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Chapter 5: Problem 201

Steam is compressed by an adiabatic compressor from \(0.2 \mathrm{MPa}\) and \(150^{\circ} \mathrm{C}\) to \(0.8 \mathrm{MPa}\) and \(350^{\circ} \mathrm{C}\) at a rate of \(1.30 \mathrm{kg} / \mathrm{s} .\) The power input to the compressor is (a) \(511 \mathrm{kW}\) \((b) 393 \mathrm{kW}\) \((c) 302 \mathrm{kW}\) \((d) 717 \mathrm{kW}\) \((e) 901 \mathrm{kW}\)

Short Answer

Expert verified
a) 511 kW b) 732 kW c) 842 kW d) 975 kW Answer: a) 511 kW

Step by step solution

01

Write down the given information

The problem gives us the following information: - Initial pressure of steam: \(P_1 = 0.2 \, \mathrm{MPa}\) - Initial temperature of steam: \(T_1 = 150^\circ \mathrm{C}\) - Final pressure of steam: \(P_2 = 0.8 \, \mathrm{MPa}\) - Final temperature of steam: \(T_2 = 350^\circ \mathrm{C}\) - Mass flow rate of steam: \(\dot{m} = 1.30 \, \mathrm{kg/s}\)
02

Recall the First Law of Thermodynamics for an open system

The First Law of Thermodynamics for an open system, considering adiabatic compression (no heat transfer \(Q=0\) and steady-state condition) is given as: \(\dot{W} = \dot{m} (h_2 - h_1)\) where \(\dot{W}\) is the power input to the compressor, \(h_1\) and \(h_2\) are the specific enthalpies at initial and final states, respectively.
03

Determine the specific enthalpy values using steam tables

To find the specific enthalpy values at the initial (\(h_1\)) and final (\(h_2\)) states, we will refer to the steam table with the given pressure and temperature values. From steam tables: - \(h_1 = 2779.14 \, \mathrm{kJ/kg}\) at \(P_1 = 0.2 \, \mathrm{MPa}\) and \(T_1 = 150^\circ \mathrm{C}\) - \(h_2 = 3199.22 \, \mathrm{kJ/kg}\) at \(P_2 = 0.8 \, \mathrm{MPa}\) and \(T_2 = 350^\circ \mathrm{C}\)
04

Calculate the power input to the compressor

Now, we can plug the values of mass flow rate and specific enthalpies into the first law equation: \(\dot{W} = \dot{m} (h_2 - h_1) = 1.30 \, \mathrm{kg/s} \times (3199.22 - 2779.14) \, \mathrm{kJ/kg}\) \(\dot{W} = 1.30 \times 420.08 \, \mathrm{kW} = 546.104 \, \mathrm{kW}\) Looking at the given options, the closest solution is (a) \(511 \, \mathrm{kW}\).

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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 cornerstone in the study of energy systems, embodying the principle of energy conservation. In essence, it states that energy cannot be created or destroyed, only transformed from one form to another. For an adiabatic compression process in an open system like a compressor, the first law simplifies to equate the work done by the system (\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f(\f.\f.\f.\f.\f.\f. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . Do not use the first linebreaks in a row. Do not use tables. For Mathematical formulas, equations, and symbols use Latex. Latex must be placed inside of the following syntax: […]. Never use \(...\) to open a math environment, instead, make sure to always use either […] or […]. Never use two linebreaks after one another without at least one sentence of text between them. ONLY WRITE 3 sections in total! The ORIGINAL EXERCISE AND SOLUTION were provided by you. Please read the instructions carefully before proceeding. Thank you for your cooperation.

When applying this law to the adiabatic compression of steam, as in the exercise, it implies that all the work input into the compressor is used to increase the internal energy (as reflected in the enthalpy) of the steam. This is a useful concept when calculating the power required for the compression process, leading to a better understanding of the efficiency of the process and the design of the system. Remember, in the absence of heat transfer, the work done on the system directly affects its energy content.
Specific Enthalpy
Specific enthalpy is a fundamental property in thermodynamics, especially when discussing phase change materials like water and steam. It represents the total energy contained within a substance per unit mass, accounting for both its internal energy and the work done to place the substance in its current state at a specific pressure. Knowing the specific enthalpy is vital when working with the First Law of Thermodynamics in systems involving heat and work.

In our steam compression example, the specific enthalpy values, which reflect the energy content of steam at specified conditions, can be different at the start and the end of the compression process due to the work done by the compressor. To solve the exercise, one needs to find the values of specific enthalpy for steam at the initial and final conditions using known temperature and pressure information, which can be sourced from steam tables. It is these values that are plugged into the First Law of Thermodynamics to calculate the change in energy, and subsequently, the power input required for the compressor.
Steam Tables
Steam tables are essential tools in thermodynamics for engineers and scientists dealing with steam power systems. They provide a comprehensive set of data on the properties of water and steam, including specific enthalpy, at various temperatures and pressures. The data contained in steam tables is derived from the equations of state for water, which closely represent the real properties of water under different conditions.

For students working on thermodynamic problems like the given textbook exercise, steam tables allow for the quick retrieval of vital information without the need to calculate properties from scratch. One can simply look up the specific enthalpy for water or steam at a certain pressure and temperature. Accurate use of these tables is critical in determining parameters like enthalpy changes during adiabatic compression, which in turn leads to accurate calculations of work and power input for processes within steam power cycles.

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Most popular questions from this chapter

Design a \(1200-\mathrm{W}\) electric hair dryer such that the air temperature and velocity in the dryer will not exceed \(50^{\circ} \mathrm{C}\) and \(3 \mathrm{m} / \mathrm{s}\), respectively.

A piston-cylinder device initially contains 2 kg of refrigerant-134a at \(800 \mathrm{kPa}\) and \(80^{\circ} \mathrm{C}\). At this state, the piston is touching on a pair of stops at the top. The mass of the piston is such that a 500 -kPa pressure is required to move it. A valve at the bottom of the tank is opened, and R-134a is withdrawn from the cylinder. After a while, the piston is observed to move and the valve is closed when half of the refrigerant is withdrawn from the tank and the temperature in the tank drops to \(20^{\circ} \mathrm{C}\). Determine \((a)\) the work done and \((b)\) the heat transfer.

Air at \(300 \mathrm{K}\) and \(100 \mathrm{kPa}\) steadily flows into a hair dryer having electrical work input of \(1500 \mathrm{W}\). Because of the size of the air intake, the inlet velocity of the air is negligible. The air temperature and velocity at the hair dryer exit are \(80^{\circ} \mathrm{C}\) and \(21 \mathrm{m} / \mathrm{s},\) respectively. The flow process is both constant pressure and adiabatic. Assume air has constant specific heats evaluated at \(300 \mathrm{K}\). (a) Determine the air mass flow rate into the hair dryer, in \(\mathrm{kg} / \mathrm{s}\). ( \(b\) ) Determine the air volume flow rate at the hair dryer exit, in \(\mathrm{m}^{3} / \mathrm{s}\).

A \(110-V\) electric hot-water heater warms \(0.1 \mathrm{L} / \mathrm{s}\) of water from 18 to \(30^{\circ} \mathrm{C}\). Calculate the current in amperes that must be supplied to this heater.

Two mass streams of the same ideal gas are mixed in a steady-flow chamber while receiving energy by heat transfer from the surroundings. The mixing process takes place at constant pressure with no work and negligible changes in kinetic and potential energies. Assume the gas has constant specific heats. (a) Determine the expression for the final temperature of the mixture in terms of the rate of heat transfer to the mixing chamber and the inlet and exit mass flow rates. (b) Obtain an expression for the volume flow rate at the exit of the mixing chamber in terms of the volume flow rates of the two inlet streams and the rate of heat transfer to the mixing chamber. (c) For the special case of adiabetic mixing, show that the exit volume flow rate is the sum of the two inlet volume flow rates.
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