Question

A \(0.18-\mathrm{m}^{3}\) rigid tank is filled with saturated liquid water at \(120^{\circ} \mathrm{C}\). A valve at the bottom of the tank is now opened, and one-half of the total mass is withdrawn from the \(\operatorname{tank}\) in the liquid form. Heat is transferred to water from a source at \(230^{\circ} \mathrm{C}\) so that the temperature in the tank remains constant. Determine \((a)\) the amount of heat transfer and (b) the total entropy generation for this process.

Short answer

Answer: (a) The amount of heat transfer is 0 kJ. (b) The total entropy generation is 0 kJ/K.

Step-by-step solution

Determine the initial properties of the system

First, let's find the initial properties of the water in the tank. The volume of the tank is given as 0.18 m³, and we know that the water is a saturated liquid at 120°C. Using a steam table or the saturated properties for water, we can find the saturation temperature for water at 120°C: \(T_s = 120^{\circ} \mathrm{C}\) Specific volume, \(v_f = 0.00106 \thinspace m^3/kg\)

Calculate the initial mass of the water in the tank

Now we can calculate the initial mass (m) of the water in the tank using the volume and specific volume: \(Initial \thinspace mass \thinspace(m_i) =\dfrac{Volume}{Specific \thinspace volume} =\) \(\dfrac{0.18 \thinspace m^{3}}{0.00106 \thinspace m^{3} / kg} \approx 169.81 \thinspace kg\)

Determine the final mass of the water in the tank

Following the process, one-half of the total mass is withdrawn from the tank. The final mass (m_f) of the water in the tank is: \(Final \thinspace mass \thinspace (m_f) = \dfrac{1}{2} \cdot (Initial \thinspace mass) = \dfrac{1}{2} \cdot 169.81 \thinspace kg \approx 84.91 \thinspace kg\)

Apply the first law of thermodynamics to the process

Let's apply the first law of thermodynamics for a closed system, considering the heat transfer (Q) and neglecting the work and kinetic and potential energy changes: \(\delta Q = m \cdot (u_2 - u_1)\) Where \(u_1\) and \(u_2\) are the initial and final internal energies of the water in the tank Since the temperature remains constant at 120°C during the process and the water is a saturated liquid, the initial and final internal energies can be found using saturated properties for water: \(u_1 = u_f(T_1) = 504.49 \thinspace kJ/kg\) \(u_2 = u_f(T_2) = 504.49 \thinspace kJ/kg\) (since the temperature remains constant) Now we can calculate Q: \(\delta Q = m_f \cdot (u_2 - u_1) = 84.91 \thinspace kg \cdot (504.49 - 504.49) \thinspace kJ/kg = 0 \thinspace kJ\) Since the internal energies are the same before and after the process, no heat transfer takes place.

Calculate the total entropy generation

Next, we'll use the second law of thermodynamics to calculate the total entropy generation (S_gen): \(\delta S_{gen} = m \cdot (s_2 - s_1) + \dfrac{Q}{T_0}\) Where \(s_1\) and \(s_2\) are the initial and final specific entropy, and \(T_0\) is the surrounding temperature. Similar to the internal energies, the specific entropies can be found using saturated properties for water: \(s_1 = s_f(T_1) = 1.4339 \thinspace \mathrm{kJ} \cdot \mathrm{kg}^{-1}\mathrm{K}^{-1}\) \(s_2 = s_f(T_2) = 1.4339 \thinspace \mathrm{kJ} \cdot \mathrm{kg}^{-1}\mathrm{K}^{-1}\) (since the temperature remains constant) Assuming the surrounding temperature (T_0) is at saturation temperature (\(T_0 = 120^{\circ} \mathrm{C}\)), we can calculate the total entropy generation: \(\delta S_{gen} = m_f \cdot (s_2 - s_1) + \dfrac{Q}{T_0} = 84.91 \thinspace kg \cdot (1.4339 - 1.4339) \thinspace\mathrm{kJ} \cdot \mathrm{kg}^{-1}\mathrm{K}^{-1} + \dfrac{0 \thinspace kJ}{120^{\circ} \mathrm{C}} = 0 \thinspace \mathrm{kJ/K}\) Since the specific entropies are the same before and after the process and no heat transfer takes place, there is no entropy generation in this process. The results are: \((a)\) The amount of heat transfer is 0 kJ. \((b)\) The total entropy generation is 0 kJ/K.

Key concepts

First Law of Thermodynamics

The first law of thermodynamics is a fundamental principle that describes how energy is conserved in a system. It states that energy cannot be created or destroyed, only transferred or transformed from one form to another. In the context of our exercise, the first law is used to analyze the heat transfer during the process of liquid withdrawal from the tank.

When applying the first law to a closed system like our water tank, we take into account the internal energy changes within the system. The formula that represents this law is given by \[\delta Q = m \cdot (u_2 - u_1)\] where \(\delta Q\) is the heat transfer, \(m\) is the mass, and \(u_1\) and \(u_2\) are the specific internal energies at initial and final states, respectively. As the exercise reveals, if the temperature remains constant and there’s no phase change or chemical reaction, the internal energy remains the same, leading to no heat transfer according to the first law. This can be confusing for students who might expect heat to always be part of such processes, but understanding the relationship between temperature, internal energy, and phase can clarify this concept.

Entropy Generation

Entropy is a thermodynamic property that measures the degree of disorder or randomness in a system. The second law of thermodynamics introduces the concept of entropy generation, which represents the amount of entropy that is produced during a process.

In any real process, entropy tends to increase, reflecting the irreversible nature of physical occurrences. In a thermodynamic process, entropy generation is calculated using the formula \[\delta S_{gen} = m \cdot (s_2 - s_1) + \frac{Q}{T_0}\] where \(s_1\) and \(s_2\) are the initial and final specific entropy values, \(Q\) is the heat transfer, and \(T_0\) is the surrounding or reservoir temperature. For students, it is important to recognize that even when no heat transfer occurs, as stated by the exercise, entropy can still be generated due to other factors like mass transfer or chemical reactions. However, in this particular exercise, no entropy is generated as the system remains at a constant temperature with no change in heat or entropy content.

Saturated Liquid Thermodynamics

Saturated liquid thermodynamics is a subfield of thermodynamics dealing with liquids at the temperature at which they begin to vaporize, known as the saturation temperature. At this point, the liquid is in equilibrium with its own vapor, and any addition of energy will cause some amount of liquid to vaporize without increasing the temperature.

In our exercise scenario, the water inside the tank is at saturation temperature of 120°C which means it is ready to vaporize if provided with enough heat. The parameters used in saturated liquid calculations are found in tables known as saturated steam tables, where properties like specific volume \(v_f\), internal energy \(u_f\), and entropy \(s_f\) are available. Understanding how to read these tables and apply them properly is crucial for working with problems involving saturated liquids. This will help students clarify why certain calculations result in no heat transfer and, consequently, no entropy generation.