List six possible valve defects that should be included in the inspection of a used valve?
1 answer:
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Answer:
Qcd=0.01507rad
QT= 0.10509rad
Explanation:
The full details of the procedure and answer is attached.
Answer:
Examples of reciprocating motion in daily life are;
1) The needles of a sewing machine
2) Electric powered reciprocating saw blade
3) The motion of a manual tire pump
Explanation:
A reciprocating motion is a motion that consists of motion of a part in an upward and downwards or in a backward and forward (↔) direction repetitively
Examples of reciprocating motion in daily life includes the reciprocating motion of the needles of a sewing machine and the reciprocating motion of the reciprocating saw and the motion of a manual tire pump
In a sewing machine, a crank shaft in between a wheel and the needle transforms the rotary motion of the wheel into reciprocating motion of the needle.
Answer:
the only one that meets the requirements is option C .
Explanation:
The tolerance of a quantity is the maximum limit of variation allowed for that quantity.
To find it we must have the value of the magnitude, its closest value is the average value, this value can be given or if it is not known it is calculated with the formula
x_average = ∑ / n
The tolerance or error is the current value over the mean value per 100
Δx₁ = x₁ / x_average
tolerance = | 100 -Δx₁ 100 |
bars indicate absolute value
let's look for these values for each case
a)
x_average = (2.1700000+ 2.258571429) / 2
x_average = 2.2142857145
fluctuation for x₁
Δx₁ = 2.17000 / 2.2142857145
Tolerance = 100 - 97.999999991
Tolerance = 2.000000001%
fluctuation x₂
Δx₂ = 2.258571429 / 2.2142857145
Δx2 = 1.02
tolerance = 100 - 102.000000009
tolerance 2.000000001%
b)
x_average = (2.2 + 2.29) / 2
x_average = 2,245
fluctuation x₁
Δx₁ = 2.2 / 2.245
Δx₁ = 0.9799554
tolerance = 100 - 97,999
Tolerance = 2.00446%
fluctuation x₂
Δx₂ = 2.29 / 2.245
Δx₂ = 1.0200445
Tolerance = 2.00445%
c)
x_average = (2.211445 +2.3) / 2
x_average = 2.2557225
Δx₁ = 2.211445 / 2.2557225 = 0.9803710
tolerance = 100 - 98.0371
tolerance = 1.96%
Δx₂ = 2.3 / 2.2557225 = 1.024624
tolerance = 100 -101.962896
tolerance = 1.96%
d)
x_average = (2.20144927 + 2.29130435) / 2
x_average = 2.24637681
Δx₁ = 2.20144927 / 2.24637681 = 0.98000043
tolerance = 100 - 98.000043
tolerance = 2.000002%
Δx₂ = 2.29130435 / 2.24637681 = 1.0200000017
tolerance = 2.0000002%
e)
x_average = (2 +2,3) / 2
x_average = 2.15
Δx₁ = 2 / 2.15 = 0.93023
tolerance = 100 -93.023
tolerance = 6.98%
Δx₂ = 2.3 / 2.15 = 1.0698
tolerance = 6.97%
Let's analyze these results, the result E is clearly not in the requested tolerance range, the other values may be within the desired tolerance range depending on the required precision, for the high precision of this exercise the only one that meets the requirements is option C .