Question

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.

Answer 1

Answer:

$h = 1429.74\,\frac{kJ}{kg}$

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:

$\nu_{f} = 0.001080\,\frac{m^{3}}{kg}$

$\nu_{g} = 0.50850\,\frac{m^{3}}{kg}$

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

$x = \frac{\nu-\nu_{f}}{\nu_{g}-\nu_{f}}$

$x = \frac{0.2\,\frac{m^{3}}{kg} - 0.001080\,\frac{m^{3}}{kg} }{0.50850\,\frac{m^{3}}{kg} - 0.001080\,\frac{m^{3}}{kg} }$

$x = 0.392$

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

$h_{f} = 589.16\,\frac{kJ}{kg}$

$h_{g} = 2733.5\,\frac{kJ}{kg}$

The specific enthalpy is:

$h = h_{f}+x\cdot (h_{g}-h_{f})$

$h = 589.16\,\frac{kJ}{kg}+0.392\cdot \left( 2733.5\,\frac{kJ}{kg} - 589.16\,\frac{kJ}{kg} \right)$

$h = 1429.74\,\frac{kJ}{kg}$

Answer 2

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 $v_f = 1.0797*10^{-3}m^3/kg$

specific volume of saturated vapor $v_g = 0.5089 m^3/kg$

specific enthalpy of saturated liquid $h_f = 589.13 kJ/kg$

specific enthalpy of saturated vapor $h_g = 2733.9 kJ/kg$

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

$x = \frac{v - v_f}{v_g-v_f}$

$x = \frac{0.2 - 1.0797*10^{-3}}{0.5089-1.0797*10^{-3}}$

$x = \frac{0.1989203}{0.5078203}$

$x = 0.3917139587$

$x = 0.390$

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

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