Question

One kilogram of water fills a 0.140 m^3 rigid container at an initial pressure of 1.8 MPa. The container is then cooled to 40°C. Determine the initial temperature and final pressure of the water.

Answer 1

Answer:

T1 = 299.18 °C

P2 = 0.00738443 MPa

Explanation:

From the data, we can get two properties for the initial condition. These are pressure and specific volume.

The pressure is 1.8 MPa and the specific volume, we can get it with the mass and volume of the container, since it’s filled this is also the volume of the water in it.

$v=\frac{vol (m^{3})}{mass (kg)} = \frac{0.140 m^{3}}{1 kg} = 0.140 \frac{ m^{3}}{kg}$

When we check in the thermodynamic tables, the conditions for saturation at 1.8 MPa we found the following:

$P^{sat} = 1.8 MPa$

$T^{sat} = 207.12 C$

$v_{g} = 0.1103\frac {m^{3}}{kg}$ specific volume for the saturated vapor  

$v_{l} = 0.001167 \frac{m^{3}}{kg}$ specific volume for the saturated liquid  

Since the specific volume in our condition is higher that the specific volume for the saturated vapor, we have a superheated steam.  

Looking in the thermodynamic tables for superheated steam we found that the temperature where the steam has a specific volume of $0.140 \frac{ m^{3}}{kg}$ at 1.8 MPa is 299.18 °C. This is the initial temperature in the container.

Since the only information that we have about the final condition is that the container was cooled. We can assume that it was cooled until a condition of saturation. So, the final pressure for the water will be the pressure of saturation for a temperature of 40°C. From thermodynamic tables we get:

$P^{sat} at 40C = 0.00738443 MPa$