Respuesta :
Answer:
ΔH+U²/2=0
and
ΔH=[tex]C{p[/tex]×ΔT
∴to get the temperature drop of air, you make ΔT subject of the formula
ΔT=-U²/2Cp
=-300²/2×[tex]\frac{7}{2}[/tex]×8.314
∴ΔT=-1546K
Explanation:
The temperature drop of air if air is assumed to be an ideal gas for which C_p = ⁷/₂R is; Δt = 1546 K
We are given;
Final velocity; v₂ = 300 m/s
C_p = ⁷/₂R
At constant pressure, the change in enthalpy is;
Δh = C_p × Δt
Now, from first law of thermodynamics;
h₂ + (v₂²/2) = h₁ + (v₁²/2)
We are told initial velocity is negligible and as such v₁ = 0 m/s
Thus;
h₂ + (v₂²/2) = h₁ + 0
(h₁ - h₂) = (v₂²/2)
Thus; Δh = v₂²/2
Finally;
C_p × Δt = v₂²/2
Δt = v₂²/2/(C_p)
Δt = (300²/2)/(⁷/₂R)
where R is ideal gas constant = 8.314 Kj/kg.mol
Thus;
Δt = (300²/2)/(⁷/₂ × 8.314)
Δt = 1546 K
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