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.
Kindly check the attached image below to see the step by step explanation to the question above.
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Answer:
Explanation:
Given that:
<u>At state 1:</u>
Pressure P₁ = 20 bar
Volume V₁ = 0.03
From the tables at saturated vapour;
Temperature T₁ = 212.4⁰ C ; = 0.0996
/ kg
The mass inside the cylinder is m = 0.3 kg, which is constant.
The specific internal energy u₁ = ug₁ = 2599.2 kJ/kg
<u>At state 2:</u>
Temperature T₂ = 200⁰ C
Since the 1 - 2 occurs in an isochoric process v₂ = v₁ = 0.099 / kg
From temperature T₂ = 200⁰ C
Since , the saturated pressure at state 2 i.e. P₂ = 15.5 bar
Mixture quality
At temperature T₂, the specific internal energy , also
Thus,
<u>At state 3:</u>
Temperature
Specific volume
Thus; ,
SInce , therefore, the phase is in a superheated vapour state.
From the tables of superheated vapour tables; at and T₃ = 200⁰ C
The pressure = 10 bar and v =0.206
The specific internal energy at the pressure of 10 bar = 2622.3 kJ/kg
The changes in the specific internal energy is:
= (2210.686 - 2599.2) kJ/kg
= -388.514 kJ/kg
≅ - 389 kJ/kg
= (2622.3 - 2210.686) kJ/kg
= 411.614 kJ/kg
≅ 410 kJ/kg
We can see the correct sketches of the T-v plot showing the diagrammatic expression in the image attached below.
Answer:
Explanation:
= Velocity of car = 65 mph =
= Density of air =
From Bernoulli's law we have
The maximum pressure on the girl's hand is
Now = 200 mph =
The maximum pressure on the girl's hand is
Answer:
a) 159.07 MPa
b) 10.45 MPa
c) 79.535 MPa
Explanation:
Given data :
length of cantilever beam = 1.5m
outer width and height = 100 mm
wall thickness = 8mm
uniform load carried by beam along entire length= 6.5 kN/m
concentrated force at free end = 4kN
first we determine these values :
Mmax = ( 6.5 *(1.5) * (1.5/2) + 4 * 1.5 ) = 13312.5 N.m
Vmax = ( 6.5 * (1.5) + 4 ) = 13750 N
A) determine max bending stress
б = =
= 159.07 MPa
B) Determine max transverse shear stress
attached below
ζ = 10.45 MPa
C) Determine max shear stress in the beam
This occurs at the top of the beam or at the centroidal axis
hence max stress in the beam = 159.07 / 2 = 79.535 MPa
attached below is the remaining solution
