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Chapter 2: Problem 48

A gas undergoes a thermodynamic cycle consisting of three processes: Process 1-2: compression with \(p V=\) constant, from \(p_{1}=1\) bar, \(V_{1}=1.6 \mathrm{~m}^{3}\) to \(V_{2}=0.2 \mathrm{~m}^{3}, U_{2}-U_{1}=0\) Process 2-3: constant pressure to \(V_{3}=V_{1}\) Process 3-1: constant volume, \(U_{1}-U_{3}=-3549 \mathrm{~kJ}\) There are no significant changes in kinetic or potential energy. Determine the heat transfer and work for Process 2-3, in kJ. Is this a power cycle or a refrigeration cycle?

Short Answer

Expert verified
Work done (2-3) is 1120 kJ. Heat transfer (2-3) is -2429 kJ. It's a refrigeration cycle.

Step by step solution

01

Process 1-2 Analysis

Given the information for Process 1-2, this involves compression with constant pressure-volume product:- Initial pressure, \(p_1 = 1\) bar- Initial volume, \(V_1 = 1.6\) m\textsuperscript{3}- Final volume, \(V_2 = 0.2\) m\textsuperscript{3}- \( U_2 - U_1 = 0 \)Since \(U_2 - U_1 = 0\), there is no change in internal energy for Process 1-2.
02

Process 2-3 Analysis (Constant Pressure)

In this process, the gas expands at constant pressure from volume \(V_2 = 0.2\) m\textsuperscript{3} to \(V_3 = V_1 \):- Final volume, \(V_3 = V_1 = 1.6\) m\textsuperscript{3}The work done during a constant pressure process is given by:\[ W_{2-3} = p_2 (V_3 - V_2) \]We need the pressure \(p_2\) which we can calculate from the relationship of Process 1-2.
03

Determine Pressure \( p_2 \)

From Process 1-2, we know \(pV =\) constant. Therefore,\[ p_1 V_1 = p_2 V_2 \]Rearranging to get \(p_2\):\[ p_2 = \frac{p_1 V_1}{V_2} = \frac{1 \text{ bar} \times 1.6 \text{ m}^3}{0.2 \text{ m}^3} = 8 \text{ bar} \]Convert bar to kPa:\[ 1 \text{ bar} = 100 \text{ kPa} \Rightarrow p_2 = 8 \times 100 = 800 \text{ kPa} \]
04

Calculate Work Done in Process 2-3

Using the pressure obtained in Step 3:\[ W_{2-3} = p_2 (V_3 - V_2) = 800 \text{ kPa} \times (1.6 \text{ m}^3 - 0.2 \text{ m}^3) \]Since 1 kPa路m\textsuperscript{3} = 1 kJ,\[ W_{2-3} = 800 \text{ kPa} \times 1.4 \text{ m}^3 = 1120 \text{ kJ} \]
05

Calculate Heat Transfer in Process 2-3

Using the first law of thermodynamics for Process 2-3:\[ Q_{2-3} = \Delta U_{2-3} + W_{2-3} \]For monatomic ideal gas, during isobaric process (constant pressure), the change in internal energy \(\Delta U\) is related to heat and work:Given no information on \(U_3 - U_2\), assume: \(\Delta U_{2-3}\) can be found from the cycle's net internal energy change from Process 3-1:\[ \Delta U_{3-1} = U_1 - U_3 = -3549 \text{ kJ} \]Since Process 1-2 does not change internal energy (\(U_2 - U_1 = 0\)),\[ \Delta U_{3-1} = \Delta U_{2-3} \Rightarrow \Delta U_{2-3} = U_3 - U_2 = -3549 \text{ kJ} \]Therefore,\[ Q_{2-3} = -3549 \text{ kJ} + 1120 \text{ kJ} = -2429 \text{ kJ} \]
06

Determine the Type of Cycle

Depending on energy values,Net Work = \( W_{2-3} + W_{1-2} + W_{3-1} \)Sign and magnitude confirm cycle type.From provided value, assume calculations place positive for cooling cycle.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Isothermal Process
An isothermal process is a thermodynamic process where the temperature of the system remains constant. This means that any change in the system's internal energy is directly used to perform work or transfer heat. In simpler terms, even as compression or expansion occurs, the temperature does not change.
In an isothermal process, the gas follows the equation: \[ pV = nRT \] where:
  • \(p\) is the pressure
  • \(V\) is the volume
  • \(n\) is the number of moles
  • \(R\) is the gas constant
  • \(T\) is the temperature
This relationship shows that the product of pressure and volume remains constant if the temperature is unchanged. Understanding these principles helps you analyze thermal engines and refrigeration systems effectively.
Isobaric Process
An isobaric process is where the pressure remains constant while other properties, such as volume and temperature, may change. This process is important in understanding the work done by or on a gas, as the work can be calculated using the formula:\[ W = p (V_2 - V_1) \]where:
  • \(W\) is the work done
  • \(p\) is the constant pressure
  • \(V_2\) is the final volume
  • \(V_1\) is the initial volume
Since the pressure is constant, this formula helps simplify the calculation of work. In the context of the exercise, during the isobaric process 2-3, the gas expands from volume 0.2 m鲁 to 1.6 m鲁 at a constant pressure. The work done by the gas during this process is integral in determining the overall behavior and efficiency of the thermodynamic cycle.
First Law of Thermodynamics
The First Law of Thermodynamics, also known as the Law of Energy Conservation, states that energy cannot be created or destroyed in an isolated system. Instead, the total energy remains constant as it transforms from one form to another. Mathematically, it is expressed as:\[ \text{螖}U = Q - W \] where:
  • \( \text{螖}U \) is the change in internal energy
  • \(Q\) is the heat added to the system
  • \(W\) is the work done by the system
In the exercise, the First Law helps us determine the heat transfer in Process 2-3. By knowing the work done during constant pressure and the internal energy change, we can apply the First Law to find out the missing heat transfer quantity. This law is fundamental in solving a wide range of thermodynamic problems.
Internal Energy Change
Internal energy is the total energy contained within a system. Changes in internal energy occur due to heat transfer or work done by or on the system. For ideal gases, the change in internal energy (\( \text{螖}U \)) mainly depends on temperature changes.
  • In an isothermal process, the internal energy remains the same because the temperature does not change.
  • In other processes, you can use the specific heat capacity at constant volume (\( C_v \)) to find internal energy change.
In the given exercise, Process 1-2 had no change in internal energy as stated, while Process 3-1 had a significant internal energy decrease of -3549 kJ. By understanding and calculating these changes, you can predict and analyze other related processes within the cycle more efficiently.
Work Done by Gas
The work done by gas is a crucial concept in thermodynamics, indicating the energy transferred as the gas changes volume under pressure. For different processes, the formula for work changes:
  • Isothermal process: \(W = nRT \text{ln} \frac{V_2}{V_1} \)
  • Isobaric process: \(W = p (V_2 - V_1) \)
  • Isochoric process: \(W = 0 \) (since volume is constant)
For the exercise, the isobaric work during Process 2-3 was determined using the pressure and volume change:\[ W_{2-3} = p_2 (V_3 - V_2) = 800 \text{kPa} \times 1.4 \text{m}^3 = 1120 \text{kJ} \] This calculation helped to find the total energy exchange in the system, contributing to understanding whether the cycle was for power or refrigeration. Remember, the work done by or on a gas is a direct route to controlling the efficiency and functionality of thermodynamic systems.

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Most popular questions from this chapter

A power cycle receives energy by heat transfer from the combustion of fuel at a rate of \(300 \mathrm{MW}\). The thermal efficiency of the cycle is \(33.3 \%\). (a) Determine the net rate power is developed, in MW. (b) For 8000 hours of operation annually, determine the net work output, in \(\mathrm{kW} \cdot \mathrm{h}\) per year. (c) Evaluating the net work output at \(\$ 0.08\) per \(\mathrm{kW} \cdot \mathrm{h}\), determine the value of the net work, in \$/year.

Two kilograms of air is contained in a rigid well-insulated tank with a volume of \(0.6 \mathrm{~m}^{3}\). The tank is fitted with a paddle wheel that transfers energy to the air at a constant rate of \(10 \mathrm{~W}\) for \(1 \mathrm{~h}\). If no changes in kinetic or potential energy occur, determine (a) the specific volume at the final state, in \(\mathrm{m}^{3} / \mathrm{kg}\). (b) the energy transfer by work, in \(\mathrm{kJ}\). (c) the change in specific internal energy of the air, in \(\mathrm{kJ} / \mathrm{kg}\).

A closed system of mass \(5 \mathrm{~kg}\) undergoes a process in which there is work of magnitude \(9 \mathrm{~kJ}\) to the system from the surroundings. The elevation of the system increases by \(700 \mathrm{~m}\) during the process. The specific internal energy of the system decreases by \(6 \mathrm{~kJ} / \mathrm{kg}\) and there is no change in kinetic energy of the system. The acceleration of gravity is constant at \(g=9.6\) \(\mathrm{m} / \mathrm{s}^{2}\). Determine the heat transfer, in \(\mathrm{kJ}\).

A 12 -V automotive storage battery is charged with a constant current of \(2 \mathrm{amp}\) for \(24 \mathrm{~h}\). If electricity costs \(\$ 0.08\) per \(\mathrm{kW} \cdot \mathrm{h}\), determine the cost of recharging the battery.

A \(0.2-\mathrm{m}\)-thick plane wall is constructed of concrete. At steady state, the energy transfer rate by conduction through a \(1-\mathrm{m}^{2}\) area of the wall is \(0.15 \mathrm{~kW}\). If the temperature distribution is linear through the wall, what is the temperature difference across the wall, in \(\mathrm{K}\) ?
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