Question

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

Answer 1

Answer:

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

$T = \frac{m_ac_{pa}T_a + m_nc_{pn}T_n}{m_ac_{pa} + m_nc_{pn}}$  for the same kind of gas cpₐ =cpₙ therefore

$T = \frac{m_aT_a + m_nT_n}{m_a + m_n}$ where mₐ and mₙ are mass flow rate

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

$T = \frac{m_a*900 + 2*m_a*400}{m_a + 2*m_a}$ = $T = \frac{900 + 800}{1 + 2}$ = 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