Respuesta :
Answer:
the maximum theoretical work is w= 1314.7 kJ/kg
Explanation:
The maximum theoretical work obtained in reversible conditions, since there is no loss of useful energy.
Since the turbine is insulated ,we can assume an adiabatic process. Then for an ideal gas under an adiabatic and reversible process the relationship between T and P is
PV^k = constant and P*V = n*R*T thus
T*P^[(1-k)/k] = constant , k=adiabatic coefficient for air
T₁*P₁^[(1-k)/k] = T₂*P₂^[(1-k)/k]
T₂ = T₁*(P₁/P₂)^[(1-k)/k]
replacing values, knowing that k= 1.4 for air
T₂ = 1400 K*(8 bar /0.8 bar)^[(1-1.4)/1.4] = 2703 K
for open systems we know that
W = ∫ V dP
since PV^k = constant = C → P= C*V^-k →dP = -kC*V^(-k-1)*dV
thus
W = ∫ V dP = -kC*∫V^(-k)*dV= -kC*[V₂^(1-k)-V₁^(1-k)]/(1-k) = k*(P₂V₂-P₁V₁)/(k-1)
= n*R*(T₂-T₁)*k*/(k-1)
W= n*k*R/(k-1) *(T₂-T₁)
this result can also be obtained from the first law of thermodynamics applied to open systems
Q + W tur = ΔH + ΔV +ΔK
since Q =0 (insulated) and ΔV=ΔK=0 (potential and kinetic energy neglected)
W tur = ΔH
for an ideal gas ΔH = n*Cp* (T₂-T₁) , and also k= Cp/Cv , Cp - Cv = R thus Cp= k*R/(k-1)
W tur = ΔH = n* Cp* (T₂-T₁) = n*k*R/(k-1)*(T₂-T₁)
since n= m/M
w = W/m = (1/M)*k*R/(k-1) *(T₂-T₁)
replacing values and knowing that M= ∑xiMi = 0.21*32+0.79*28 = 28.84 g/mol
w =(1/M)*k*R/(k-1) *(T₂-T₁) = 1/ (28.84 g/mol) *1.4 * 8.314 J/molK /(1.4-1) *(2703 K- 1400K)
w = 1314.7 J/g = 1314.7 kJ/kg
