Using the tables for water determine the specified property data at the indicated state. For H2O at T = 140 °C and v = 0.2 m3/kg, find h in kJ/kg. Note: Round up the answer to 2 decimal places.

Respuesta :

Answer:

[tex]h = 1429.74\,\frac{kJ}{kg}[/tex]

Explanation:

The determination of any further properties requires the knowledge of two independent properties. (Temperature and specific volume in this case). The specific volumes for saturated liquid and vapor at 140 °C are, respectively:

[tex]\nu_{f} = 0.001080\,\frac{m^{3}}{kg}[/tex]

[tex]\nu_{g} = 0.50850\,\frac{m^{3}}{kg}[/tex]

Since [tex]\nu_{f} < \nu < \nu_{g}[/tex], it is a liquid-vapor mixture. The quality of the mixture is:

[tex]x = \frac{\nu-\nu_{f}}{\nu_{g}-\nu_{f}}[/tex]

[tex]x = \frac{0.2\,\frac{m^{3}}{kg} - 0.001080\,\frac{m^{3}}{kg} }{0.50850\,\frac{m^{3}}{kg} - 0.001080\,\frac{m^{3}}{kg} }[/tex]

[tex]x = 0.392[/tex]

The specific enthalpies for saturated liquid and vapor at 140 °C are, respectively:

[tex]h_{f} = 589.16\,\frac{kJ}{kg}[/tex]

[tex]h_{g} = 2733.5\,\frac{kJ}{kg}[/tex]

The specific enthalpy is:

[tex]h = h_{f}+x\cdot (h_{g}-h_{f})[/tex]

[tex]h = 589.16\,\frac{kJ}{kg}+0.392\cdot \left( 2733.5\,\frac{kJ}{kg} - 589.16\,\frac{kJ}{kg} \right)[/tex]

[tex]h = 1429.74\,\frac{kJ}{kg}[/tex]

Answer:

h = 1425.59 kJ/kg

Explanation:

when temperature of water is = 140° C

specific volume (v) = 0.2 m³/kg

We Obtain the properties of water from the table properties of saturated water, at a temperature of  T = 140° C

specific volume of saturated liquid [tex]v_f = 1.0797*10^{-3}m^3/kg[/tex]

specific volume of saturated vapor [tex]v_g = 0.5089 m^3/kg[/tex]

specific enthalpy of saturated liquid [tex]h_f = 589.13 kJ/kg[/tex]

specific enthalpy of saturated vapor [tex]h_g = 2733.9 kJ/kg[/tex]

Let's first find the quality (x) of the liquid by using the expression

[tex]x = \frac{v - v_f}{v_g-v_f}[/tex]

[tex]x = \frac{0.2 - 1.0797*10^{-3}}{0.5089-1.0797*10^{-3}}[/tex]

[tex]x = \frac{0.1989203}{0.5078203}[/tex]

[tex]x = 0.3917139587[/tex]

[tex]x = 0.390[/tex]

To determine the specific enthalpy h; we use the relation:

[tex]h = h_f + x (h_g -h_f)\\h = 589.13 + 0.390(2733.9 -589.13)\\h = 589.13 + 0.390(2144.77)\\h = 589.13 + 836.4603\\h = 1425.5903 \\h = 1425.59 \ kJ/kg[/tex]