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
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.
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 longIn 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
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
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.
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]
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 :-)
Mr. 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.
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
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
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
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
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.
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)
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.Vapor lock occurs when the gasoline is cooled and forms a gel, preventing fuel flow and
engine operation. TRUE or FALSE
Answer:
True
Explanation:
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]
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
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
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: