Complete question:
The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.
Answer:
The exit velocity is 629.41 m/s
Explanation:
Given;
initial temperature, T₁ = 1200K
initial pressure, P₁ = 150 kPa
final pressure, P₂ = 80 kPa
specific heat at 300 K, Cp = 1004 J/kgK
k = 1.4
Calculate final temperature;
k = 1.4
Work done is given as;
inlet velocity is negligible;
Therefore, the exit velocity is 629.41 m/s
Answer:
It's only 1.11 m/s2 weaker at 400 km above surface of Earth
Explanation:
Let Earth radius be 6371 km, or 6371000 m. At 400km above the Earth surface would be 6371 + 400 = 6771 km, or 6771000 m
We can use Newton's gravitational law to calculate difference in gravitational acceleration between point A (Earth surface) and point B (400km above Earth surface):
where G is the gravitational constant, M is the mass of Earth and r is the distance form the center of Earth to the object
So the gravitational acceleration at 400km above surface is only 0.885 the gravitational energy at the surface, or 0.885*9.81 = 8.7 m/s2, a difference of (9.81 - 8.7) = 1.11 m/s2.
An equation can be dimensionally correct but that does not mean that it is physically correct. Explanation is given in the photo.
