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Chapter 4: Problem 49

Air is compressed at steady state from \(1 \mathrm{bar}, 300 \mathrm{~K}\), to 6 bar with a mass flow rate of \(4 \mathrm{~kg} / \mathrm{s}\). Each unit of mass passing from inlet to exit undergoes a process described by \(p v^{1.27}=\) constant. Heat transfer occurs at a rate of \(46.95 \mathrm{~kJ}\) per \(\mathrm{kg}\) of air flowing to cooling water circulating in a water jacket enclosing the compressor. If kinetic and potential energy changes of the air from inlet to exit are negligible, calculate the compressor power, in \(\mathrm{kW}\).

Short Answer

Expert verified
The compressor power is approximately 366.5 kW.

Step by step solution

01

- Understand the Given Data

Identify and write down the data:- Initial pressure, \(p_1 = 1 \mathrm{bar}\)- Initial temperature, \(T_1 = 300 \mathrm{K}\)- Final pressure, \(p_2 = 6 \mathrm{bar}\)- Mass flow rate, \(\dot{m} = 4 \mathrm{kg / s}\)- Heat transfer rate, \(\dot{Q} = 46.95 \mathrm{kJ/kg}\)- Process equation: \(p v^{1.27} = \text{constant}\)
02

- Calculate Work Done per Unit Mass

Use the polytropic process formula to find the work done per unit mass.For a polytropic process:\[ W = \frac{p_2 v_2 - p_1 v_1}{1 - n} \]Given that \(n = 1.27\) and using the ideal gas law \( v = \frac{RT}{p} \), we need to find the specific volumes \(v_1\) and \(v_2\) at the given states.
03

- Apply Ideal Gas Law to Find Specific Volumes

Using the ideal gas law at state 1:\[v_1 = \frac{RT_1}{p_1}\]\(R\) for air is approximately \(287 \mathrm{J/(kg \, K)}\). Plug in the values:\[v_1 = \frac{287 \times 300}{100000} = 0.861 \frac{\mathrm{m^3}}{\mathrm{kg}}\]Using the process relationship \(p v^{1.27} = \text{constant}\), find \(v_2\):\[v_2 = v_1 \left(\frac{p_1}{p_2}\right)^{1/1.27} = 0.861 \left(\frac{1}{6}\right)^{1/1.27} = 0.330 \frac{\mathrm{m^3}}{\mathrm{kg}}\]
04

- Calculate Work Done per Unit Mass

Insert the specific volumes into the work formula:\[ W = \frac{p_2 v_2 - p_1 v_1}{1 - 1.27} = \frac{6 \times 10^5 \times 0.330 - 1 \times 10^5 \times 0.861}{1 - 1.27} \]Simplify the expression to find the work done per kg.
05

- Calculate Compressor Power

Use the mass flow rate to determine the total power.\[ \dot{W} = \dot{m} \left( W + \dot{Q} \right) \]Insert the mass flow rate and the work done per kg calculated earlier along with the heat transfer rate.
06

- Find Final Compressor Power in kW

Combine everything together:\[ \dot{W} = 4 \left( \text{Work per unit mass} + 46.95 \right) \]Calculate the final result for the compressor power in \(\mathrm{kW}\).

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Key Concepts

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

Polytropic Process
A polytropic process is described by the equation: \(p v^n = \text{constant}\). Here, \(p\) is the pressure, \(v\) is the specific volume, and \(n\) is the polytropic index. In our exercise, we have \(n = 1.27\), indicating a process where both temperature and heat transfer affect the gas behavior. To find the work done during such a process, we use the formula: \[ W = \frac{p_2 v_2 - p_1 v_1}{1 - n} \] This involves calculating the specific volumes at the initial and final states, and then substituting these values into the equation.
Ideal Gas Law
The ideal gas law is crucial in solving problems involving gasses. It is given by: \(\ pV = nRT\), where \(p\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is the temperature. For specific calculations, we often transform it into: \(v = \frac{RT}{p}\), where \(v\) is the specific volume (volume per unit mass). By using this formula, we calculate the specific volumes at different states in the process. In the exercise, we used: \[v_1 = \frac{287 \times 300}{100000} = 0.861 \ \text{m^3/kg}\]
Specific Volume
Specific volume is a core concept when dealing with gasses. It is the volume occupied by a unit mass of the gas, given in units of \(\text{m^3/kg}\). In our example, we determine the specific volume at the initial and final states using the ideal gas law for the initial state and the polytropic relationship for the final state. The calculations are as follows:
  • Initial specific volume, \(v_1 = 0.861 \ \text{m^3/kg}\)
  • Final specific volume, \(v_2 = v_1 \left( \frac{p_1}{p_2} \right)^{1/n} = 0.330 \ \text{m^3/kg}\)
Understanding these volumes is essential for computing the work done.
Steady-State Compression
Steady-state compression implies that the process is continuous and without fluctuations in the conditions over time. This assumption simplifies the analysis since we can ignore transients and focus on the inlet and exit conditions under equilibrium. In this scenario, we assume mass flow rate (rate at which mass enters and exits) is constant and given by \(\dot{m} = 4 \ \text{kg/s}\). This indicates that four kilograms of air per second is undergoing compression. Ignoring potential and kinetic energy changes ensures our focus remains on pressure, volume, and temperature changes.
Heat Transfer Calculation
Heat transfer in thermodynamic processes is the energy exchange due to temperature differences. In this exercise, the heat transfer rate (\(\dot{Q} = 46.95 \ \text{kJ/kg}\)) represents energy removed by the cooling water. When calculating compressor power, the heat removed needs to be added to the work done because it influences the total energy balance. The total power equation is:\(\dot{W} = \dot{m} \left( W + \dot{Q} \right)\)This equation combines both the work done on the gas and the heat transfer rate. It is essential for evaluating the compressor's energy requirements accurately.

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Most popular questions from this chapter

Steam enters a well-insulated turbine operating at steady state at \(5 \mathrm{MPa}\) with a specific enthalpy of \(3000 \mathrm{~kJ} / \mathrm{kg}\) and a velocity of \(9 \mathrm{~m} / \mathrm{s}\). The steam expands to the turbine exit where the pressure is \(0.1 \mathrm{MPa}\), specific enthalpy is \(2540 \mathrm{~kJ} / \mathrm{kg}\), and the velocity is \(100 \mathrm{~m} / \mathrm{s}\). The mass flow rate is \(12 \mathrm{~kg} / \mathrm{s}\). Neglecting potential energy effects, determine the power developed by the turbine, in \(\mathrm{kW}\).

Refrigerant \(134 \mathrm{a}\) enters a compressor operating at steady state as saturated vapor at \(0.14 \mathrm{MPa}\) and exits at \(1.4 \mathrm{MPa}\) and \(80^{\circ} \mathrm{C}\) at a mass flow rate of \(0.106 \mathrm{~kg} / \mathrm{s}\). As the refrigerant passes through the compressor, heat transfer to the surroundings occurs at a rate of \(-0.40 \mathrm{~kJ} / \mathrm{s}\). Determine at steady state the power input to the compressor, in \(\mathrm{kW}\).

A large pipe carries steam as a two-phase liquid-vapor mixture at \(1.0 \mathrm{MPa}\). A small quantity is withdrawn through a throttling calorimeter, where it undergoes a throttling process to an exit pressure of \(0.1 \mathrm{MPa}\). For what range of exit temperatures, in \({ }^{\circ} \mathrm{C}\), can the calorimeter be used to determine the quality of the steam in the pipe? What is the corresponding range of steam quality values?

Carbon dioxide gas is heated as it flows steadily through a 2.5-cm-diameter pipe. At the inlet, the pressure is 2 bar, the temperature is \(300 \mathrm{~K}\), and the velocity is \(100 \mathrm{~m} / \mathrm{s}\). At the exit, the pressure and velocity are \(0.9413\) bar and \(400 \mathrm{~m} / \mathrm{s}\), respectively. The gas can be treated as an ideal gas with constant specific heat \(c_{p}=0.94 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\). Neglecting potential energy effects, determine the rate of heat transfer to the carbon dioxide, in \(\mathrm{kW}\).

A rigid tank of volume \(0.75 \mathrm{~m}^{3}\) is initially evacuated. A hole develops in the wall, and air from the surroundings at 1 bar, \(25^{\circ} \mathrm{C}\) flows in until the pressure in the tank reaches 1 bar. Heat transfer between the contents of the tank and the surroundings is negligible. Determine the final temperature in the tank, in \({ }^{\circ} \mathrm{C}\).
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