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Chapter 4: Problem 46

At steady state, a well-insulated compressor takes in air at \(15^{\circ} \mathrm{C}, 98 \mathrm{kPa}\), with a volumetric flow rate of \(0.57 \mathrm{~m}^{3} / \mathrm{s}\), and compresses it to \(260^{\circ} \mathrm{C}, 827 \mathrm{kPa}\). Kinetic and potential energy changes from inlet to exit can be neglected. Determine the compressor power, in \(\mathrm{kW}\), and the volumetric flow rate at the exit, in \(\mathrm{m}^{3} / \mathrm{s}\).

Short Answer

Expert verified
Compressor power is 167.55 kW, and exit volumetric flow rate is 0.1238 m^3/s.

Step by step solution

01

- Identify Known Values

List all the known values provided in the problem: - Inlet temperature, \(T_1 = 15^{\circ} \text{C} = 288 \text{K}\)- Inlet pressure, \(P_1 = 98 \text{kPa}\)- Inlet volumetric flow rate, \( \dot{V}_1 = 0.57 \text{m}^3/ \text{s}\)- Exit temperature, \(T_2 = 260^{\circ} \text{C} = 533 \text{K}\)- Exit pressure, \(P_2 = 827 \text{kPa}\)
02

- Apply Ideal Gas Law at Inlet and Exit

Use the ideal gas law \(P V = n R T\). For an ideal gas, the number of moles can be calculated as \(n = \frac{P V}{R T}\). This can be rewritten for mass flow rate and specific volume:\[ \dot{m} = \frac{P_1 \dot{V}_1}{R T_1} = \frac{98 \times 10^3 \times 0.57}{287 \times 288} \] Solve to find mass flow rate \(\dot{m}\).
03

- Calculate Mass Flow Rate

Determine mass flow rate, \(\dot{m}\), using the values:\[ \dot{m} = \frac{98 \text{kPa} \times 0.57 \text{m}^3/ \text{s}}{287 \text{J/(kg.K)} \times 288 \text{K}} \approx 0.677 \text{kg/s} \]
04

- Calculate Compressor Work Using Steady-State Energy Equation

The power required for the compressor can be determined using the steady-state energy equation for an adiabatic process: \[ W_{\text{comp,in}} = \dot{m}c_p \(T_2 - T_1\)\] where \(c_p\) is the specific heat at constant pressure, typically \(c_p \approx 1005 \text{J/(kg路K)}\).Substitute the known values:\[ W_{\text{comp,in}} = 0.677 \text{kg/s} \times 1005 \text{J/(kg路K)} \times (533 \text{K} - 288 \text{K})\] and convert \(W_{\text{comp,in}}\) to kW.
05

- Calculate Volumetric Flow Rate at the Exit

Use the ideal gas law again for exit conditions to find volumetric flow rate at the exit \( \dot{V}_2 \): \[ \dot{V}_2 = \frac{\dot{m}RT_2}{P_2} \]Substitute the known values:\[ \dot{V}_2 = \frac{0.677 \text{kg/s} \times 287 \text{J/(kg路K)} \times 533 \text{K}}{827 \text{kPa}}\] and solve for \( \dot{V}_2\).

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Key Concepts

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

steady state thermodynamics
In thermodynamics, steady-state refers to a condition where the system properties do not change over time. This means that the inflow and outflow of energy and matter are constant. For a compressor operating at steady-state, the amount of air entering and leaving the system is balanced, ensuring that temperature, pressure, and other properties remain stable over time.
This concept simplifies calculations since transient effects (changes over time) can be ignored.
Understanding steady-state allows us to use energy balance equations effectively, such as the steady-state energy equation, which is crucial for calculating compressor work.
ideal gas law
The Ideal Gas Law is a fundamental equation in thermodynamics that describes the behavior of ideal gases. It is expressed as: \[ PV = nRT \] Where:
  • P is the pressure
  • V is the volume
  • n is the number of moles of gas
  • R is the universal gas constant (approximately 8.314 J/(mol路K))
  • T is the temperature in Kelvin

To adapt this to flows in the compressor problem, we use mass flow rate and specific volume, focusing on how pressure and temperature changes affect volume and density.
The ideal gas law helps us relate the volumetric flow rates at the inlet and outlet, critical for determining the mass and behavior of the gas under compression.
mass flow rate calculation
The mass flow rate (\dot{m}) is a measure of the amount of mass passing through a given system per unit of time. It is a crucial parameter in many thermodynamic processes, especially in compressors. It can be calculated using the ideal gas law at the inlet as: \[ \dot{m} = \frac{P_1 \dot{V}_1}{R T_1} \]
Here, \dot{V}_1 is the volumetric flow rate at the inlet, T_1 is the temperature at the inlet, and P_1 is the pressure at the inlet.
The mass flow rate gives insight into how much air is being compressed, enabling us to subsequently calculate the required compressor power. It's a fundamental step for understanding the system's energy balance.
adiabatic process
An adiabatic process is a type of thermodynamic process where no heat is transferred to or from the system. This means that all changes in the system's internal energy are due to work done by or on the system. For a compressor with no heat exchange due to insulation, we assume an adiabatic process.
The steady-state energy equation for an adiabatic compressor becomes: \[ W_{\text{comp,in}} = \dot{m} c_p (T_2 - T_1) \]
Here, T_1 and T_2 are the temperatures at the inlet and exit, c_p is the specific heat at constant pressure, and W_{\text{comp,in}} is the compressor work or power.
This equation allows us to calculate the work input necessary for the compression process, considering just the temperature change and mass flow rate.

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Most popular questions from this chapter

Liquid water at \(20^{\circ} \mathrm{C}\) enters a pump though an inlet pipe having a diameter of \(160 \mathrm{~mm}\). The pump operates at steady state and supplies water to two exit pipes having diameters of \(80 \mathrm{~mm}\) and \(120 \mathrm{~mm}\), respectively. The velocity of the water exiting the \(80 \mathrm{~mm}\) pipe is \(0.5 \mathrm{~m} / \mathrm{s}\). At the exit of the \(120 \mathrm{~mm}\) pipe the velocity is \(0.1 \mathrm{~m} / \mathrm{s}\). The temperature of the water in each exit pipe is \(22^{\circ} \mathrm{C}\). Determine (a) the mass flow rate, in \(\mathrm{kg} / \mathrm{s}\), in the inlet pipe and each of the exit pipes, and (b) the volumetric flow rate at the inlet, in \(\mathrm{m}^{3} / \mathrm{s}\).

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