Question #175244

An adiabatic expansion of air occurs through a nozzle (W1=0) from 850 kPa and 75C to 140 kPa. For an isentropic expansion find: specific volume(m/kg), temperature(C) and velocity(m/s) at the exit section assuming initial kinetic energy is negligible


1
Expert's answer
2021-04-01T01:10:21-0400

The inlet temperature(T1)= 75 C=75+273=348 K

Inlet Pressure(P1)=850 x 103 Pa

The exit Pressure(P2)=140x103Pa

Assuming air as ideal gas having following properties:

Cp = 1.004 kJ/kg.K, Cv = 0.718 kJ/kg.​K, k = 1.4 and R=0.287 kJ/kg K


Since the expansion is adiabatic,

T2T1=P2P1K1K\frac{T_2}{T_1}={\frac{P_2}{P_1}}^{\frac{K-1}{K}}

T2348=[140850]1.411.4\frac{T_2}{348}={[\frac{140}{850}]}^{\frac{1.4-1}{1.4}}


T2=207.865KT_2=207.865 K

specific volume at exit is given by,

v2=RT2P2v2=287×207.86140000=0.426kg/m3v_2=\frac{{RT_2}}{P_2}\\v_2=\frac{{287\times 207.86}}{140000}=0.426kg/m^3


In an adiabatic turbine there is no Heat or work transfer between system and surrounding,

W=0, Q=0

Applying steady flow energy equation between inlet and exit. h1+V122=h2+V222\ h_1+\frac{V^2_1}{2}=h_2+\frac{V^2_2}{2}

At inlet velocity of flow is negligible,

=> V1=0

The velocity at exit is given by,

V2=h1h2{V_2}=\sqrt{h_1-h_2}

V2=Cp(T1T2)=1.004×(348207.865)×1000=375.09 m/s{V_2}=\sqrt{C_p(T_1-T_2)}=\sqrt{1.004\times (348-207.865)\times1000}=375.09\space m/s


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