Question

1-m^{3}\( of saturated liquid water at \)200^{\circ} \mathrm{C}$ is expanded isothermally in a closed system until its quality is 80 percent. Determine the total work produced by this expansion, in kJ.

Short answer

Question: Calculate the work produced when saturated liquid water with an initial volume of 1 m³ at 200°C expands isothermally in a closed system until it reaches a quality of 0.8 (80% vapor, 20% liquid). Answer: To calculate the work produced during the expansion process, follow these steps: 1. Identify the initial state (saturated liquid at 200°C and 1 m³) and final state (same temperature, quality of 0.8). 2. Determine the properties of water in the initial and final states using steam tables or software (specific volume and internal energy). 3. Calculate the mass of the water in the system using the initial volume and specific volume. 4. Calculate the initial and final internal energy of the system using the mass and specific internal energies. 5. Apply the First Law of Thermodynamics for a closed system (ΔU = W, since Q = 0). 6. Calculate the work produced by subtracting the initial internal energy from the final internal energy (W = U₂ - U₁). The total work produced during the expansion process will be equal to the calculated value of W in kJ.

Step-by-step solution

Identify the initial state and final state of the system

Initially, the water is in a saturated liquid state at \(T_1 = 200 ^\circ \mathrm{C}\) and a given volume \(V_1 = 1 \thinspace m^3\). The final state occurs at the same temperature \(T_2=T_1=200^{\circ} \mathrm{C}\), but with a quality \(x = 0.8\). Our goal is to find the work done during this expansion.

Determine the properties of water in the initial and final states

We can use the steam tables or software to find the specific properties of water in the initial and final states. Since the water is initially in the saturated liquid state, we need the specific volume \(v_f\) and specific internal energy \(u_f\) at the given temperature at the initial state. Similarly, for the final state, we need the specific volume \(v_g\) and internal energy \(u_g\) of the saturated vapor at the given temperature. Then, we calculate the specific volume and internal energy of the final state using the quality. At 200°C, from steam tables: \(v_f = 0.001156 m^3/kg\) \(u_f = 850.6 kJ/kg\) \(v_g = 0.12736 m^3/kg\) \(u_g = 2584.0 kJ/kg\) Since the quality is 0.8 at the final state: \(v_2 = v_f + x (v_g - v_f)\) \(u_2 = u_f + x (u_g - u_f)\)

Calculate the mass of the water in the system

Knowing the total volume and the specific volume of the initial state, we can find the mass of the water in the system. \(m = \frac{V_1}{v_f}\)

Calculate the initial and final internal energy of the system

Multiply the mass of the water with the respective specific internal energies to get the initial and final internal energies. \(U_1 = m * u_f\) \(U_2 = m * u_2\)

Apply the First Law of Thermodynamics

Since the expansion is isothermal, we have no heat transfer \((Q = 0)\). Apply the First Law of Thermodynamics for a closed system. \(\Delta U = W\) Which can also be written as: \(U_2 - U_1 = W\)

Calculate the work produced

Now, plug in the values for \(U_2\) and \(U_1\) from step 4 and solve for the work \(W\). \(W = U_2 - U_1\) The total work produced during the expansion process is equal to the calculated value of \(W\) in kJ.