Air enters a turbine at an inlet Ti 1000 K pi 10 bar and l

Air enters a turbine at an inlet (T_i = 1,000 K, p_i = 10 bar) and leaves the turbine at an exit (T_e = 500 K, p_e = 1 bar). We also know the velocities of air at both inlet (V_i = 10 m/s) and exit (V_e = 90 m/s). The turbine operates at steady state and develops a stable power output (W_cv = 3,000 kW). At the same time, there is a heat transfer happening from the turbine to surroundings (Q_cv = -235 kW). R_air = 0.287 kJ/(kg middot K). The potential energy is negligible at both inlet and exit, please Determine the term 1/2 (V_i^2 - V_e^2), in kJ/kg; Read the Table 22 to determine the term (h_i - h_e), in kJ/kg; Determine the term dE_CV/dt in kW; Write the equation of energy rate balance at steady state for one-inlet one-exit control volume; Use the equation above to determine the mass flow rate, m, in kg/s; Use the Ideal Gas Model to calculate specific volumes at inlet and exit, v_1 and v in m^3/kg; Calculate the volumetric flow rates at inlet and exit, (AV)_i and (AV)_e in m^3/s; Calculate the cross-sectional areas of inlet and exit, A_i and A_e in m^3.

Solution

a) 0.5 * (Vi^2 - Ve^2 )   = 0.5 * (10^2 - 90^2 )    = - 4 kJ/kg

b) (hi - he)   = 1046.04   - 503.02 = 543.02 KJ/kg

c) dEcv / dt = Q - W

                   = -235 - 3000   = -3235 kW

d) at steady state dEcv / dt = 0

    Q   - W   = m * ( (he-hi) + 0.5*(Ve^2 - Vi^2) )

e)    Q   - W   = m * ( (he-hi) + 0.5*(Ve^2 - Vi^2) )

       3235   = m * (543.02 - 4 )

       m = 6 kg/s

f) vi   = R Ti / Pi    = 287 * 1000 / 10 * 10^5    = 0.287 m^3/kg

    ve = R Te / Pe    = 287 * 500 / 1 * 10^5    = 1.435 m^3/kg