Question

Air enters the compressor of a gas-turbine plant at ambient conditions of \(100 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\) with a low velocity and exits at \(1 \mathrm{MPa}\) and \(347^{\circ} \mathrm{C}\) with a velocity of \(90 \mathrm{m} / \mathrm{s}\). The compressor is cooled at a rate of \(1500 \mathrm{kJ} / \mathrm{min}\), and the power input to the compressor is \(250 \mathrm{kW}\). Determine the mass flow rate of air through the compressor.

Short answer

Answer: The mass flow rate of air through the compressor is approximately \(0.0735 \mathrm{kg/s}\).

Step-by-step solution

List known and unknown variables

We are given the following information: - Inlet pressure: \(P_1 = 100 \mathrm{kPa}\) - Inlet temperature: \(T_1 = 25^{\circ} \mathrm{C}\) - Outlet pressure: \(P_2 = 1 \mathrm{MPa}\) - Outlet temperature: \(T_2 = 347^{\circ} \mathrm{C}\) - Outlet velocity: \(v_2 = 90 \mathrm{m/s}\) - Compressor cooling rate: \(\dot{Q} = -1500 \mathrm{kJ/min}\) - Power input to the compressor: \(\dot{W} = 250 \mathrm{kW}\) Our goal is to find the mass flow rate (\(\dot{m}\)) of air through the compressor.

Find the change in specific kinetic energy

Since the inlet velocity is said to be low, we can assume \(v_1 \approx 0 \mathrm{m/s}\). The change in specific kinetic energy in the compressor can be calculated as follows: \(\Delta \mathrm{KE} = \frac{v_2^2}{2} - \frac{v_1^2}{2} = \frac{v_2^2}{2}\) Substitute the given outlet velocity: \(\Delta \mathrm{KE} = \frac{90^2}{2} = 4050 \mathrm{J/kg}\)

Use the energy equation to solve for the mass flow rate

We will apply the energy equation (conservation of energy) for a steady-flow system: \(\dot{m}(\Delta h + \Delta \mathrm{KE})=-\dot{Q}+\dot{W}\) In this case, the change in specific enthalpy can be represented as follows: \(\Delta h = C_p (T_2 - T_1)\) where \(C_p\) is the specific heat at constant pressure for air, approximately \(1005 \mathrm{J/kgK}\). Now, substitute the given values and solve for \(\dot{m}\): \(\dot{m}(1005 (347 - 25) + 4050) = -(-1500 \cdot 10^3 / 60) + 250 \cdot 10^3\) \(\dot{m}(1005(322) + 4050) = 24000\) \(\dot{m} = \frac{24000}{1005 \cdot 322 + 4050}\) \(\dot{m} = 0.0735 \mathrm{kg/s}\)

Final Answer

The mass flow rate of air through the compressor is approximately \(0.0735 \mathrm{kg/s}\).

Key concepts

Air Compressor

An air compressor is a device that converts power into potential energy stored in pressurized air. By increasing the pressure of a gas, typically air, the compressor serves various applications in industrial, commercial, and personal settings. It works on the principle of forcing air into a chamber, where the volume is decreased to compress the air. \(
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For our exercise, we looked at a gas-turbine air compressor that intakes air at ambient conditions and discharges it at a significantly higher pressure and temperature. The compressor is powered by external work, quantified as power input, and there's also a mention of the cooling rate, which is a form of heat exchange often necessary to dissipate the heat generated during compression, ensuring the system's efficiency. Understanding the operation of an air compressor is essential as it directly influences the mass flow rate calculation. \(
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In the context of our problem, the parameters such as inlet and outlet pressures, inlet and outlet temperatures, and power input are essential pieces of data to determine the mass flow rate. The mass flow rate is crucial as it indicates the amount of air processed by the compressor in a given period, thus defining its performance.

Specific Kinetic Energy

Specific kinetic energy refers to the kinetic energy per unit mass. In the context of fluid flow, it is expressed as \(\frac{v^2}{2}\), where \(v\) is the velocity of the fluid. \(
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Given that kinetic energy depends on the square of velocity, even small increases in fluid velocity can lead to significant changes in kinetic energy. While solving for our mass flow rate, the change in specific kinetic energy from the inlet to the outlet of the compressor is a critical factor in the energy balance equation, especially since the velocities at these points differ. \(
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In the provided exercise, we assumed the inlet velocity to be negligible compared to the outlet velocity. Thus, the focus was placed on the kinetic energy at the outlet, derived from the outlet velocity, which was squared and divided by two to find the change in specific kinetic energy.

Steady-Flow Energy Equation

The steady-flow energy equation is a form of the First Law of Thermodynamics, applied to systems where fluids flow through a control volume stably over time. This equation takes into account energy transfers like work and heat, along with changes in kinetic and potential energies. \(
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It's expressed as \(\dot{m}(\Delta h + \Delta KE + \Delta PE) = \dot{Q} - \dot{W}\), where \(\dot{m}\) is the mass flow rate, \(\Delta h\) is the change in specific enthalpy, \(\Delta KE\) and \(\Delta PE\) are changes in specific kinetic and potential energy, \(\dot{Q}\) is the heat transfer rate, and \(\dot{W}\) is the work transfer rate. For most air compressor problems, unless specified, the change in potential energy is often neglected due to small elevation differences. \(
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In our exercise, this equation is used by equating the energy added to the system through power input and subtracting the energy lost through cooling to the changes in enthalpy and kinetic energy. By rearranging the equation, we solved for the mass flow rate.

Specific Enthalpy Change

Specific enthalpy change, denoted as \(\Delta h\), is the difference in enthalpy per unit mass between two points in a system. Enthalpy itself is a thermodynamic property that represents the total heat content of a system, comprising internal energy plus the energy associated with pressure and volume. \(
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In the context of compressors, it's critical to estimate the change in enthalpy across the compressor to understand the energy transformation involved. Specific enthalpy change is instrumental in calculating the mass flow rate through the steady-flow energy equation and is dependent on the specific heat capacity at constant pressure (\(C_p\)) and temperature difference between the outlet and inlet. \(
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In our calculation, recognizing the value of \(C_p\) for air is vital since this specific heat value allows us to translate temperature differences into enthalpy changes, which, combined with kinetic energy changes, gives us the energy balance needed to find mass flow rate.

Conservation of Energy in Thermodynamics

The principle of conservation of energy in thermodynamics states that energy cannot be created or destroyed, only transformed from one form to another. This is the core of the First Law of Thermodynamics, which applies to all thermodynamic processes. \(
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In the context of our exercise, the application of this principle means that the work input and heat extracted from the air compressor must be accounted for as changes in the internal energy of the air – reflected as changes in specific enthalpy and specific kinetic energy. It is this foundational concept that underpins the steady-flow energy equation used to solve for the mass flow rate of air through the compressor. It ensures that all energy entering and leaving the control volume of the compressor is accounted for, enabling us to deduce the performance and efficiency of the compressor as well as to calculate the mass flow rate, which is a measure of the effectiveness of the compressor in processing the air. \(
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Recognizing that all energy in the system is conserved, it must all be accounted for, which gives us a methodical approach to solving complex problems involving energy transformations in systems like air compressors.