Question

A \(4-L\) pressure cooker has an operating pressure of 175 kPa. Initially, one- half of the volume is filled with liquid and the other half with vapor. If it is desired that the pressure cooker not run out of liquid water for \(1 \mathrm{h}\), determine the highest rate of heat transfer allowed.

Short answer

Answer: The maximum heat transfer rate allowed is 0. This means that no additional energy is required in the system to maintain the initial liquid level since the initial conditions are already at saturation levels.

Step-by-step solution

List the given information and assumptions

: 1. Total volume of the pressure cooker, \(V_t = 4 L\) 2. Operating pressure, \(P = 175 kPa\) 3. Volume filled with liquid water initially, \(V_l = V_t /2 = 2 L\) 4. Volume filled with water vapor initially, \(V_v = V_t /2 = 2 L\) 5. The pressure cooker should not run out of liquid water for 1 hour. So, we need to determine the highest heat transfer rate that maintains the liquid level at least at \(2 L\).

Find initial mass of liquid water and water vapor

: We know that the pressure and temperature of the pressure cooker represent saturation conditions. Thus the saturation temperature is $$ T = T_{sat}(P) = 401.93 K $$ Now, the mass of liquid water and water vapor can be calculated using specific volumes: 1. The specific volume of liquid water at given pressure and saturation temperature, \(v_{l} = 0.0010483 m^3/kg\) 2. The specific volume of water vapor at given pressure and saturation temperature, \(v_{g} = 0.19444 m^3/kg\) Now, the mass of liquid water, \(m_l\), and the mass of water vapor, \(m_v\), can be calculated using the formula: $$ m = \frac{V}{v} $$ \(m_l = \frac{V_l}{v_l} = \frac{0.002 m^3}{0.0010483 m^3/kg} = 1.907 kg\) \(m_v = \frac{V_v}{v_g} = \frac{0.002 m^3}{0.19444 m^3/kg} = 0.01028 kg\)

Apply energy balance equation

: To find the maximum heat transfer, we need to use the energy balance equation over the system boundary. The energy balance equation is given by: $$ Q_{in} = W_{out} + \Delta E $$ Here, \(Q_{in}\) is the heat transfer rate into the system, \(W_{out}\) is the work done by the system (negligible since the pressure cooker is sealed), and \(\Delta E\) is the change in total energy of the system. Since the liquid water should not run out in 1 hour, the final mass of liquid water, \(m_{l_f}\), should be equal to the initial mass of liquid water. $$ m_{l_f} = m_{l} $$ The final mass of water vapor, \(m_{v_f}\), is the sum of the initial mass of water vapor and the mass of water vapor evaporated from the liquid during heating. Thus, we can write: $$ m_{v_f} = m_{v} + m_{v_{evap}} $$ To find \(\Delta E\) over a period of 1 hour, we need to consider the change in internal energy, \(\Delta U\), and the change in potential and kinetic energies (negligible). Therefore, the \(\Delta E\) can be represented as: $$ \Delta E = \Delta U = m_l (u_{l_f} - u_{l}) + m_v(u_{v_f} - u_{v}) + m_{v_{evap}}(u_{v_f}-u_{l_f}) $$ Where \(u_{l}\), \(u_{v}\), \(u_{l_f}\), and \(u_{v_f}\) are specific internal energies of the initial and final states of the liquid and vapor, respectively. Now, we can express the energy balance equation as: $$ Q_{in} = m_l (u_{l_f} - u_{l}) + m_v(u_{v_f} - u_{v}) + m_{v_{evap}}(u_{v_f}-u_{l_f}) $$

Calculate the heat transfer rate

: Using the specific internal energies at the initial and final states of the liquid and vapor, we can calculate the heat transfer rate: 1. Initial specific internal energy of liquid water, \(u_{l} = 755 kJ/kg\) 2. Initial specific internal energy of water vapor, \(u_{v} = 2594.34 kJ/kg\) 3. Final specific internal energy of liquid water, \(u_{l_f} = u_l = 755 kJ/kg\) 4. Final specific internal energy of water vapor, \(u_{v_f} = u_v = 2594.34 kJ/kg\) Now, we can plug these values into the energy balance equation to find the heat transfer rate, \(Q_{in}\), over a period of 1 hour: $$ Q_{in} = m_l (u_{l_f} - u_{l}) + m_v(u_{v_f} - u_{v}) + m_{v_{evap}}(u_{v_f}-u_{l_f}) $$ $$ Q_{in} = 1.907(755-755) + 0.01028(2594.34-2594.34) + m_{v_{evap}}(2594.34-755) $$ $$ Q_{in} = 1839.34 m_{v_{evap}} $$ Now, let's calculate the evaporation rate value over the period of 1 hour: $$ m_{v_{evap}} = \frac{m_{v_f}-m_v}{1 h} $$ We have already found the value of \(m_v\), so we just need to find \(m_{v_f}\). We can use specific volume for this: $$ m_{v_f} = \frac{V_t-V_l}{v_g} $$ $$ m_{v_f} = \frac{0.004 m^3 - 0.002 m^3}{0.19444 m^3/kg} = 0.01028 kg $$ So, the evaporation rate over the period of 1 hour is zero. Thus, the heat transfer rate, \(Q_{in}\), is also zero.

Interpret the result

: The calculated maximum heat transfer rate allowed for the pressure cooker to not run out of liquid water for 1 hour is zero. This means that no additional energy is required in this system to maintain the initial liquid level since the initial conditions are already at saturation levels.

Key concepts

Understanding Saturation Conditions

When discussing thermodynamics and heat transfer, particularly in contexts involving phase change materials like water, the term 'saturation conditions' frequently arises. Saturation conditions occur when a substance is at a temperature and pressure at which its liquid and vapor phases can coexist in equilibrium. This is the point on a temperature-pressure diagram where the phase change happens, such as the boiling point of water at atmospheric pressure.

Visualize a closed container half-filled with liquid water and half with its vapor. At equilibrium, the liquid won't turn into vapor or vice versa unless the pressure or temperature changes. When we say a system, like our pressure cooker example, is under saturation conditions, we understand that the pressure and temperature of the water won't allow more liquid to evaporate or vapor to condense without an alteration of system conditions.

For students working on thermodynamics problems, recognizing saturation conditions helps narrow down the properties needed for calculations, like specific volumes and internal energies, which are often found on saturation tables. Recognizing that a system is at saturation allows us to make accurate assumptions and simplifications for our computations, leading to proper implementation of the energy balance equation.

The Role of Specific Volume in Calculations

Specific volume is a crucial concept in thermodynamics related to the density of substances. Defined as the volume that a unit mass of a substance occupies (\( v = V/m \)), specific volume is especially significant when dealing with changes in phase, such as in our pressure cooker. This property varies greatly between the liquid and vapor phases, with the vapor phase occupying much more space for the same mass of a substance than the liquid phase.

Understanding specific volume enables you to determine the mass of a substance in a particular phase given the volume it occupies. In the pressure cooker example, we have initial specific volumes for both the liquid and vapor phases given saturation conditions. Using these values, calculating the mass of each phase within the container becomes straightforward with the formula \( m = V/v \).

Specific volume also plays an indispensable role in energy calculations. For instance, when exploring how much vapor can be produced without changing the container's pressure, we rely on specific volume to maintain the balance between phases. Employing the correct specific volume values in problems about heating and work done by systems gives students the ability to extend these calculations into more complex scenarios like engine cycles or refrigeration processes.

Applying the Energy Balance Equation

The energy balance equation is the cornerstone of many thermodynamics problems. It is the mathematical expression of the First Law of Thermodynamics, which states that energy cannot be created or destroyed, only transferred or converted from one form to another. The basic form of the energy balance equation is \( Q_{in} = W_{out} + \Delta E \), where \( Q_{in} \) is the heat added to the system, \( W_{out} \) is the work done by the system, and \( \Delta E \) represents the change in energy of the system.

In the context of the pressure cooker, we're primarily concerned with the change in internal energy (\( \Delta U \) part of \( \Delta E \)). Because it is a sealed container with water in liquid and vapor forms, we disregard the work done (\( W_{out} \) is negligible), and kinetic and potential energy changes are minimal, so we focus on the heat transfer and the internal energy. We aim to find how much heat can be added without losing all the liquid, which depends on the internal energy changes of the existing liquid and vapor.

Using specific internal energy values for water's liquid and vapor at saturation conditions, students can calculate the heat transfer required for phase changes. In our example, where we wish to maintain liquid water for a specific time, the equation assists in determining the maximal heat transfer rate while ensuring we don't deplete the liquid water in the pressure cooker. This application of the energy balance equation is a practical exercise that mirrors real-life engineering systems such as steam generators or cooling systems.