Estimate the uncertainty in a 22 m/sec air velocity measurement using a Pitot tube at 20C. Assume the atmospheric pressure is 1
00 kPa and that R and the density of water, H O2 are known with high enough accuracy to consider them as constants (zero uncertainty). The pressure for each tap is measured with separate manometers that have a resolution of 0.5 mm. The manometers use water as the fluid and each are oriented vertically. The absolute pressure (which gives you the density of air) is known with an uncertainty of 0.1 kPa. The temperature is known with an uncertainty of 1C. Assume the stagnation pressure, pt, is indicated by a static height of 7 mm.
1 answer:
Answer:
Check the explanation
Explanation:
In calculating the second version of velocity that is expected velocity, which involves the division of the overall amount of estimated story points by the amount of sprints. Take for instance, if the development team estimates a total of 150 points over five sprints, then we can say that the team's expected velocity would be 30 points per sprint.
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Answer:
Q=0.000604 m³/s
Explanation:
Given that
d₁=5 cm
d₂=1 cm
P= 30 KPa
Density of water ,ρ=1000 kg/m³
As we know that volume flow rate Q given as
A₁=0.0019 m²
A₂=0.000078 m²
Q=0.000604 m³/s
Answer:
1) A=282.6 mm
2)
3)T=0.42 sec
4)f= 2.24 Hz
Explanation:
Given that
V=3.5 m/s at x=150 mm ------------1
V=2.5 m/s at x=225 mm ------------2
Where x measured from mid position.
We know that velocity in simple harmonic given as
Where A is the amplitude and ω is the natural frequency of simple harmonic motion.
From equation 1 and 2
------3
--------4
Now by dividing equation 3 by 4
So A=0.2826 m
A=282.6 mm
Now by putting the values of A in the equation 3
ω=14.609 rad/s
Frequency
ω= 2πf
14.609= 2 x π x f
f= 2.24 Hz
Maximum acceleration
Time period T
T=0.42 sec