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Chapter 7: Problem 63

Air enters a compressor operating at steady state at \(T_{1}=\) \(320 \mathrm{~K}, p_{1}=2\) bar with a velocity of \(80 \mathrm{~m} / \mathrm{s}\). At the exit, \(T_{2}=550 \mathrm{~K}, p_{2}=6\) bar and the velocity is \(180 \mathrm{~m} / \mathrm{s}\). The air can be modeled as an ideal gas with \(c_{p}=1.01 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\). Stray heat transfer can be ignored. Determine, in kJ per \(\mathrm{kg}\) of air flowing, (a) the power required by the compressor and (b) the rate of exergy destruction within the compressor. Let \(T_{0}=\) \(300 \mathrm{~K}, p_{0}=1\) bar. Ignore the effects of motion and gravity.

Short Answer

Expert verified
The power required by the compressor is 232.3 kJ/kg. The rate of exergy destruction is 53.1 kJ/kg.

Step by step solution

01

- Apply the First Law of Thermodynamics for Steady-Flow Devices

For a compressor, the first law of thermodynamics in steady state is given by: -
02

- Simplify the Energy Equation by Ignoring Motion and Gravity

Since the problem states to ignore the effects of motion and gravity, we can simplify the energy equation as: Applying the conditions (constant cp, ideal gas): -
03

- Use the Given Cp Value to Find the Power Required by the Compressor

The power required by the compressor per unit mass of air is calculated as: The values provided: -
04

- Calculate The Change in Exergy (Irreversibility)

Exergy destruction is calculated by the change in specific exergy (availability function): For ideal gas: When ignoring kinetic potential energy differences: -
05

- Substitute the Given Values to Find Exergy Destruction

Substitute values: -

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Key Concepts

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

First Law of Thermodynamics
The First Law of Thermodynamics is a fundamental principle. It states that energy can neither be created nor destroyed. Instead, energy is conserved.
Specifically, it highlights that the change in energy of a system equals the heat added to the system minus the work done by the system.
For steady-state devices like compressors, this law translates to considering the energy balance between the inlet and outlet of the system.
In our exercise, the energy equation at steady state can be represented as: \(\frac{dE_{system}}{dt} = \frac{dE_{in}}{dt} - \frac{dE_{out}}{dt}\). This equation can be simplified to mostly involve enthalpy when ignoring motion and gravity.
steady-flow devices
Steady-flow devices are systems where matter continuously enters and exits. Consequently, these devices operate in a steady-state.
Compressors, turbines, and nozzles are common examples.
For these devices, it鈥檚 crucial to analyze the flow rates and energy changes.
Steady-state means that properties like mass flow rate, pressure, and temperature remain constant over time.
Compressors particularly elevate the pressure of incoming gases (like air).
In our exercise, we assume a steady-state operation of the compressor to apply the simplified energy equation.
exergy destruction
Exergy destruction refers to the loss of potential work due to irreversibilities.
It quantifies how much potential work is wasted during a process.
In thermodynamic processes, exergy destruction is often due to inefficiencies like friction and heat loss.
The rate of exergy destruction provides insight into system performance.
For our compressor, we calculate exergy destruction using the difference between the initial and final specific exergies.
This calculation involves the availability function: \(\text{Exergy} = (h_2 - h_1) - T_0 (s_2 - s_1)\)
ideal gas model
The ideal gas model is a simplified version of gas behavior.
This model assumes that the gas particles do not interact and occupy no volume.
The equation of state for an ideal gas is \(PV = nRT\), where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.
It鈥檚 useful for calculations under certain conditions.
In the exercise, air is modeled as an ideal gas to simplify computations.
We use the specific heat capacity (\(c_p\)) and apply it in our energy equation to find the power requirement.
specific heat capacity
Specific heat capacity (\(c_p\)) defines how much heat energy is needed to raise the temperature of a unit mass of a substance.
For gases, we use constant pressure specific heat capacity (\(c_p\)).
This is crucial for understanding energy changes in the gas as it passes through devices like compressors.
For our exercise, \(c_p = 1.01 \frac{kJ}{kg \bullet K}\) for air is given. \(Q = m \bullet c_p \bullet (T_2 - T_1)\) calculates the heat added.
This helps determine the power required to elevate the air temperature in the compressor.

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Most popular questions from this chapter

At steady state, hot gaseous products of combustion from a gas turbine cool from \(3000^{\circ} \mathrm{F}\) to \(250^{\circ} \mathrm{F}\) as they flow through a pipe. Owing to negligible fluid friction, the flow occurs at nearly constant pressure. Applying the ideal gas model with \(c_{p}=0.3 \mathrm{Btu} / \mathrm{lb} \cdot{ }^{\circ} \mathrm{R}\), determine the exergy transfer accompanying heat transfer from the gas, in Btu per lb of gas flowing. Let \(T_{0}=80^{\circ} \mathrm{F}\) and ignore the effects of motion and gravity.

Three pounds of carbon monoxide initially at \(180^{\circ} \mathrm{F}\) and \(40 \mathrm{lbf}\) in. \({ }^{2}\) undergo two processes in series: Process 1-2: Constant pressure to \(T_{2}=-10^{\circ} \mathrm{F}\) Process 2-3: Isothermal to \(p_{3}=10 \frac{\mathrm{lbf}}{\mathrm{in}^{2}}\) Employing the ideal gas model (a) represent each process on a \(p-v\) diagram and indicate the dead state. (b) determine the change in exergy for each process, in Btu. Let \(T_{0}=77^{\circ} \mathrm{F}, p_{0}=14.7 \mathrm{lbf} / \mathrm{in}^{2}\) and ignore the effects of motion and gravity.

Argon enters an insulated turbine operating at steady state at \(1000^{\circ} \mathrm{C}\) and \(2 \mathrm{MPa}\) and exhausts at \(350 \mathrm{kPa}\). The mass flow rate is \(0.5 \mathrm{~kg} / \mathrm{s}\) and the turbine develops power at the rate of \(120 \mathrm{~kW}\). Determine (a) the temperature of the argon at the turbine exit, in \({ }^{\circ} \mathrm{C}\). (b) the exergy destruction rate of the turbine, in \(k W\). (c) the turbine exergetic efficiency. Neglect kinetic and potential energy effects. Let \(T_{0}=20^{\circ} \mathrm{C}\), \(p_{0}=1\) bar.

Carbon monoxide \((\mathrm{CO})\) enters an insulated compressor operating at steady state at 10 bar, \(227^{\circ} \mathrm{C}\), and a mass flow rate of \(0.1 \mathrm{~kg} / \mathrm{s}\) and exits at 15 bar, \(327^{\circ} \mathrm{C}\). Determine the power required by the compressor and the rate of exergy destruction, each in \(\mathrm{kW}\). Ignore the effects of motion and gravity. Let \(T_{0}=17^{\circ} \mathrm{C}, p_{0}=1\) bar.

Water vapor at \(4.0 \mathrm{MPa}\) and \(400^{\circ} \mathrm{C}\) enters an insulated turbine operating at steady state and expands to saturated vapor at \(0.1 \mathrm{MPa}\). The effects of motion and gravity can be neglected. Determine the work developed and the exergy destruction, each in \(\mathrm{kJ}\) per \(\mathrm{kg}\) of water vapor passing through the turbine. Let \(T_{0}=27^{\circ} \mathrm{C}, p_{0}=0.1 \mathrm{MPa}\).
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