#### Open System. Conservation of energy. Entropy

Steam is streaming out of a tank through a nozzle in an **adiabatic,
reversible** fashion and is expanding to a pressure of 1 MPa.

The conditions inside the tank are :

T = 700^{o}C , p = 6 MPa , v = 0.07352 m^{3}/kg

u = 3453.1 kJ/kg , h = 3894.2 kJ/kg , s = 7.4234 kJ/(kg K)

Assume that the steam enters the nozzle with negligibly small velocity
and that changes in potential energies can be neglected.

**Properties of superheated steam at 1 MPa**

T [^{o}C] |
p [MPa] |
v [m^{3}/kg] |
u [kJ/kg] |
h [kJ/kg] |
s [kJ/(kg K)] |

320 | 1.0 | 0.2678 | 2826.1 | 3093.9 |
7.1962 |

360 | 1.0 | 0.2873 | 2891.6 | 3178.9 |
7.3349 |

400 | 1.0 | 0.3066 | 2957.3 | 3263.9 |
7.4651 |

Based on the information given above determine :

- The enthalpy and temperature of the exiting steam
- The exit velocity

Linear interpolation is required. Note : 1kJ/kg = 1000 Nm/kg = 1000 (m/s)²