Thermodynamics

Analysis We denote the entrance, throat, and exit states by 1, t, and 2,respectively, as shown in Fig. 17–60.(a) Since the inlet velocity is negligible, the inlet stagnation and static statesare identical. The ratio of the exit-to-inlet stagnation pressure isThen the throat velocity is determined from Eq. 17–3 to beV t 22 1h 01 h t 2 32 13248.4 3076.82 kJ/kg4 a 1000 m2 >s 2b 585.8 m/sB 1 kJ>kgThe flow area at the throat is determined from the mass flow rate relation:A t m# v tV tAt state 2s,The enthalpy of the steam at the actual exit state is (see Chap. 7)Therefore,P t 1.09 MPas t 7.1292 kJ>kg # Kf h t 3076.8 kJ>kgv t 0.24196 m 3 >kg 12.5 kg/s2 10.2420 m3 /kg2585.8 m/sP 2s P 2 300 kPas 2s s 1 7.1292 kJ>kg # Kf h 2s 2783.6 kJ>kgh N h 01 h 2h 01 h 2s0.93 3248.4 h 23248.4 2783.6P 2 300 kPah 2 2816.1 kJ>kg fv 2 0.67723 m 3 >kgs 2 7.2019 kJ>kg # KThen the exit velocity and the exit area become 10.33 10 4 m 2 10.33 cm 2¡ h 2 2816.1 kJ>kgV 2 22 1h 01 h 2 2 32 13248.4 2816.12 kJ>kg4 a 1000 m2 >s 2b 929.8 m>sB 1 kJ>kgP 1 = 2 MPaT 1 = 400°CChapter 17 | 871h N = 93%m · = 2.5 kg/sV 1 ≅ 0STEAMP 2 300 kPaP 01 2000 kPa 0.15 ThroatIt is much smaller than the critical-pressure ratio, which is taken to behP*/P 01 0.546 since the steam is superheated at the nozzle inlet. Therefore,the flow surely is supersonic at the exit. Then the velocity at the throatis the sonic velocity, and the throat pressure is1P t 0.546P 01 10.5462 12 MPa2 1.09 MPatAt the inlet,P 1 P 01 2 MPaT 1 T 01 400°C fh 1 h 01 3248.4 kJ>kgs 1 s t s 2s 7.1292 kJ>kg # K22sAlso, at the throat,sFIGURE 17–60Schematic and h-s diagram forExample 17–16.P tP 1 = P 01 = 2 MPaP 2 = 300 kPaA 2 m# v 2V 2 12.5 kg>s2 10.67723 m3 >kg2929.8 m>s 18.21 10 4 m 2 18.21 cm 2