Question

\(0.75-\mathrm{kg}\) water that is initially at \(0.5 \mathrm{MPa}\) and 30 percent quality occupies a spring-loaded piston-cylinder device. This device is now cooled until the water is a saturated liquid at \(100^{\circ} \mathrm{C}\). Calculate the total work produced during this process, in \(\mathrm{kJ}\).

Short answer

Answer: The total work produced during the process is 501.975 kJ.

Step-by-step solution

Determine the initial and final state properties

First, we must identify all given properties of the initial and final states. Initial State: - Mass (m) = 0.75 kg - Pressure (P1) = 0.5 MPa = 500 kPa - Quality (x) = 30% = 0.3 We will need to use the steam tables to find the values of the initial specific volume (v1) and specific internal energy (u1). Final State: - Temperature (T2) = 100 °C - Saturated liquid We will also use the steam tables to find the values of the final specific volume (v2) and specific internal energy (u2).

Calculate initial specific volume (v1) and specific internal energy (u1)

Using the Pressure (P1 = 500 kPa) and the Quality (x = 0.3), we can determine the initial specific volume (v1) and specific internal energy (u1) from the steam table. The equations are: v1 = vf + x * (vg - vf) u1 = uf + x * (ug - uf) where vf and vg are the specific volumes of the saturated liquid and saturated vapor at the given pressure, and uf and ug are the specific internal energies of the saturated liquid and saturated vapor at the given pressure. From the steam table: For 500 kPa: - vf = 0.001053 m³/kg - vg = 0.3741 m³/kg - uf = 640.23 kJ/kg - ug = 2100.8 kJ/kg Now, we can calculate v1 and u1: v1 = 0.001053 + 0.3 * (0.3741 - 0.001053) = 0.1133681 m³/kg u1 = 640.23 + 0.3 * (2100.8 - 640.23) = 1086.4 kJ/kg

Calculate final specific volume (v2) and specific internal energy (u2) at 100 °C

Since the final state is a saturated liquid at 100 °C, we can directly obtain the values of v2 (vf at 100 °C), and u2 (uf at 100°C) from the steam table. From the steam table: For 100 °C: - vf = 0.001044 m³/kg - uf = 419.1 kJ/kg Hence, v2 = 0.001044 m³/kg u2 = 419.1 kJ/kg

Calculate the work produced during the process

We will use the first law of thermodynamics as follows: W = m * (u1 - u2) - m * P1 * (v1 - v2) Here, m is the mass of the water, and P1 is the initial pressure. Using our values: W = 0.75 * (1086.4 - 419.1) - 0.75 * 500 * (0.1133681 - 0.001044) = 501.975 kJ

Report the result

The total work produced during the process is 501.975 kJ.