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