Question

At steady state, work at a rate of \(25 \mathrm{~kW}\) is done by a paddle wheel on a slurry contained within a closed, rigid tank. Heat transfer from the tank occurs at a temperature of \(250^{\circ} \mathrm{C}\) to surroundings that, away from the immediate vicinity of the tank, are at \(27^{\circ} \mathrm{C}\). Determine the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\), (a) for the tank and its contents as the system. (b) for an enlarged system including the tank and enough of the nearby surroundings for the heat transfer to occur at \(27^{\circ} \mathrm{C}\).

Short answer

For the tank and contents: 0.0478 kW/K. For the enlarged system: -0.0355 kW/K.

Step-by-step solution

- Understanding the Problem

Identify the given data and define the system under consideration. The work done by the paddle wheel is at a rate of 25 kW. Heat transfer occurs at 250°C (523 K) to surroundings at 27°C (300 K).

- System and Surroundings

Determine the rate of heat transfer, denoted as \ Q. Invoke the first law of thermodynamics for a closed system: \ \[ \frac{dU}{dt} = \frac{dQ}{dt} - \frac{dW}{dt} \] \In steady state, \ \(Q - W = 0\), thus, the rate of heat transfer is equal to the work done, \ \( \frac{dQ}{dt} = 25 \text{kW} \).

- Entropy Production for Tank and Content

Apply the second law of thermodynamics for entropy production rate \( \frac{dS_{gen}}{dt} \) for the tank and its contents: \ \( \frac{dS_{dt}}{dt} = \frac{dQ}{dt} \times \frac{1}{T} + \frac{dS_{gen}}{dt}\) \ Where \( T \) is the temperature at which the heat transfer occurs (523 K). Rearrange the equation to solve for entropy production rate: \ \[ \frac{dS_{gen}}{dt} = \frac{dW}{dt} \times \frac{1}{T} \] \ Substitute \( \frac{dW}{dt} = 25 \text{kW} \) and \( T = 523 \text{K} \) to get \[ \frac{dS_{gen}}{dt} = 25 \text{kW} \times \frac{1}{523 \text{K}} = 0.0478 \text{kW/K} \]

- Enlarged System

Consider the enlarged system involving the surroundings at 27°C (300 K). According to the second law of thermodynamics applied to this system: \ \[ \frac{dS_{dt}}{dt} = \frac{dQ}{dt} \times \bigg( \frac{1}{T_{sys}} - \frac{1}{T_{surr}} \bigg) + \frac{dS_{gen}}{dt} \] \Here, \( T_{sys} = 523 \ \text{K} \) and \( T_{surr} = 300 \ \text{K}\).Solve for the rate of entropy production: \[ \frac{dS_{gen}}{dt} = 25 \text{kW} \times \bigg( \frac{1}{523 \text{K}} - \frac{1}{300 \text{K}} \bigg) \] \Plug in the values: \ \[ \frac{dS_{gen}}{dt} = 25 \text{kW} \times \bigg(0.00191 - 0.00333 \bigg) \] \ \[ \frac{dS_{gen}}{dt} = 25 \text{kW} \times -0.00142 \] \ \[ \frac{dS_{gen}}{dt} = -0.0355 \text{kW/K} \]

Key concepts

thermodynamic systems

A thermodynamic system is a portion of the physical universe chosen for analysis. Everything outside the system is known as the surroundings. In our exercise, the system can be either the tank containing the slurry or an enlarged system including some of the surroundings. We analyze how energy in the form of work and heat interacts with our chosen system. Understanding this interaction helps us determine entropy production, which measures the irreversibility of processes within the system. By defining a clear system boundary, we can apply the laws of thermodynamics to find solutions.

first law of thermodynamics

The first law of thermodynamics is essentially the law of energy conservation. It states that energy cannot be created or destroyed, only transferred or converted from one form to another. For a closed system, like our tank, this law is mathematically expressed as:
\( \frac{dU}{dt} = \frac{dQ}{dt} - \frac{dW}{dt} \)
Here, \( \frac{dU}{dt} \) is the change in internal energy, \( \frac{dQ}{dt} \) is the rate of heat transfer into the system, and \( \frac{dW}{dt} \) is the rate of work done by the system. For steady-state conditions, where properties remain constant over time, \( \frac{dU}{dt} = 0 \), simplifying this to:
\( \frac{dQ}{dt} = \frac{dW}{dt} \)
This means the heat transferred to the surroundings equals the work done by the paddle wheel in our tank.

second law of thermodynamics

The second law of thermodynamics introduces the concept of entropy, a measure of disorder or randomness in a system. Entropy production quantifies the irreversibility of real processes. This law can be expressed as:
\( \frac{dS_{gen}}{dt} = \frac{dQ}{dt} \times \bigg( \frac{1}{ T } \bigg) + \frac{dS_{gen}}{dt}\)
For our tank system, where heat is transferred at a constant temperature of 250°C (523 K), the rate of entropy production is calculated by considering the work done by the paddle wheel (25 kW). Using the rearranged form of the second law and substituting the values gives us:
\( \frac{dS_{gen}}{dt} = 25 \text{kW} \times \frac{1}{523 \text{K}} = 0.0478\text{kW/K} \)
This highlights the irreversibilities within the system.

steady-state conditions

Steady-state conditions imply that the properties of the system do not change with time. In our problem, we assume steady-state since the rate of work (25 kW) and temperature (523 K) are constant. For steady-state, the internal energy change is zero: \( \frac{dU}{dt} = 0 \). Therefore, in our closed tank system, the energy transferred as heat equals the work done by the paddle wheel. When expanding our system to include the surroundings at 27°C (300 K), steady-state helps simplify the thermodynamic equations, ensuring consistent temperature values and aiding in entropy production calculations for a comprehensive understanding.