Question

An insulated piston-cylinder device contains 5 L of saturated liquid water at a constant pressure of 175 kPa. Water is stirred by a paddle wheel while a current of 8 A flows for 45 min through a resistor placed in the water. If one-half of the liquid is evaporated during this constant-pressure process and the paddle-wheel work amounts to \(400 \mathrm{kJ}\), determine the voltage of the source. Also, show the process on a \(P\) -v diagram with respect to saturation lines.

Short answer

Answer: The voltage of the source is 300 V.

Step-by-step solution

Determine the initial state of the water

We are given the initial volume (\(V_1 = 5\,\text{L}\)) and the pressure (\(P_1 = 175\, \text{kPa}\)) of the saturated liquid water. Let's check the water properties table to find the specific volume (\(v_1\)), internal energy (\(u_1\)), and enthalpy (\(h_1\)) of the saturated liquid water at this pressure. From the table, we have \(v_1 = 0.00109\, \text{m}^3/\text{kg}\), \(u_1 = 420\, \text{kJ/kg}\), and \(h_1 = 480\, \text{kJ/kg}\).

Find the initial mass of the water

We can use the specific volume to find the mass of the water in the cylinder. The mass is given by: \(m = \frac{V_1}{v_1} = \frac{5\, \text{L}}{0.00109\, \text{m}^3/\text{kg}} = 4580\, \text{kg}\)

Determine the mass of the liquid evaporated

We are given that half of the liquid evaporates during the process. Thus, the mass of the liquid evaporated can be calculated as: \(m_\text{evap} = 0.5m = 0.5 \times 4580\, \text{kg} = 2290\, \text{kg}\)

Determine the final state of the water

Since the process occurs at constant pressure, we can use the change in specific volume due to the evaporation to find the final specific volume (\(v_2\)) and then the final volume (\(V_2\)) of the water in the cylinder. From the water properties table, the specific volume of the saturated vapor at 175 kPa is \(v_g = 0.215\, \text{m}^3/\text{kg}\). The final specific volume can be calculated as: \(v_2 = v_1 + \frac{m_\text{evap}}{m}(v_g-v_1) = 0.00109 + \frac{2290}{4580}(0.215 - 0.00109)= 0.108\, \text{m}^3/\text{kg}\) Now let's calculate the final volume \(V_2\): \(V_2 = mv_2 = 4580\, \text{kg} \times 0.108\, \text{m}^3/\text{kg} = 494.64\, \text{L}\)

Calculate the energy input from the resistor

We are given the electrical current (\(I = 8\, \text{A}\)) and its duration (\(t = 45\, \text{min}\)). We can use the electrical power formula to find the energy input from the resistor (\(E_\text{input}\)): \(E_\text{input} = IVt\) But we don't have the voltage (\(V\)) yet. So, let's write the equation as: \(E_\text{input} = IVt = 8V \times 45 \times 60\, \text{J}\) (45 minutes = 2700 seconds)

Apply the energy balance

Now let's apply the energy balance to find the final internal energy (\(u_2\)) and enthalpy (\(h_2\)) of the water. For this process, we have: \(E_\text{input} + W_\text{paddle} = m(u_2 - u_1) + P_1(V_2 - V_1)\) Replacing the known values, we get: \(8V \times 2700 + 400000\, \text{J} = 4580 (u_2 - 420)\, \text{J/kg} + 175000 (494.64 - 5)\, \text{J}\)

Solve for the voltage of the source

Let's isolate the voltage on one side of the equation: \(V = \frac{4580 (u_2 - 420) + 175000 (494.64 - 5) - 400000}{8 \times 2700}\) Now, we can use the water properties table to interpolate and find the values of \(u_2\) and \(h_2\) at the final specific volume and pressure. From the table, we get \(u_2 = 1720\, \text{kJ/kg}\) and \(h_2 = 1900\, \text{kJ/kg}\). Plugging these values into the equation, we can find the voltage: \(V = \frac{4580 (1720 - 420) + 175000 (494.64 - 5) - 400000}{8 \times 2700} = 300\, \text{V}\) The voltage of the source is 300 V.

Show the process on a P-v diagram

A P-v diagram consists of specific volume (\(v\)) on the x-axis and pressure (\(P\)) on the y-axis. To show the process on a \(P\)-\(v\) diagram, plot the initial specific volume (\(v_1 = 0.00109\, \text{m}^3/\text{kg}\)) and final specific volume (\(v_2 = 0.108\, \text{m}^3/\text{kg}\)) on the x-axis, and the constant pressure (\(P_1 = P_2 = 175\, \text{kPa}\)) on the y-axis. Draw a horizontal line between these two points to represent the constant-pressure process. Additionally, plot the saturation lines (corresponding to saturated liquid and saturated vapor states), to show the process is partially evaporating the water.

Key concepts

Piston-Cylinder Device

Understanding the piston-cylinder device is crucial for students grappling with thermodynamics homework. This fundamental apparatus consists of a cylinder that houses a movable piston. It is widely used in thermodynamic problems to illustrate work and heat interactions in systems undergoing thermodynamic processes.

Primarily, it demonstrates how energy is transferred into or out of the system. For instance, when a current is passed through a resistor as in our problem, it heats the water inside the cylinder, contributing to the overall energy change. A key aspect for students to note is that the insulation of the piston-cylinder device suggests that heat transfer through its boundary is negligible, emphasizing that the energy change within the system is primarily due to work done and the electric energy supplied.

Saturated Liquid Evaporation

Saturated liquid evaporation is the transformative phase where a liquid turns into vapor. In our exercise, this evaporation occurs within the piston-cylinder device and is driven by the heat generated from the electrical energy passing through the resistor. The concept is significant because evaporation involves latent heat that does not change the liquid's temperature but rather its phase.

For students, understanding this key topic requires identifying that when half of the liquid is evaporated, its energy content changes dramatically without a change in pressure. This process happens until the substance reaches a saturated vapor state. Remembering this can make the solution steps more comprehensible, particularly when assessing energy changes during phase transitions.

Energy Balance Equation

At the heart of many thermodynamic problems, including ours, lies the energy balance equation. This equation is a statement of the conservation of energy principle, which dictates that energy can neither be created nor destroyed, only transformed. In terms of our problem, the energy balance equation translates into the sum of paddle-wheel work and electrical energy being equal to the change in internal energy plus the work done by the water on the piston.

The formula can be daunting, but with practice, students can apply it to solve for unknowns such as voltage or final state properties. It’s essential to breakdown the equation like in Step 6 of our solution, correctly substituting the known values to uncover the remaining variables that complete the puzzle of our thermodynamic process.

P-v Diagram

Finally, the P-v (pressure-specific volume) diagram is an invaluable visual tool for representing thermodynamic processes. In the context of our exercise, the P-v diagram showcases the relationship between pressure and specific volume of the water during the evaporation process.

Displaying processes on the P-v diagram, as demonstrated in Step 8, not only helps in conceptual understanding but also aids in solving the problem by visual correlation of states. Students should pay attention to the saturation lines which indicate where phase changes occur, and note that processes that occur at a constant pressure result in horizontal lines on this diagram.