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
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Answer:
is the minimum gap between the slabs
Explanation:
Given:
- length of the concrete highway,

- coefficient of thermal expansion,

- range of temperature variation,

<u>Now form the equation of thermal expansion:</u>



Since each slab of the highway expands by the above length so the minimum gap between the slabs to prevent buckling:


is the minimum gap between the slabs
-1.1m/s
Explanation:
Work out which of the displacement (S), initial velocity (U), acceleration (A) and time (T) you have to solve for final velocity (V). If you have U, A and T, use V = U + AT. If you have S, U and T, use V = 2(S/T) - U.