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:
The answers to the question are as follows
(a) W = -175.6 MJ
(b) W = -329.256 MJ
The peak temperature of the isentropic compression process is 886.974 K
Explanation:
(a) We are given the initial conditions as
v₂ = 10 m³
T₂ = 15 °C
p₂ (gauge) = 4.5 MPa gauge → 4.5 MPa + 1 atm = 4.5 MPa + 101325 Pa = 4.601 MPa
p₁ = 1 atm
Therefore isothermal compression we have the work done given by
per unit mass of the given gas, hence
From the relation
p₁·v₁ =p₂·v₂ therefore v₁ = p₂·v₂/p₁ = 4.6 MPa× 10 m³/(1 atm) = 4.6 MPa× 10 m³/(101325 Pa) = 454.115 m³
Therefore W₁₂ = 101325 Pa × 454.11 m³× ㏑((10 m³)/(454.115 m³)) = 46013250×(-3.82) = -175575813.855 J = -175.6 MJ
W = -175.6 MJ
(b) For isentropic compression we have
W = m×cv×(T₂ -T₁)
for air we put K = 1.4
therefore we have from which
v₁ = 152.65 m³
We also have or
from which we find the value of T₂ as
= 886.974 K (peak temperature)
Therefore from pv = RT and R =cp -cv = 1.005 -0.718 = 0.287 kJ/kg·K
Therefore number of moles = pv/(RT) = (4601325×10)(287×288.15) = 556.394 kg
m = 556.394 kg
Therefore work done at constant pressure = m·cp·(T₂-T₁) gives
556.394 kg × 1.005 kJ/kg⋅K×(298.15 K-886.974 K ) = -329256.19 kJ or -329.256 MJ
The peak temperature of the isentropic process = 886.974 K
Answer:
a)
b) The flow would be going from section (b) to section (a)
Explanation:
1) Notation
For above conversions we use the conversion factor
head loss from section
2) Formulas and definitions
For this case we can apply the Bernoulli equation between the sections given (a) and (b). Is important to remember that this equation allows en energy balance since represent the sum of all the energies in a fluid, and this sum need to be constant at any point selected.
The formula is given by:
Since we have a constant section on the piple we have the same area and flow, then the velocities at point (a) and (b) would be the same, and we have just this expression:
3)Part a
And on this case we have all the values in order to replace and solve for
4)Part b
Analyzing the value obtained for is a negative value, so on this case this means that the flow would be going from section (b) to section (a).