Answer:
combining scientific knowledge, careful reasoning, and artistic invention in a flexible approach to problem-solving
Explanation:
Answer:
a.) a component item is coded at the lowest level at which it appears in the BOM structure is the correct answer.
Explanation:
- Low-level coding is a kind of programming language used in BOM structures and it carries basic commands that are identified by a computer.
- The two types of low-level coding are
- Assembly language.
- machine language.
- The advantages of using low-level coding are programs develop by using low-level code are very memory effective and quick and there no need to use interpreters for the conversion of the source to machine code.
The statement that best describes the lower vc-turbo engine compression ratios is that its generate an increased amount of power with the average fuel consumption.
<h3>
Vc-turbo engine</h3>
In the engine, lower temperatures allow to run more ignition timing advance which makes significantly more power than increasing the compression ratio in a turbo engine.
Hence, the turbo engines run lower static compression ratios to increase the amount of power they can reliable generate on pump gas.
Therefore, the statement that best describes the lower vc-turbo engine compression ratios is that its generate an increased amount of power with the average fuel consumption.
Read more about turbo engines
<em>brainly.com/question/26409491</em>
Answer:
0.0432 m^3/s
Explanation:
Internal diameter of smaller pipes = 2.5 cm = 0.025 m
pipa wall thickness = 3 mm = 0.003 m
internal diameter of larger pipes = 8 cm = 0.8 m
velocity of region between smaller and larger pipe = 10 m/s
Calculate discharge in m^3/s
First we calculate the area of the smaller pipe
A = 
= 
= 0.00023571 m^2
next we calculate area of fluid between the smaller pipes and larger pipe
A = ![[\frac{\pi }{4} D^{2} _{L} ] - 3(A_{s})](https://tex.z-dn.net/?f=%5B%5Cfrac%7B%5Cpi%20%7D%7B4%7D%20D%5E%7B2%7D%20_%7BL%7D%20%20%5D%20-%203%28A_%7Bs%7D%29)
= ![[ \frac{\pi }{4} (0.08 )^2 - 3 ( 0.00023571 )]](https://tex.z-dn.net/?f=%5B%20%5Cfrac%7B%5Cpi%20%7D%7B4%7D%20%280.08%20%29%5E2%20-%203%20%28%200.00023571%20%29%5D)
= [ 0.00502857 - 0.00070713 ]
= 0.00432144 m^2
hence the discharge in m^3/s
Q = AV
= 0.00432144 * 10
= 0.0432 m^3/s
Answer:
A. Overall efficiency =83.95%
B. 87.45%
C.120.465kw
Explanation:
H=75m
Flow rate= 120L/s=1.02m^2/s
Po=Output power= 630kw
As power input Pi= Total power available
=gQH/1000
Where
g=9.81
Q=120L/s=1.02m^2/s
H=75m
= (9.81m/s^2 × 1.02m^2/s×75m)/1000
750.465kw
A. Overall efficiency of generator turbine unit (Po/Pi)
= Output power/ power input
= 630/750.465= 0.8395
Overall efficiency of the unit= 0.8395 × 100
=83.95%
B. Turbine efficiency if generator efficiency is 96%
no × nt= 0.96
nT= 87.45%
C. Power losses= Power input - Power output
Pi - Po
=750.465- 630
=120.465kw
Power losses due to inefficiencies in turbine of the generator is 120.465kw
