Question

Superheated steam is stored in a large tank at 6 MPa and 800°C, The steam is exhausted isentropically through a converging-diverging nozzle. Determine the velocity of the steam flow when the steam starts to condense, assuming the steam to behave as a perfect gas with γ = 1.3.

Answer 1

1542.9 m/s

Step-by-step explanation:

The table of thermodynamic properties of steam (Steam Table) is required to solve this question. Using steam table:

At p = 6000 kPa and T = 800°C, then entropy
`(s_(1))` = 7.6554 kJ/(kg*K), internal energy
`(u_(1))` = 3641.2 kJ/kg, and
`(v_(1))` = 0.08159 m^3/kg. Thus:

`h_(1) = u_(1) + p_(1)v_(1)` = 3641.2 + 6000*0.08159 = 3641.2 + 489.54 = 4130.74 kJ/kg

For condensation of steam to occur,
`s_(2) = s_(g) = s_(1)`

Thus,
`p_(2) = 42 kPa`,
`T_(2)` = 77°C,
`h_(2)` = 2638.8 kJ/kg

If we assume that steam is a perfect gas, we have:

`(p)/(p_(1)) = (42)/(6000) = 0.007`,
`M_(e) =3.7794`,
`T_(e) = 273 + 600 = 873K`

`T_(2) = 873(0.3182) = 277.79 K`

`v_(2) = 3.7794√(1.3(461.5)277.79) = 1542.9`