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

Answer:

1542.9 m/s

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:

$\frac{p}{p_{1}} = \frac{42}{6000} = 0.007$, $M_{e} =3.7794$, $T_{e} = 273 + 600 = 873K$

$T_{2} = 873(0.3182) = 277.79 K$

$v_{2} = 3.7794\sqrt{1.3(461.5)277.79} = 1542.9$