5. A stream of air at 12 bar and 900 K is mixed with another stream of air at 2 bar and 400 K with 2 times the mass flowrate. If this process were accomplished reversibly and adiabatically, what would be the temperature and pressure of the resulting air stream

The anserrs to the question are

(a)  The temperature would be 566.67  K and

(b) The pressure of the resulting air stream is 14 bar

Explanation:

Here the two streams of gases meet ad form a single stream

The steady-flow energy equation can be implemented at the mixing point of the to streams as follows

mₐhₐ₁ + mₙ hₙ₁ + Q° + W° = mₐhₐ₂ + mₙhₙ₂

Where the flow is adiabatic, we have

Q = 0 and  W = 0  hence

mₐhₐ₁ + mₙ hₙ₁ = mₐhₐ₂ + mₙhₙ₂ where h = cp×T we have

mₐcpₐ×Tₐ + mₙcpₙ×Tₙ  = mₐcpₐ×T + mₐcpₙ×T

Therefore the final temperature T is given by

 for the same kind of gas cpₐ =cpₙ therefore

where mₐ and mₙ are mass flow rate

Therefore we have where Tₐ = = 900 K and Tₙ = 400 K and mₙ = 2mₐ gives

= = 1700/3 = 566.67 K  

From Dalton's law, the total pressure of a mixture of gases is equal to the sum of the partial pressures of the components of the mixture

Therefore the total pressure of the combined stream = pₐ + pₙ = p

= 12 bar + 2 bar = 14 bar

Stream pressure = 14 bar

when heated, the valance electrons increased their speed and transferred the heat from electron to electron.   i just took the test.