## Steam Tables

Saturated water at 115°C flows into a throttling valve, and exits the valve at 0.10135 MPa. At exit, what is the quality of the water?

##
__
__**Hint**

**Hint**

Quality, x, (for two-phase liquid-vapor systems at saturation) is defined as

$$$h=xh_g+(1-x)h_f\:or\:h=h_f+xh_{fg}$$$

where
$$h_{fg}$$
is the enthalpy change for phase transitions (evaporation),
$$h_f$$
is the enthalpy of saturated liquid,
$$h_g$$
is the enthalpy for saturated vapor, and
$$h$$
is specific enthalpy.

##
__
__**Hint 2**

**Hint 2**

Quality, x, (for two-phase liquid-vapor systems at saturation) is defined as

$$$h=xh_g+(1-x)h_f\:or\:h=h_f+xh_{fg}$$$

where
$$h_{fg}$$
is the enthalpy change for phase transitions (evaporation),
$$h_f$$
is the enthalpy of saturated liquid,
$$h_g$$
is the enthalpy for saturated vapor, and
$$h$$
is specific enthalpy.

From the steam table at 115°C, the saturation pressure is 0.16906 MPa, and the enthalpy,
$$h_1$$
, is 482.48 kJ/kg (the problem states saturated water at 115°C). Because enthalpy does not change as liquid passes through a throttling valve,
$$h_1=h_2=482.48\:kJ/kg$$
. In the steam table again, 0.1035 MPa correlates to
$$T=100°C$$
,
$$h_f=419.04\:kJ/kg$$
and
$$h_{fg}=2257\:kJ/kg$$
.

Thus, for our problem statement:

$$$h_1=h_f+xh_{fg}$$$

$$$482.48\frac{kJ}{kg}=419.04\frac{kJ}{kg}+x(2257\frac{kJ}{kg})$$$

$$$x=\frac{63.44}{2257}=0.028$$$

0.028