Question 7.10: Problem: Steam enters an adiabatic nozzle at 1 MPa and 400 °...
Problem: Steam enters an adiabatic nozzle at 1 MPa and 400 °C with negligible velocity. It exits at a pressure of 0.6 MPa. Assuming that the expansion is reversible, calculate the exit temperature and velocity.
Find: Exit temperature T_2 and velocity \pmb{V_2}.
Known: Inlet pressure P_1 = 1 MPa, inlet temperature T_1 = 400 °C, exit pressure P_2 = 0.6 MPa, nozzle is adiabatic.
Diagram:
Assumptions: Expansion is reversible, and since the nozzle is adiabatic, the process is isentropic.
Governing Equation:
Adiabatic nozzle \pmb{V_2}=\sqrt{2(h_1-h_2)}
At the inlet we know the pressure and temperature, which fixes the state. At the outlet we know the pressure, and we can obtain a second property by assuming that the expansion is isentropic so that s_1= s_2. The inlet conditions are P_1 = 1 MPa and T_1 = 400 °C. From the superheated steam tables (Appendix 8c) specific entropy s_1 = 7.4651 kJ / kgK and specific enthalpy h_1 = 3263.9 kJ / kg. For saturated vapour at P_2 = 0.6 MPa (Appendix 8b) specific entropy of vapour s_g = 6.7600 kJ / kg K. Since s_2> s_g , the steam is superheated at the outlet. We can represent the process on a T‐ν diagram.
We know the pressure and specific entropy at the outlet. To find the temperature we must interpolate in the superheated steam table at 0.6 MPa. The value of s_2 = 7.4651 kJ / kgK lies between those for s at 300 and 350 °C. From the superheated steam tables, at 0.6 MPa:
| T(°C) | h(kJ / kg) | s(kJ / kgK) |
| 300 | 3061.6 | 7.3724 |
| T_2=? | h_2=? | s_2 = 7.4651 |
| 350 | 3165.7 | 7.5464 |
Interpolating for temperature,
\frac{T_2-300 \ °C}{350 \ °C-300 \ °C}=\frac{7.4651 \ kJ/kgK-7.3724 \ kJ/kgK}{7.5464 \ kJ/kgK-7.3724 \ kJ/kgK} , \\ T_2=326.638 \ °C.
Interpolating for enthalpy,
\frac{h_2-3061.6 \ kJ/kg}{3165.7 \ kJ/kg – 3061.6 \ kJ/kg}=\frac{7.4651 \ kJ/kgK-7.3724 \ kJ/kgK}{7.5464 \ kJ/kgK-7.3724 \ kJ/kgK} , \\ h_2=3117.06 \ kJ/kg.
From an energy balance for an adiabatic nozzle,
\pmb{V_2}= \sqrt{2(h_2-h_1)}=\sqrt{2(3263.9 \ kJ/kg-3117.1 \ kJ/kg)10^3 \ J/kJ}= 541.849 \ m/s.
Answer: The exit temperature is 327 °C and the exit velocity is 541.8 m / s.