Answer:
The speed of shaft is 1891.62 RPM.
Explanation:
given that
Amplitude A= 0.15 mm
Acceleration = 0.6 g
So
we can say that acceleration= 0.6 x 9.81

We know that

So now by putting the values



We know that
ω= 2πN/60
198.0=2πN/60
N=1891.62 RPM
So the speed of shaft is 1891.62 RPM.
both b and c are the right
Answer:
Part A: (d/D=0.1)
DeltaV percent=42.6%
Part B:(d/D=0.01)
DeltaV percent=21.7%
Explanation:
We are going to use the following volume flow rate equation:

Above equation can be written as:


First Consider no wire i.e d/D=0
Above expression will become:


Part A: (d/D=0.1)




DeltaV percent=42.6%
Part B:(d/D=0.01)




DeltaV percent=21.7%
Answer:
peak flow and any engineering considerations related thereto
Explanation:
It should be no surprise that a peak flow meter will report peak flow, sometimes with important maximum-value, time-constant, or bandwidth limitations. There are many engineering issues related to flow rates. A peak flow meter can allow you to assess those issues with respect to the flows actually encountered.
Peak flow can allow you to assess adequacy of flow and whether there may be blockages or impediments to flow that reduce peak levels below expected values. An appropriate peak flow meter can help you assess the length of time that peak flow can be maintained, and whether that delivers sufficient volume.
It can also allow you to assess whether appropriate accommodation is made for unexpectedly high flow rates. (Are buffers or overflow tanks of sufficient size? Is there adequate protection against possible erosion? Is there adequate support where flow changes direction?)