Air enters a compressor operating at steady state at T1 = 320 K, p1= 2 bar with a velocity of 80 m/s. At the exit, T2 = 550 K, p
2 answers:
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Answer:
T = 15 kN
F = 23.33 kN
Explanation:
Given the data in the question,
We apply the impulse momentum principle on the total system,
mv₁ + ∑ = mv₂
we substitute
[50 + 3(30)]×10³ × 0 + FΔt = [50 + 3(30)]×10³ × ( 45 × 1000 / 3600 )
F( 75 - 0 ) = 1.75 × 10⁶
The resultant frictional tractive force F is will then be;
F = 1.75 × 10⁶ / 75
F = 23333.33 N
F = 23.33 kN
Applying the impulse momentum principle on the three cars;
mv₁ + ∑ = mv₂
[3(30)]×10³ × 0 + FΔt = [3(30)]×10³ × ( 45 × 1000 / 3600 )
F(75-0) = 1.125 × 10⁶
The force T developed is then;
T = 1.125 × 10⁶ / 75
T = 15000 N
T = 15 kN
Answer:
Explanation:
From the information given:
Life requirement = 40 kh = 40
Speed (N) = 520 rev/min
Reliability goal = 0.9
Radial load = 2600 lbf
To find C10 value by using the formula:
where;
The Weibull parameters include:
∴
Using the above formula:
Recall that:
1 kN = 225 lbf
∴