pump inlet conditions : Pump exit to Boiler pressure:
P1 = 50 KPa, sat liquid P2 = P3 = 2 MPa
h1 = 340.54 KJ/kg s2 = s1 = 1.0912 KJ/kg.K
s1 = 1.0912 KJ/kg.K h2 = 908.47 KJ/kg
- Apply energy balance for the pump and determine the work input ( Win ) required by the pump:
Win = h2 - h1
Win = 908.47 - 340.54
Win = 567.93 KJ/kg
Isentropic expansion of steam in turbine:
Turbine inlet conditions : Turbine exit to condenser pressure:
P3 = 2MPa, T3 = 1100°C P4 = P1 = 50 kPa
h3 = 4889.1 KJ/kg s4 = s3 = 8.7842 KJ/kg.K .. superheated
s3 = 8.7842 KJ/kg.K h4 = hg = 2665.4 KJ/kg
- Apply energy balance for the turbine and determine the work output ( Wout ) produced by the turbine:
Wout = h3 - h4
Wout = 4889.1 - 2665.4
Wout = 2223.7 KJ/kg
- The net work-output obtained from the cycle ( W-net ) is governed by the isentropic processes of pump and turbine.
W_net = Wout - Win
W_net = 2223.7 - 567.93
W_net = 1655.77 KJ/kg ... Answer
- The fraction of work generated by turbine is used to operate the pump. The a portion of Wout is used to drive the motor of the pump. The pump draws ( Win ) amount of work from pump. The ratio of work extracted from turbine ( n ) would be:
n = Win / Wout
n = 567.93 / 2223.7
n = 0.25539 ... Answer ( 25.5 % ) of work is used by pump
- The amount of heat loss in the condenser ( consider reversible process ). Apply heat balance for the condenser, using turbine exit and condenser exit conditions:
Ql = h4 - h1
Ql = 2665.4 - 340.54
Ql = 2324.86 KJ/kg ... Answer
- The amount of heat gained by pressurized water in boiler ( consider reversible process ). Apply heat balance for the boiler, using pump exit and boiler exit conditions:
Qh = h3 - h2
Qh = 4889.1 - 908.47
Qh = 3980.63 KJ/kg ... Answer
Vapor lock occurs when the gasoline is cooled and forms a gel, preventing fuel flow and
engine operation. TRUE or FALSE
Answer:
True
Explanation:
Consider flow in between two parallel plates located a distance H from each other. Fluid flow is driven by the bottom plate moving to the right with a velocity of U (note, NO pressure gradient). The top plate has a fixed temperature of TT and the bottom plate has a temperature of Tb. Starting with the governing equations for incompressible flow, find the velocity profile and the temperature profile for the flow in between the plates. In addition, what is an expression for the heat flux at the bottom plate (assuming a thermal conductivity of k)
B. Is the “Loading Time” of any online application a functional or a non-functional requirement? Can the requirement engineers specify this property before the system is actually implemented, how?
Answer:
non-functional requirement,
Yes they can.
The application loading time is determined by testing system under various scenarios
Explanation:
non-functional requirement are requirements needed to justify application behavior.
functional requirements are requirements needed to justify what the application will do.
The loading time can be stated with some accuracy level after testing the system.
Un grupo de trabajadores tenia un rendimiento poco satisfactorio. Elena trabajo varias noches para poner por escrito las metas que deberian cumplir sus subordinados. Les especifico a cada uno de ellos su tarea de produccion y les conmino a cumplir a como diera lugar. que tipo de lider es?
Answer:
Elena es una líder de tipo autoritario.
Explanation:
El liderazgo autoritario representa el control individual por parte del líder de la toma de decisiones y el proceso de elección y planificación en una determinada organización.
En este liderazgo no se promueve la participación efectiva del equipo en los proyectos, solo el líder toma todas las decisiones necesarias y generalmente oprime a sus subordinados. Así, genera muchas veces situaciones de tensión y agotamiento entre los empleados y demás subordinados.
B. Is the "Loading Time" of any online application a functional or a non-functional requirement? Can the requirement engineers specify this property before the system is actually implemented, how?
Answer:
Non functional
Explanation:
Loading time is a requirement that does not work in applications. There are many non-functional requirements, such as the usability, performance, and reliability of the application. The loading time falls into the display category. Generally, the loading time limit is specified in the application, and the application is exempt from the application for exit if the application loading time is too longMr. auric goldfinger, criminal mastermind, intends to smuggle several tons of gold across international borders by disguising it as lumps of iron ore. he commands his engineer minions to form the gold into little spheres with a diameter of exactly and paint them black. however, his chief engineer points out that customs officials will surely notice the unusual weight of the "iron ore" if the balls are made of solid gold (density ). he suggests forming the gold into hollow balls instead (see sketch at right), so that the fake "iron ore" has the same density as real iron ore one of the balls of fake "iron ore," sliced in half. calculate the required thickness of the walls of each hollow lump of "iron ore." be sure your answer has a unit symbol, if necessary, and round it to significant digits.
Answer:
The thickness of the walls of each hollow lump of "iron ore" is 2.2 cm
Explanation:
Here we have that the density of solid gold = 19.3 g/cm³
Density of real iron ore = 5.15 g/cm³
Diameter of sphere of gold = 4 cm
Therefore, volume of sphere = 4/3·π·r³ = 4/3×π×2³ = 33.5 cm³
Mass of equivalent iron = Density of iron × Volume of iron = 5.15 × 33.5
Mass of equivalent iron = 172.6 cm³
∴ Mass of gold per lump = Mass of equivalent iron = 172.6 cm³
Volume of gold per lump = Mass of gold per lump/(Density of the gold)
Volume of gold per lump = 172.6/19.3 = 8.94 cm³
Since the gold is formed into hollow spheres, we have;
Let the radius of the hollow sphere = a
Therefore;
Total volume of the hollow gold sphere = Volume of gold per lump - void sphere of radius, a
Therefore;
[tex]33.5 = 8.94 - \frac{4}{3} \times \pi \times a^3[/tex]
[tex]\frac{4}{3} \times \pi \times a^3 = 33.5 - 8.94[/tex]
[tex]a^3 = \frac{24.6}{\frac{3}{4} \pi } = 5.9[/tex]
a = ∛5.9 = 1.8
The thickness of the walls of each hollow lump of "iron ore" = r - a = 4 - 1.8 = 2.2 cm.
Steam at a pressure of 0.08 bar and a quality of 93.2% enters a shell-and-tube heat exchanger where it condenses on the outside of tubes through which cooling water flows, exiting as saturated liquid at 0.08 bar. The mass flow rate of the condensing steam is 3.43 x 10^5 kg/h. Cooling water enters the tubes at 15.8°C and exits at 35.8°C with negligible change in pressure.
1. Neglecting stray heat transfer and ignoring kinetic and potential energy effects, determine the mass flow rate of the cooling water, in kg/h, for steady-state operation.
Answer:
The answer is "[tex]\bold{9.09\times 10^6 \frac{kg}{hour}}[/tex]".
Explanation:
For the reference table we get:
[tex]h1 = 2410 \frac{kJ}{kg} \ , at \ \ \\\\ \ P = 0.08 \ bar \ and \ \ quality = 0.932[/tex]
Through steam tables, they get:
[tex]\ h2 = 173.9 \frac{kJ}{kg} \ on\\\\ \ P = 0.08 \ bars \ \ \ but \ quality = 0 (sat.liquid),[/tex]
Water power transfer = [tex]\ 3.4 \times 10 ^ 5 \times (2410-173.9)\\[/tex]
It should be comparable to the water enthalpy:
[tex]m_{water}\times Cp\times (T2-T1)\\\\For \ eg:\\\\ = 3.4 \times 10 ^ 5\times (2410-173.9) \\\\ = m_{water}\times4.18\times(35-15)\\[/tex]
[tex]m_{water}=9.09\times 10^6 \frac{kg}{hour}[/tex]
Indicate the correct statement about the effect of Reynolds number on the character of the flow over an object.
If Reynolds number is high enough the effect of viscosity is negligible and the fluid flows over the plate without sticking to the surface.
If Reynolds number increases the extent of the region around the object that is affected by viscosity decreases.
If Reynolds number is low enough the effect of viscosity is so high that there is a region near the plate where the fluid is stationary.
If Reynolds number increases the size of the region around the object that is affected by viscosity increases.
Answer:
If Reynolds number increases the extent of the region around the object that is affected by viscosity decreases.
Explanation:
Reynolds number is an important dimensionless parameter in fluid mechanics.
It is calculated as;
[tex]R_e__N} = \frac{\rho vd}{\mu}[/tex]
where;
ρ is density
v is velocity
d is diameter
μ is viscosity
All these parameters are important in calculating Reynolds number and understanding of fluid flow over an object.
In aerodynamics, the higher the Reynolds number, the lesser the viscosity plays a role in the flow around the airfoil. As Reynolds number increases, the boundary layer gets thinner, which results in a lower drag. Or simply put, if Reynolds number increases the extent of the region around the object that is affected by viscosity decreases.
A spherical ball of solid, nonporous naphthalene, a "moth ball," is suspended in still air. The naphthalene ball slowly sublimes, releasing the naphthalene into the surrounding air by a diffusion limited process.
1. Estimate the time required to reduce the diameter from 2 cm to 0.5 cm. when the surrounding air is at 347 K and 1.013 x 10^5 Pa. Naphthalene has a molecular weight of 128 g/mole, a solid density of 1.145 g/cm^3, a diffusion coefficient in air of 8.19 x 10^-6 m^2/s, and exerts a vapor pressure of 5 torr (670 Pa) at 347 K.
Answer:
61.6 hours will be needed to reduce the diameter of solid spherical ball from 2 cm to 0.50 cm.
Explanation:
Find the given attachments
Under normal conditions, gauge pressure is ______ absolute pressure.
A. Higher than
B. Lower than
C. Equal to
D. Cannot say
Answer:
Option B Lower than
Explanation:
Gauge pressure is a relative measurement based on atmospheric pressure. Gauge pressure can be positive if it is above atmospheric pressure or it can also be negative it is below. On another hand, absolute pressure is an actual pressure in a space and its value has always to be zero or above. Basically absolute pressure is zero if it is in a perfect vacuum. So the measurement of absolute pressure is gauge pressure + atmospheric pressure. This is the reason in normal condition the gauge pressure = absolute pressure - atmospheric pressure and therefore is lower than absolute pressure
The steady-state data listed below are claimed for a power cycle operating between hot and cold reservoirs at 1200 K and 400 K, respectively. For each case, evaluate the net power developed by the cycle, in kW, and the thermal efficiency. Also in each case apply the equation below on a time-rate basis to determine whether the cycle operates reversibly, operates irreversibly, or is impossible.
(a) Qh(dot)=600 kW, Qc(dot)=400 kW
(b) Qh(dot)=600 kW, Qc(dot)=0 kW
(c) Qh(dot)=600 kW, Qc(dot)=200 kW
∮ (δQ/T)_b = -σ_cycle
Answer:
(a) Qh(dot)=600 kW, Qc(dot)=400 kW is an irreversible process.
(b) Qh(dot)=600 kW, Qc(dot)=0 kW is an impossible process.
(c) Qh(dot)=600 kW, Qc(dot)=200 kW is a reversible process.
Explanation:
T(hot) = 1200k, T(cold) = 400
efficiency n = (Th - Tc ) / Tc
n = (1200 - 400) / 1200 = 0.667 (this will be the comparison base)
(a)
Qh = 600 kW, Qc = 400 kW
n = (Qh - Qc) / Qh ⇒ (600 - 400) / 600
n = 0.33
0.33 is less than efficiency value from temperature 0.67
∴ it is irreversible process
(b)
Qh = 600 kW, Qc = 0
n = (Qh - Qc) / Qh ⇒ (600 - 0) / 600 = 1
efficiency in any power cycle can never be equal to one.
∴ it is an impossible process.
(c)
Qh = 600 kW, Qc = 200 kW
n = (Qh - Qc) / Qh = (600 - 200) / 600
n = 0.67 (it is equal to efficiency value from temperature)
∴ it is a reversible process
What can you do to protect your hands from any hazards on your worksite? Select the 2 answer options that apply. Use your gloves correctly Avoid jobs that could hurt your hands Always wear gloves Inspect your gloves for any damage before wearing
Answer:
"Use your gloves correctly" and "Inspect your gloves for any damage before wearing."
Explanation:
If you don't use your gloves correctly you could risk your hands being infected/burned. If your gloves are damaged they will provide minimum protection but there is still a potential safety hazard.
Answer:
Use your gloves correctly
Inspect your gloves for any damage before wearing
A heat engine operates between 2 reservoirs at TH and 18oC. The heat engine receives 17,000 kJ/h from the high temperature reservoir, and delivers half of its power output to drive a Carnot heat pump. The Carnot heat pump removes heat from the cold surroundings at 0oC and transfers it to a house which is maintained at 24oC. If the house is losing heat at a rate of 80,000 kJ/h, determine the temperature TH of the heat engine reservoir.
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem
Aerotron Electronics is considering purchasing a water filtration system to assist in circuit board manufacturing. The system costs $32,000. It has an expected life of 7 years at which time its salvage value will be $5,000. Operating and maintenance expenses are estimated to be $2,000 per year. If the filtration system is not purchased, Aerotron Electronics will have to pay Bay City $13,000 per year for water purification. If the system is purchased, no water purification from Bay City will be needed. Aerotron Electronics must borrow 1/2 of the purchase price, but they cannot start repaying the loan for 2 years. The bank has agreed to 3 equal annual payments, with the 1st payment due at the end of year 2. The loan interest rate is 8.5 % compounded annually. Aerotron electronics’ MARR is 12.5 % compounded annually.
Required:
a. What is the present worth of this investment? (Carry all interim calculations to 5 decimal places and then round your final answer to the nearest dollar. The tolerance is £10.)
b. What is the decision rule for judging the attractiveness of investments based on present worth?
c. Should Aerotron Electronics buy the water filtration system?
Answer:
Explanation:
a) Present worth of the system:
First step :
Calculation of bank installment:
We are given:
·nitial costs = $32,000
Borrow amount = ½ of purchase price
Payment of borrow amount = EOY 2 to EOY 4 (3 Equal installments)
Bank loan interest = 8.5% = 8.5/100 = 0.085
Assume the installment amount is F. They will be paid at end of year 2 to end of year 4. Their present value must be equal to borrow amount.
Present value of cost to be incurred in future can be calculated by below formula:
F P (1 + i)
F= Future cost
i = Rate of interest
n = time (in years)
Therefore,
F F $32,000 2 F (1 +0.085)2 (1 +0.085) 3 (1 +0.085) + +
.: 2.35394F = 16,000 or F = 16,000 2.35394 - $6,797.11
Step 2: Present worth of the system:
Given Data:
Initial costs = $32,000
Expected life = 7 years
Salvage value = $5,000
O&M Costs = $2,000 per year
MARR = 12.5%
= 12.5/100
= 0.125
Present value of uniform recurring payments is given by below formula:
P=A (1 + i) - 1 i(1+i)n 72
Where,
P = Present Value
A = Recurring payments per annum
i = rate of interest
n = time (in years)
Hence present value of O&M costs,
P1 = -2,000 x (1 + 0.125) 7-1 0.125 x (1 + 0.125) 7 -$8,984.60
Present worth of the system calculated in below table:
Description
F ($)
MARR (i)
per year
n (years)
P ($)
Initial investment (1/2 of purchase price)
-16,000.00
0.125
0
-16,000.00
Bank installment EOY2
-6,797.11
0.125
2
-5,370.56
Bank installment EOY3
-6,797.11
0.125
3
-4,773.83
Bank installment EOY4
-6,797.11
0.125
4
-4,243.40
Salvage value
5,000.00
0.125
7
2,192.31
O&M Costs
-8,984.60
0.125
0
-8,984.60
The current worth of the new system
-37,180.07
Part b) Decision rule of judgment:
Assuming current value of costs is lower than current value of benefit, an alternative is known to be economic to use based on current worth analysis.
Part c) Decision for the water filtration system:
Given Data:
· Annual savings from filtration system (A) = $13,000 per year
· Expected life (n) = 7 years
· MARR = 12.5%
= 12.5/100
= 0.125
Present value of benefits = 13,000 X (1 + 0.125)7 - 1 0.125 x (1 + 0.125) 7 $58,399.91
Since current value of costs is less than current value of benefit, this is an worthwhile system and has to be be purchased.
A structural component is fabricated from an alloy that has a plane-strain fracture toughness of 62 MPa√m. It has been determined that this component fails at a stress of 250 MPa when the maximum length of an internal crack is 1.6 mm. What is the maximum allowable internal crack length (in mm) without fracture for this same component exposed to a stress of 250 MPa and made from another alloy that has a plane strain fracture toughness of 40 MPa√m?
Answer:
0.67 mm
Explanation:
Solution:
We find the dimensionless parameters by applying the critical stress crack propagation formula stated below:
σс= Klc/Y√πa
Y = Klc/σс √πa
σс = this is the critical stress needed for initial cracking propagation
Klc = the plain stress fracture toughness
a = surface length of the crack
Y = the dimensionless parameter
Now, we substitute the values 62MPa√m for Klc, 250 MPa for σс and 1.6 * 10 ^⁻3 for a in the dimensionless parameter equation.
Thus,
Y = Klc/σс √πa
= 62/250(√π * 1.6* 10 ^⁻3)
= 3.492
The next step is to find the maximum permitted surface crack length by applying the critical stress crack propagation equation given below:
σс= Klc/Y√πa
a= 1/π (Klc/Yσс)²
Now, substitute 40 MPa√m for Klc, 250 MPa for σс and 3.492 for surface length crack equation
So,
a= 1/π (Klc/Yσс)²
= 1/π[40/3.492 * 250]²
=1/π[40/873]²
=1/π[0.0458]²
0.318[0.0458]²
=0.318[0.00209]
= 0.0066
0.67* 10 ^⁻3 m
= 0.67 mm
Therefore the maximum surface crack length produced is 0.67 mm
Makine yüzeysel gemi ünvan değişikliği teknolojisi
I inferred you meant emerging technologies we see today.
Explanation:
1. 3D Printing
A three dimensional printing allows a digital model to be printed (constructed) into a physical object.
A box, an industrial design and many more could be printed within minutes.
2. AI voice recognition devices
Another trend in the tech world is the rise in artificial intelligence been used in voice recognition devices that not only recognize one's voice but also take commands from the user.
This technology allows one to listen to the internet, such as listening to music online.
The tech world is ever changing with new technology beyond one's consumption on the increase daily.
Kirby is conducting a literature review in preparation for his study of “expectations regarding the sharing of financial and practical responsibilities among married and cohabiting couples in which both partners are between the ages of 20 and 29.” Conducting a keyword search on “couples” and “responsibility,” Kirby has generated a lengthy list of research articles. He decides to shorten the list of potential articles by eliminating all articles that were not published in prestigious research journals. He will include all the remaining articles in his literature review. What is your opinion of Kirby’s approach to selecting articles for the literature review?
Answer:
My opinion towards Kirby's approach in choosing articles for literature reviews is that, it is not the considered a good approach because when choosing articles based only on Journal it can't be considered the best.
Various methods needs to be considered by Kirby's before selecting articles, which are stated in the explanation section below
Explanation:
Solution:
Kirby’s method in choosing articles is not regarded to be a better proposal because choosing articles with regards to the journal can’t be seen as good. There are other things that should to be taken into consideration by Kirby which is explained below:
It is also important to confirm the editors who are in charge of the journals. It is advisable to view the profile of the editors in various links such as LinkedIn, Google scholar, before choosing their articles.It is very important to stay away from people who might find a way to exploit this situation. some research articles may be produced just for the aim of making money & there might exist no good quality information that is needed by the researcher for conducting his research.A proper journal is the one that produces work on the journal that the paper addresses & it is the one that presents the researcher’s needs through its authenticity and aspirations.Some particular journals are regarded to offer good source of information for research. examples are Thomson Reuters website etc.The various information produced aside from quality, is also important to consider when choosing the source of information that the article presents.It is proposed to absorb acetone from air using water as a solvent. Operation is at 10 atm and is isothermal at 20°C. The total flow rate of entering gas is 10 kmol /h. The entering gas is 1.2 mol% acetone. Pure water is used as the solvent. The water flow rate is 15 kmol/h. The desired outlet gas concentration should be 0.1 mol % acetone. For this system, Henry's law holds and Ye = 1.5 X where Ye is the mol fraction of acetone in the vapour in equilibrium with a mol fraction X in the liquid.
KGa = 0.4 kmol*m^-3*s^-1
1. Draw a schematic diagram to represent the process.
2. Determine the mole fraction of acetone in the outlet liquid.
Answer:
The meole fraction of acetone in the outlet liquid is [tex]x_1 = 0.0072[/tex]
Explanation:
1.
The schematic diagram to represent this process is shown in the diagram attached below:
2.
the mole fraction of acetone in the outlet liquid is determined as follows:
solute from Basis Gas flow rate [tex]G_s = 10(1-0.012) =9.88 kmol/hr[/tex]
Let the entering mole be :[tex]y_1 = 1.2[/tex] % = 0.012
[tex]y_1 =(\dfrac{y_1}{1-y_1})[/tex]
[tex]y_1 =(\dfrac{0.012}{1-0.012})[/tex]
[tex]y_1 =0.012[/tex]
Let the outlet gas concentration be [tex]y_2[/tex] = 0.1% = 0.001
[tex]y_2 = 0.001[/tex]
Thus; the mole fraction of acetone in the outlet liquid is:
[tex]G_s y_1 + L_s x_2 = y_2 L_y + L_s x_1[/tex]
[tex]9.88(0.012-0.001)=15*x_1[/tex]
[tex]9.88(0.011) = 15x_1[/tex]
[tex]x_1 = \dfrac{0.10868}{15}[/tex]
[tex]x_1 = 0.0072[/tex]
The mole fraction of acetone in the outlet liquid is [tex]x_1 = 0.0072[/tex]
a) Describe the operation of a heat pump operating on the theoretical reversed Carnot cycle, with a neat sketch of the layout.
b) What modifications are required in order to convert a steam power plant working on the ideal Carnot cycle to a plant operating on the Rankine cycle? Explain briefly why these modifications are necessary to enable the operation of a practical cycle? Illustrate your answer with sketches using appropriate property diagrams (p-v and T-s diagrams).
Answer:
a) The operation of a heat pump involves the extraction of energy in the form of heat Q₁ from a cold source
b) The modifications required to convert a plant operating on an ideal Carnot cycle to a plant operating on a Rankine cycle involves
i) Complete condensation of the vapor at the condenser to saturated liquid for pumping to the boiler
ii) Heating of the pumped, pressurized water to the boiler pressure
Explanation:
a) 1 - 2. Wet vapor enters compressor where it undergoes isentropic compression to state 2 by work W₁₂
2 - 3. The vapor enters the condenser at state 2 where it undergoes isobaric and isothermal condensation to a liquid with the evolution of heat Q₂
3 - 4. The condensed liquid is expanded isentropically with the work done equal to W₃₋₄
4 - 1. At the state 4, with reduced pressure from the previous expansion, the liquid makes its way to the evaporator where it absorbs heat, Q₁, from the body to be cooled.
b. i) Complete condensation of the vapor at the condenser to saturated liquid for pumping to the boiler
Here the condensation process is modified from partial condensation to complete condensation at the same temperature which reduces the size of the pump required to pump the liquid water as opposed to pumping steam plus liquid
ii) Heating of the pumped, pressurized water to the boiler pressure
The pumped water at state 4 will be required to be heated to saturated water temperature equivalent to the boiler pressure, hence heat will need to be added at state.
Sketches of the schematic of a Basic Rankine cycle is attached
define sheer stress
Every one deals with stress. Stress or anxiety is usually when your scared or anxious because you know you could have done better then what you did. For example on a test, you could be scared about what grade you get. You can also have stress when you feel something bad is going to happen.
Answer:
it is the component of stress coplanar with a material cross section. It arises from the shear force, the component of force vector parallel to the material cross section
Explanation:
hope this helps and have a good day :-)
Steam enters an adiabatic condenser (heat exchanger) at a mass flow rate of 5.55 kg/s where it condensed to saturated liquid water at P = 20 kPa. The change in enthalpy of the steam is – 2,491 kJ/kg. The steam is cooled by water from a river. Environmental regulations require that the maximum increase in the river water temperature is 10°C.
a) What is the minimum mass flow rate of river water through the condenser to cool the steam? The Cp for water = 4.184 kJ/kg/K.
Answer:
The minimum mass flow rate will be "330 kg/s".
Explanation:
Given:
For steam,
[tex]m_{s}=5.55 \ kg/s[/tex]
[tex]\Delta h=2491 \ kg/kj[/tex]
For water,
[tex]\Delta T=10^{\circ}C[/tex]
[tex](Cp)_{w}=4.184 \ kJ/kg^{\circ}C[/tex]
They add energy efficiency as condenser becomes adiabatic, with total mass flow rate of minimal vapor,
⇒ [tex]m_{s}\times (\Delta h)=M_{w}\times(Cp)_{w}\times \Delta T[/tex]
On putting the estimated values, we get
⇒ [tex]5.55\times 2491=M_{w}\times 4.184\times 10\\[/tex]
⇒ [tex]13825.05=M_{w}\times 41.84[/tex]
⇒ [tex]M_{w}=330 \ kg/s[/tex]
why are apartments called apartments if there together? and why are buildings called buildings if there already built? hmmmm
Answer:huh
U right Even though all the apartments within a single building are indeed stuck together, they are also apart from each other.
For you’re second question That means when the building was not built, at that time it was building, now when it is built we remember the past and give it respect of being built - so a building is called a building when it is already built - well it was building back at that time when it was not built !!!
Explanation:
42. A vehicle has sagged rear springs and reduced rear curb riding height. This problem results
in?
A. Excessive positive camber on the front wheels
B. Excessive toe-out on the front wheels
C. Excessive positive caster on the front wheels
D. Excessive toe-in on the front wheels
43. On a vehicle equipped with rear parallel leaf springs and a solid rear axle, customer is
complaining that the vehicle reacts erratically (it darts) during turns. What is the most likely
cause of this complaint?
A. Incorrect ride height
B. Incorrect driveline angle
C. Loose rear axle U-bolts
D. Missing jounce/rebound bumpers
44. A power steering pump is being tested with a pressure gauge for maximum output pressure.
Which of the following statements is correct?
A. The pressure gauge should be attached to the pump return port
B. The steering wheel should be held in the right lock position
C. A maximum output pressure dead heading test should last no longer than 5 seconds
D. One should check output pressure with engine speed above 4000rpm
Answer:1. Driving under the influence of any drug that makes you drive unsafely is:
a. Permitted if it is prescribed by a doctor
b. Against the law
c. Permitted if it is a diet pill or cold medicine
2. Which fires can you put out with water:
a. Tire fires
b. Gasoline fires
c. Electrical fires
3. How far should a driver look ahead of the vehicle while driving:
a. 9-12 seconds
b. 12-15 seconds
c. 18-21 seconds
4. To prevent shifting, there should be at least one tie-down for ever ____feet of cargo: a. 10
b. 15 c. 18
5. Which of these statements about downshifting is true:
a. When you downshift for a curve, you should do so before you enter the curve
b. When you downshift for a hill, you should do so after you start down the hill
c. When you downshift for a curve, you should do so after you enter the curve
6. How do you test hydraulic brakes for a leak:
a. Move the vehicle slowly and see if it stops when the brake is applied
b. With the vehicle stopped, pump the pedal three time, apply pressure then hold
For five seconds and see it the pedal moves.
c. Step on the brake pedal and accelerator at the same time and see if the vehicle moves
7. For an average driver, driving 55 MPH on dry pavement, it will take about _____ to bring The vehicle to a stop:
a. Twice the length of the vehicle
b. Half the length of a football field
c. The length of a football field
8. You are driving a vehicle with a light load, traffic is moving at 35 MPH in a 55 MPH zone. The safest speed for your vehicle in this situation is most likely:
a. 30 MPH
b. 35 MPH
c. 40 MPH
9. Which of these is a good rule to follow when driving at night:
a. Keep your speed slow enough to stop within the range of your headlights
b. Look directly at oncoming headlights
c. Keep your instrument lights bright
10. A moving vehicle ahead of you has a red triangle with an orange center on the rear. What does this mean?
a. The vehicle is hauling hazardous materials
b. It may be a slow-moving vehicle
c. It may be oversized
11. You wish to turn right form a two lane two way street to make the turn. Which of these Drawings show how the turn should be made.
12. You are driving a heavy vehicle and must exit a highway using an offramp that curves downhill:
a. Use the posted speed limit for the offramp
b. Slow down to a safe speed before the turn
c. Wait until you are in the turn before downshifting
13. Which of these statements about using mirrors is true:
a. You should look at a mirror for several seconds at a time
b. There are “blind spots” that your mirror cannot show you
c. A lane change requires you to look at the mirrors twice
14. You must park on the side of a level, straight, two-lane road. Where should you place the three reflective triangles?
a. one within 10 feet of the rear of the vehicle, one about 100 feet to the rear and one about 200 feet to the rear.
b. One with 10 feet of the rear of the vehicle, one about 100 feet to the rear and one about 100 feet from the front of the vehicle
c. One about 50 feet from the rear of the vehicle, one about 100 feet to the rear and one about 100 feet from the front of the vehicle
15. Your vehicle is in a traffic emergency and may collide with another vehicle if you do not take action. Which of these is a good rule to remember at such a time?
a. Stopping is always the safest action in a traffic emergency
b. Heavy vehicles can almost always turn more quickly than they can stop
c. Leaving the road is always more risky than hitting another vehicle
16. The most important reason for being alert to hazards is:
a. Law enforcement personnel can be called
b. You will have time to plan your escape if the hazard becomes an emergency
c. You can help impaired drivers
17. You are traveling down a long, steep hill. Your brakes begin to fade and then fail. What should you do?
a. Downshift
b. Pump the brake pedal
c. Look for an escape ramp or escape route
18. The most common cause of serious vehicle skids is:
a. Driving too fast for road conditions
b. Poorly adjusted brakes
c. Bad tires
19. To avoid a crash, you had to drive onto the right shoulder. You are now driving at 40 MPH on the shoulder. How should you move back onto the pavement?
a. If clear, come to a complete stop before steering back onto the pavement
b. Brake had to slow the vehicle, then steer sharply onto the pavement
c. Keep moving at the present speed and steer very gently back onto the pavement
20. If
a. Slide sideways and spin out
b. Go straight ahead but will turn if you turn the steering wheel
c. Go straight ahead even if the steering wheel is turned
A 50-cm x 50-cm circuit board that contains 121 square chips on one side is to be cooled by combined natural convection and radiation by mounting it on a vertical surface in a room at 25 °C. Each chip dissipates 0.18 W of power, and the emissivity of the chip surfaces is 0.7. Assuming the heat transfer from the back side of the circuit board to be negligible, and the temperature of the surrounding surfaces to be the same as the air temperature of the room, determine the surface temperature of the chips. Evaluate air properties at a film temperature of 30 °C and 1 atm pressure. Is this a good assumption?
Answer:
Ts = 311.86 K = 38.86°C
Explanation:
The convection heat transfer coefficient for vertical orientation of the board is given by the formula:
[tex]h = 1.42(\frac{T_{s} - T_{f} }{L})^{0.25}[/tex]
where,
h = heat transfer coefficient
[tex]T_{s}[/tex] = surface temperature
[tex]T_{f}[/tex] = Temperature of fluid (air) = 30°C + 273 = 303 K
L = Characteristic Length = 50 cm = 0.5 m
Since the heat transfer through convection is given as:
[tex]Q_{conv} = hA_{s}(T_{s} - T_{f})[/tex]
using value of h, we get:
[tex]Q_{conv} = 1.42(\frac{T_{s} - T_{f} }{L})^{0.25} A_{s} (T_{s} - T_{f} )[/tex]
[tex]Q_{conv} = 1.42 A_{s} \frac{(T_{s} - T_{f} )^{1.25} }{L^{0.25} }[/tex]
where,
[tex]A_{s}[/tex] = Surface Area = (0.5 m)(0.5 m) = 0.25 m²
Now, the radiation heat transfer is given by:
[tex]Q_{rad} =[/tex] εσ[tex]A_{s} [(T_{s})^{4} - (T_{surr})^{4}][/tex]
where,
ε = emissivity of surface = 0.7
σ = Stefan Boltzman Constant = 5.67 x 10⁻⁸ W/m².k⁴
[tex]T_{surr}[/tex] = Temperature of surroundings = 25°C +273 = 298 k
Now, the total heat transfer rate will be:
[tex]Q_{total} = Q_{conv} + Q_{rad}[/tex]
using values:
[tex]Q_{total} =[/tex] [tex]1.42 A_{s} \frac{(T_{s} - T_{f} )^{1.25} }{L^{0.25} } +[/tex] εσ[tex]A_{s} [(T_{s})^{4} - (T_{surr})^{4}][/tex]
we know that the total heat transfer from the board can be found out by:
[tex]Q_{total} = (0.18 W) (121) = 21.78 W[/tex]
using values in the equation:
21.78 = (1.42)(0.25)[tex](T_{s} - 303)^{1.25}/0.5^{0.25}[/tex] + (0.7)(5.67 x 10⁻⁸)(0.25)[tex][(T_{s})^{4} - 298^{4}][/tex]
21.78 = (0.4222)[tex](T_{s} - 303)^{1.25}[/tex] + 9.922 x 10⁻⁹[tex](T_{s} )^{4}[/tex] - 78.25
100.03 = (0.4222)[tex](T_{s} - 303)^{1.25}[/tex]+ 9.922 x 10⁻⁹[tex](T_{s} )^{4}[/tex]
Solving this equation numerically by Newton - Raphson Method (Here, any numerical method or an equation solver can be used), we get the value of Ts to be:
Ts = 311.86 K = 38.86°C
The film temperature is the average of surface temperature and surrounding temperature. Therefore,
Film Temperature = (25°C + 38.86°C)/2 = 31.93°C
Since, this is very close to 30°C.
Hence, the assumption is good.
Design a rectangular metallic waveguide to be used for transmission of electromagnetic power at 2.45 GHz. This frequency should be at the middle of the operating frequency band. The design should also allow maximum power transfer without sacrificing the operating bandwidth. In each case, you should use a safety factor of 10 and neglect ohmic loss in the conductors.
a. Design an air-filled guide to meet the given specifications. Find the maximum power the waveguide can transmit. The breakdown electric field in air is assumed to be 2 x10^6 V/m
b. Now assume that a electric material is used to fill the waveguide. The material is characterized by ε = 2.5 εo, μ = μ o and σ = 0. The breakdown field in the dielectric is 10^7 V/m. How many times more power can be transmitted by this waveguide?
Answer:
A) 1.4 *10^11 watts
B) 41.42 ≈ 41 TIMES
Explanation:
Designing a rectangular metallic wave guide using the given data
Electromagnetic power = 2.46 GHz also at the middle of operating frequency
A) Design an air-filled guide to meet the given specifications.
operating frequency range = C / αa < f < C / a
2.45 GHz = [tex]\frac{\frac{C}{ba}+ \frac{c}{a} }{2}[/tex]
The given frequency middle at the middle of operating frequency range
= 4.9 GHz = [tex]\frac{c + 2c }{ba}[/tex] = 3C / βa
α = [tex]\frac{3*3*10^{10} }{2*4.9*10^9}[/tex] = 45/4.9 = 9.18 cm
note: to operate in dominant mode aspect ratio should be b = α/2
therefore b = 4.59 cm
Also Maximum power can be carried by wave guide only in dominant mode
i.e TE10 mode
power carried = I E I^2ab / 4Zte using this formula
ZTE = impedance when operated in TE mode = [tex]\sqrt[n]{1-(\frac{Fc}{f} )^{2} }[/tex]
Fc = cutoff frequency = (3*10^16) / (2*9.18) = 1.6GHz
F = operating frequency = 2.45 GHz
n = freespace impedance = 377 ohms
input all the given values back to ZTE equation
ZTE = 285 ohms
power carried = [tex]\frac{|2*10^6|^{2}* 9.18 * 4.59 }{4 * 285}[/tex] = 4*10^12 * 0.036
THEREFORE power carried 1 = 1.4 *10^11 watts
B) The dielectric materials given data/parameters
∈ = 2.5 ∈o ∪ = ∪o
breakdown field = 10^7
free space impedance n = [tex]\sqrt{\frac{u}{e} } = \sqrt{\frac{UoUr}{EoEr} }[/tex]
therefore for the given dielectric n = [tex]\sqrt{\frac{Uo}{Eo} } \sqrt{\frac{1}{2.5} } = \frac{377}{\sqrt{2.5} }[/tex] n = 238.43
ZTE = [tex]\sqrt[n]{1-(\frac{1.6}{2.45} )^{2} }[/tex]
therefore ZTE = 180.56 ohms
power carried 2 = [tex]\frac{|10^7|^2*9.18*4.59}{4*180.56} = 58*10^{11} N[/tex]
To calculate the number of time power can be transmitted by the waveguide = power carried 2 / power carried 1
= 58*10^11 / 1.4*10^11 = 41.42 ≈ 41
Consider a sphere made of stainless steel with diameter of 25 cm. It is heated to temperature of 300°C for some chemical tests. After finishing the tests, the sphere is cooled by exposing it to a flow of air at 1 atm pressure and 25°C with a velocity of 3 m/s. By the end of cooling process, the sphere's temperature drops to 200°C. The rate of heat transfer loss due to convection is closest to:__________.
a) 485 W
b) 513 W
c) 88 W
d) 611 w
Answer:
263.69 W.
(None of the option).
Explanation:
So, from the question above we are given the following parameters or data or information which is going to allow us to solve this question and they are;
(1). diameter of 25 cm.
(2). Initial temperature of 300°C.
(3).temperature drops to 200°C = final temperature.
Step one: Calculate the Reynolds number.
Reynolds number = 3 × 0.25/1.562 × 10^-5 = 48015.365.
Step two: Calculate average heat transfer coefficient.
The average heat transfer coefficient = k/D { 2 + (0.4Re^1/2 + 0.06Re^2/3} px^0.4 × (u/uz)^1/4.
The average heat transfer coefficient = 0.10204 × [ 2 + (87.65 + 79.26) (0.8719) × 0.8909.
average heat transfer coefficient = 0.20204 ( 2 + 129.652).
average heat transfer coefficient = 13.43/m^2.k.
Step three: The rate of heat transfer loss due to convection = (average heat transfer coefficient ) × πD^2 × ( T1 - T2).
The rate of heat transfer loss due to convection= 13.43 × π(0.25)^2 × (300 - 200).
=>The rate of heat transfer loss due to convection = 263.69 W.
A particular DSL modem operates at 768 kbits/sec. How many bytes can it receive in 1 minute? USB 3.0 can send data at 5 Gbits/sec. How many bytes can it send in 1 minute?
Answer:
a. 1.6 Kbytes/min
b. 10.417 Mbytes/min
Explanation:
a. The DSL modem operates at 768 Kbits/sec.
But,
8 bits = 1 byte and 1 Kbit = 1 000 bits, so that:
= [tex]\frac{768 000}{8}[/tex]
= 96 000 bytes
Therefore, the modem operates at 96 Kbytes/sec.
The byte to be received in 1 minute can be calculated thus;
Since 60 seconds = 1 minute, then:
= [tex]\frac{96000}{60}[/tex]
= 1600
= 1.6 Kbytes/min
The modem receives 1.6 Kbytes/min
b. The USB sends 5 Gbits/sec.
But, 8 bits = 1 byte and 1 Gbit = 1000000000 bits so that:
= [tex]\frac{5000000000}{8}[/tex]
= 625000000
= 0.625 Gbytes
The USB sends 0.625 Gbyte/sec.
Since 60 seconds = 1 minute, then:
= [tex]\frac{625000000}{60}[/tex]
= 10416666.67
= 10.417 Mbytes/min